First Edition, 2009
ISBN 978 93 80168 57 9
© All rights reserved.
Published by: Global Media 1819, Bhagirath Palace, Chandni Chowk, Delhi-110 006 Email:
[email protected] Table of Contents 1. Theoretical Representations 2. Variation Method 3. The Solids 4. Transition Elements 5. Non-transition Elements 6. Lanthanides and Actinides 7. Statistical Thermodynamics 8. Non-equilibrium Thermodynamics 9. Molecular Orbital and Valance Bond 10. The Solutions
Theoretical Representations 1 Theoretical Representations Great Orthogonality Theorem Explained : The great orthogonality theorem can be stated as
=
... (1)
where `h' is the order of the group li is the dimension of the ith representation lj is the dimension of the jth representation R denotes the various operations in the group, éi(R)mn denotes ith irreducible representation of an element in the mth row and the nth column and [é j(R)m ¢n¢]* denotes the complex conjugate of the factor on the left hand side. From the above equation, it is clear that in the set of matrices constituting any one irreducible representation, one from each matrix behaves as the components of a vector in h-dimensional space in such a way that all these vectors are mutually orthogonal and each is normalized so that the square of its length equals . For example equation (1) can be split into three equations. Since the equations are split, the complex conjugate can be omitted.
Chemistry : Practical Application éi(R )mn éj(R)mn = 0, if i ¹ j ... (2)
éi(R )mn éj(R)m ¢n¢ = 0, if m ¹ m¢ and/or n = n¢ ... (3)
éi(R )mn éj(R)mn =
... (4)
Equation (2) indicates that the vectors differ only in the fact that they are chosen from matrices of different representations and are orthogonal. Equation (3) indicates that the vectors from the same representation but from different sets of elements in the matrices of this representation are orthogonal.
Equation (4) indicates the fact that the square of the length of any such vector equals The operators
and
.
form a group :
Taking water molecule as the object of operations, let sv be the reflection in the molecular plane and s¢v the reflection in the plane bisecting the H—O—H angle. Then constructing the multiplication table by investigating the effect of two successive operations. The effect of \
followed by another
is to bring the water molecule back to the original position.
=
The result of
followed by
is the same as the rotation by 180°. Hence
= By proceeding along these lines, we get the following group multiplication table Theoretical Representations First Operation Second Operation
From this table we see that the four conditions for group multiplication are met. Hence the above
operations form a group. In this particular case the operations also commute. A hypothetical molecule A3 with the geometry of an equilateral triangle. The basis set consisted of one function from each atom. Symmetry adapted functions (SAF), constructed :
The Molecule belongs to the C3 group f2 f3 Let us denote the three basis functions as f1, f2 and f3 and the SAFs by y(a), y(e) and y¢(e). From the projection operator theorem y(a) =
= f1 + f2 + f3
The E representation is doubly degenerate.
Chemistry : Practical Application The functions belonging to it are y (e) =
= 2f1 _ f2 _ f3
By starting with the function f2 we have y¢ (e) = 2f2 _ f3 _ f1 By starting with f3, we obtain (2f3 _ f1 _ f2). However, this function is not independent of the other two functions, as may be seen by adding them. Instead of the above set we may work with
y1 = y(e) + y¢ (e) = 2f3 _ f1 _ f2 and y2 = y(e) _ y¢ (e) = f1 _ f2 Since any linear combination of degenerate eigen functions is also an eigen function. The 1Ag ® 1B3u transition in pyrazine is allowed: For a transition to be allowed the transition moment integral must have a non zero value. The components of the transition moment integral are ò y0 xy1 dv º x01, òy0 yy1 dv = y01 ò y0 zy1 dv º z01 The representation for each component is obtained by referring to the D2h group table. Since y0 É A1g and y1 É B3u é (x01) = A1g é(x) B3u = A1g B3u B3u = A1g é (y01) = A1g é(y) B3u = A1g B2u B3u = B1g é (z01) = A1g é(z) B3u = A1g B1u B3u = B2g \ x01 ¹ 0 and y01 = z01 = 0 We see from this that the transition is allowed in the direction perpendicular to the molecularplane. Irreducible Representations The important rules about irreducible representations : Theoretical Representations (1) The sum of the squares of the dimensions of the irreducible representations of a group is equal to the order of the group, that is Sl12 = l12 + l22 + l32 + ...... = h
(2) The sum of the squares of the characters in any irreducible representation equals h, that is
=h (3) The vectors whose components are the characters of two different irreducible representations are orthogonal, thatis = 0 when i ¹ j (4) In a given representation (reducible or irreducible) the characters of all matrices belonging to operations in the same class are identical. (5) The number of the irreducible representations of a group is equal to the number of classes in the group. The energy level diagram of the molecular orbitals in the octahedral symmetry and forming only a bonds, drawn : Since only particular kinds of orbitals lead to octahedral symmetry i.e., dx2 _ y2, dz2, s, px, py, pz so the dxy, dyz and dxz will be unaffected by the bonding as these orbitals are not involved in the bond formation. The D, splitting between T2g & eg* depends upon the strength of the ligands.
Chemistry : Practical Application The molecular orbital diagram for [FeF6]3_ complex, drawn : Since F_ ion is a weak ligand, the electrostatic field splitting D will be small and will lead to high spin
complex. The molecular orbital energy level diagram will be t*1u — — — a*1g — e*g - t2g - - eg xx xx a1g xx t1u xx xx xx Note : xx ® Indicates the electrons of six ligands. The selection rule for IR spectra : For a fundamental transition to occur by absorption of Infrared radiation, it is necessary that one of integrals of equation (1) given below be non-zero.
...(1)
Let us consider the integral consider yi in this integral is totally symmetric. The product representation of X and yi will be totally symmetric only if yj has the same symmetry as X. Hence we can have the rule for the activity of fundamentals in IR absorption. A fundamental will be infrared active only if the normal mode which is excited belongs to the same representation as any one or several of the Cartesian coordinates. The HOMO in CO is : (a) p bonding (b) s bonding
(c) p antibonding (d) s antibonding Theoretical Representations `d' because MO configuration of CO is (1s)2 (2s)2 (1p)4 (3s)2. The HOMO is 3s orbital which is antibonding. The rules of irreducible representation for C2v point group : The C2v consist of 4 element and each is in a separate class. Hence according to rule there will be four irreducible representations for the group and according to rule the sum of the squares of dimensions of these representations equals to g. Thus l12 + l22 + l32 + l42 = 4 Thus l1 = l2 ± l3 = l4 = 1 Thus C2v has four one-dimensional irreducible representation. On working out the characters of these four irreducible representations on the basis of vector properties and rules of irreducible representations one can write, C2v E C2 sv s¢v é1 1 1 1 1 This is a suitable vector in 4-space which has a component of 1 corresponding to E. From this S[X, (R)]2 = 12 + 12 + 12 + 12 = 4 thus satisfying the rule. Now all other representations are to be such that
=4 which will be true for X1 (R) = ±1
According to rule in order for each of the other representation to be orthogonal to é1 there must be two + 1 and two _ 1. Thus (1)(_1) + (1) + (_1) + (1) (1) + (1) (1) = 0
Chemistry : Practical Application The total can be written as C2v E C2 sv s¢v é1 1 1 1 1 é2 1 _1 _1 1 é3 1 _1 1 _1 é4 1 1 _1 _1 Trigonal planar molecule such as BF3 cannot have degenerate orbitals : BF3 belongs to D3h point group. The maximum number in the column headed by the identity E is the maximum orbital degeneracy possible in a molecule of that symmetry group. The character table of D3h shows that maximum degeneracy is 2, as no character exceeds 2 in the column headed E. This means, the orbitals cannot be triply degenerate. The multiplication table of the Pauli Spin matrices and the 2 × 2 unit matrix.
sx =
sz =
sy =
E=
The multiplication table is—
The matrices do not form a group since the product 1sz, isy, isx and their negatives are not among the four given matrices. Taken 1s orbitals as a basis of the two hydrogen and the four valence orbitals of the oxygen atom in H2O molecule to
Theoretical Representations set up 6 × 6 matrices and then confirm by explicit matrix multiplication the group multiplication (i) C2TV = T¢v and (ii) sv s¢v = C2 . H2O belongs to C2v point group. Places orbitals h1 and h2 on the H atoms and s, px, py and pz on the O atom. The z-axis is the C2 axis, X lies perpendicular to s¢v, y lies perpendicular to sv. Then draw up the following table of the effect of the operations on the basis—
Express the column headed by each operation R in the form (new) = D(R) original, where D(R) is the 6 × 6 representative of the operation R. We use the rules of matrix multiplication. E : (h1, h2, s, px, py, pz) ¬ (h1, h2, s, px, py, pz) is reproduced by the 6 × 6 unit matrix.
C2 : (h2, h1, s, _ px, _py, pz) ¬ (h1, h2, s, px, py, pz) is reproduced by
D (C2) = sv = (h2, h1, s, px, _ py, pz) ¬ (h1, h2, s, px, py, pz) is reproducedby
Chemistry : Practical Application
D (sv) = s¢v = (h1, h2, s, _ px, py, pz) ¬ (h1, h2, s, px, py, pz) is reproducedby
D (s¢v) = (i) To confirm the correct representation of C2sv = s¢v, we write D (C2) D(sv)
=
=
= D (s¢v)
(ii) Similarly, to confirm the correct representation of sv s¢v = C2, we write Theoretical Representations
=
= D(C2)
The irreducible components of representations generated by a set of a type atomic orbitals in XY3 molecules of C3v and D3h symmetry.
C3v:
In case methane molecule is distorted to (a) C3v, point group due to bond lengthening and (b) to C2v point group due to scissor action of molecular vibration. More d-orbitals would become available for bonding. (a) In C3v symmetry the H1s orbitals span the same irreducible representations as in NH3, which is A1 + A2 + E. There is an additional A1 orbital because a fourth H atom lies on the C3 axis. In C3v, the d orbitals span A1 + E + E. Therefore, all five d orbitals may contribute to the bonding. (b) In C2v symmetry the H1s orbitals span the same irreducible representations as in H2O, but one `H2O' fragment is rotated by 90° with respect to the other. Therefore, whereas Chemistry : Practical Application in H2O the H1s orbitals span a1 + B2 [H1 + H2, H1 _ H2] in the distorted CH4 molecule they span A1 + B2 + A1 + B1 [H1 +H2,H1_ H2, H3 + H4, H3 _ H4]. In C2v the d-orbitals span 2A1 + B1 + B2 + A2, therefore, all except A2 (dxy) may participate in bonding. The ground state of NO2 (C2v) is A1. The excited states may be excited by electric dipole transition and whose polarization of light is it necessary to use : Ans. The electric dipole moment operator transforms as x (B1), y (B2) and z (A1) [C2v character table]. Transitions are allowed if òyf m yi dt is non-zero and hence are forbidden unless éf × é(m) × éi contains A1. Since éi = A1, this requires éf × é(m) = A1 Since B1 × B1 = A1, and B2 × B2 = A1 and A1 × A1 = A1 x-polarized light may cause a transition to a B1 term, y-polarized light to a B2 term and z-polarized light to an A1 term. The benzene may be reached by electric dipole transition from their (totally symmetrical) ground states
: The point group of benzene is D6h, where m spans E1u (x,y) and A2u (z) and the group term is A1g. Then using A2u × A1g = A2u, E1u × A1g = E1u, A2u × A2u = A1g and E1u × E1u = A1g + A2g + E2g we conclude that the upper term is either E1u or A2u. The symmetry elements listed and named the point groups to which following molecule belongs : (i) Staggered CH3CH3, (ii) Chair and boat cyclohexane, (iii) B2H6, (iv) [Co(en)3]3 +, (v) S8 which of them molecules can be polar and chiral. (i) Staggered CH3CH3 : E, C3, C2, 3sd ; D3d (ii) Chair C6H12 : E, C3, C2, 3sd ; D3d Boat C6H12: E, C2, sv, s¢v, C2v (iii) B2H6 : E, C2, 2 C¢2; sn; D2h (iv) [Co(en)3]3+ : E, 2C3, 3C2; D3 (v) Crown S8 : E, C4, C2,4C¢2, 4sd, 2S8; D4d Theoretical Representations Only boat C6H12 may be polar, since all the others are D point groups. Only [Co(en)3]3+ belongs to a group without an improper rotation axis (S1 = s) and hence is chiral. A weak infrared band is seen at 2903 cm_1 in the IR spectrum of BF3. The feature assigned and deduced its symmetry properties : The band can be assigned to 2v3, the first overtone of the E¢ B—F stretch (2 × 1453). To determine the symmetry species of this overtone use the equation written below
c2(R) = For D3h, we have
The resulting representation is reducible to A1 + E¢. Hence, the first overtone of the E¢ B—F stretch of BF3 has symmetry species A¢1 and E¢. The transition A1 ® A2 is forbidden for electric dipole transmission is C3v molecule : Considering all three components of the electric dipole moment operator, m. Component of m: X Y Z A1 1 1 1 1 1 1 1 1 1 é(m) 2 _1 0 2 _1 0 1 1 1 A2 1 1 _1 1 1 _1 1 1 _1 A1é(m)A2 2 _1 0 2 _1 0 1 1 _1 E E A2 Since A1 is not present in any product, the transition dipole moment must be zero.
Chemistry : Practical Application Raman Spectra The selection rule for Raman spectra :
For Raman scattering it is necessary that atleast one of the integrals of the type be nonzero. In these types of integrals P is one of the quadratic function of the cartesian coordinates namely, x2, y2, z2, xy, yz, zx all of which are listed opposite to the representations that they generate in the character
tables. These P's are components of the polarizability tensor. The above integral will become non-zero only if there is a change in polarizability of the molecule during transition. A fundamental transition will be Raman active only if the normal mode involved belongs to the same representation as one or more of the components of the polarizability tensor of the molecule. For example for NH3 molecule the charactertable for C3v group is used to obtain the following irreducible representations for the quadratic and binary cartesian coordinates. Quadratic and Binary Cartesian Coordinates Representation Z 2 A1 (x2 _ y2, xy) (xz, yz) E The representations obtained for the quadratic and binary coordinates [Z2; (x2 _ y2, xy); (xz, yz)] correspond to the symmetry species of the vibrational modes for NH3 molecule. Therefore, the vibrational modes of NH3 molecule are Raman active. Quantum Chemistry The principles of symmetry and group theory used in the area of quantum chemistry: Principles of symmetry and group theory find applications in several areas of quantum chemistry like chemical bonding, molecular spectroscopy, ligand field theory, crystal field theory etc. The procedure in all these cases involves— (i) Generating a reducible representation ér of the symmetry group to which the molecule belongs using a set of atomic orbitals as basis. Theoretical Representations (ii) Resolving the ér to irreducible representation éi's using formula.
ai = ai ® number of irreducible representations (symmetry types) of one kind i in Tr, n is the total number of symmetry operations in the symmetry group. c(R) is character of the operation R in the reducible representation, ci (R) is the character of the same operation R in the irreducible representation i, and n is number of the operations in one class.
(iii) Constructing SALC's (symmetry adapted linear combinations) corresponding to the Ti's of the group. Energy Level Diagrams for : (a) p orbitals of napthalene, (b) Ferrocene. (a) Napthalene belongs to the point group D2h.
Energy level diagram for n orbitals of naphthalene.
Chemistry : Practical Application (b) Ferrocene belongs to D5d group.
An energy level diagram for ferrocene.
Variation Method 2 Variation Method An anhormonic oscillator has a potential energy given by :
U(x) =
=
The first order perturbation correction to the harmonic oscillator energy level, calculated : The wave functions corresponding to the energy En are given by yn (x) = Nn Hn (a1/2 x) e_ax2/2
where a = and the Normalization constant Nn is given by
Nn =
Since DE = and the perturbation term of the Hamiltonian is
Chemistry : Practical Application =
We have DE =
The integrand here is overall an odd function. The integral therefore vanishes. i.e., DE = 0 corresponding to y = y0 + Dy we have E = E(0) + DE where DE is the first order correction to E(0). The energy is E = E(0) + òy(0)*.H¢.y(0) dt+ higher order terms The unperturbed energies are
E(0) = The first terms correction to the harmonic oscillator energy are therefore,
E=
+ higher order terms
Frequency of Radiation The frequency of radiation emitted by a bound systems is given by :
w=
= 1, 2, .... ... (A)
where Dn is the change in quantum number n and En is the energy associated with the change in quantum number. If the WKB (Wentzel, Kramers & Brillion) energy levels En are determine by:
f (En) =
... (B)
Variation Method where f is the phase integral, shown that the radiation emitted by the bond system obeys the correspondence principle : Differentiation of phase integral can be expressed as [by differentiating equation (B)]
=
=p
Since by WKB
we get
=
= 2ph/ v ... (1)
Combining equation (A) and (1) gives w = Dn (2pv), Dn = 1, 2, 3, ... ... (2) This is analogous to the equation w = Dn Wn ... (3) where Wn is the angular velocity of the electron in its orbit. The classical angular frequency is given by wC1= KWn, K = l, 2, 3... ... (4) Comparison of equations (2) and (4) show that the radiation from the bond state described by WKB obeys the correspondence principle. A system consists of a particle in a one dimensional well shown below C 0 ¬¾¾¾ x ¾¾¾¾® a
Chemistry : Practical Application The potential energy inside the well is given by U (x) = C(x/a) Using perturbation theory, the first order estimate of the energy of this systems, determined: The Hamiltonian for the unperturbed problem is
= This has the solution
y0n =
with AA* = 2/a and En0 = for 0 < x < a the Hamiltonian for the perturbed system is
= where
= = C (x/a)
Now we have
= For m = n
=
=
= The value of the integral
= Variation Method
Hence H¢nn =
=
=
Setting l = 1, signifying that the perturbation is fully applied, the first order estimate of energyis
En =
=
=
The two normalized eigen vectors are C(1) =
,
ya(x) = e_ax2 (a > 0), C(2) =
. Shown that C(1) and C(2) are orthogonal.
Two column vectors b and c that each have n elements are said to be orthogonal if
=0 There are two column vectors C(1) and C(2)
C(1)* =
C(2) =
.
so that
=0
Chemistry : Practical Application
=
=
=0
A trial function e_ax2/2 used to calculate the ground state energy of a quartic oscillator whose potential is given by V (x) = Cx4
We have to use the equation
Ef =
... (1)
We have
=
...(2)
which gives
=
... (3)
=
Also ò f*f dt =
...(4)
The value of Ef is therefore,
Ef =
...(5)
Now according to the variational principle we differentiate Ef with respect to the variational parameter a and set the result to zero. On doing this we get
a=
... (6)
Substituting value of a from equation (6) into equation (5) and simplifying we get
Emin = Variation Method
Variation Principle The variation principle is used when the Schrodinger equation cannot be solved. However let us assume that we know the eigen function of the correct Hamiltonian. Let us label these functions y0, y1, y2,... These form the bases for a function space. Since the eigen functions form a complete set, any function we might guess must be either : (a) One of the basis functions or (b) a linear sum of the bases. Let us consider both possibilities. (a) If we happen to guess the lowest energy function, we will have the minimum possible energy. Any other guess will give us a higher energy. (b) Suppose we pick a function say c that is not one of the bases, then it must be a linear sum of the bases: c = C0y0 + C1y1 + C2y2 + ... ... (1) Where y represents the eigen functions and c is the chosen function. The calculated energy of the system E (trial) = òc * Hc dV = C02 E0 + C12 E1 + C22 E2 +... ... (2) is the weighted average of the lowest as well as the higher energies. Since an average value can never be lower than the lowest of the values, E(trial) must be greater than E0. The wave function equation y11 = ls (1) . ls (2) (Lowest energy level of the helium atom) would be an eigen function of the Hamiltonian for a two electron system, under the following conditions : Let us consider the Hamiltonian in equation which is for a two electron atom
H^ =
0
Chemistry : Practical Application without the last term. In that case the Hamiltonian may be written as = where case
depends only on the coordinates of electron 1, and
on the coordinates of electron 2. In that
=
= = (E1 + E2) [1s(1) 1s (2)] Therefore, equation y11 = 1s (1) 1s (2) is an eigen function of the Hamiltonian. Second, suppose the effect of electron repulsion is simply a reduction in nucleus-electron attraction. In that case we will have Z¢e2/4pe0 r1 (Z¢ is some number smaller than Z) and another term like that for electron 2 in the Hamiltonian, instead of Ze2/4pe0r1 . In that case the Hamiltonian will be the sum of two one electron Hamiltonian and equation y11 = 1s (1) 1s (2) will be its eigen function. Shown that the probability of having three electrons in a 1s orbital is zero : Since there are only two spin states for an electron, atleast two of the three electrons must have the same spin. Therefore, the determinant form of the wave function is
where we have assigned lsa for two electrons. There are two identical rows in the above determinant hence it must vanish. Variation Method Therefore, the probability of finding three electrons in 1s (or any other orbital) must be zero. The Slater determinants for `Be' in their lowest energy configuration, written : Be has the closed shell configuration 1s2 2s2 and can be represented by a single slater determinant
y= The diagonal of the determinant is Be = 1s a (1) 1s b (2) 2s a (3) 2s b (4) Spectroscopic terms which arise from the 1s2 2s2 2pl 3pl configuration of the carbon atom or the N+ ion are : We need to consider only the electrons beyond the closed shall part of the configuration 2p13p1. The electrons in these functions are said to be non equivalent since they have different n quantum numbers. We can obtain the allowed ML and Ms values by addition as indicated below :
Chemistry : Practical Application Only if L is allowed the values are 2, 1 and 0 and if S is allowed the values are 1 and 0 we can account for all the components of angular momenta in the above diagram. Hence we conclude that the 2p13p1 configuration yields the terms 3D, 1D, 3P, 1P, 3S, 1S. The variation method used to obtain an upper bound to the ground state energy of a particle in a one dimensional box and compare the result with the true value given in equation
E=
n = 1, 2, ... ... (A)
Normalized trial variation function that satisfies the boundary conditions that y = 0 at x = l is :
y=
.
=
=
=-
=
> E0
The true value is
% error =
= 1.3 %
\ an approximate wave function always yields a higher energy than the true wave function. Variation Method The atomic term symbol for Lithium and Boron is : Since Li has a 2s electron with l = 0, its magnetic quantum number m must also be zero. m ms ml MS MJ 0 ± 1/2 0 ± 1/2 ± 1/2
Thus L=0, S = and J =
so the term symbol is 2S1/2.
Boron has a p electron so that m ms ml MS MJ 0 ±1/2 0 ±1/2 ±1/2 ±1 ±1/2 ± 1 ± 1/2 ± 1/2 ± 3/2
Thus L = 1, S =
,J=
,
, so that there are two possible term symbols 2P3/2 and 2P1/2.
Application of the Variation Method The steps involved in the application of the variation method are : The application of the variation method involves the following steps : (a) Choose a trial function y with some variable parameters. (b) Calculate the integral < y | H | y >. (c) Since this integral always gives an upper bound to the true energy unless the chosen function y happens to be the exact one, minimise the integral with respect to the variable parameters. (d) The function y with the optimum value of the parameters is the best approximation of its class to the true wave function and the lowest value of < y | H | y > is the nearest approximation to the true energy, for all trial wave functions belonging to the same class.
Chemistry : Practical Application The spectroscopic term symbol for the electronic configuration 2p1 3p1 3d1 system, calculated: First take 2pl 3pl, l1 & l2 for these electrons are 1 each, \ L = 1 + 1 = 2, 1 + 1 _ 1 = 1, 1 _ 1 = 0
S=
_
+
= 1, 2S + 1 = 3
= 0 2S + 1 = 0
The term symbols due to coupling of 2p1 3p1 electrons can be 3D, 3P, 3S and 1D, 1P and 1S. Now each of six states can couple with 3d1 electron to give the resultant L and S.
Now each of six states can couple with 3d1 electron to give the resultant L and S. For e.g., 3D 3d1 gives, L = 2 + 2 = 4, 2 + 2 _ 1 = 3, 2 + 2 _ 2 = 2, 2 + 2 _ 3 = 1, 2 _ 2 = 0
S=1+
=
1_
, 2S + 1 = 2
=
, 2S + 1 = 4,
\ The term symbols for the coupling of 3D 3d1 configuration are 4G 4F 4D 4P 4S and 2G 2F 2D 2P 2S. For 3P 3d1 the terms are 4F 4D 4P and 2F 2D and 2P For 3S 3d1 terms are 4D and 2D For lD 3dl terms are 2G 2F 2D 2P 2S For 1P 3d1 terms are 2F 2D 2P and For 1S 3d1 the term is 2D Variation Method It can be seen that some of the term symbols occurs more than once in different combinations and they can combined to give as 4G 4F(2) 4D(3) 4P(2) 4S 2G(2) 2F(4) 2D(6) 2P(4) 2S(2)
The number in the brackets tells us how many times the particular state is formed by different combinations.
The term symbols for (1) L = 2, S =
(2) L = 1, S =
, determined :
; 2S + 1 = 2
(1) J =
\ terms are 2D5/2 and 2D3/2.
(2) J =
; 2S + 1 + 4
\ terms are 4P5/2, 4P3/2, 4P1/2 The terms symbols for the ground and first excited state of He, determined : L = l1 + l2 = 0
S1 = S2=
,
S = 1 (spins, parallel) S = 0 (spins, paired) Ground State _ Spins are paired, i.e., S = 0, J = 0 \ term symbol 1S0 Excited State— (a) When spins are paired (S = 0), J = 0, Term symbol: 1S0 (singlet)
Chemistry : Practical Application
(b) When spins are parallel (S = 1), J = 1 Term symbol: 3S1 (triplet) Slater Rules Stated : (1) The effective principal quantum number n¢ is related to the quantum number n as follows: n:123456 n¢ : 1 2 3 3.7 4.0 4.2 (2) The effective nuclear charge Z¢ is related to the actual atomic number Z as Z¢ = Z _ S, where S is the screening constant. The value of S is calculated by the following rules. The orbitals are first grouped as : (1s); (2s, 2p); (3s, 3p); (3d); (4s, 4p); (4d); (4f) The screening constant (S) is then calculated by adding the contributions from electrons as follows : (a) No contribution from electrons in groups higher than (i.e., to the right of) the group in question is included. (b) Each electron in the same group contributes 0.35 the electron in Is group however contributes 0.30 only. (c) In case of s or p group, each electron in the next lower group contributes 0.85 from an electron in d or f group the contribution from each inner electron is 1.00 (d) Each electron in groups lower by two or more numbers from the one in question contributes 1.00. Effective Nuclear Charge The effective nuclear charge for the 1s electron in He and 2s electron in C, determined : He (2s2); S = 0.30
\ Z¢ = 1.70 Variation Method C (1s2 2s2 2p2); For each electron in 2s or 2p groups S = 3 × 0.35 (due to the remaining three electrons in that group) + 2 × 0.85 (due to the two electrons in the next lower group i.e., 1s) = 2.75 \ Z¢ = Z _ S = 3.25 The trial function y = (1 _ x2) xa used to calculate the ground state energy of a particle in a one dimensional box of unit length and having infinite potential walls at the two ends; a is a variable parameter :
=
=
E=
=
=
=
=
=0
Chemistry : Practical Application or 8a3 + 8a2 _ 14a + 1 = 0, this is a cubic equation having three roots. Numerical solution yieldsone of the roots (a = 0.862) which is relevant to the present problem. On substituting this value we get,
E = 5.13 in a.u where
=m=1
The exact value of
E=
=
=
= 4.94 (in a. u, n & L being equal to 1)
The trial function y = exp (_ar2) used (without the normalisation factor) for the ground state of the hydrogen atom : (a) Variation method applied to find the value of a that would give the minimum energy : (b) That energy :
E=
=
= Variation Method From the table of integrals
=
=
and
=
\E=
=
=
From which a =
=0
\E=
=_
= _ 0.4244 a.u. The perturbed Hamiltonian =
where
operation. Shown that if
is given by the relation
is that for the unperturbed system and and
are Hermitian,
due to a small perturbation
must also be Hermitian :
=
=
... (i)
Chemistry : Practical Application =
and
=
... (ii)
Since
and
and Hermitian,
=
Therefore, i.e.,
must be equal to
is Hermitian.
The following is wrong with the following atomic term symbols 4S1 and 0D1 : 4S
: 2S + 1 = 4; \ S = 3/2. For the S atomic state, L = D, hence J must be 3/2 and not 1. 0D1 : 2S + 1 cannot be equal to zero. 1
The Solids 3 The Solids Crystallography The branch of science which deals with the study of geometry, properties and structure of crystals and crystalline substances is known as crystallography. Three fundamental laws of crystallography are (1) The Law of constancy of interfacial angles. (2) The Law of rotationality of indices. (3) The Law of symmetry.
A crystal plane has intercepts on the three axes of crystal in the ratio of of the plane are :
: 2 : 1. The Miller indices
The reciprocal of the ratio of intercepts : Multiply by the LCM of 3 and 2 i.e., 6 4:3:6 The Miller indices of the plane are 436.
Chemistry : Practical Application Density of NaCl = 217 gm cc_1 (at 20° C), molar mass = 58.45. The interplanar distance (d200)calculated. The first order reflection maximum of the Ka rays of palladium form the 200 planes of NaCl occurs at a glancing angle 5.90°. Calculate the wavelength of the X-radiation :
d200 =
= = 2.82 × 10_8 cm = 2.82 Å n l = 2 d sin q; n = 1, d = 2.82 Å. and sin q = sin 5.9° = 0.1028 l = 2 × 2.82 × 0.1028 = 0.5797 Å Caesium bromide has CsCl structure (body centered cubic type of lattice). Its density is 4.49 gm cc_1. The side of the unit cell calculated : Unit cell of CsBr contains effectively/one molecule of CsBr. Molar mass of CsBr is 212.8 (i.e., 212.8 gms mol-1)
Molar volume = N molecules (6.023 x 1023) correspond to 47.39 cm3 \ 1 molecule will be corresponding to
= i.e., the volume of the unit cell = 7.868 × 10-23 cm3 \ Side of the unit cell = (7.868 × 10-23)1/3 = 4.285×10-8cmor 4.285Å. The Solids
The crystal structure of zinc blende, ZnS is related to the diamond structure with zinc and sulphur atoms occupying the carbon sites. If the coordinates of one of the zinc sites are (0 0 0), give the fractional coordinates of the remaining Zn and S sites in the unit cell. If covalent radii of Zn and S are 1.32 Å and 1.04 Å respectively, the unit cell length of zinc blende calculated: Zinc blende has face centered cubic lattice in Zn atoms and again face centred lattice in S atoms. The coordinates of Zn atoms and S atoms in a unit cell are Zn atoms :
S atoms :
The interatomic distance r = (1.32 + 1.04)Å
=
where `a' is the unit cell length. Thus
= 2.36 Å
a = 5.45 Å The mole fractions of Schottky and Frenkel defects in a NaCI crystal at 1000 K estimated. The energies of formation of these defects are 2 eV and 3 eV respectively given : 1 eV = 1.602 × 10-19J, K = 1.38 × 1023 JK-1 Schottky defects is given by n/N = exp (_ E/2KT)
= = exp (_ 11.6) = 9.17 × 10-6
Chemistry : Practical Application Frenkel Defects : According to the structure of the unit cell of NaCI crystal, the number of inter-stitial spaces is twice the number of Na + ions present in the unit cell, that is, Ni = 2N. The Frenkel defects are formed only by the migration of the smaller Na+ ion. They are given by the equation n = (NNi)1/2exp (_E/2KT) Since Ni = 2N, hence n/N = 21/2 exp (_ E/2KT)
= The given examples into metal excess defects or metal deficiency defects which are generally found in them : (a) NaCl (b) FeO (c) FeS (d) NiO (e) KC1 (a) NaCl ® Metal excess defect (b) FeO ® Metal deficiency defect (c) FeS ® Metal deficiency defect (d) NiO ® Metal deficiency defect (e) KC1 ® Metal excess defect Imperfections Due to Transient Atomic Displacement.
The vibrations of a crystal can be linked to the vibrations of a simple harmonic oscillator having three normal modes of vibration which in the simplest case, correspond to one longitudnal mode and two mutually perpendicular transverse modes. When a transition takes place from a higher vibrational to a lower vibrational state, a quantum of thermal energy is emitted. In analogy with a photon of radiation energy, the quantum of thermal energy is termed as phonon. Absorption of phonons by a crystal can produce atomic displacement leading to imperfections. Unlike point defects and line defects, phonons produce atomic displacements that are time dependent and hence transient i.e., short lived. The Solids The conduction of electricity in Be and Li metals with the help of a band theory. The electronic configuration for Li is Is2 2sl so there will be 1s and 2s bands as shown in figure.
Since only half of 2s orbitals are occupied by the electrons, the conduction will take place. The electronic configuration for Be is 1s2 2s2 and the bands are formed due to Is and 2s electrons. Both the bands are completely filled but since 2p band which is completely empty, overlaps the 2s band
There is sufficient space for electrons to move and explains the conduction part. Conduction Phenomenon in Semiconductors : In semiconductors, the gap between the filled and unfilled band is reduced by the presence of electrons due to impurities and the electrons of the impurities are promoted to the empty conduction band either by increasing the voltage or by shining light on the semiconductor. The fraction (n/N) of the lattice sites are vacant at 298 K fora crystal: l eV = 1.602 × 10-19 J
=
Chemistry : Practical Application The schematic band models of solids i.e. insulator, metal, intrinsic semiconductor and impurity semiconductor which are classified according to electronic properties, drawn :
Solids, classified on the basis of their electrical conductivity. They are of three types (a) Conductors or metals, which offer a low resistance to the flow of electrons when a potential difference is applied. The resistivities of metals lie m the range 10-6 to 1020 W. m at room temperature and increases with the temperature. (b) Insulators which have high resistivities from 108 to 1020 W m at room temperature. (c) Semiconductors, the resistivities of which are intermediate between those of typical metals and typical insulators and decrease with rising temperature. The conductivity decreases in case of metals on increasing the temperature. The conductance of metal decreases with an increase in temperature this is due to the fact that vibratory motion of the
The Solids atoms increases with temperature which interferes with the free movement of electrons. Matched : (1) MgAI2O4 (a) Perovskite structure. (2) FeTiO3 (b) Inverse spinel structure. (3) CaTiO3 (c) Spinel structure. (4) Zn2TiO4 (d) Rutile structure. (5) TiO2 (e) Ilmenite structure. (l)_(c), (2)_(e), (3)_(a), (4)_(b), (5)_(d). The type of conduction is observed in compounds like Fe1 — YO and Fe1 _ yS : Semiconductor behaviour is to be found in compounds Fe1 _ YO and Fe1 _ YS where electron transfer from Fe+2 to Fe+ 3 causes a virtual migration of Fe3 + ions and hence p-type conduction. The two consequences of metal excess defects : (1) Crystals having this defect can conduct electricity to some extent. Such crystals are generally termed as semiconductors. (2) Crystals showing metal excess defects are generally coloured. This is due to the presence of free electrons. Intrinsic and Extrinsic Semiconductors Intrinsic Semiconductors : Semiconductors due to thermal defects. At OK, pure Si or Ge act as insulators, however at higher temperatures some of the covalent bonds are `broken and the electrons so released become free to move in the crystal and thus conduct electric current. This type of conduction is known as intrinsic conduction as it can be introduced in the crystal without adding an external substance.
Extrinsic Semiconductors : Semiconductors due to impurity defects. Si and Ge (group 14 elements) in pure state has very low electrical conductivity, it can be enhanced by the addition of traces of elements belonging to group 13 or group 15. Chemistry : Practical Application Introduction of group 13 and group 15 elements to the crystal lattice of group 14 elements produces ptype semiconductors' and n-type semiconductors. This process of addition of impurities is called doping. The Superconductivity Ordinary metals conduct electricity and their conductivity is around 106 ohm-1 cm-1. Materials are said to be superconducting when they offer no resistance to the passage of electricity. In this state the materials become diamagnetic and are repelled by magnets. This phenomenon was discovered by Kammerlingh Onnes in 1913 when he found that Mercury became superconducting around 4 K. Most metals become superconducting at very low temperatures (2K5 K). Many metal oxides have been found to become superconducting at fairly high temperatures. For e.g., YBa2Cu3O7 at 90 K, Bi2Ca2Sr2Cu3O 10 at 105 K etc. Possible applications of this phenomenon are in electronics, building magnets, leviation transportation and power transmission. A plot of resistivity (P) versus temperature (T) showing the drop to zero at the critical temperature (Tc) for a superconductor and the finite resistivity of a normal metal at absolute zero, drawn :
Critical Field Strength Superconducting material can be made non-superconducting by the application of large magnetic field. The Solids The minimum value of the field strength required to bring about this change is called the critical field strength (Hc); its value depending on the material and on the temperature. Similarly, if the current in the superconductor exceeds a critical current, the superconductivity is
destroyed this is known as Silsbee effect, the size of the critical current is dependent on the nature and geometry of the particular sample. The full form of SQUIDs. The two uses of these devices : SQUIDs : Superconducting Quantum Interference Devices. This device is extremely sensitive to changes in magnetic field and can measure voltages as small as 1018 V, current of 10-18 A and magnetic field of 10 -14 T. Squids are being used in medical research to detect small changes in magnetic field in the brain. Geologists employ these devices in prospecting for minerals and oil where deposits can cause small local changes in the earths magnetic field. The Quantum Mechanical Aspect of the Band Theory : The potential energy will vary periodically as an electron moves through the crystal. The Schrodinger equation for a free electron moving in X direction is
= 0 …(1) This, as we know leads to the solution y = …(2)
Let us now define, wave number K = Since the kinetic energy (E _ V) = p2/2m
applying de Broglie relation, we get `p =
E_V=
this gives
…(3)
Chemistry : Practical Application Substituting this in the Schrodinger equation, we obtain the general solution for a free electron moving at constant potential as
y(x) = Ce±iKx …(4) According to the band theory, however the potential energy V is not a constant but it is a function of x and the Schrodinger equation is written as
…(5) The solution of this equation was obtained by Bloch y(x) = uK (x) e±iKx …(6) where uK is a periodic function. When the electron is moving in the x direction the magnitude of uK (x) changes but has the same value at two points separated by the lattice spacing. This means uK (x + a) = uK (x) …(7) where a is the lattice constant. y (x) is known as Bloch function. The form of uK (x) depends on the direction in the crystal. Kronig and Penney made the assumption that potential energy is zero near the nucleus and is maximum Vo at the point half way between two neighbouring nuclei. Further, they assumed the product of the potential height and the width (Vow) as constant. They obtained the relation
cos Ka =
…(8)
where P = Now cos Ka can vary only between + 1 and - 1 if K is real. This implies that only certain values of a are allowed which in turn means that all values of energy E are not allowed. Graphical solution of eq. (8) is possible. It is seen that as energy increases, the width of the allowed energy zone increases. The width depends on the magnitude of P too. If P ® ¥, the allowed zone narrows to become discrete levels at aa = np where n is a positive integer when P ® 0, the forbidden region disappears and the electron behaves like a free electron.
The Solids Matched The graphs showing the fraction of the solid decomposed versus time :
(1)
(2)
(b) Initial slow gas evolution.
(3)
(c) Self catalysed.
(4)
(d) No induction period.
(1)(c), (2)(a), (3)(d), (4)(b). The Reactivity in Solids can be Improved : Increasing the Surface Area : This gives a larger area of contact between reactants and it reduces the diffusion path. Preparation of ternary oxides through single phase precursor : By using single phase precursor travel distance is brought down from several thousand Angstroms to about10A.
Addition of impurities : They introduce a distortion in the crystal structure thereby weakening the chemical bonds.
Chemistry : Practical Application Coupling effect : This term stands for combining the effect of one transformation in the solid with another reaction. Suppose instead of mixing two metal oxides and heating them to obtain a new single phase mixed oxide we heat their carbonates or the oxalates. The carbonates or oxalates will initially decompose to the respective oxides. These instantaneously formed oxides will react much more rapidly giving the product. Effect of high energy radiation : This effect is particularly important in the case of organic solids due to the formation of free radicals. Radiation induced polymerisation is also known. An intrinsic semiconductor has a band gap of 1. 5 eV. The wavelength of the electromagnetic radiation required to cause the material photoconducting, calculated :
Band gap = 1.5 × 1.602 × 10-12 erg =
l= = 8.272 × 10_5 cm = 8272 Å. Frenkel and Schottky point defects in crystals defined. The effects they have on the density of a crystal : Schottky Defects : These defects arise if some of the lattice points in a crystal are unoccupied. The points which are unoccupied are called lattice vacancies. These defects appear generally in ionic crystals in which the positive and the negative ions do not differ much in size, example sodium chloride and cesium chloride. Frenkel Defects : These defects arise when an ion occupies an interstial position between the lattice points. It appear in crystals in which the negative ions are much larger than the positive ions, example AgBr and ZnS crystals. Schottky defects lower the density of the crystal but Frenkel defects do not change the density significantly.
Transition Elements 4 Transition Elements The structures of the products formed in various reactions, keeping in view the 18-electron rule. h4C4H6Fe(CO)3 + HC1 ¾® h4C4H6HFe(CO) 3Cl
(h5cp)2Fe + HBF4 ¾® (h5cp)2FeH+ + BF4_ Addition of H+ does not involve any addition of electrons. Hence the 18 electron rule is obeyed. The structure of the product can be represented as
Chemistry : Practical Application From among the following reactions, the type of reaction involved viz., oxidative-addition, reductiveelimination, insertion or addition, identified : (i) [RhI3(CO)2CH3] _ ¾® [RhI3CO(solvent)(COCH3)] (ii) Co2 (CO)8 + H2 ¾® 2 [CoH(CO)4] (iii) Mn2 (CO)10 + Br2 - ¾® 2 MnBr (CO)5 (i) [RhI3(CO)2CH3] _ ¾® [RhI3CO(solvent)COCH3]_
It is a type of insertion reaction in which solvent is inserted while methyl migrates, (ii) Co2(CO)8 + H2 ¾® 2[CoH(CO)4] It is a type of oxidative addition reaction. In these reactions a low valent transition metal complex reacts with a molecule to give a product in which both the oxidation number and coordination number of the metal are increased. In the given reaction H has entered as the hydride ion and hence the oxidation number of cobalt has increased from 0 to + 1. Mn2(CO)10 + Br2 ¾® 2MnBr(CO)5 It is also an oxidation addition reaction, where Br_ has increased the oxidation state of Mn from 0 to + 1. The preparation and electrophilic substitution reactions of ferrocene. Preparation of Ferrocene : (i) 2C5H6 + 2N(Et)2H + FeCl2 ¾® Fe(C5H5)2 + 2[N(Et)2H2] +Cl_ Cyclopentadiene Ferrocene (in excess) (ii) Fe + R3NH +C1- ¾® FeCl2 + 2 R3N + H2 FeCl2 + 2C5H6 + R3N ¾® Fe(C5H5)2 + 2R3NH + C1_ Fe + 2C5H6 ¾® Fe(C5H5)2 + H2 Electrophilic Substitution Reactions The mechanism of electrophilic substitution reactions in ferrocene is illustrated below Transition Elements The attacking electrophile E + first interacts with Fe atom to give a p diene cationic intermediate which then loses a proton to give the final product, as shown
This mechanism explains why the rate of electrophilic substitution in ferrocene derivatives is proportional to the ease of oxidation of Fe atom. Nitration and Halogenation Direct nitration and halogenation leads to the decomposition of ferrocene. The processes are, therefore, carried out indirectly as described below Nitration:
Bromination :
Nitroferrocene on reduction with Pd/HCl gives amino ferrocene.
Chemistry : Practical Application Carboxylation : Ferrocene can also be carboxylated as shown
Vilsmeir Reaction : Ferrocene undergoes Vilsmeir reaction to yield ferrocene carboxyaldehyde which is a useful starting material for the preparation of other ferrocene derivatives.
Mannich Condensation : Ferrocene also undergoes Mannich condensation as shown below
The Vilsmeir reaction and the Mannich condensation are reactions which are shown by aromatic systems that are more reactive than benzene. It follows, therefore, that ferrocene is more aromatic than benzene. Alkyiation of ferrocene can also be carried out as shown below
Transition Elements Since the presence of an alkyl group increases the electron density on the C5H5- ring, hence, in the
presence of excess of ligand, the disubstitution of alkyl group also occurs in the same ring, as shown above. Friedel-Crafts Reaction : Ferrocene readily undergoes acylation as illustrated below
Ferrocene can also be alkylated in a similar manner. Structure of Ferrocene Ferrocene has a structure in which the iron atoms is sandwiched between two C5H5 organic rings. It has been observed by X-ray diffraction that in gas phase the structure of ferrocene is eclipsed while at low temperature the structure of ferrocene is staggered. The staggered configuration in the solid phase exists because of the crystal packing forces so that CC and HH repulsions between the two rings are minimum.
Chemistry : Practical Application The five pp AOs of the C5H5 ring combine linearly to give an equal number of MOs having symmetry a (i.e., one MO with no planar node), symmetry ep (i.e., two degenerate MOs having one planar node) and symmetry e2 (i.e., two degenerate MOs having two planar node). The five MOs of C5H5 system are presented in Fig.
The n molecular orbitals formed from the set of pn orbitals of the C5H5 ring. The MOs i.e., y1y2 etc. of one (i.e., upper) C5H5 ring then linearly combine with MOs of the same symmetry i.e., y'i, y'2 etc. of the lower C5H5 ring to given 10 MOs of resulting two C5H5 ring system. The ten MOs which are linear combinations of ligand orbitals of the two C5H5 ring ligand system then combine with orbitals of the metal of the same symmetry to give bonding and antibonding MOs of ferrocene. For (hC5H5)2 Fe, there are 18 valence electrons to be accommodated; 5p-electrons from each C5H5 ring and 8 valence shell electrons from the iron atom. These 18 electrons are arranged in the nine bonding and non-bonding MOs. As a result the 10th antibonding MOs remain vacant. Transition Elements
An approximate MO diagram for ferrocene In the ferrocene molecule, the most significant bonds are the bonds of p-symmetry formed between Fe atom and the two C5H5 rings. These are obtained by the overlap of dxz and dyz AOs of Fe with appropriate ligand group MOs of C5H5 ring. Ferrocene is diamagnetic in nature. Ferrocene has p-delocalised MOs and shows aromatic character. (If we consider the formation of ferrocene from Fe2+ and two C5H5_ anions containing six n-electrons which obeys the 4n+ 2 rule of aromaticity and not the radical.) Fe(C5H5)2 is more stable than Co(C5H5)2 or Ni(C5H5)2. This is due to the stability of Fe(C5H5)2 with 18 electrons compared to Co(C5H5)2 with 19 electrons and Ni(C5H5)2 with 20 electrons. Cobalocene and nikelocene have one and two electrons, respectively, in their ABMO whereas ferrocene has none. The electrons in ABMO are easier to remove so that cobalocene and nickelocene are easily oxidisable compared to ferrocene.
Chemistry : Practical Application Structure of Ruthenocene
For ferrocene the symmetry of the space group requires the two five membered rings to have staggered conformation. In ruthenocene and osmocene the rings adopt the eclipsed conformation.
[Co(H)(N2)(PPh3)3 ] :
[IrCl(CO)(O2) (PPh3)2] :
Structure of Fe2(CO)9 : This carbonyl is formed by the photolysis of Fe(CO)5. In the structure of this compound there is a weak coupling of the electron spins of Fe and Fe shown by dotted lines. This type of weak coupling of electron spins gives rise to a fractional single bond between two Fe atoms. These atoms are at a larger distance from each other than Transition Elements expected in a normal FeFe single bond. The complex is diamagnetic and obeys the inert gas rule. Each Fe
atom appears to have 36 electrons around itself (26e_ from Fe + 3e_ from three bridging COs + 6e_ from 3 terminal COs.
Structure of Mn2(CO)10 : In the structure of this complex each metal atom has an octahedral environment and the two Mn(CO)5 units are joined solely by an MM bond as shown. The equatorial CO groups of the two square planes are slightly bent towards each other. The Mn atom in half of the molecule has an effective atomic number of 35 (paramagnetic). One electron to become diamagnetic is provided by the MM bond. Thus, diamagnetism of bi-metal (or bi-nuclear) carbonyls can be explained on the basis of the formation of an MM bond in these carbonyls.
Structure of Zeise's Salt : It is a complex between platinum metal and ethylene, obtained when ethylene is passed through aqueous solution of potassium tetra chlroplatinate. [PtCl4]2_ + C2H4 ¾® [Pt(C2H4)Cl3] _ + Cl_ Zeise's salt In the structure of Zeise's salt, the ethylene occupies the fourth coordination site of the square planar complex with the CC axis perpendicular to the platinum-ligand plane. In this compound, the dsp2 hybridised s orbital of Pt overlaps with p-bonding molecular orbitals of ethylene. Simultaneously, the filled dp orbital of Pt overlaps with p* orbital of C2H4.
Chemistry : Practical Application
The structure of the anion in Zeise's salt, trichloro (ethylene) platinate(II) ion. Structure of Re2Cl82_ : The complex ion octachlorodirhenate(III) ion has short ReRe distance and eclipsed configuration of the chlorine atom Re(III) is a d4 species. The ReCl bonds may be considered tcrbe dative bonds from Cl_ ions to Re3+. The formation of one sigma bond, two pi bonds and one delta bond causes the pairing of four electrons in quadruple bond, hence the complex is diamagnetic.
Structure of [Mo(C5H5)3NO] : This molecule could be named (monohaptocyclopentadienyl) (trihaptocyclopentadienyl) (Pentahapto focyclopentadienyl) nitrosylmolybdenum or formulated as (h1—C5H5 )(h3—C5H5 )(h5C5H5)MoNO.
Transition Elements
Explain the nature of metal-olefin interaction : Organometallic complexes containing unsaturated ligands such as alkenes are considered olifinic complexes. Ziese characterised the pale yellow crystalline solid of composition K[PtCl3 (C2H4)] which is now known as Zeise salt. This compound was prepared by passing ethylene gas through aqueous solution of potassium tetracholoro platinate. PtCl42_ + C2H4 ¾® [PtCl3(C2H4)] _ + Cl_ In the olifin ligand, in addition to a s bond formed due to overlap of sp2 hybrid orbitals of two carbon atoms, the pz orbitals of the two carbon atom overlap to form a p-bonding and a p-antibonding molecular orbital. The p-bonding molecular orbital is occupied whereas p* antibonding molecular orbital is vacant. This type of bonding is shown in Fig.
Stepwise bonding in metal-olefin complex. The occupied rc;bonding molecular orbital of olefin then overlaps with empty s orbital of metal while one of the filled dn orbitals of the metal interacts with the empty p*-antibonding molecular orbital of olefin resulting in a p-symmetry about the Z-axis. Thus, the metal-olefin bond consists of two compounds Chemistry : Practical Application viz., s-donation from ligand to metal and simultaneous p-donation from metal to ligand. The metal carbon bond is, thus similar to metal-carbon bond in metal carbonyls. Both olefin and CO are weak s-donors but the presence of low energy empty p* MO on these ligands makes the dpp* bonding possible. The transfer of electrons from the ligand to metal during the formation of s bond is enhanced by simultaneous removal of charge from the metal through dpp* back donation. This effect is known as `synergic effect', would strengthen the MC bond.
The structure of K + [PtCl3(C2H4)] _ has been established to approximately square planar. The olefinmetal bond lies perpendicular to the plane of the molecule. In this compound dsp2 hybridised s-orbital of Pt overlaps with p-bonding MO of C2H4. Simultaneously, the filled dp orbital of Pt overlaps with p* orbital of C2H4. The two carbon atoms of C2H4 are equidistant from Pt. The ethylene group lies perpendicular to the PtCl bond which is trans to the C2H4 ligand. The entire Cl3PtC2H4 group in Zeise salt has C2v symmetry.
Transition Elements Alkyne Complexes Alkynes can be bound to metals using one or both sets of p-bonds, as 2, 3 or 4e_ donors. Acetylene bridges of several types are known. In acetylene there are two p-bonds at 90° to each other and both can be bound to a metal. The Co atoms and the acetylene carbon atoms forms a distorted tetrahedron and the C6H5 or other groups [Co2(CO)8(C2R 2)] on the acetylene are bent away as shown
There are also complexes where the alkyene is coordinated to only one metal atom and serves simply as the equivalent of the olefin or carbon monoxide ligand. Thus we have thereactions —
A third way of bonding, notably in Pt, Pd and Ir complexes is that shown below. In these the CC stretching frequency is lowered considerably, to the range 1750 to 1770 cm "` indicative of a CC doublebond.
Chemistry : Practical Application
The structure of (Ph3P)2 Pt (PhC2PH), in which diphenylacetylene is most simply formulated as a divalent, bidentate ligand. Many important reactions of acetylenes, especially with metal carboynyls, involve incorporation of the acetylenes into rings, thus producing species with new organic ligands bound to the metals. Some
examples are the following
Hydrogenation The first rapid and practical system for the homogenous reduction of alkenes, alkynes and other unsaturated substances at 25° C and 1 atm pressure used the complex RhCl(PPh3)3 known as Wilkinson catalyst. It dissociates to only a small extent at 25° C. RhCl(PPh3)3
RHCl(PPh3)2 + PPh3
Transition Elements Under hydrogen RhCl (PPh3)3 solution rapidly becomes yellow and 1H and 3IP NMR studies show that an octahedral dihydride is first formed but, due to the strong trans effect of H, this rapidly dissociates at room temperature to give a fluxional five-coordinate rhodium(III) species.
This species then coordinates alkene.
Simplified catalytic cycle for hydrogenation of C = C bonds by species derived from RhCl(PPh3)3 or from [(alkene)2RhCl]2 + PR3. Possible solvent coordination is disregarded. Cycle shows only major species involved for millimolar rhodium concentrations with large alkene concentrations under ambient conditions. The 14-electron species RhCl(PPh3)2 adds a molecule of hydrogen oxidatively to form the 5-coordinate, 16-electron dihydrocomplex. This in turn adds a molecule of olefin to form a 6-coordinate, 18-electron complex. Transfer of a hydrogen atom to the (b-carbon atom yields a 5-coordinate alkyl intermediate which then undergoes another rearrangement to give hydrogenated product and to reform the RhCl(PPh3) species.
Chemistry : Practical Application The process is very slow because of the formation of stable rhodium ethylene complex, which does not readily undergo reaction with H2. (Ph3P)3RhCl + H2C = CH2 ¾® (Ph3P)2RhCl(n2C 2H4) + Ph3P Catalytic Cycle for the Production of Acetaidehyde Wacker Process : The Wacker process is primarily used to produce acetaldehyde from the oxidation of ethylene by palladium(II)Copper(II) chloride solution. C2H4 + PdCl2 + H2O ¾® CH3CHO + Pd + 2HC1
Pd + 2CuCl2 ¾® PdCl2 + CuCl
2CuCl + 2HC1 +
C2H3 +
O2 ¾® 2CuCl2 + H2O
O2 ¾® CH3CHO
Oxidation of alkenes of the type RCH=CHR´ or RCH=CH2 gives ketones, for example, acetone from propene. The overall catalytic cycle is shown below
Proposed mechanism for the oxidation of ethylene to acetaldehyde in the Wacker process. Chloride ligands have been omitted. The oxidation number of palladium is + 2 at all stages of this cycle except the upper left where reductive elimination of acetaldehyde gives Pd (0), which is oxidized by Cu (II). The complete cycle for the reoxidation of Cu (I) is not shown. Transition Elements The hydration of alkenePd(II) complex (a) occurs by the attack of H2O from the solution on the concentrated ehtylene rather than the insertion of coordinated OH. Hydration, to form (b), is followed by two steps that isomerized the coordinated alcohol. First (b-hydride eliminatiorf occurs with the formation of (c), and then migratory insertion results in the formation of (d). Elimination of the acetaldehyde and a hydrogen ion then leaves Pd (0). The Pd (0) is
converted back to Pd(II) by the auxiliary Cu(II)-catalysed air oxidation. Acetone is prepared similarly from propene
CH3CH=CH2+PdCl2 +H2O
CH3_
CH3+Pd+2 HC1
Ziegler-Natta Polymerization The polymerization of olefins by catalysts of the Ziegler-Natta type represents a most important example of the insertion reaction. The Ziegler catalyst formed from TiCl4 and aluminium trialkyl. These component may react in a following way TiCl4 + A1(C2H5)3 ¾® A1(C2H5)2C1 + TiCl3C2H5 TiCl3C2H5 ¾® HC1 + C2H4 + TiCl2 TiCl3C2H5 + HCl ¾® C2H6 + TiCl4 TiCl2 + TiCl4 ¾® 2TiCl4 In the inert hydrocarbon solvents used an insoluble mixed halide-alkyl complex of aluminium and titanium is formed of variable composition. This material is the active catalyst for the polymerization of ethylene, persumably acting as a heterogenous catalyst. Natta and co-workers had produced stereospecific polymers. For example, olefins like propylene have been Chemistry : Practical Application polymerized in such a way as to yield long linear head to tail chains consisting of sequences of monomer units having the same steric structures. These polymers are called isotactic polymers and they crystallize easily, whereas those monomeric units of different steric arrangement phased at random do not crystallize well. These polymers are called atactic. Polymers of regular, alternating structure are called syndyotactic polymers.
Different types of polymers The Ziegler-Natta system is not homogenous, it is heterogenous and the active metal species is fibrous form of TiCI3 formed in situ from TiCl4 and AlEt3 but performed TiCl3 can be used. According to mechanism proposed by Cossee, the metal ion active site on the catalyst surface is octahedrally coordinated and which has a growing polymer chain R at one position and one position is vacant. Transition Elements The alkene is coordinated to vacant site and an insertion reaction occurs. After promotion of an electron from the Tialkyl bond to a molecular orbital of the complex, a four centre transition state is produced which enables an alkyl group to transfer to coordinated alkene. A further molecule of alkene is then bound to vacant site and the process is repeated. The mechanism is then as follows
The functions of aluminium alkyl are mainly (i) The formation of active metal species TiCl3. (ii) The replacement of one of the chloride ions in a Ti3+ ion at the surface by an alkyl radical derived from it. The stereoregular polymerization of propene may arise because of the nature of the sterically hindered surface sites on the TiCl3 lattice. An important extension of Ziegler-Natta polymerization is the copolymerization of styrene, butadiene and a third component such as dicyclopentadiene or 1, 4-hexadiene to give synthetic rubbers. Vanadyl halides rather than titanium halides are then used as the metal catalyst. Wilke and his coworkers have shown that zera-valent complexes, especially of nickel, obtained by reduction with aluminium alkyls can be used in a wide variety of polymerizations such as trimerization of butadiene to trans,tran, trans-cyclododecatriene.
Chemistry : Practical Application Hydroformylation Hydroformylation of alkenes (the `oxo' reaction) is their conversion into aldehydes by the action of hydrogen and carbon monoxide at 90°-200° C temperature and 100-400 atm pressure in the presence of Co2(CO)g.
Under these conditions Co2(CO)8 is converted into HCo(CO)4 and strong evidence for this as the important cobalt containing species in the overall reaction is provided by its ability to bring about hydroformylation at ordinary temperature and pressure. The main steps in the mechanism are as follows
HCo(CO)4
HCo(CO)3 + CO ...(i)
RCH = CH2 + HCo(CO)3
RCH2CH2Co(CO)3 ...(ii)
RCH2CH2Co(CO)3 + CO
RCH2CH2COCo(CO) 3 ...(iii)
RCH2CH2COCo(CO)3 + H2
RCH2CH2CHO + HCo(CO)3 ...(iv)
The eq. (ii) involves the b-hydrogen transfer to coordinated olefin, eq. (iii) the insertion of CO to form acyl and eq. (iv) the acyl cleavage by hydrogen. This process is frequently followed by reduction of the aldehyde to an alcohol. The cobalt system is difficult to study. The complex RhH(CO)(PPh3)3 is catalytically active even at 25° C and 1 atm pressure and in contrast to cobalt system, produces only aldehyde. The catalyst cycle is shown in Fig. The initial step is attack of the alkene on the 16-electron species RhH(CO)PPh3 (A) which leads to an alkyl complex (B) which then undergoes CO addition and insertion to form acyl derivative (C), which undergoes oxidative addition of molecular hydrogen to give the dihydrodoacyl complex (D). The high concentration of PPh3 that are essential to provide high yield of linear aldehyde are probably required in order to suppress dissociation and the formation of monophosphine species and to force the associative attack of olefin on species (A). Transition Elements
Catalytic cycle for the hydroformylation of alkenes involving triphenylphosphine rhodium complex species. Note that the configurations of the complexes are not known with certainty. A wide variety of unsaturated substance can be hydroformylated by cobalt or rhodium catalysts but conjugated alkenes (e.g., butadiene) may give a number of products including hydrogenated monoaldehydes. The mechanism is different, since addition of MH to. dienes leads to alicyclic species which may be present as a-bonded intermediate or as h3allyls.
Hydroformylation can also be achieved complexes such as Ru(CO)3(PPh3)2 and PtH(CO)(SnCl3) (PPh3)2.
Chemistry : Practical Application Isomerization and Racemization of Tris-chelate Complex Tris-chelate complexes exist in
enantiomeric configuration ^ and D about the metal atom, and when the chelating ligand is unsymmetrical, there are also geometrical isomers, cis and trans. Each geometrical isomer exists in enantiomeric forms; thus there are four different molecules. In the case of tris complexes with symmetrical ligands, the process of inversion (interconversion of enantiomers) is important. When the metal ions are of the inert type, it is often possible to resolve the complex; then the process of race-mization can be followed by measurement of optical rotation as a function of time. Possible pathways for racemization fall into two broad classes; those without bond rupture and those with bond rupture. There are two pathways without bond rupture. One is the trigonal or Bailor twist and other is rhombic, or Ray-Dutt, twist shown in Fig. Some cis-M(CO)4(PR3)2 complexes are believed to isomerize in this way.
Three possible modes of intramolecular racemization of a tris-chelate complex, (a) trigonal twist, (b) rhombic twist, (c) one of several ring-opening paths. Transition Elements There are several possible dissociative pathway, in which a five-coordinated intermediate of either tbp or sp geometry are formed. One of them is shown below. There are no evidence for the actual occurence of associative pathway through seven coordinate intermediate.
It was found that both the isomerization and the racemization are intramolecular processes, which occur at approximately the same rate and with activation energies that are identical within experimental error. It thus appears likely that the two processes have the same transition state. This exclude a twist mechanism as the principal pathway for racemization. Moreover, it was found that isomerization occurs mainly with inversion of configuration. This imposes a considerable restriction on the acceptable pathways. Detailed consideration of the stereochemical consequences of the various dissociative pathways, and combinations thereof, leads to the conclusion that for this system the major pathway is through a tbp intermediate with the dangling ligand in an axial position as in Fig. (a) Isomerization There are many instances of the isomerization of alkenes or alkenilic compounds under the influence of transition metal compound catalyst. Some examples arepent-1-ene into cis- and trans-pent-2-ene (RhCl3 in methanal or H2PtCl6 + SnCl2); hex-1-ene into hex-2-enes and hex-3-enes (Fe 3(CO)12); penta-l,4-diene into trans-penta-l,3-dieneFe(CO)3Fe(CO)5 ; allyl alcohol into propionaldehyde (Fe(CO)5 or HCo(CO)4) propylene oxide into acetone (Co2(CO)8 in methanal) and isopropyl magnesium bromide into «-propyl compound (TiCl4). In these processes it is believd that the double bond reacts with transition metal complex, giving an intermediate which decomposes with the regeneration of the double bond in a new position. Possible mechanism of such changes are shown below
Chemistry : Practical Application
Isomerization normally leads to the isomer or mixture of isomers predicted by thermodynamic considerations. For the isomerization of allyl alcohol to propionaldehyde by HCo(CO)4 the following
mechanism has been suggested
CH2 = CHCH2OH + HCo(CO)4 ¾®
¾® Evidence in support of this mechanism is provided by using the deuteride DCo(CO)4 as catalyst, when CH2DCH2CHO is the only deuterium-containing organic product of the reaction. In the isomerization of isopropylmagnesium bromide the mechanism appears to be (CH3)2CHMgBr
(CH3)2CHTiCl 3 ¾® TiCl3H + CH3CH = CH2
CH3CH = CH2 + TiCl3H
CH3CH2CH2TiCl 3
Transition Elements CH3CH2CH2TiCl 3 + (CH3)2CHMgBr
(CH3)2CHTiCl3
+ CH3CH2CH2MgBr Whether the True or False "The NiC bond length in nickelocene is longer than the FeC bond length in ferrocene." Statement is True : Ferrocene [(cp)2Fe] has 18 electrons in the molecular orbitals : six electrons are from Fe2 + and six electrons from each of the two cyclopentadiene anions. Nickel has two electrons more than iron so nickellocene [(cp)2Ni] has 20 electrons. The additional electron in nickellocene are in an antibonding orbital. The bond order of nickellocene is higher than that of ferrocene hence NiC bond length in nickellocene is larger (2.20 A) than the FeC bond length in ferrocene (2.06A).
Structure of Co4(CO)12 : One metal atom has three terminal CO groups, but the remaining nine CO groups occupy both symmetrical bridging positions and terminal positions around the triangle formed by the other three metal atoms. The molecule has C3V symmetry.
Isolobal Relationship : Two molecular fragments are isolobal if the number, symmetry properties, shapes and approximate energies of their frontier orbitals are the same. They may or may not also be isoelectronic. If the two fragments are isoelectronic, in the sense that the ratio of the number of electrons in frontier orbitals to the number of frontier orbitals is same, then the net charges on the two species will be the Chemistry : Practical Application same. The orbitals whose similarity is critical in determining isolobality are called frontier orbitals. Cobalt fragment Co(CO)3 is isolobal with CH.
The cluster formed after replacing Co(CO)3 fragment by isolobal CH is Co3(CO)9CH, with the structure:
n-acidity : The ability of a ligand to accept electron density into vacant n orbitals is called n-acidity. Complexes of metals with carbon monoxide are called metal carbonyls. Oxidation state of metal in these complexes is either zero or a low positive value. There are no attractive interactions between the metal and the ligands as is possible with the positively charged metal ion. It is the main characteristic of CO ligand that it can stabilize low oxidation states. This is due to the fact that it possesses vacant p orbitals in addition to lone pairs. The formation of a sigma bond by the donation of a lone pair of electrons into the suitable vacant metal orbitals leads to excessive negative charge on the metal (in zero or negative oxidation state). To counter the accumulation of negative charge on the metal, a p-bond is Transition Elements formed by the back donation of electrons from the filled metal orbitals into the vacant p-type orbitals on the ligand. This also supplements the s-bond. This ability of the ligand (CO) to accept electron density into vacant p orbitals is called p-acidity. Therefore, CO is also called a p-acceptor ligand and the metal carbonyls are referred as complexes of p-acceptor (or p-acid) ligands. Vibrational Spectra IR has given valuable support for bonding in metal carbonyls. These studies provide information regarding bond orders of MC and CºO bonds. The decrease in CO bond order or force constant is estimated by studying the CO stretching frequency in IR spectroscopy. The IR spectra is characterised by frequency of vibration, which is related to force constant (k) as
where m is reduced mass of bonded atoms with mass m1 and m2 and is given by The force constant is a measure of bond strength. The larger the force constant, the stronger the bond and higher the vibrational frequency. In IR spectrum CO frequencies are generally very strong and, therefore, can be used to study the force constant or bond strength of CO bond in different bonding situation. Free CO has IR stretching frequency of 2143 cm-1. In the case of terminal CO groups in neutral molecules, the CO frequency has been found to be 2125-1850 cm-1. This suggests that the bond strength of CO has decreased or MC bond strength has increased. This is only possible if there is back bonding from filled metal d-orbitals to the antibonding orbitals of CO.
Chemistry : Practical Application
Possible Isomers of cp2Fe2(CO)4 Geometrical Isomers of cp2Fe2 (CO)4 :
=17. Thus the effective atomic number of each Mo The total number of electrons per Mo atom atom is one short of 18. So there is an MoMo bond and the bond order is one. Assuming the coordination number of Mo as 6, Mo2(cp)2(CO)6 has bridge structure, the cis and trans isomers can be drawn as
Propose a Structure
Transition Elements
(i) Treatment of Cr(CO)6 with LiCH3, followed by [(CH3)3O]BF4, yields the carbene complex
. Propose a mechanism for this synthesis. (ii) Ethylenecanoxidisedby aqueous solution of PdCl42-: C2H4 + PdCl42- + H2O —® Pd + 2H+ + 4C1- + CH3CHO The rate law is
Propose a plausible mechanism for the reaction.
(i) (ii) PdCl42-+C2H4 +H2O —® PdCl2(C2H4)OH -+2Cl-+H+
The slow step may involve the formation of an unstable intermediate of the type C12(H2O) PdCH2CH2OH-. Reasonable product for the following reactions, predicted:
(i) (ii) Co(CO3)NO + PPh3 (iii) Mo(CO)6 + CH3CN
Chemistry : Practical Application (iv)
(v)
(vi) (vii) Mn(CO)5Br + [Mn(CO)5]-
(i)
(ii) Co(CO)2(NO)(PPh3)
(iii) Mo(CO)3(CH3CN)3 (iv)
(v)
(vi)
(vii) Mn2(CO)10 + BrThe product and state which reactions are oxidative addition and which are insertion reaction, predicted : (i) cp(CO)3Re + Br2 —® (ii) trans-(PPh3)2Ir(CO)Cl + H2 —® (iii) cp (CO)2FeCH3 + PPh3 —® (iv) EtRe(CO)5.+ CH3CN —® (i) cp(CO)2ReBr2 + CO — Oxidative addition Re (I) ® Re (III) Transition Elements
(ii) (iii) cp (CO)(PPh3)FeCOCH3 Insertion reaction (iv) (cis)-(CH3CN)Re(CO)4(COEt) Insertion reaction The formal oxidation state and d-electron count of the metal in the following complexes (a) (h6C6H6)2 Mo, (b) cp2ZrCl (OMe) (c) [(PMe)3Pd(h3C3H 3)] +, (d) (CO)5ReEt
(a) Mo(O), d6; (b) Zr(IV), d°; (c) Pd(II), d8; (d) Re(I), d6 Favourable Factors for the Oxidative-addition Reactions In general, the tendency of a complex to undergo oxidative-addition reactions is governed by the amount of electron-density at the metal i.e., by the ease with which the metal can be oxidized. Thus, several general trends have been observed (i) The presence of electron rich ligarids in the coordination sphere of the metal increases the rate of oxidative-addition. (ii) A low initial oxidation state of the metal is more favourable for oxidative-addition to occur e.g., all other factors being equal, Fe (0) is easier to oxidize than Co (I), which is easier to oxidize than Ni (II), even though the d-electron configuration is same for these metals. (iii) The tendency to oxidative-addition to occur increases down a given group e.g., Ir (II) is easier to oxidize than Rh (II), which is easier to oxidize than Co (II). (iv) Oxidative-addition occurs more readily in coordinatively unsaturated systems.
Chemistry : Practical Application The total number of electrons, determine the number of MM bonds present, and predict a structure, indicated : (a) mCO[(h4C4H4)Fe(CO)] 2, (b) mCOmCRR´ [cp*Rh]2, (c) [mXmCH2(Os3(CO)10)] - where X = halide, (d) (mBr)2[Mn(CO)4]2. (a) mCO 2 electrons neutral 2 (h4C4H4) 8 electronsneutral 2 CO 4 electrons neutral
2Fe(0) 16 electrons neutral Total 30 electrons MM bonds = number of metals (n) × 2 - total no. of electron in complex/2 = [18(2) - 30]/2 = 3 FeFe bonds
(b) mCO 2 electrons neutral mCRR' 2 electrons neutral 2[cp*]_ 12 electrons -2 2 Rh (I) 16 electrons +2 Total 32 electrons MM bonds = [18(2) _ 32]/2 = 2 RhRh bonds
Transition Elements (c) [mX]- 4 electrons - 1 mCH2 2 electrons neutral
10 CO 20 electrons 3 Os (0) 24 electrons neutral Total 50 electrons MM bonds = [18(3) - 50]/2 = 2 OsOs bonds
(d) 8 CO 16 electrons neutral 2 mBr 8 electrons - 2 2 Mn (I) 12 electrons +2 Total 36 electrons MM bonds = [2(18) _ 36]/2 = 0 MnMn bonds
For each of the following metal and ligand combinations, the simplest neutral compound that conforms to the 18-electron rule. Draw a reasonable structure, formulated : (a) Fe, CO, COT = cyclooctatetraene; (b) Fe, cp, CO. (a) (CO)3Fe(COT) 3 CO 6 electrons neutral Fe (0) 8 electrons neutral
COT 4 electrons neutral Total 18 electrons
Chemistry : Practical Application
(b) [cp (CO)2Fe]2 4 CO 8 electrons neutral 2 cp- 12 electrons -2 2 Fe(H) 12 electrons +4 Total 34 electrons plus one Fe—Fe bond
The structures of isomers of complex dibromodicarbonyl cyclopentadienyl-rhenium drawn: This is an example of geometrical isomerism in five coordinate complex, having square pyramid structure.
Transition Elements Factors affecting the CO stretching frequency in the infra-red spectrum of metal carbonyl derivatives : The bonding in metal carbonyl complex is depicted below
The pi bond between the metal and carbon atom arises as a result of a backbonding interaction between a filled metal rf-orbital of the appropriate symmetry and the empty pi star orbital of the carbonyl. Any factor that increases the electron density at the metal will increase the strength of the MC pi bond, and thus the contribution from resonance form b. The result is a decrease in the CO bond order and a resulting decrease in the CO stretching frequency in the infra-red spectrum of the complex. Aspects which affect the DR. spectrum of metal carbonyl derivatives include the cf-electron count of the metal, the donor properties of the ligands and the charge on the complex. For example, in case of [W(CO)5C1]- and Re(CO)5Cl, both metals are d6 but since [W(CO)5C1]- is anionic, it will have a lower CO stretching frequency. Similarly, in case of Fe(CO)5 and Fe(CO)4Br2, Fe(CO)5 will have a lower CO stretching frequency because being d8 it has more electron density at the metal centre than the d6 in another because of the lower oxidation state of the metal. In the NMR spectrum of a metal complex at - 97° C, there are two signals, one at 182 ppm and 178 ppm. At 58° C the signals have just coalesced into a broad signal at 180 ppm. The rate constant for exchange of the two carbon atoms that produce these signals is :
Chemistry : Practical Application At the coalescence temperature, the equation for the rate constant is
The difference in chemical shift of the two signals is 4 ppm, but the formula requires that this shift difference be in Hz. This value depends on the strength of the magnet used for the measurement and can be obtained by multiplying the operating frequency of the magnet (in MHz) and the difference in chemical shifts in ppm. Thus, for this example, we obtain a value of Dn = 4 ppm (25 MHz) = 100 Hz and the value of the rate constant at _ 58° C is calculated tobe
k= k = 2.2 × l02s_1 If a magnet of different field strength were used for this same measurement, the coalescence temperature would also be different. Transition Elements The fact that D0 increases in the order CrCl33_, Cr(NH3)63+, Cr(CN)63_, rationalised : Inspire of negative charge of Cl_, D0 for Cl_ is less than D0 for NH3 because of the repulsive interaction between the pn electrons of Cl_ and the d p electrons of Cr3+. Do for CN_ is very high because of back bonding ie., the shift of electron density from the dp orbitals of Cr3+ to the antibonding p* orbitalsofCN_. The latter interaction stabilizes the dp orbitals and increasesD0. (A ligand such as H_, NH3 or CH3_ can engage only in a bonding. However, a ligand such as Cl_, O2_ or CO32_ has electrons in essentially non-bonding p-orbitals of the donor atoms which can interact with appropriate d-orbitals of the metal atom. Similarly, a ligand such as CN_, CO or PR3 has empty p-orbitals which can interact with d-metal orbitals. In Transition Elements the case of CN_ and CO, the empty p-orbitals are the antibonding p-MOs of these diatomic species.)
Molecular orbital energy level diagram for an octahedral complex with only o bonding.
Energy level diagram showing the interaction of the t2gd orbitals with filled p-orbitals of the ligands. The non-bonding ligand p-orbitals are not shown. Notice that the interaction tends to decrease D0.
Energy level diagram showing the interaction of the t2gd orbitals with empty p-orbitals of the ligands. The ligand p-orbital combinations of other symmetries are not shown. Notice that the interaction tends to increase D0.
Chemistry : Practical Application An explanation for the fact that MnO (NaCl structure) has no net magnetic moment : ZrCI4, Diamagnetic NiCl42_, Ti(CO)44+, Cd(CO)3, Fe3(CO)12, AuF52_ ? Why ? MnO is antiferromagnetic. The crystal contains two interpenetrating sets of Mn2+ ions. In one set, all the magnetic moments are aligned in one direction, in the other set, the magnetic moments are aligned in the opposite directions. Spin polarization of the oxide ions is believed to aid the magnetic coupling between the two sets of Mn2+ ions. Diamagnetic NiCl42_ could be square planar; such a structure would be unstable with respect to tetrahedral form, probably because of repulsions between the chlorines. Ti(CO)42_ is unstable because there is so dn electrons to engage in back bonding. Cd(CO) 3 would be expected to be unstable because delectrons are held so tightly (high nuclear charge of Cd) that they cannot engage in back bonding to the CO group. AuF5 is unstable with respect to AuF4_ because fluoride ions cannot engage in back bonding and cannot remove the negative charge which builds up on the Au atom when five ligands are coordinated to it. The d-orbital splitting diagram for a square pyramidal and a trigonal bipyramidal complex:
Transition Elements d-orbital Splitting in Square Pyramidal Complex : Square pyramidal has C4V geometry.
In this pyramid base in xy plane and energies of various orbitals in Dq units are (_0.86),dxz and dyz(_ 4.57).
(9.14), (0.86), dxy
d-orbital Splitting in Trigonal Bipyramidal Complex: The trigonal bipyramidal has D3n symmetry.
Pyramid base in xy plane and energies of various orbitals in Dq units are (7.07), and dyz (_2.72).
and dxy (_0.82), dxz
The complex [Fe(H2O)6]2+ displays two overlapping absorption bonds at _1000 nm. An explanation. [Fe(H2O)6]2 + is a high spin d6 ion, the ground state of d6 ion is 5D. The ground state energy level 5T2g originates from the configuration f2g4 eg2. The excited energy level 5Ef originates from t2g3 eg3. The absorption band observed is, therefore, assigned to sT2g ¾® 5Eg. The band appears to be composite one due to splitting of excited state energy level 5Eg into two non-degenerate energy levels due to John Tellar distortion. The explanation of quenching of orbital angular momentum on the basis of crystal field theory: Quenching of the Orbital Angular Momentum : As electrons will have orbital momentum around the given axis (say az-axis), if the orbital that it occupies can be transformed into an entirely equivalent orbital (which should be degenerate also with it) by a simple rotation around the axis. The electron in the orbital in a free ion will have a orbital angular magnetic momentum of ±2h/2p units about the z-axis as rotation by 45° will carry it into an equivalent dxy orbital. Similarly, a 90° rotation of the da orbital around the z-axis will transform the orbital into the degenerate dyx orbital. Hence, a dzx or dyx orbital has an angular momentum of ± 1 (h/2p) units around the z-axis. An electron in the any orbital
orbital will not have
Chemistry : Practical Application momentum along the z-axis as it cannot be transformed into
or dzx etc.orbitals by simple rotation.
In an octahedral ligand field, only t2g orbitals remain degenerate and rotationally related. The Eg orbitals get separated by 10 Dq.
Hence, orbital momentum due to the should apply.
orbital electron get quenched, and the spin-only formula
It can be seen that the orbital angular momentum formula should be important for the high spin d1, d2, d6 and d7 ion complexes and low spin d4, d5 ions in octahedral field. In the tetrahedral field, the high spin d3,
d4, d8 and d9 ions should have a significant contribution from the orbital angular momentum. The magnetic moment of [CoCl4]2_ [4.4 BM] and that of [Co(H2O)6]+2 (5.0) confirm the above statements, the orbital moment contributes for the high spin octahedral, but not for the tetrahedral complexes. Even for the other ions, where no orbital moment is expected, the observed values significantly depart from the spin-only formula (though the differences are small). This is attributed to the spin-orbital interactions which oppose the quenching of the orbital moments by mixing the orbitals. This mixing (or hybridization) of orbitals, separated by energy difference D, changes the magnetic momentto
s is a constant depending on the spectroscopic state and the no. of d electrons, being two for 2D or 5D; four for 3F or 4F and zero for 6S. l is the spin-orbit coupling constant for different J states and related to the spin-orbit coupling constant as given (it is positive for less than half filled subshell and negative for more than half filled subshell). This explains the generally, lower meff values obtained for Cr2+, Cr3+, V3+ and V+ and higher values for spin free Fe2+, Co2+, Ni2+ and Cu2+ complexes. Greatest Transition Elements deviations occur for the Co2+ and Fe2+ complexes, for which unquenched orbital moments contribute significantly. The observed magnetic moments for the metals in t2g ground state are temperature dependent and usually depart from the ms values due to probably the t2g electron delocalizations and lower symmetry ligand field components. The structures of : (i) Perovskite, CaTiO3, (ii) Rutile, (iii)CuCI. (i) Perovskite : The mineral perovskite, CaTiO3, has a structure in which the oxide ions and the large cations (Ca2+) forms a ccp array with the smaller cation (Ti4+) occupying those octahedral holes formed exclusively by oxide ions. This structure is adopted by many ABO3 oxide ion.
(ii) Rutile : The rutile structure, named after mineral TiO2, is very common among oxides and fluorides of the MF2 and MO2 types, where the radius ratio favours coordination number 6 for the cation. Each Ti4+ is octahedrally surrounded by six O2_ ions and each O2_ ion has three Ti4+ ions round it in a plane triangular arrangement.
(iii) CuCl : CuCl has a covalent, tetrahedral zinc blende [ZnS] structure with the metal atom tetrahedrally surrounded Chemistry : Practical Application by halogen atoms. Cu+ ions occupy half of the suitable tetrahedral sites in a face centered cubic lattice formed by Cl_ ions. Thus each Cl_ ion is also surrounded tetrahedrally by four Cu+ ion.
In vapour state, CuCl is polymeric and contains a cyclic ring of alternating Cu and Cl atoms. The formation of cadmium (II) complexes with chloride anion exhibits the successive equilibrium constant K1 = 1.56, K2 = 0.54, K3 = 0.05 and K4 = 0.46, an explanation. The equilibrium constant for the formation of this complex can be represented as follows Cd2+ + Cl_ CdCl+ + Cl_
CdCl+ K1 = 1.56 CdCl2 K2 = 0.54
CdCl2 + Cl
CdCl3_ K3 = 0.05
CdCl3_ + Cl_
CdCl42_ K4 = 0.46
From these data it can be seen that CdCl2 is very stable with respect to disproportionation of CdCl+ and CdCl3_. The reason may be an abrupt change in coordination number and hybridization in the sequence of complexes. 10 ml of a 0.001 M metal solution and 10 ml of a 0.003 M ligand solution were mixed. The final concentration of the metal complex ML3 was 0.000499 M. The stability constant of the complex, calculated : Transition Elements The reaction can be written as Mn+ + 3L
[ML3]n+
Initial 0.001 M 0.003 M 0M Equilibrium 0.001-0.000499 0.003 _ 3 × 0.000499 0.000499M = 0.000501 M 0.003 _ 0.001497 = 0.001503 The stability constant of the complex is given by
K=
= = 0.29335 × 109 = 2.9335 × 108 The concentration of [Cu(H2O)4]2+ (represented by [Cu2 +]) in 1.0 L of a solution made by dissolving
0.10 Cu2+ in 1.00 M aqueous ammonia Ks = l × 1012, calculated : Cu2+ + 4NH3
[Cu(NH3)]2+
The reaction between Cu2+ and NH3 goes almost to completion as is suggested by large magnitude of the formation constant. Thus the equilibrium concentration of tetraamine complex is approximately 0.10 M (If x mol/L Cu2+ ion still remain uncoordinated, the concentration of the complex will be 0.10 _ x). With the formation of 0.10 M Cu(NH3)42+, which requires 4 mol NH3 per mol complex, the ammonia concentration becomes 1.00 _ 4(0.10) = 0.60 M (Dissociation of the complex provides 4x mol NH3 per x mol Cu2+).
Chemistry : Practical Application initial used up produced equilibrium [Cu2+J 010 0.10 - x x [NH3] 1-00 4(0.10 - x) 0-60 + 4.x [Cu(NH3)42+] 0.00 0.10 - x 0.10 - x
x = 8 × 10_13 M The ferromagnetism and antiferromagnetism, explained : Ferromagnetism and Antiferromagnetism : Ferromagnetism and antiferromagnetism are cooperative phenomenon and are, therefore more likely to occur where there is a high concentration of paramagnetic species. A distinction is therefore, made between magnetically concentrated and magnetically dilute materials. In magnetically dilute substances i.e., in which the metal ion containing unpaired electrons are isolated from one another by a cluster of ligands, each paramagnetic metal ion will act independent of other such ions because the distance between the metal ions is too large to allow the paramagnetic spin of one to interact with that of the other ion. But, this is not the case in magnetically concentrated substances
i.e., in which the paramagnetic ions allow their total electron spins to interact with one another through exchange interaction. If it is assumed that only the spin magnetic moment contributes towards meff of a magnetically concentrated substances, then DE, the energy of spin interaction of paramagnetic ion is given by DE = 2J (S)i(S)k where (S)i and (S)k are the total electron spin of the ith and £th metal ion of the substance. J is known as exchange coupling constant. J is a measure of the strength of interaction between the total electron spin of the metal ions. If J is positive, then the total electron spin (S)s of all the paramagnetic ions in the magnetically concentrated substance in ground state get aligned in the same direction. This is called ferromagnetism. Transition Elements If J is negative, then total electron spins (S)s of all the paramagnetic ions in magnetically con-centrated substance in ground state are paired up. This is called antiferromagnetism. The ferromagnetic alignments of spins produce c greatly in excess of that produced by normal paramagnetic alignments. c in ferromagnetic substances is dependent on the strength of magnetic field applied. The ferromagnetic behaviour exists upto curie temperature or curie point. The total electron spin vector (S)s of atom or ions in a ferromagnetic material are all coupled parallel to one another in groups or domains. In presence of magnetic field the magnetic moments of all the domains start aligning parallel to the direction of the applied field. This tendency is opposed by thermal agitation which tends to randomise the magnetic moments of domains. However, the thermal energy is not always able to randomise the magnetic moment once these have been aligned. This results in the retention of magnetic moment of domains in a ferromagnetic substance even after the external field is removed. This is known as hysteresis. Antiferromagnetic spin exchange interactions cause a lowering of c. In most of the antiferromagnetic substances, the electron spin interactions leading to the pairing of total electron spins of neighbouring metal ions occurs through the intervening diamagnetic atom or ions. This type of total electron spin exchange occurs through linear O2_ or F_ bridges lying between the paramagnetic metal ions of a ferromagnetic substance (Fig.)
Antiferromagnetism is very much temperature dependent. For antiferromagnetic substances there is a characteristic temperature, known as Neel temperature or Neel point above which the antiferromagnetic substance behave as a normal paramagnetic substance.
Chemistry : Practical Application The magnetic moments from magnetic susceptibilities : When a substance is subjected to magnetic field H, and a magnetization I is induced. The ratio I/H is called the volume susceptibility k and can be measured by a number of techniques, including Gouy balance method, Faraday method and an nmr method. The volume susceptibility is related to the gram susceptibility c and the molar susceptibility cm.
…(1) where d and M are the density and mol. wt. of the substance. For a paramagnetic substance k, c and cm are positive quantities. The effective magnetic moment meff is calculated from the relation
meff = 2.83
...(2)
where T is the absolute temperature and the constant 2.83 is obtained from quantum mechanical calculations cpara = NA m2eff/3 kT ...(3) NA = Avogadro number, k = Boltzmann constant. On substituting the values we get the equation(2). Magnetic susceptibility would be independent of the strength of the external field applied. However, magnetic susceptibility would depend on temperature since the thermal agitation would tend to randomise the atomic or molecular magnetic dipoles which get aligned in the direction of the magnetic field applied. The magnetic susceptibility of a paramagnetic substance, cpara would, therefore, decrease with increase in temperature. The variation of cpara with temperature is expressed in the form of Curie law as
cpara =
where c = Curie constant
The Curie law is obeyed with great accuracy by some systems such as [FeF6]3_. However, many paramagnetic material deviate slightly from the ideal behaviour, and obey Curie-Weiss law
Transition Elements
cpara =
where q = Weiss constant
The origin of the intense colour : (i) [Cu(b py)] bpy = 2, 2´ = bipyridine (ii) K3[RuCl6] (i) [Cu (bpy)] : bpy or 2, 2' bipyridine is a p-acceptor ligand. The origin of intense colour of the complex is Metal ® Ligand charge transfer spectra as the ligands have empty p-antibonding orbitals. Thus the bonds responsible are mainly dp* charge transfer bands. Charge transfer processes from metal to ligand are favoured in complexes that have occupied metal-centred orbitals and vacant low lying ligand-centred orbitals. (ii) K3[RuCl6] : The origin of the intense colour in this complex is Ligand ® Metal charge transfer spectra charge transfer transitions tn1u, nt2u ® t*2g occur along with other charge transfer transition tn1u, tn2u ® eg. A charge transfer process is similar to redox process. The energy of transition measures the tendency of the redox reaction to proceed. As the charge transfer bands are neither multiplicity forbidden, nor Laporte forbidden, they have high absorption intensities. The unusual positive oxidation states of Mn : Manganese shows a number of oxidation states due to negligible energy difference between ns and (n _ 1)d orbitals, so electrons from both the orbitals can participate in chemical reaction. The outer electronic configuration of Mn is 3d5 4s2. The most stable oxidation states of Mn are + II, + IV and +VII. The unusual positive oxidation states shown by Mn are + I, + III, + IV and + VI. Other than this Mn shows oxidation states 0, _1, _II and _III. Mn shows (_ III) oxidation in the compound [Mn(NO)3CO]. The examples of compounds in various oxidation are given below in table
Chemistry : Practical Application Oxidation States and Sterochemistry of Manganese Oxidation Coordination Geometry Examples state number
Mn_III 4 Tetrahedral Mn(NO)3CO Mn_II 4 or 6 Square [Mn(phthalocyanine)]2_ Mn_I 5 tbp Mn(CO)5_, [Mn(CO)4PR3]_ 4 or 6 Square [Mn (phthalocyanine)]_ Mn° 6 Octahedral Mn2(CO)10 Mn1, d6 6 Octahedral Mn(CO)5Cl, K5[Mn(CN)6], [Mn(CNR)6] + MnII, d5 2 Linear Mn[C(SiMe3)2]2 4 Tetrahedral MnCl42_, MnBr2(OPR3)2, [Mn(CH2SiMe3)2] n 4 Square [Mn(H2O)4]SO4.H 2O, Mn(S2CNE2)2 5 Distorted tbp MnBr2[(MeHN)2CO]3 6 tbp Octahedral [Mn(trenMe6)Br]Br 7 NbF72_ structure [Mn(H2O)6]2 +, [Mn(NCS)6]4_ Pentagonal [Mn(EDTA)H2O)2 _ d bipyramidal MnX2(N5 macrocycle) 8 Dodecahedral (Ph4As)2Mn(NO3) 4 Transition Elements Contd... Oxidation Coordination Geometry Examples state number
MnIII, d4 4 Square [Mn(S2C6H3Me) 2]_ 5 sp MnX sal2en, [bipyH2]MnCl5 5 tbp MnI3(PMe3)2 f 6 Octahedral Mn(acac)3, [Mn(ox)3]3_, MnF3(distorted), 7 Mn(S2CNR2)3 [Mn(EDTA)H2O]_, MnH3(dmpe)2 MnIV, d3 4 Tetrahedral Mn (l-norbornyl)4 6 Octahedral MnO2, MnMe4(dmpe), MnCl62- Mn(S2CNR2)3 Mnv, d2 4 Tetrahedral MnO43_ MnVI, dl 4 Tetrahedral MnO42_ MnVII, d° 3 Planar MnO3+ 4 Tetrahedral MnO4_, MnO3F
Chemistry : Practical Application Zinc sulphide forms a cubic unit cell of length 6 × 10_10 m. Zinc ions form a face centred cubic lattice and sulphide ions occupy the centre of the alternate small cubes : (a) find the position of zinc and sulphide ions, (b) assign the coordinates, (c) effective number of Zn and S ions and formula of zinc sulphide, (d) density of ZnS if the atomic weight of Zn and S are 65.4 and 32 g mol-1, (e) Smallest distance between zinc and sulphide ion,
(f) coordination number of each ion. The structure of zinc sulphide crystal is face centred cubic lattice.
(a) The • represents zinc ion and x represents sulphideions. (b) The cordinates of zinc ions are
For sulphide ions are
(c) Effective number of zinc ions in unit cell are
=4 Effective number of sulphide ions are 4 × 1 = 4 Transition Elements So the molecular formula is ZnS. (d) Density is given by the formula where,
P= where, Z = Number of particles per unit cell M = Atomic mass in kg mol-1 a = Edge of the unit cell N0 = Avogadro number
= = 2.995 × 103 kg m3
(e) The smallest distance between zinc and sulphide ions is abcdefgh
of the body diagonal of small cube
(fb)2 = (db)2 + (df)2
= (3×10-I0)2+[(3×10 _I0)2+(3+10_10)2 ] = 27 x 10_20 m2
=
=
= 3 × 1.732 × 10_10 m
= 2.598 × 10_10 m
i.e. smallest distance between zinc and sulphide ions is 2.598 × 10_10 m. (f) Each sulphide ion has four zinc as the nearest neighbours arrange tetrahedrally and each zinc ion is surrounded by four sulphide ions tetrahedrally. The chemistry of isopoly and heteropoly acids/salts of molybdenum and tungsten :
The poly acids of molybdenum and tungsten are of two types
Chemistry : Practical Application (a) the isopoly acids and their related apions which contain only molybdenum or tungsten along with oxygen or hydrogen; (b) the heteropoly acids and anion which contain one or two atoms of another elements in addition to molybdenum, tungsten, oxygen and hydrogen. The polyanions are built primarily of MO6 octahedra, but they are prepared by starting with MO42_ ions. The polymolybdates are formed by the condensation of MoO42_ unit and their composition depends upon the pH of the solution.
The formation of isopolytungstate is similar, the simple tungstate WO42_ exists in strongly basic solution. Acidification results in the formation of polymers built up from WO6 octahedra. The important species formed are W7O246_ and W12O42I2_. Heteropoly acids and their salts are formed when molybdate and tungstate solutions containing other oxoanions (e.g., PO43_ and SiO44_) or metal ions are acidified. At least 35 elements are known to be capable of functioning as the hetero atoms.
Some Representative Heteropoly Salts and their Nomenclature Formula IUPAC names Na3[PvMoI204o] Sodium 12-molybdophosphate; sodium dodecamolybdophosphate H3[PvMo[204o] 12-Molybdophosphoric acid; dodecamolybdophosphoric acid K3[Co2IIW12 O42] Dimeric potassium 6-tungstocobaltate; dimeric potassium hexatungstocobaltate (II)
Na3[CeIVMo12C 42] Sodium 12-molybdocerate (IV); sodium dodecamolybdocerate (IV) Transition Elements In the structure of isopoly and heteropoly anions the tungsten and molybdenum atoms lie at the centres of octahedra of oxygen atoms and the structures are built up of these octahedra by means of shared corners and shared edges (but not shared faces).
(a) Diagrammatic representation of MoO6 and WO6 octahedra used in showing structures of some isopoly and heteropoly anions. (b) The structure of the para-molybdate anion, [Mo7O24l6_. (c) The structure of the octamolybdate anion, [Mo8O26]4_ (note that one MoO6 octa-hedron is completely hidden by the seven that are shown). (d) The structure of the [W12O42]12_ unit in the paratungstate ion.
The structure of the series A 12-molybdo and 12-tungs o-heteropoly anions of general formula [Xn +Mo O ] (8_n)_. 12 40 The spin orbit coupling for p2 system : Spin Orbit Coupling : When several electrons are present in a sub-shell, the overall effect of the individual orbital angular momenta l is given by resultant quantum number L, and the Chemistry : Practical Application overall effect of the individual spin ms is given by the resultant spin quantum number S. In an atom, the magnetic effects of L and S may interact or couple, giving a new quantum number J called total angular momen-tum quantum number which results from vectorial combination of L and S (Russell - Saunders or LS coupling). Spin Orbit Coupling of p2 : With a p2 arrangement, resultant orbital quantum number values of L = 2, 1 and 0, and also resultant quantum number values S = 1 and 0 are obtained. These may be coupled to give the total angular quantum number J. Each of these arrangement corresponds to an electronic arrangement called a spectroscopic state.
Each of these arrangements corresponds to an electronic arrangement sometimes called a spec-troscopic state, which is describe by a full term symbol. The letter D indicates that the L quantum number has a value of two, P indicates that L = 1 and S denotes a value of L = 0 as described previously. The lower right hand subscript denotes the value of the total quantum number J, and the upper left superscript indicates the multiplicity, which has the value of 2S + 1 (where S is the resultant spin quantum number). The relation between the Transition Elements
number of unpaired electrons, the resultant spin quantum number S, and the multiplicity is given in table. Unpaired electrons S Multiplicity Name of state 0 0 1 Singlet 1
2 Doublet
2 1 3 Triplet 31
4 Quartet
4 2 5 Quintet Thus, the symbol 3D2 (pronounced triplet D two) indicates a D state, hence L = 2; the multiplicity is three, hence S = 1 and the number of unpaired electrons is 2; and the total quantum number J =2. All of the spectroscopic terms derived above for a p2 configuration would occur for an excited state of carbon Is2, 2s2, 2pl, 3p1. However, in the ground state of the atom Is2, 2s2, 2p2 the number of states is limited by the Pauli exclusion principle since no two electrons in the same atom can have all four quantum numbers the same. In the ground state configuration, the two p electrons both have the same values of n = 2 and l = 1 so they must differ in at least one of the remaining quantum numbers m or ms. This restriction reduces the number of terms from 3D, 3P, 3S, 1D, 1P and 1S to 1D, 3P and1S. This can be shown by writing down only those electronic arrangements of m and ms which do not violate the Pauli exclusion principle. For p electrons, the subsidiary quantum number l = 1, and the magnetic quantum number m may have values from + l ® 0 ® l, giving in this case values of m = + 1, 0 and - 1. There are 15 possible combinations. The rules for determining the order of energy for different terms : The rules are as under
Chemistry : Practical Application 1. The terms are arranged depending upon their spin multiplicities, e.g., their S-values. The most stable state has the highest S-value and stability decreases as the value of S decreases. This means that the most stable state (ground state) has maximum unpaired electrons because, this gives the minimum electrostatic repulsion. This is in accordance with Hund's rule.
2. For a given value of S, the state with the highest value of L is the most stable state. This means that if two or more terms have the same value of S (same spin multiplicity), the state with highest value of L will have the lowest energy. 3. For a given value of S and L (i.e., having same S and L values), the terms with smallest J value is most stable if the subshell is less than half filled and the term with maximum J value is most stable if the subshell is more than half filled. Using crystal field model, explained why Fe3O4 has an inverse spinel structure while Mn3O4 has a normal spinal structure : Normal spinels have the formula AIIB2IIIO4 where AII can be a group IIA metal or a transition metal in the +2 oxidation state and BIII is a group III A metal or a transition metal in the +3 oxidation state. The oxide ions form a close packed cubic lattice with eight tetrahedral holes and four octahedral holes per molecule of AB2O4. Mn3O4 has normal spinel structure as it is a mixed oxide of the formula MnII(Mn2III)O 4 corresponding to the formula AB2O4. AII ions occupy of the tetrahedral holes and BIII ions occupy one half of the available octahedral holes. In the inverse spinel structure the AII ions and one-half of the BIII ions have exchanged place that is the AII ions occupy octahedral holes along with one-half of the BIII ions while half of the BIII ions are in tetrahedral holes. Fe3O4 (magnetite) is an example of an inverse spinel structure. Although both A and Transition Elements B ions in this case are iron, some are ferrous and others ferricFeIII[FeIIFeIII ]O4. CFT helps in predicting the structures of spinels. CFSE values in octahedral and tetrahedral fields have been used for interpretation. For this it is assumed that the oxide ion O2_, like water molecule produces weak field. CFSE values (in terms of D0) for Mn3+ (d4), Fe3+ (d5), Mn2+ (d5) and Fe2+ (d6) in octahedral and tetrahedral weak field (high spin) are Mn3+ Mn2+ Fe3+ Fe2+ (d4) (d5) (d5) (d6) CFSE (Oh weak field) 0.60 0 0 0.40 CFSE (Th weak field) 0.18 0 0 0.27 For Mn3+ (d4) and Fe2+ (d6) ions the CFSE values are greater for octahedral than tetrahedral sites. Thus
Mn3+ and Fe2+ ions will preferentially occupy the octahedral sites, maximising the CFSE values of the system. Hence in Mn3O4 all the Mn3+ ions occupy octahedral sites and all Mn2 + ions are in the tetrahedral sites i.e., normal spinel structure MnII[Mn2III]O 4. In Fe3O4 all the Fe2+ ions and half of the Fe3+ ions are in the octahedral sites, while the remaining half of Fe3+ ions occupy tetrahedral sites. Thus, it is an inverse spinel structure FeII[FeIIFeIII]O 4. The spectrum of Nickel(II) octahedral complex ion : Ni(II) in octahedral complexes such as [Ni(H2O)6]2+ has d8 configuration. Their spectra with three spin allowed transition are shown in Fig.
Chemistry : Practical Application 3A ® 3T 9000 cm_1 2g 2g 3A 2g
® 3T1g (F) 14,000 cm_1 for [Ni (H2O)6]2+
3A 2g
® 3T1g(P) 25,000 cm_1
The bands are quite weak as expected. The absorption spectrum is shown in Fig. b-elimination in transition metal alkyls ? Explained : In higher alkyl organometallic compounds, the hydrogen present on carbon atom at b-position to the metal is labile. It brings about decomposition into metal hydride and an alkene. This is called belimination. The b-elimination is responsible for the instability of alkyl organometallic compounds. Therefore, the blockage of b-elimination reaction can bring about remarkably stability in metal alkyls. Therefore, the alkyls which do not contain replaceable hydrogen atom on b-carbon are highly stable, for example,
MCH2CH3, MCH2Ar, MCH2NR2 etc. High spin-low spin cross overs. The pre-requisities for a coordination compound to be amenable to such cross oversare: Spin-state Cross Overs : In the majority of complexes the ligand field is either weak enough to conclude that a high spin (HS) ground state prevails at all associable temperatures or strong enough to assure that a low spin (LS) ground state exists at all acessible temperatures. There are, however, cases where a high spin and a low-spin states are separated by only about the thermal energy prevailing at or below room temperature. The magnetic properties of the complex, therefore, change as a function of temperature. In the following types of octahedral systems spin-state cross overs are commonly observed dn LS configuration and state HS configuration and state Example
d5
FeIII
d6
FeII, CoIII
d7
CoII
Transition Elements The origin of spin state cross overs is related to magnitude of the orbital splitting (D0) and the spinpairing energy P. When P and D0 are about equal, the high spins and low spin states will have very similar energies. Because of the change in M-L bond lengths, which decrease in the LS state (by 0.1 _ 0.2 Å) the behaviour of the system is strongly dependent on the pressure as well as temperature. HS complexes are associated with the condition D0 < P and low spin complexes with D0 > P. For complexes in which the energy difference between A0 and P is relatively small, an intermediate field situation is possible for the two spin states to coexist in equilibrium with each other. Consider the Fe2+ ion. At the two extremes, it forms high spin paramagnetic [Fe(H2O)6]2+ (S = 2) and low spin diamagnetic [Fe(CN)6]4_ (S=0). The Tanabe-Sugano diagram pertaining to these d6 complexes shows that near the cross over point between weak and strong fields the difference in energy between the spin-free (5T2g and spin paired (1Ag) ground states becomes very small (Fig.).
Variation in energies of the 5T2g and 1A1g terms with increasing D0 for d6 octahedral complexes. At weak fields (high spin complexes) the ground term is 5T2g, while at strong fields (low spin complexes) it is 1A1g. Note that in the region immediately on each side of the spin cross over point, the energy difference between the two terms is small; thus high and low spin complexes may coexist. Within this region, it is reasonable to expect that both spin states may be present simultaneously and the degree to which Chemistry : Practical Application each is represented will depend on temperature D0 _ P = kT. For complex [Fe(phen)2(NCS)2] illustrating these effects is shown in Fig. At high temperatures a moment coexistent with four unpaired electrons is observed, but as the temperature is decreased, a sharp drop in magnitude is observed at 175 K where the low spin form becomes dominant.
In a broader sense the spin cross-over phenomenon are of two types Continuous : Occurring smoothly, with little or no hysteresis over a wide range temperature.This behaviour is shown when HS and LS molecules can form solid solutions within a common packingpattern.
Discontinuous : Occurring in a narrow range of temperature, often with a hysteresis. In these cases there are strong cooperative intermolecular interactions and first-order phase transition occurs. Nephelauxetic Effect Electrons in the partly filled d-orbitals of a metal ion repel one another and give rise to a number of energy levels depending upon the arrangement of these electrons in the d-orbitals. The energy of each of these levels can be expressed in terms of some inter-electronic repulsion parameters. It is observed experimentally that magnitude of these inter-electronic repulsion parameters always decreases on the Transition Elements complexation of the metal ion. This is possible only if the magnitude of inter-electronic repulsion between the d-electrons of the metal ion decreases on complexation. The magnitude of inter-electronic repulsion is inversely proportional to the distance r between the regions of maximum charge density of d-orbitals which are occupied by electrons. This repulsion would decrease only if the distance r increases or the lobes of d-orbitals extend in space. The extention ofthe lobes of the af-orbitals which means the expansion of d-electron charge cloud is known as nephelauxetic effect. Thus in the complex the electron charge cloud is more diffuse than in the free ion, hence, the average electronelectron distance is greater in the complex.
The extension of d-orbitals in complex occurs to maximise the overlap of ligand and metal orbitals, which is an essential condition of for covalent bonding. Thus, it is the covalent bonding which decreases the inter-electronic repulsions parameters when a free metal ion gets complexed. The larger the decrease in inter-electronic repulsion, the greater is the extent of covalent bonding in complex. The capability of dorbitals of different metal ions to extend themselves forms the basis of nephelauxetic effect. Nephelauxetic effect for a fixed ligand and different metal ions is Pt(IV) ~ Mn (IV) > Co (III) > Rh (III) ~ Ir (III) ~ Fe (III) > Cr (III) > Mo (III) > Ni (II) > V (II) > Mn(II).
Chemistry : Practical Application For a given metal ion and different ligands the nephelauxetic effect is (C2H5O)2PSe 2_ > (C2H5O)2PS2 _ > I_ >Br_ > CN_ ~ Cl_ > C2O42_ ~ en > NH3 > (NH2)2CO > H2O>F_ Since the inter-electronic repulsion parameters almost always decrease on complexation, the metal-ligand bonds in all the complexes must contain some contribution from covalent bonding. The most electronegative metal ions, Pt (IV) and Mn (IV) would form the most covalent bond with the least electronegative ligands. The ratio of a given inter-electronic repulsion parameter for the metal ion in a complex to its value in the gaseous ion is expressed as nephelauxetic ratio (b). Spectrochemical Series The variation of the magnitude of the crystal field splitting (D) with the nature of the ligand follows a regular order, known as spectrochemical series. This series is given below in the order in which they produce increasing value of D. I_ < Br_ < S_2 < Cl_ < NO3_ < F_ < OH_ < OX_2 < H2O < NCS_ < CH3CN < NH3 < en < dipy < phen < NO2_ < phosph < CN_ < CO. The idea of this series is that the d-orbital' splitting and hence the relative frequencies of visible absorption bands for two complexes containing the same metal ion but different ligands can be predicted. The ligands which are n-back bonders such as CN_, CO etc. will produce large splitting. The ligands which are only p-donors will give small splitting. When sulphur is in positive oxidation state, it can act as a p-back bonder (electron acceptor) and be high in spectrochemical series e.g., SO32_. When sulphur is negative, as in RS _, it is a p-donor primarily and will be low in the series. A plausible explanation for the above series of the ligands Transition Elements can be offered on the basis of their electronegativity, polarisibility, polarising powers, permanent dipole moments and p-bonding properties. Jorgensen has developed a means of estimating the value of A for an octahedral complex by treating it as the product of two independent factors D0 = f × g × 10_3 cm
The quantity f describes the field strength of ligand relative to water, which is assigned a value of 1.00. Values ranges from about 0.7 for weak field Br_ ions to about 1.7 for the very strong field CN_ ion. The g factor is characteristic of the metal ion and varies from 8000 to 36000 cm_1. Charge Transfer Spectra : Charge transfer spectra originate from a redistribution of electron density within the molecule. The charge transfer process is a photochemical oxidation reduction reaction (redox spectra). The appearance of charge transfer bands is connected with electron donating and electron accepting properties of ligands and metal ions. The greater the oxidizing power of the metal ion and the reducing power of the ligand, the lower is the energy and hence the frequency at which the charge transfer band appear. The charge transfer within a complex can occur from the metal ion to the ligand as also from ligand to metal ion. Ligand to Metal (L ® M) Charge Transfer Spectra : When the electron transition takes place from a MO located primarily on the ligand (ML bonding s or p orbitals) to a non-bonding or antibonding MO located primarily on the metal atom, the L ® M charge transfer bands are observed. The LM CT bands are shown by the oxide, chloride, bromide and iodide complexes. For d0 ions, where the d-d transitions are not possible, the CT bands are used for the determination of 10 Dq values. MnO4_ shows the L ® M CT transition. Mn is in a formal oxidation state of + 7 and combined with four oxide ions. From the MO diagram for tetrahedral complexes we can identify possible LMCT transitions.
Chemistry : Practical Application
Molecular orbital diagram for a tetrahedral ML4 complex, showing possible ligand-to-metal charge transfer (LMCT) transitions. In any tetrahedral complex, the four lowest energy s-bonding orbitals will be primarily ligand in character. Next there are two sets a-non-bonding MO's one Iigand centered and one metal centered. In MnO4_, these orbitals would correspond to filled oxygen pp-orbitals and empty manganese 3d-orbitals, respectively. All of the higher energy antibonding MO's would be unoccupied for Mn (VII) complex. Hence there are four possible L ® M transitions L(t1) ® M(e) L(t1) ® M(t*2) L(t2) ® M(e) L(t2) ® M(t*2) Transition Elements For MnO4_ all four of these transitions have been observed, but only the absorption at 17,700 cm_1 (tl ® e) falls within the visible range (14,000 _ 28,000 cm_1) and is responsible for deep purple colour of MnO4_. Metal to Ligand (M ® L) Charge Transfer Spectra : The charge transfer spectral of this type can be observed only when there are vacant orbitals localised on the ligands which are higher in energy than the filled MO's of the metal atoms. Prime examples are complexes in which the ligands have empty pantibonding orbitals, like CO, CN, py, bipyridine, pyrazine and o-phenanthroline.
Simplified molecular orbital diagram for an octahedral ML6 complex showing possible metal toligand charge transfer (MLCT) transitions when both the t2g and e*g orbitals are occupied and the ligands have empty orbitals. As the charge transfer bands are neither multiplicity forbidden nor Laporte forbidden, they have high absorption intensities. The electronic spectrum of KMnO4 shows a broad band at 18000 cm_1, while in K2CrO4 the band is observed at a higher frequency 26000 cm_1.
Chemistry : Practical Application The colours of MnO4_ and CrO42_ are due to charge transfer phenomenon, here transfer of electron takes place from oxygen to the transition metal ion (L ® M type). The greater the oxidising power of the metal ion, the lower the energy at which the charge transfer (L ® M) occurs. Here Mn is in + VII oxidation state in MnO4_ has a greater oxidising power than Cr6+. Therefore, MnO4_ shows the charge-transfer band at a lower frequency or wave number than CrO42_. Orgel diagram drawn and explained the spectrum of [CoF6]3_ which occurs as a broad and split band in the visible region:
Orgel diagram are used for the interpretation of the spin allowed absorption bands of crystal field or d-d origin in the electronic spectra of tetrahedral and octahedral transition metal complexes. If we plot the magnitude of splitting of the energy levels with the increasing ligand field for a dn system and take into consideration spin-orbit coupling and mixing of the different energy states, especially under a strong field, the Orgel diagram for the ion is obtained. Orgel Diagram of [CoF6]3_ : [CoF6]3_ is a high spin octahedral complex Co (III) has d6 configuration for which ground state term is 5D. Due to ligand 5D term splits into two states 5T2g and 5Eg. Ground state energy level 5T2g originates from the configuration t2g4 eg2 which has three degenerate levels corresponding, to three orbital arrangements of equal energy viz. dxz1 dyz1 dxy2 and dxz1 dyz2dxy 1
dz1; dxz2dyz 1dxy1
.
The excited energy level 5Eg originates from t2g3 eg3 configuration containing two degenerate energy levels corresponding to the arrangements dxy1 dxz1 dyz1
The absorption band
observed is, therefore, assigned to 5T2g ® 5Eg. The band appears to be Transition Elements composite are due to splitting of the excited energy level 5Eg into two non-degenerate energy levels due of John-Teller distortion.
On the basis of Orgel diagram, given the expected electronic transitions of the species [V(H2O)6]3+ : [V(H2O)6]3 + : In [V(H2O)6]3+, vanadium is in + III oxidation state with d2 configuration. For the free d2 ion, the energy states are Ground state : 3F Excited State : 3P, 1G, 1D and 1S In a multielectron system, the energy levels depend upon (i) ligand field strength 10 Dq and (ii) inter-electronic interactions. The ground state of a d2 ion can be dxy1 dyzl, dyz1 dxz1 or dxzl dxy1. This is the 3T1g (F) state. When the electron is excited, the configuration t2g1 eg1 gives two probabilities. If the dxz or dyz Chemistry : Practical Application electron is promoted, it goes to the present
orbital, where it experience's less repulsion due to the dxy electron
then it would have experience had it been in the orbital.
Similarly, an electron promoted from dxy orbital will go to
orbital leaving the other electron in
either dyz or dxz orbital. Thus three degenerate excited states exist (3T2g (F)) : . The other degenerate arrangement
and
produces 3T1g (P)
and
state. Other arrangement requires reversing the spin of one of the electron, so that spin multiplicity changes. Hence this transition will be multiplicity forbidden. Lastly, the two electrons may give eg2 state which is, the singly degenerate. Thus we actually expect three absorption bands due to 3Tlg (F) ® 3T2g (F), 3T1g (F) ® 3Tlg (P) and 3T1g (F) ® 3A2g electronic transitions. But actually only two absorption bands at 17,200 cm_1 and 25,600 cm_1 are observed due to 3Tlg (F) ® 3T2g and3Tlg (F) ® 3T1g (P). The third electronic transition 3Tlg (F) ® 3A would 2g
involve excitation of both the electrons to higher energy level (t2g to eg) and hence would not be likely to be observed (forbidden).
Transition Elements
Jahn-Teiler Distortion : Non-linear molecular system possessing degenerate electronic state will be unstable and will undergo distortion to form a system of lower symmetry and lower energy and thus will remove degeneracy. Distortion is due to unsymmetrical t2g or eg orbitals.
[CuCl6]4_ : In this complex Cu2+ ion has d9 configuration. In strong as well as weak field the ion has the configuration t2g6 eg3. Two possible arrangement of electrons in t2g and eg orbitals are
Distortion arises from the repulsion of ligands by the electrons occupying eg orbitals (unsymmetrical eg orbitals in both the cases). Thus d9 configuration in both the fields show strong distortion. [Cr(acac)3] : Cr3+ has d3 configuration. The distribution of d-electrons in t2g and eg sets in stronger and weaker octahedral ligand field is t2g eg0. The t2g and eg orbitals which are empty (t2g0 and eg0), half-filled (t2g3 and eg0) or completely filled (t2g6 eg4) are said to be symmetrical. Thus t2g3 eg0 lead to perfectly symmetrical octahedral complex and in this case there is no distortion. [Co(CN)6I4 Low Spin : In this case Co (II) has d7 configuration. The complex is low spin the distribution of Chemistry : Practical Application d-electron in t2g and eg orbital of low spin i.e., strong field complex is as t2g6 egl, which leads to dissymmetry of the eg-orbitals. Thus this complex show strong Jahn-Teller distortion. Jahn-Teller Effect : It explain why certain six coordinated complexes undergo distortion to assume distorted octahedral or tetragonal geometry. The d-orbitals which have both t2g and eg sets as symmetrical orbitals lead to perfect i.e., regular octahedral complexes. The d-orbitals d0 (t2g0 eg°), d3 (t2g3 eg°), d5 (t2g3 eg2), d8 (t2g6 eg2) and d10 (t2g6 eg4) of High spin-octahedral complexes and d0 (t2g0 eg0), d3 (t2g0 e 0), d6
(t2g6 e g0) and d10 (t2g6 eg4) of low spin-octahedral complexes give perfectly regular octahedral complexes i.e., there is no distortion. g
Condition for Slight Distortion : The high spin complexes of d1 (t2gl eg0), d2 (t2g2eg0 ), d6 (t2g4 eg2) and low spin complexes of d1 (t2gl eg0), d2 (t2g2 eg0), d4 (t2g4 eg0) and d5 (t2g5 eg0) ions undergo slight distortion. Thus whenever the t2g orbitals which do not come directly in the path of the ligands disposed octahedrally around the central metal ion but point between the ligands contain 1, 2, 4 or 5 electrons, we shall expect slight distortion.
Condition for Strong Distortion : The high spin complex of d4 (t2g3 egl), d9 (t2g6 eg3) and low spin complexes b (t2g6 eg1), d7 (t2g6 eg2) and d9 (t2g6 eg3) lead to strong distortion. Distortion in d4 High Spin Octahedral Complex : High spin octahedral complexes of d4 ion have any of the following configuration I ® t2g3(dz )1
II ® t2g3 (dz2) 0
When the configuration has one electron in orbital and the is empty (I), cation-anion interaction along the z-axis is less than that along the x and y-axes leading to a larger interionic distance along the z-axis and hence a tetragonal structure. Transition Elements Distortion in d9 Octahedral Complex : Complexes of d9 configuration such as aqueous solution of [Cu(NH3)4]2+ in which trigonal distortion is so strong that a square planar complex results. The two possible arrangement of t2g6 .eg3 in both the fields are I ® t2g6(
)2
II ® t2g6 (dz2)1
As t2g is completely filled, distortion or asymmetry is caused by incomplete filling of eg-orbitals i.e., from the repulsion of ligands by the electrons in eg-orbitals. In configuration (I)
orbital points at the ligand on the z-axis offers greater shielding of the Cu2+
nucleus than the half-filled orbital which points towards the ligands in the xy plane. Thus the ligands on the x and y axes experience a higher effective nuclear charge, while those on z-axis experience a lower effective nuclear charge. Consequently the ligands on the x and y axes are drawn closer to Cu2+ nucleus and those, as z-axis move further out. Thus four short and two long bonds are observed, i.e., the ligands existing along the z-axis would be at a greater distance from the central metal ion and the remaining four co-planar ligands would be at a shorter distance. With configuration (II) exactly opposite distortion is expected. Table showing : (i) Distribution of electrons in t2g- and eg-orbitals in HS and LS-octahedral complexes.
(ii) Symmetrical [abbreviated as (sym)] and unsymmetrical [abbreviated as (unsym)] t2g- and eg-orbitals. (iii) Predicted distortion in octahedral complexes. Note : The orbitals that are unsymmetrical produce distortion.
Chemistry : Practical Application HS-octahedral complexes : weak ligand field LS-octahedral complexes : strong ligand (spin-free) field (spin-paired) No. of Distribution of Predicted Distribution of Predicted d- electrons in t2g and eg- Distortion electrons in t2g- and eg- Distortion electrons orbitals and symmetrical orbitals and symmetrical and unsymmetrical orbitals and unsymmetrical orbitals d0 t2g0 (sym) eg0 (sym) No distortion t2g0 (sym) eg0 (sym) No distortion dl t2g1 (unsym) eg0 (sym) Slight distortion t2gl (unsym) eg0 (sym) Slight distortion d2 t2g2 (unsym) eg0 (sym) Slight distortion t2g2 (unsym) eg0 (sym) Slight distortion d3 t2g3 (sym) eg0 (sym) No distortion t2g3 (sym) eg0 (sym) No distortion d4 t2g3 (sym) eg1 (unsym) Strong distortion t2g4 (unsym) eg0 (sym) Slight distortion d5 t2g3 (sym) eg2 (sym) No distortion t2g5 (unsym) eg0 (sym) Slight distortion d6 t2g4 (unsym) eg2 (sym) Slight distortion t2g6 (sym) eg0 (sym) No distortion d7 t2g5 (unsym) eg2 (sym) Slight distortion t2g6(sym) eg1 (unsym) Strong distortion d8 t2g6 (sym) eg2 (sym) No distortion t2g6 (sym) eg2 (unsym) Strong distortion
leading to square planar d9 t2g6 (sym) eg3 (unsym) Strong distortion t2g6 (sym) eg3 (unsym) Strong distortion d10 t2g6 (sym) eg4 (sym) No distortion t2g6 (sym) eg4 (sym) No distortion Transition Elements The d-orbital energy level change in Cu2+ (d9) ion when the regular octahedral distorts : In the diagram below the splitting of the more stable octahedral distortion corresponding to configuration has been considered, d1 and d2, which represent the splitting of the eg3 and t2g6 levels t2g6 ( )2 respectively, both are much smaller than D0 and d2 is much smaller than d1 (D0 >>> d1 > d2).
The two eg orbitals separate so that goes up as much as the goes down; the t2g orbitals separate so that dyz and dxz goes down only half as far as the single orbital dxy goes up. For t2g electrons there is no net energy change since four dxy and dxz electrons are stabilised by electrons are destabilized by However, in the splitting of eg level
while two dxy
Thus in the splitting of t2g level net energy gain is zero.
Net energy gain = energy gain + energy loss
=
Chemistry : Practical Application =
=
Thus the net lowering of the electronic energy in case of eg level is d1/2, equal to _ d1/2, might be called the Jahn-Teller stabilisation energy and provides the driving force for the distortion. Drawn a crystal field splitting diagram for a complex in a linear field. Assume that the ligands lie on the z-axis : orbital will be highest in energy. An In a linear field with two negative charges on the z-axis, the electron in this orbital will be located in a region of space directly adjacent to the negative charges. The dxz and dyz orbitals are degenerate and are next highest in energy. The dxy and degenerate (both lying in the xy plane) and lowest in energy.
orbitals are
Lower the symmetry of a complex, lower is its magnetic moment, to the spin only value, explained : The orbital angular momentum is reduced from what it would be in the free metal ion by the presence of ligands. The orbital contribution to the magnetic moment is said to be quenched if the symmetry of the complex is lowered as in Jahn-Teller distortion. Thus the value of magnetic moment approaches the spin only value m = [n (n + 2)]1/2 If t2g orbitals are not completely filled or not exactly half-filled, symmetry will be lowered in an octahedral geometry field.
Transition Elements The effect of addition of Co2+(aq) to the equilibrium reaction given : Co2+(aq) + 4NCS_ ® [Co(NCS)4]2_ (Pale Pink) (Colourless) (Blue) The addition of Co2+ to the equilibrium mixture, will raise the concentration of complex ion in the mixture. Thus the percentage of absorption by the complex increases and transmittance decreases. Comparatively more intense d-d-transition : (i) trans-[CrCl2(H2O)4 ]+ or cis-[CrCl2(H2O)4] + (ii) Ni(CO)4 or Fe(CO)5 (i) The cis isomer has no centre of symmetry, the trans isomer has a centre of symmetry, therefore, Leporte selection rule is better applicable to the trans isomer. As a result, the crystal field electronic transition of d-d origin (i.e., g-g type) are more Leporte forbidden in trans isomer than similar transition in cis isomer (Due to lack of symmetry in the cis isomer, the energy levels cannot be termed g or u if we apply symmetry rules strictly). In other words, the intensity of absorption band is lower in trans isomer than the intensity of bands in the cis isomer. Thus, the lowering of molecular symmetry relaxes Leporte selection rule and increases the intensity of absorption of crystal field transition. (ii) Nickel carbonyl; Ni(CO)4, having Ni(O) in sp3 configuration is tetrahedral and diamagnetic. Fe(CO)5 having Fe(O) in dsp3 configuration is trigonal bypyramid and diamagnetic. Since in a tetrahedral complex the metal ion is not at the centre of symmetry the mixing of d and p orbitals is much favoured and as such the intensity of Leporte forbidden d-d absorption bands in tetrahedral complex is much higher than in trigonal bipyramid or octahedral complex.
Chemistry : Practical Application For the [Cr(H2O)6]2+ ion, the mean pairing energy P is found to be 23500 cm_1. The magnitude of D is 13900 cm_1. Calculate the crystal field stabilization energy for the complex in configuration corresponding to high spin and low spin states. More stable : Cr2+ had d4 configuration d4 ion in a high spin state (weak field) t2g3 egl
CFSE = _0.6 D0 = _0.6 × 13,900 cm_1 = _8340 cm_1 d4 ion in a low spin rate (strong field) CFSE = _1.6 D0 + P = _1.6 (13900 cm_1) + 23500cm_1 = + 1260 cm_1 Since D0 < P, the high spin configuration would be more stable. The enthalpy of hydration of Cr2+ is _ 460 k cal/mol. In the absence of crystal field stabilisation energy, the value of DH would be _ 435 k cal/mol. The value of D for [Cr(H2O)6]2+, estimated : The enthalpy of hydration of Cr2 + is 0.6 D higher for this d4 weak field ion than it would be in the absence of CFSE. Thus, _0.6 D = (_460 k cal/mol) _ (435 k cal/mol) = _25 k cal/mol
D= = 14600 cm_1 The actual value of D is 13900 cm_1. (i) The [Mn(H2O)6]2+ ion has an extremely pink colour attributable to transitions which are formally forbidden. Why there are no completely allowed d-d transitions, explained : (ii) The fact that a number of tetrahedral Co (II) complexes are stable, whereas the corresponding Ni (II) complexes are not, suggested a reason : Transition Elements (i) The Mn(H2O)62+ ion is a weak field, high spin d5 complex with the spectroscopic symbol 6A1. There is only one combination of quantum numbers corresponding to five unpaired electrons. In as much as transition involving a change in multiplicity are forbidden, there are no allowed transitions. The pale
colour of Mn(H2O)62+ is due to weak forbidden transitions. (ii) Co(II) has d1 and Ni(II) has d8 configurations. The ligand field stabilization energy of a d7 tetrahedral complex is greater than that of a d8 tetrahedral complex and the ligand field stabilization energy of a d8 square planar complex is greater than that of a d7 square planar complex. The dimeric structure of copper(II) acetate, explained : Cu(II) acetate is dimeric in both solid and solution state, [Cu(CH3COO)2].2H2O. Each Cu(II) is coordinated in a square plane by four oxygens of bridging acetato groups. The two H2O molecules are attached to each Cu(II). The CuCu distance is short and this bond results from a lateral overlap of singly filled orbitals of two Cu(II) ions. Such type of metal-metal bond is called d-bond and is different from a s and a p-bond. Such overlap leads to a partial queueing of the magnetic moment (m = l.43 BM).
Chemistry : Practical Application (a) Whether the reactants or products are favoured in the following equilibria, predicted: Cul2 + Cu2O
2CuI + CuO
(b) The complex [CuCl4]2_ exists, but [Cul4]2_ does not, explained : (c) The fact that among the complexes [Ni(OH2)6]2+, [Ni(en)3]2+ and [Ni(NH3)6]2+, [Ni(en)3]2+ is most stable, accounted :
(a) In the reaction Cul2 + Cu2O
2CuI + CuO
products are favoured. Thus the equilibrium lies more towards the right hand side. Cu2I2 is stabler than Cul2, similarly CuO is stabler than Cu2O. (b) Cu (II) is reduced to Cu (I) by I_ but not by Cl_, that is why [Cul4]2_ does not exist. (c) Of the three complex given [Ni(en)3]2+ is most stable. Since en = ethylene diamine being a bidentate ligand occupies two coordination position in the metal ion and ring type complex i.e., chelate is formed. Chelate ring complexes are usually more stable complexes than complexes containing no rings. The chelated complex dissociates less than a non-chelated complex and hence will register a higher stability constant. Tris (trifluoroacetylacetonato) cobalt(III) is formed from a bidentate ligand trifluoro-acetylacetone {CF3C(OH) = CHCO-CH3}. The structure of the complex drawn. The isomerism predicted and drawn the structures of isomers if isomerism is possible. Trifluoroacetylacetonato ligand may be represented as Transition Elements
It is a bidentate ligand and coordinate with cobalt to form complex like
The complex Co(acac)3 complex would show optical isomerism
A stepwise ligand displacement reaction of hexaaquonickel (II) ion with ammonia considered, and given the composition and structure of all the possible geometrical isomers of this system :
Among these compounds formed the compounds [Ni(H2O)6]2+, [Ni(H2O)5NH3] 2+, [Ni(H2O) (NH3)5] and Chemistry : Practical Application [Ni(H2O)6]2+ will have no geometrical isomsers. Their structures are as—
(a) In the reaction [Co(Cl)2(NH3)4] + + Cl_ ® [Co(CI)3(NH3)3] + NH3, only one isomer of the complex product is obtained. The initial complex whether cis or trans : (b) When [Ni(NH3)4]2+ is treated with concentrated HCl, two compounds having the formula Ni(NH3)2Cl2 (designated I and II) are formed. I can be converted into II by boiling in Transition Elements dilute HCl. A solution of I reacts with oxalic acid to form Ni(NH3)2 (C2O4). II does not react with oxalic acid. The configuration of I and II and the geometry of nickel (II) complexes, deduced : (c) The ion [Co(en)Cl2Br2]_ can have many different isomers. The structures of these isomers, drawn : (d) Whether [Ir(en)3]3+ should exhibit optical isomerism. Proved by diagram.
(e) Many geometrical isomers of [Cr(en)2Cl2]+ exist and isomer(s) might display optical activity, decided : (a) The original complex must have the two chloride ions in the trans position, since then all the other four positions which could be replaced are identical. If the original isomer were cis, the third chloride could replace an ammonia cis to both or trans to one, and two isomers would be expected. (b) Compound I is the cis isomer, which can easily form a chelate ring with oxalate group. Trans- isomer cannot form a chelate ring. I and II are square planar. (c) [Co(en)Cl2Br2]_ ion has four different isomers
(d) There are two optical isomers of [Ir(en)]3+
Chemistry : Practical Application (e) [Cr(en)2Cl2]+ has two geometrical isomers, cis and trans. Only the cis isomer can be optically active, the trans isomer is its own mirror image.
(i) Complexes in order of increasing liability arranged : Fe(H2O)63+, Cr(H2O)63+, Y(H2O)63+, AI(H2O)63+, Sr(H2O)62+, K(H2O)6+. (ii) There are two isomers of Pt(NH3)2Cl2, A and B. If A is treated with thiourea (tu), Pt(tu)42+ is formed. If B is treated with thiourea, Pt(NH3)2(tu)22+ is formed. The isomers and explain the data, identified : (iii) The reaction of CrCl3 with liquid ammonia gives principally [Cr(NH3)5Cl]Cl2, but when trace of KNH2 is present the main product is [Cr(NH3)6]Cl3, explained : (iv) The oxidation of Sn (II) by Fe3+ is first order in each reagent. Whether Sn (III) is involved in the mechanism, explained : (i) Cr(H2O)63+ < A1(H2O)63+ < Fe(H2O)63+ < Y(H2O)63+ < Sr(H2O)62+ < K(H2O)6+ (ii) A is the cis-complex; B is the trans-complex. In the cis-complex, the ammonia molecule is labilized by the trans-chloride ions; replacement of ammonia molecule by thiourea causes the trans-chloride to labilized. Hence all positions became occupied by thiourea. In the trans-complex, the chloride ions labilize each other; only these ligands are placed by thiourea molecules. Transition Elements (iii) The NH2_ ions catalyse the displacement of the last coordinated chloride ion by ammonia. [Cr(NH3)5Cl]2+ + NH2
Cr(NH3)4(NH2)Cl + + NH3
Cr(NH3)4(NH2)Cl + ¾® Cr(NH3)4(NH2 )2+ + Cl_ Cr(NH3)4(NH2) 2+ + 2NH3 ¾® [Cr(NH3)6]3+ + NH2_
(iv) If the reaction rate is reduced by the addition of Fe2+ to the system, the formation of Sn (III) in a reversible primary step would be strongly indicated. Sn (II) + Fe3+
Sn (III) + Fe2+
Sn (III) + Fe3+ ¾® Sn (IV) + Fe2+ (a) The trans-effect, with its utility, explained : (b) [Pt(NO2)Cl3]2_ is treated with NH3, and the isomer of Pt(NO2)Cl2NH3 is formed : (c) The product cis-[PtCl2(NH3)2] + Py given : (a) Trans-effect : The ability of an attached group to direct substitution into position trans (i.e., opposite) to itself is called the trans-effect or the labilization of ligands trans to other, trans directing ligands. By comparison of a large number of reactions, it is possible to set up a trans-directing series. The order of ligands in this series is as follows CN_ ~ CO ~ NO ~ H_ ~ > CH3 ~ SC(NH2)2 ~ SR2 ~ PR3 > SO3H > NO2_ ~ I_ ~ SCN_ > Br_ > Cl ~ > Py > RNH2 ~ NH3 > OH_ > H2O. The trans-effect series has proved useful in rationalising known synthetic procedures and devising new ones. Let us consider the synthesis of cis and trans isomers of [Pt(NH3)2Cl2]. The cis isomer is synthesized by the treatment of [PtCl4]2_ ion with NH3.
Chemistry : Practical Application In the first step [PtNH3(Cl3)]_ is formed. In the second step since Cl_ has a greater trans-directing influence than does NH3, substitution of NH3 into [Pt(NH3)Cl3]_ is least likely to occur in the position trans to the NH3 already present and thus the cis-isomer is favoured. The trans isomer is made by treating [Pt(NH3)4]2 + with Cl_.
Here the superior trans-directing influence of Cl_ causes the second Cl_ to enter trans to the first one producing trans [Pt(NH3)2Cl2]. (b) The complex ion [Pt(NO2)Cl3]2_ when treated with NH3 give [Pt(NO2)Cl2(NH3 )]_. The two possible isomers are
Experiments have shown that it is only trans-isomer which is formed by the replacement of Cl lying trans to NO2 i.e., since NO2 has greater trans effect than Cl_, the Cl_ trans to NO2 and not one trans to Cl_ is replaced by NH3.
Transition Elements Py occupies the position trans to Cl ~ because trans directing effect of Cl . is greater than that of NH3. The reaction :
[Co(NH3)5CI]2+ + OH_
[Co(NH3)sOH]2+ + Cl_
obeys a rate law which is compatible with SN1CB mechanism. Give the steps in the mechanism. SN1CB Mechanism : This is known as SN1 transformation involving the conjugate base. Here the conjugate base (CB) reacts by a dissociation (SN1) process. For the given reaction following mechanism was suggested (i) [(NH3)5CoCl]2 + + OH_ (ii) [(NH3)4Co(NH2)Cl] +
[(NH3)4Co(NH2)Cl] + + H2O [(NH3)4Co(NH2)] 2 + + Cl_
(iii) [(NH3)4Co(NH2)] 2+ + H2O
[(NH3)5Co(OH)]2+
In the reaction (i) OH_ ions abstract a proton from a ligand in N5 group giving the conjugate base of the ligand. CB produced in reaction (i) contains a coordinated :NH2- ligand which is capable of pi-bonding with Co(III) more efficiently than in the seven coordinated intermediate, the better pi-stabilised five coordinated intermediate was proposed to be formed from the conjugate base in the second step. The pibonding :NH2_ ligand also helps in the dissociation of Cl_ from Co(III). The five coordinated intermediate [Co(NH3)4NH2] 2+ then quickly reacts with H2O to give final product of hydration. The reaction (ii) is supposed to be slower than reaction (i) and (iii), thus it constitute the rate determining step of the reaction. The detailed mechanism of the reaction between [Co(NH3)5Cl]+2 and [Cr(H2O)6]+2 to form [Co(H2O)6]2+ and [Cr(H2O)5Cl]2+ : The required reaction is [CoIII(NH3)5Cl] 2+ + [CrII(H2O)6]2+ ® [CoII(H2O)6] 2+ + [CrIII(H2O)5Cl]2+ This is an example of inner sphere redox reaction. The reaction Chemistry : Practical Application proceeds by ligand bridging mechanism. This reaction involve the reduction of Co (III) (in [Co(NH3)5Cl]2+) by chromium II (in [Cr(H2O)6]2+). The mechanism is based on the following facts
(i) Both Co (III) and Cr (III) form inert complexes. (ii) The complexes of Co (II) and Cr (II) are labile. Under these circumstances the chlorine atom, while remaining firmly attached to the inert Co (III) ion, can displace a water molecule from the labile Cr (II) complex to form a bridged intermediate. [CoIII(NH3)5Cl] + [CrII(H2O)6] ® [(H3N)5CoIIIClCr III(OH2)5]4+ + H2O
[CrIII(H2O)5Cl] 2+ + [Co(NH3)5(H2O)] 2+ ¬ [(H3N)5CoIIClCr(OH 2)]4+
[CoII(H2O)6]2+ + 5 NH4+ The redox reaction takes place within this dinuclear compiex with the formation of reduced Co (III) and oxidised Cr (III). The latter species form an inert chloroaqua complex but the Co (II) complex is labile, so the intermediate dissociates with the'chlorine atom remaining with the chromium. The mechanism of the reaction : [Fe(CN)6]4_ + [Mo(CN)8]3- ® [Fe(CN)6]3_ + [Mo(CN)8]4 _ This reaction proceed through the outer sphere reaction mechanism in which the coordi-nation sphere of the ions undergoing redox reaction is not altered. The reaction involve different central metal ions with different oxidation states occurs at a faster rate. For such reaction when excited state of the products are converted into ground states the decrease in energy can appear as the free energy of the reaction (DG° must be negative). This means that for such reaction the structure of transition state is more like that of the reactants and consequently the activation energy is decreased and the rate is increased. Such a reaction may be considered to approximate a simple collision model. The rate of reaction is faster than cyanide Transition Elements exchange for either reactant so we consider the process consist of electron transfer from one stable complex to another with no breaking of FeCN or MoCN bonds. For this process to occur there is a critical requirement which is given by Frank-Condon principle which states that there can be no appreciable change of atomic arrangement during the time of an electronic transition.
Marcus has derived a relationship to calculate rate constant for outer sphere reaction k12 = (k11, k22, k12f12)1/2 Rate constant, k12 for the cross reaction 1. [Fe(CN)6]4_ + [Mo(CN)8]3- ® [Fe(CN)6]3_ + [Mo(CN)8]4_ is calculated as compared to a self-exchange reaction ku is the rate constant for the reaction. 2. [Fe(CN)6]4_ + [Fe*(CN)6]3_ ® [Fe(CN)6]3_ + [Fe*(CN)6]4_ k22 is rate constant for the reaction. 3. [Mo(CN)8]3_ + [Mo*(CN)8]4_ ® [Mo(CN)8]4_ + [Mo*(CN)8]3_ kl2 is the equilibrium constant for the overall reaction 1. The factor f12 has been described as a correction for tha difference in the free energies of the two reactants and is often close to unity. Fluxional Molecules No molecule is strictly rigid even at 0 K due to interatomic vibrations, but the amplitude of vibration about the mean position is about 10-30 pm. However, some molecules like PF5, some metal carbonyls and organometallic compounds, called the fluxional molecules, undergoes deformational rearrangements in which the atoms actually interchanges their places with one another. A class of fluxional molecule of great importance are those with trigonal bipyramid configuration. When all five appended groups are identical single atoms, as in AB5, the symmetry of the molecule in D3n. In case of PF5, for example, rapid exchange Chemistry : Practical Application of axial and equatorial fluorine atoms in trigonal bipyramid structure through the square pyramid (> 105 s') makes the NMR spectrometer record all the F atoms in an identical environment which is the weighted average of that prevailing in the separate environment.
Rapid exchange of the axial and the equatorial fluorine atoms through the intermediate formation of the square pyramidal intermediate. The species before and after arrangement in case of PF5 is chemically identical, so that DG° = DH0 = DS0 = 0. The process in which one of the more symmetrical forms (TBP for the 5-coordinated complex) changes to another (square pyramid) and then back to an equivalent form in which the interchange of ligands takes place-is called polytopal arrangement. Because of the small differences in energies, the 5-coordinated and the 8-coordinated complexes (square antiprism ® dodecahedron ® square antiprism) are generally fluxional even at low temperature. The octahedral complexes, in general, are not fluxional. Organometallic compounds which show fluxional behaviour are as
Transition Elements Most molecules are rigid because their atoms execute approximately harmonic vibration about their equilibrium positions. But in a number of cases in which molecular vibrations or intramolecular rearrangements carry a molecule from one nuclear configuration into another. When such process occur at a rate permitting detection by at least same physical or chemical method, the molecules are designated as stereochemically non-rigid. In some cases two or more configurations are not
chemically equivalent and the process of interconversion is called isomerization or tautomerization. In other cases the two or more configurations are chemically equivalent, and this type of stereochemically non-rigid molecule is called fluxional. The NMR spectroscopy is most commonly used to reveal the occurrence of stereochemical non-rigidity, since its time scale is typically in the range 10_2 to 10_5 s. The rearrangement involved in stereochemically nonrigid behaviour are rate processes with activation energies. When these activation energies are in the range of 25 to 100 kJ mol_1, the rates of the arrangements can be brought into the range 102 to 105 s_l at temperatures between + 150 to _ 150° C. Thus by proper choice of temperature, many such rearrangements can be controlled so that they are slow enough at low temperature to allow detection of individual moleculesor environment within the molecules and rapid enough at higher temperature for the signals from the different molecules or environments to be averaged into a single line at the mean position. Thus by studying inner spectra over a suitable range, the rearrangement process can be examined in much detail. An example of this type of study of interconverting tautomers is provided by the systems of cis- and trans- isomers of (h3 _ C5H5)2 Fe (CO)4.
Chemistry : Practical Application The proton NMR spectrum of (C5H5)4Ti showed two peaks at _ 30° C and one peak at 62° C, explained : The structure of (C5H5)4Ti-tetracyclo-pentadienyl titanium is as follows
Its formula can be formulated as [Ti(hi-C5H5)2(h 5-C5H5)2]
It is a fluxional molecule in which the averaging of two different nuclear environments occurs as a result of intermolecular arrangement is provided by (C5H5)4Ti. Even at a temperature as low as _ 30° C, the five protons in the b and b'rings are indistinguishable owing to a rapid internal rotation which is characteristic of rings bonded in this way to metal atoms. Thus at _ 30° C there are, effectively two types of protons in the molecule, ten of one kind in the a and a´ rings and ten of a second kind in the b and b' rings and the rate at which these two ring type interchange their role is slow enough for two separate, sharp signals to be observed. As the temperature is raised, a process in which rings of type a and b exchange roles becomes more and more rapid and eventually rapid enough for the two types of proton environment to be no longer distinguishable in the proton NMR spectra. The most common polyhedral arrangements encountered in metal atom clusters: Metal Atom Cluster : In the clusters there is no central atom and the essential feature is a system of bonds connecting each atom directly to its neighbours in the polyhedron. Polyhedra found in cages and clusters are the same as those adopted by complexes e.g., tetrahedron, tbp and octahedron but there are others especially the polyhedra with six vertices. Transition Elements A metal atom cluster may be defined as a group of two or more metal atoms in which there are substantial and direct bonds between the metal atoms. Metal cluster compounds can be grouped into two classes (i) polynuclear carbonyls, nitrosyls and related compounds. These are known as low valence type, contain metal atoms in formal valence states _1 to +1 have carbonyl groups to stabilize low oxidation states and they contain relatively long MM bond. (ii) lower halides and oxides. These are known as high valence type. These are formed by earlier transition metals, and are restricted to the metals of second and third transition series (i.e., Nb, Ta, Mo, W and Re). The tendency to retain cluster of metal atoms will predominate in those metals with very large energies of atomization (and hence very high melting and boiling points). Thus the most refrectory metals (Zr, Nb, Mo, Tc, Ru, Rh, Hf, Ta, W, Re, Os, Ir and Pt) have the greatest tendency to form metal clusters. A second factor necessitating low oxidation states in the nature of the d-orbitals. The size of the d-orbitals is inversely related to the effective nuclear charge. Since effective overlap of d-orbitals appears to stabilize metal clusters, excessive contraction of them will destabilize the metal cluster. Hence large charges resulting from high oxidation states are unfavourable.
Representative cluster compounds, (a) Os5(CO)16 a typical metal carbonyl cluster,
Chemistry : Practical Application [Nb6Cl12]2 +, a typical halide cluster containing a metal from the left of the d-block, a mixed Co-Sb cluster, Sb4[Co(CO)3]4. Carbonyl clusters of different types are Triatomic Cluster : M3(CO)12 clusters (M = Fe, Ru and Os).
Four-atom Clusters : M4(CO) I2 clusters (M = Co, Rh andIr) Heteronuclear analogues of this type with tetrahedral geometry are
Besides the tetrahedral geometry, four atom clusters are known with butterfly and planar structure
Transition Elements Halide clusters of different types are Dinuclear Compounds : [Re2X8]2_
2ReO4
[Re2X8 ]2_ X = Cl, Br, I, NCS
Trinuciear Clusters : [(ReCl3)3]
Rhenium(IIl) clusters : (a) The structural unit present in a Re(III) trinuciear cluster. The positions marked O are empty in the trihalides in the gas phase but have coordinating groups in other situations. (b) The structure of solid (ReCl3)x (c) The [Re3Cl12] anion. Tetranuclear Clusters : Although common among carbonyl clusters, far fewer examples of tetranuclear clusters are found among the halides and oxides. One example noted previously is W4(OR)12 which forms by dimerization of W2(OR)6. The tetrameric W4(OR)16 (i has also been synthesized. Whereas W2(OR)6 and W4(OR)12 may be viewed as unsaturated, W4(OR)16 is saturated, containing WW single bonds.
Chemistry : Practical Application
Hexanuclear Clusters : Clusters of six molybdenum, niobium, or tantalum atoms have been known for many years, predating the work with rhenium. There two types : In the first, an octahedron of six metal atoms is coordinated by eight chloride Iigands, one on each face of the octahedron. This is found in `molybdenum dichloride', Mo6Cl12, better formulated as [Mo6Cl8]Cl4. Each Mo (II) atom can use its four electrons to form four bonds with adjacent molybdenum atoms and can receive dative bonds from the four chloride Iigands.
Non-transition Elements 5 Non-transition Elements The Structural Features of : (a) B4H10 (b) B5H9(c) B5H11 (a) Tetraborane, B4 HI0: In this molecule four B-atoms may be regarded as a portion of slightly distorted octahedron. The structure of this molecule contains : (i) Fourt bridging (3c2e) BHB bonds (ii) One direct (2c2e) BB bond (iii) Six terminal (2c2a) BH bonds The structure can be represented as follows :
The structure and line representation of the bonding in tetraborane, B4H10.
Chemistry : Practical Application Thus six BH electron pair, one BB electron pair and four BHB bridge bonds together account for twentytwo valence electrons contributed by four boron and ten hydrogen atoms. Thus skeletal electrons are 22. (b) Pentaborane-9, B5H9 : In this molecule five B-atoms form a square-based pyramid. Four B-atoms located in the base of the square-pyramid are bonded to each other by four (3c _ 2e) BHB bonds while the B-atom located at the apex of the pyramid is bonded to the two basal B-atoms (2c _ 2e) BB bond. This molecule also contains one closed (3c - 2e) BBB bond. Thus B5 H9 molecule contains :
(i) Five terminal BH bonds (ii) Four bridging BHB bonds (iii) Two BB bonds (iv) One closed BBB bond Total 24 skeletal electrons are involved in various bonds formation in the molecule.
The geometrical structure of B5H9, and one of its four-bond resonance structures. (c) Pentaborane-11, B5H11 : This molecule has an unsymmetrical square pyramidal structure. Five Batoms occupy the five comes of a square pyramid. This molecule contains : (i) Eight terminal BH bonds (ii) Three bridging (3c - 2e) BHB bonds (iii) Two closed (3c - 2e) BBB bonds Non-transition Elements The structure is shown below :
Structure of pentaborane-11, B5H11 Total number of skeletal electrons are 26. The wave functions of molecular orbitals of diborane in terms of orbitals The hydrogen bridge structure of diborane (B2H6) is shown in Fig.
It contains two type of bonds : (i) Four terminal (2c - 2e) BH bonds (normal covalent a bonds) (ii) Two bridging (3c - 2e) BHB bonds In terms of molecular orbitals, the three centre BHB orbital may be considered to result from the combination of one sp3 orbital from each borane and the j-orbital of the hydrogen. Thus two bridging (3c 2e) BHB bonds are Chemistry : Practical Application formed by two of the four sp3 hybrid orbitals, one of them is empty and other is singly filled. Other two singly filled
sp3 hybrid orbitals of a boron atom form two terminal BH bonds. A three centre bond has a banana shape (banana bond). This is due to the repulsion between positive charges on the two bridge hydrogen atoms, causing the three centre bonds to be bent away from each other in the middle.
(A) Qualitative picture of bonding in diborane. (B) A common method of depicting BHB bridges Wave mechanical picture of diborane : Consider each borane atom to be sp3 hybridised. Two terminal BH bonds are cr bonds involving a pair of electron each. This accounts for eight of the total twelve electrons available for bonding. Each of the bridging BHB linkage then involves a delocalized or three centre bond as follows. The appropriate combination of three orbital wave functions, fB1, fB2 (approximately sp3 hybrids) and fH (an s-orbital) results in three molecular orbitals.
Bonding yb =
Non-bonding yn =
Antibonding yn = The diagrammatically possibilities of overlap together with the resulting MOs and their energies are given in Fig : Non-transition Elements
Qualitative description of atomic orbitals (left), resulting three-centre molecular orbitals (right), and the approximate energy level diagram (centre) for one BHB bridge in diborane. The preparation, properties and structure of S4N4 : S4N4, Tetrasulphur Tetranitride : It is prepared as follows : 6SC12 + 16NH3 ¾® S4N4 + 2S + 14NH4C1 6S2C12 + I6NH3
S4N4 + 8S + I2NH4CI
6S2C12 + 4NH4C1 ¾® S4N4 + 8S + 16HC1 The compound is also formed when sulphur reacts with anhydrous liquid ammonia. 10 S + 4 NH3
S4N4 + 6 H2S
S4N4 is solid, m.p. 178°C. It is thermochromic, that it is changes colour with temperature. At liquid nitrogen temperatures it is almost colourless, but at room temperature it is orange-yellow, and at 100°C it is red. S4N4 is very slowly hydrolysed by water, but reacts rapidly with warm NaOH with the break-up of the ring : S4N4 + 6NaOH + 3H2O ¾® Na2S2O3 + 2Na2SO3 + 4NH3
Chemistry : Practical Application
Structure of S4N4 The structure of S4N4 is given in figure. The structure is an eight membered heterocyclic ring and cradle shaped. Since decomposition of tetrasulphur tetranitride with alkali gives ammonia, and reduction with stannous chloride followed by decomposition with alkali also gives ammonia, there canbe no NN links in molecule. If the group NN were present, direct decomposition would yield ammonia and reduction followed by decomposition would yield hydrazine or its derivatives. S4N4 is diamagnetic SN bond lengths are 1.74Å, SS distance = 2.63Å, NN distance = 1.47. Band angles are SNN = 110°, SNS = 98°, NSN = 76°. B10C2H12 is isostructural and isoelectronic with what borane ion BxHy2_ : B10C2H12 is isostructural and isoelectronic with B12H122_ borane ion. The boron atom each have one fewer electron than a carbon atom. To keep the total number of electrons the same in B10C2H12 and the borane ion, the replacement of two carbon atoms with boron atoms must be accompanied by addition of two extra electrons.
Non-transition Elements
Framework of theB12H122_ ion Carboranes The carboranes are mixed hydrides of carbon and boron having both carbon and boron atoms in electron deficient skeletal framework. The polyhedral carboranes may be considered as formally derived from the BnHn2_ ions on the basis that the CH group is isoelectronic and isostructural with, and may thus replace, the BH_ group. Geometrically carboranes are classified into two types : (i) Those in which the boron atom framework closes in on itself to form a polyhedron. These are closo (cage) compounds. These have the general formula C2Bn _ 2 Hn (n = 5 _ 12). (ii) Those which have the open cage structures, derived formally from one or other of several boranes and containing from one to four carbon atoms in the skeleton. These are nido (nest) compounds. The boranes most commonly used in making the smaller carboranes are B4H10, B5H9 and B5H11. For example, B5H9 and C2H2 react in gas phase at 215°C to give mainly the nido carborane 2, 3 -C2B4H8. The same reactants at 450° or in an electric discharge give the closo carborane 1,5- C2B3H5 1, 6-C2B4H6 and 2, 4-C2B5H7. The nido carborane 2, 3-C2B4H8 is converted to the coloso-carboranes C2B3H5, C2B4H6 and C2B5H7 on pyrolysis orultraviolet irradiations.
Chemistry : Practical Application The closo-carboranes of the CnBn _ 2Hn series are isoelectronic with corresponding [BnHn]2_ ions and have the same closed polyhedral structures with one hydrogen atom bonded to each carbon and boron.
The nido -carboranes formally related to B6H10. All have eight pairs of electrons bonding the six cage atoms together. Hydrogen bridges are represented by curved lines.
An account of phosphazenes with their structural aspects: Phosphazenes or Phosphonitrilic Compounds : Phosphazenes are a group of compounds represented by the general formula (NPX2)n where X = F, Cl, Br, SCN, CH3, C6H5 etc. All these compounds are polymeric, in these compounds P atom is in oxidation state (+V) and N is in the (+ III) state. The compounds are formally unsaturated. Phosphonitrilic chlorides are important phosphazenes with the general formula (NPC12)j where x ranges from 3 to 7 and Non-transition Elements upwards. The monomer (x = 1) and dimer (x = 2) are not known. These compounds were originally called phosphonitrilic halides, but are now named systematically poly (chlorophosphazenes). These can be prepared as follows : By the ammonolysis phosphorus pentachloride
3PC15 + 3NH3 ¾¾® (NPC12)3 + 9HC1 4PC15 + 4NH3 ¾¾® (NPC12)4 + 12HC1 By the reaction of ammonium chloride with phosphorus pentachloride xPCl5 + xNH4C1
(NPC12)X + 4xHC1
The chlorine atoms are reactive, and most reactions of chlorophosphazenes involve replacement of Cl by groups such as alkyl, aryl, OH, OR, NCS, or NR2. Alkyl or aryl groups may be introduced using lithium or grignard reagents. (NPCl2)3 + 6CH3MgI ¾¾® [NP(CH3)2]3 + 3MgCl2 + 3MgI2 (NPCl2)3 + 6C6H5Li ¾¾® [NP(C6H5)2]3 + 6LiCl (NPCl2)3 + 6NaOR ¾¾® [NP(OR)2]3 + 6NaCl On hydrolysis chlorine atoms of chlorophosphazenes can replaced by OH groups. Structure : X-ray examination reveals that the trimer and tetramer chlorophosphazenes are having the six and eight membered rings and composed of alternate nitrogen and phosphorus atoms. Thus these compounds may be regarded as the phosphorus-nitrogen analogues of benzene and cyclo-octatetraene (NPC12)3 has almost planar ring. There is an approximately tetrahedral distribution of valencies round phosphorus corresponding with sp3 hybridisation with the result that the chlorine atoms in each pair lie on opposite side of the rings. All the NP bonds in the ring are also of equal length i.e., 1.59 ± 0.02Å.
Chemistry : Practical Application
The Structure of N3P3Cl6 trimer. In the above structure it is evident that two chlorine atoms are attached to each phosphorus atom. This is confirmed by two chemical experiments such as : (i) Benzene reacts with N3P3CI6 in the presence of AICI3 thus:
The product (III) is hydrolysed by water to (C6H5)2 PO.OH. Hence both the chlorine atoms originally have been attached to the same phosphorus atom. (ii) Phenyl magnesium bromide at 115° also reacts with N3P3C16 to produce diphenyl derivatives. When molecular orbital theory is applied to the structure of N3P3C16, it is observed that there are normal localised a bonds through the ring formed by overlap of sp3 hybrid orbitals of phosphorus with sp2 orbital of nitrogen. In addition, each introgen atom has an electron in a pz orbital and each phosphorus atom in a d-orbital. These orbitals combine to give delocalised pp - dp orbitals which extend over the whole ring structure. The presence of negative chlorine atoms on Non-transition Elements phosphorus makes the diffuse phosphorus d-orbitals more compact and favours their overlap. As shown below, resonance structures can be drawn analogous to those for benzene indicating aromaticity
in the ring
The resonance in the trimer is justified by the fact that all the NP distances are the same (1-67 Å) which are less than that expected for a PN single bonds (1.78°A) as in sodium phosphoramidate. The tetrameric chloride (NPC12)4 has a remarkable puckered ring structure of alternate phosphorus and nitrogen atoms, with two chlorines on each phosphorus atom.
The Structure of trimer and tetramer of phosphonitrilic chlorides. [NPC12]3 is flat and [NPC12]4 exists in the chair and boat conformation.
Flat structure of (NPCl2)3; Chair and bot form of (NPCl2)4
Chemistry : Practical Application The reaction completed and given the structure of the main product :
[PNCl2]3 + excess (CH3)2 NH ® [PNC12]3 + excess (CH3)2 NH ® [P3N3{(CH3)2 N}6]
Balanced Reactions (i) Reaction of potassium superoxide with carbon dioxide (ii) Hydrolysis of phosphorus sulphide (iii) Reaction of sodium chlorite with nitrogen trichloride (iv) Hydrolysis of calcium cyanamide (v) Reaction of borontrioxide with cobalt oxide (vi) Hydrolysis of phosgene (vii) Reaction of carbon tetrachloride with hydrogen fluoride in anhydrous conditions. (viii) Reaction of hydrazine with zinc in acidic medium (ix) Diborane reacts with ammonia (x) Silica reacts with carbon in an electric furnace (xi) Reaction of ammonia with disulphur dichloride (xii) Reaction of bromate and chloride in acid medium (xiii) Hydrolysis of nitrogen trichloride (xiv) Reaction of bromate and bromide in acid solution
(xv) Hydrolysis of borazine (xvi) Hydrolysis of silicon tetrafluoride (xvii) Hydrolysis of tetrasulphur tetranitride Non-transition Elements (i) KO2 + 2CO2 ¾¾® K2CO3 + O2 This reaction make the use of KO2 in space capsules, submarines and breathing marks, because i both produces oxygen and removes carbondioxide. Both functions are important in life support system. (ii) P4SI0+16H2O ¾¾® 4H3PO4 + 10H2S (iii) 6NaClO2 + 2NC13 ¾¾® 6ClO2 + 6NaCl + N2 and NCI3 + 3H2O + 6NaC1O2 ¾¾® 6ClO2 + 3NaCl + 3NaOH + NH3 (iv) CaNCN + 3H2O ¾¾® 2NH3 + CaCO3 (v) CoO + B2O3 ¾¾® Co(BO2)2 Cobalt metaborate Blue colour (vi) COC12 + H2O ¾¾® 2HC1 + CO2 Phosgene
(vii) CCl4 + 2HF
CCl2 + F2 + 2HCl
Feron (viii) N2H4+ Zn + 2HCl 2NH3 + ZnCl2
Hydrazine (-II)
(ix)
B2H6.2NH3
(x) SiO2 + 3C ¾¾® Si + 2CO Si + C ¾¾® SiC
(xi) 6S2C12 + I6NH3
Chemistry : Practical Application (xii) 2BrO3_ + 10C1_ + 12H+ ¾¾® Br2 + 5C12 + 6H2O (xiii) NCl3 + 4H2O ¾¾® NH4OH + 3HOC1 (xiv) BrO3_ + 5Br_ + 6H+ ¾¾® 3Br2 + 3H2O (xv) B3N3H6 + 9H2O ¾¾® 3NH3 + 3H3BO3 + 3H2 (xvi) 3SiF4 + 4H2O ¾¾® H4SiO4 + 2H2SiF6 (xvii) S4N4 + 6OH_ + 3H2O ¾¾® S2O32_ + 2SO32_ + 4NH3 The reaction steps and condition for the following conversions : (i) PC15 to poly-dichlorophosphazene (ii) CO to Urea (i) Conversion of PC15 to Poly-dichlorophosphazene : It takes place at 120° - 150°C in an inert solvent
like tetrachloroethane. xPC15 + xNH4C1 ¾¾® (NPCl2)x + 4x HCl The reaction steps may be (1) NH4C1 + PC15 ¾¾® NH4PC16 unstable (2) NH4PC16 ¾¾® HNPC13 + 3HC1 (3) xHNPC13
(NPC12)x + xHC1
(ii) Conversion of CO to Urea : It takes place in following steps : CO + C12 ¾¾® COC12 Carbonylchloride (phosgene)
or
¾¾® CO2
Non-transition Elements
CO2 + 2NH3
The reactions supplying the missing reactions and products: (i) 3BC13 + 3NH4C1 ¾¾® (ii) A12(CH3)6 + 2H2O ¾¾® (in) NO + O3 ¾¾® (iv) n[(CH3)2 SiO4] + (CH3)3 SiOSi (CH3)3 (v) [3Ca3(PO4)2] CaF2 + 7H2SO4 ¾¾® (vi) PtF6 + O2 ¾¾® (vii) SbF5 + BrF3 ¾¾® (viii) PI3 + 3H2O ¾¾®
(i) 3BC13 + 3NH4C1 ¾¾® (ii) A12(CH3)6 + 2H2O ¾¾® 2Al(OH)3 + CH4 Correct balanced equation is A12(CH3)6 + 6H2O ¾¾® 2Al(OH)3 + 6CH4 (iii) NO + O3 ¾¾® NO2 + O2 (iv) n[(CH3)2SiO4 ] + (CH3)3 SiOSi (CH3)3 (CH3)3SiO[Si(CH3 )2O]Si(CH3)3 (v) [3Ca3(PO4)2] CaF2 + 7H2SO4
(vi) PtF6 + O2 ¾¾® O2+ [PtF6]_
Chemistry : Practical Application (vii) SbF5 + BrF3 ¾¾® SbF3 + BrF5 (viii) PI3+3H2O ¾¾® H3PO3 + 3HI The balanced equation for the reactions of the following with water under ambient conditions: (i) PC13(ii) NC13(iii) BrF3(iv) Al4C3(v) SF6
(i) PCI3 + 3HOH ¾¾®
(ii) NCl3 + 4H2O ¾¾®
(iii) 2BrF3 + 3H2O ¾¾®
(iv) Al4C3 + 12H2O ¾¾® (v) SF6(g) + H2O(g) ¾¾® SO3(g) + 6HF; DG = _200 kJ Explained : (i) BF bond is larger in BF4_ than in BF3 molecule. (ii) The order of Lewis acid strength of different halides of boron is BF3 < BC13 < BBr3 or, BBr3 is a stronger Lewi's acid than BF3. (iii) N(CH3)3 is pyramidal in shape while N(SiH3)3 has planar triangular arrangement. or
(SiH3)3N is a weaker base than (CH3)3N. (iv) Dipole moment of NH3 molecule is larger than that of NF3. (v) PF3 can act as donor molecule while NF3 show little tendency to act as donor. (vi) C1F3 exists whereas FC13 does not. Non-transition Elements (vii) Ba(OH)2 is fairly soluble in water but Mg (OH)2 is not. (viii) Dipole moment of CH3CI is greater than that of CH3F. (ix) HClO4 is an acid and an oxidising agent whereas H2C2O4 is an acid and a reducing agent. (i) BF bond is larger in BF4_ than in BF3 molecule. This is due to the fact that in BF4_, boron atom is sp3 hybridised while in BF3, boron atom is sp2 hybridised and bond length decreases with increase in scharacter since s-orbital is samller than a p-orbital. In case of sp3 hybridisation 25% s-character is there while in sp2 hybridisation 33-3% s-character is there. (ii) The order of Lewis acid strength of different halides of boron is BBr3 is a stronger'Lewis acid than BF3. This order is just reversed than that expected on the basis of electronegativity values of halogens. The electron density from the filled orbitals on halogens is transferred to the empty orbital present on the boron atom. This is known as back donation. This is explained on the basis of overlapping of the 2p-filled orbital of halogens sidewise with the empty 2p-orbital of boron atom forming pn-pn back bonding
Due to back bonding, the electron deficiency in the boron atom is decreased, consequently, its Lewis character or electron accepting tendency decreases. Since back bonding is maximum in case of fluorine because of its small size, BF3 is least acidic. The tendency of back bonding decreases as BF3 > BC13 > BBr3, the acidic character falls as BBr3 > BC13 > BF3.
Chemistry : Practical Application The tendency of back bonding decreases as BF3 > BC13 > BBr3 because the overlapping of 3p and 4p filled orbitals of Cl and Br atoms with empty 2p-orbitals of B atoms does not take place effectively due to difference in the energy state of the orbitals involved. (iii) The geometry around the nitrogen atom in trimethyl amine N(CH3)3 is pyramidal (sp3 hybridisation)
due to lone pair-bond pair repulsion. In case of similar silicon compound, N(SiH3)3 called trisilylamine, it is planar triangular arrangement of its three bonds (sp2 hybridisation N atom). In this case the lone pair on nitrogen present in 2p orbital is transferred to the empty d-orbital of silicon forming dp-pp bond. Moreover, this makes (SiH3) N a weaker base than (CH3)3 N. (iv) The dipole moment of NH3 is more than that of NF3. This is due to different directions of the bond moments of NH and NF bonds. In the first case N is more electronegative but in second case F is more electronegative as shown below
Non-transition Elements Thus in NH3, the dipole moments of NH bonds are in the same direction as that of the lone pair but in NF3, the dipole moments at NF oppose that of the lone pair. (v) Both NF3 and PF3 are weak Lewis bases or donors due to the presence of a lone pair of electron on the central atom. PF3 acts as a donor while NH3 shows little tendency to act as donor is because phosphorus is less electronegative than nitrogen and phosphorus has a larger size than nitrogen. Due to greater electronegativity N cannot easily part with the lone pair of electrons. In metal complexes, the a donation by PF3 to metal atom is stabilized by back 7t donation from the filled orbital of the metal to the empty orbital of F. (vi) Outer electronic configuration of Cl is 3s2 3p5. An electron from 3p can jump to 3d orbitals, so it can show an oxidation state of +3 and combine with more electronegative fluorine. Outer electronic configuration of F is 2s22p5. No d orbital is available (when n = 2) for excitation of electron. Moreover, fluorine being the most electronegative element, it shows .an oxidation state of _ l only. (vii) Ba(OH)2 is fairly soluble in water but Mg(OH)2 is not. Both Ba and Mg are the members of IIA or 2 group (alkaline earth metals). Thus in general the solubility of the alkaline earth metal hydroxides in water increases with increase in atomic number down the group. This is due to the fact tot lattice energy decreases down the group due to increase in size of alkaline earth metal cation whereas the hydration energy of the cations remains almost unchanged. Thus AH solution becomes more negative as we move from Be(OH)2 to Ba(OH)2 which accounts for increase in solubility or Ba(OH)2 is fairly soluble Mg (OH)2 is not. (viii) Dipole moment of CH3F is less than that of CH3C1: No doubt CF bond is more polar than CCl bond due to greater electronegativity of fluorine atom than chlorine but actually CF bond length is much smaller than CCl bond length, that is why CH3C1 has greater dipole moment.
Chemistry : Practical Application (ix) HClO4 is a strong acid and an oxidising agent. Whereas H2C2O4 is also an (weak) acid but a reducing agent. Chlorine is more electronegative than carbon and hence can accept electron and get converted to Cl_. In HClO4, the chlorine is in its maximum oxidation state so cannot act as a reducing agent. Recovery of elemental silver from silver resides from photographic processing (AgCl) is achieved by converting it into A, using common ionic compound B. The compound A upon heating decomposes to give an intermediate compound C before giving metallic silver as the end product. A, B and C by giving equations for the reaction involved, identified :
(a) When AgCl is fused with Na2CO3, Ag2CO3 is formed. 2AgCl + Na2CO3 ¾¾® Ag2CO3 + 2NaCl (B) (A) Ag2CO3
Ag2O + CO2
(C) 2Ag2O3
4Ag + O2
AlBr3 dimerises to Al2Br6, while BCI3 is monomeric : AlBr3 in the vapour state as well as in inertorganic solvents exists as dimer. In the dimeric structure, each aluminium atom forms one coordinate bond by accepting a lone pair of electrons from bromine atom of another AlBr3 molecule and thus completes an octet of electrons.
Both boron and aluminium halides are Lewis acids but only aluminium halides exist as dimer whereas boron halides exists as monomers. This is due to the reasons that boron atom is so small that it cannot accommodate four large sized halides Non-transition Elements ions around it. Moreover, in case of boron halides there is some back donation (p-back bonding) from halide to boron. Experimentally found value of SiF distance in SiF4 is less than that of the theoretical value, reason : Experimental Values of Si : F bond distance is less than theoretical values because of a dative p-back bonding arising from the presence of filled p-orbital in fluorine and empty `d' orbital in silicon. HC1 is not a suitable acid medium permanganimetric titrations :
In permanganimetric titration KMnO4 is always acidified with H2SO4 because H2SO4 is not self oxidising. HCl and HNO3 are never used as KMnO4 reacts with HC1 liberating chlorine (KMnO4 is oxidised) liberating chlorine 2KMnO4 + 16HC1 ¾¾® 2KC1 + 2MnCl2 + 5C12 + 8H2O HNO3 is not used because it is itself an oxidising agent. The structure of oxides of nitrogen and phosphorus. Structure of Oxides of Nitrogen Nitrous Oxide (N2O) : A symmetrical linear molecule, polar structure
Nitric Oxide (NO) : Odd electron molecule, paramagnetic in gaseous state but diamagnetic in solid state due to dimeric structure, Bond order 2.5
NO : KK s (2s)2 (2s)2 s (2pz )2 p(2px)2 p(2py)2
Chemistry : Practical Application Nitrogen Sesquioxide (N2O3)
(2px)1
The oxide exists in two different forms. These may be intercon verted by irradiation with light of appropriate wavelength.
Nitrogen Dioxide (NO2) and Dinitrogen Tetraoxide (N2O4) The NO2 is obtained as a brown liquid and on cooling becomes colourless solid due to dimerization of NO2 into N2O4
NO2 is an odd electron angular molecule O = ® O N2O4 has planar structure
Non-transition Elements Nitrogen Pentoxide (N2O5)
Solid N2O5 is ionic NO2+ NO3_. It is covalent in solution and in gas phase and has the structure.
Structure of Oxides of Phosphorus : Phosphorus (or other elements of group VA) form fewer oxides than does nitrogen, presumably because of the inability of these elements to form pp-pp double bonds. Phosphorus Trioxide (P4O6) : It has four P atoms at the corners of a tetrahedron with six O atoms along the edges. Each P being covalently bonded to three O atoms, each of which is bonded to two phosphorus atoms. There exists a considerable double bond character in the PO bonds because of the formation of a pp-dp dative bond with oxygen. Phosphorus Pentoxide (P4O10): In this each P atom forms three bonds to the oxygen atoms and also an additional coordinate bond with an oxygen atom. Terminal PO bond is much shorter than PO bond. It shows that there is a considerable pp-dp back bonding because of the lateral overlap of full p-orbitals on oxygen with empty dp-orbitals on phosphorus.
Chemistry : Practical Application Structure of Phosphorus Halides : The halogens react .with white phosphorus to form two types of halides, PX3 and PX5. Iodineis an exception, since its compounds with phosphorus are PI3 and P2I4. PC13 is rapidly hydrolysed to H3PO3. PCI3 + 3H2O ¾¾® H3PO3 + 3H+ + 3C1_ The trihalides all have pyramidal geometry similar to NH3 and PH3, the XPX bond angle is near102°.
Pyramidal structure of Trigonal bipyramidal structure PCI3 (sp3 hybridization) of PC15 (sp3d hybridization) In the vapour phase the pentahalides PF5 PC15 and PBr5 are discrete molecules that have trigonal bipyramidal structure. Each PCl bond is a sigma bond formed by sp3d-p overlap. The d-orbital used in sp3d hybridisation is 3dz2. In this three bond angles are of 120°, the two bond angles are 90° each. In the solid phase, however PCI5 exists as an ionic solid consisting of PC14+ and PCl6_ which have the following structures,
The phosphoryl halides (POX3) have a slightly distorted tetrahedral structure, with XPX bond angles of approximately 103°. Non-transition Elements The Structure of Compounds : N3H, TeCl4, H4P2O7, SO32-, Borax N3H : (Hydrogenazide)It has a bent structure.
The bond angle HNN is 112° and two NN bonds are of different lengths. TeCl4 (Tellurium Tetrachloride) : The structure of TeCl4 is established as distorted trigonal bipyramidal as shown below with one equatorial position occupied by a lone pair.
H4P2O7 (Pyro or Dipolyphosphoric Acid) : It is a tetrabasic acid but gives rise to only two series of salts e.g., Na2H2P2O7 and Na4P2O7. It also responds to ammonium molybdate test which is characteristic for ortho-phosphates. The acid has been assigned the following structure
SO32_ (Sulphite Ion) : The sulphite ion exists in crystals and has a pyramidal structure, that is tetrahedral with one position Chemistry : Practical Application occupied by a lone pair. The bond angles OSO are slightly distorted (106°) due to the lone pair, and the bond lengths are 1-51 Å. The n bond is delocalized, and hence the SO bonds have a bond order of 1.33.
Borax (Na2B4O7.10H2O) : It is made up of two triangular and two tetrahedral units. The ion is [B4O5(OH)4]2_ , remaining eight water molecules are associated with two sodium ions. Thus borax should be better formulated as Na2 [B4O5(OH)4].8x 2O.
Non-transition Elements The Structures of Sulphides of Phosphorus : Sulphides of Phosphorus : Sulphides of the formula P4Sn (where n = 3, 5, 7, 8, 9, or 10) have been described. These may be obtained by heating red phosphorus with sulphur in appropriate amount.
Structure of P4 and its sulphides. Similar to phosphorus oxides, the sulphides may be regarded as derivatives of the P4 tetrahedral structure. There Chemistry : Practical Application are three sulphur bridges in P4S3, four in P4S5, five in P4S7 and 6 in P4S9. Non-bridged sulphur atoms occupy terminal positions. The structure of P4S10 is exactly analogous to that of P4O10. The Structures of SO2, SeO2 and TeO2, Compared :
SO2 gas forms discrete V-shaped molecules and this structure is retained in the solid state. The bond angle is 119-30°, corresponding to a planar trigonal structure involving sp2 hybridisation. Out of the three sp2 hybrid orbitals of sulphur, one is occupied by a lone pair of electron. The remaining two half filled orbitals overlap with one of the p orbitals of each of the two oxygen atoms forming sigma bonds. This gives rise to planar trigonal structure, the bond angle OSO is slightly reduced from 120° to 119.5° due to repulsion between lone pair and a bond pair.
Sulphur atom has still two half-filled orbitals i.e., one p and one d-orbital. Each oxygen atom has one halffilled p-orbital. So p-orbital of one of the O-atom overlaps with p-orbital of S-atom forming pppp bond. Similarly, p-orbital of the second O atom forming pp_dp(3dxz) bond. Both the SO bonds have exactly the same length because of resonance.
Multiple bonding explains the shorter length of SO bond (SO3 has planar trigonal structure involving sp2 hybridisation of S-atom, three s bond, one pppp and 2ppdp bonds are formed. SeO2 is a white volatile solid. In the gaseous state, it exists in the form of discrete molecules, just like SO2 but in the solid Non-transition Elements state it has a polymeric structure comprising of infinite chains which are not planar.
TeO2 is non-volatile crystalline ionic solid and occurs in two crystalline forms. The Structure of Halides of Sulphur :
SF6: It has an octahedral structure, as shown in fig.
sp3d2 hybridization in SF6 molecule. Dotted arrows represent electrons supplied by fluorine atoms. SF4: It has trigonal bipyramidal in structure. However, due to the presence of a lone pair of electrons, there is distortion in the molecule and the bond angles are 89° and 177° instead of normal bond angles of 90° and 180°, respectively.
Chemistry : Practical Application SC12 : The formation of SC12 molecule is illustrated in figure.
S2F2 : Monohalides of sulphur are dimeric like S2Cl2, S2F2. The electron diffraction studies have shown that the structure of S2F2 is nonplanar and is similar to that of H2O2, as shown in figure : S2Cl2 has got similar structure having bond angle of 104°.
Various Types of oxoacids of sulphur, formed : Oxoacids of Sulphur: Sulphur forms a number of oxoacids, many of the oxoacids of sulphur do not exist free acids, but are known as anions. Oxoacids of sulphur may be classified into following four series: 1. Sulphurous acid series 2. Sulphuric acid series 3. Thionic acid series 4. Peroxoacid series Non-transition Elements Sulphurous Acid Series
H2SO3 sulphurous acid
S(+IV)
H2S2O4 dithionous acid
S(+III)
S(+IV)
S(+VI) Thionic Acid Series
H2S2O6 dithionic acid
S(+V)
Chemistry : Practical Application
Peroxoacid Series
S(+VI)
S(+VI) The Silicones The silicones are a group of organosilicon polymers. The complete hydrolysis of SiCl4 gives SiO2 which has a stable three dimensional structure. Hydrolysis of alkyl substituted chlorosilane yield long chain polymers called silicones (not the silicon compounds analogous to ketone)
Non-transition Elements Since an active OH group is left at each end of chain, polymerisation reaction continues and the length of the chain continue to increase. The starting materials for the manufacture of silicones are alkyl or aryl substituted chlorosilanes. Methyl compounds are mainly used, though some phenyl derivatives are used as well. Hydrolysis of dimethyldichlorosilane (CH3)2SiCI2 gives rise to straight chain polymers and, as an active OH group is left at each end of the chain, polymerization continues and the Chain increases in length. (CH3)2SiCl2 is therefore a chain building unit. Normally, high polymers are obtained.
Hydrolysis under carefully controlled conditions can produce cyclic structures, with rings containing three, four, five or six Si atoms.
Hydrolysis of trimethylmonochlorosilane (CH3)3SiCl yields (CH3)3SiOH trimethylisilanol as a volatile liquid, which can condense, giving hexamethyldisiloxane. Since this compound has no OH groups, it cannot polymerize any further.
Chemistry : Practical Application If some (CH3)3 SiCl is mixed with (CH3)2 SiCl2 and hydrolysed, the (CH3)3SiCl will block the end of the straight chain produced by (CH3)2SiCl2, Since there is no longer a functional OH group at this end of the chain, it cannot grow any more at this end. Eventually the other end will be blocked in a similar way. Thus (CH3)3SiCl is a chain stopping unit, and the ratio of (CH3)3SiCl and (CH3)2SiCl2 in the starting mixture will determine the average chain size. The hydrolysis of methyl trichlorosilane RSiCl3 gives a very complex cross-linked polymer.
It may be noted that unlike carbon which cannot hold more than one OH group, the silicon atom in the compound RSi(OH)3 can hold three OH groups. It is this property of silicon that makes the formation of organosilicon polymers possible. The strength and inertness of silicones are related to two factors. 1. Their stable silica-like skeleton of SiOSiOSi. The SiO bond energy is very high (502 kJ mol_1). 2. The high strength of the SiC bond. Silicones were originally developed as electrical insulators, because they are more stable to heat than are organic polymers, and if they do break down they do not produce conducting materials as carbon does. They are resistant to heat, oxidation and most chemicals. They are strongly water repellent, are good electrical insulators, and have non-stick properties and anti-foaming properties. Their strength and inertness are related to two factors. Non-transition Elements
Graphite Compounds A very loose, layered structure of graphite makes it possible for many molecules or ions to penetrate between the layers, forming interstitial or lamellar compounds. These are of two basic types: those in which the graphite, which has good conducting properties, becomes non-conducting and those in which high electrical conductivity remains and is enhanced. Only two substances of first types are known, namely graphite oxide and graphite fluoride. Graphite oxide is obtained by treating graphite with strong aqueous oxidising agents such as fuming HNO3 or KMnO4. Its composition is not fixed but approximates to C2O, the layer separation increases to 67Å and it is believed that oxygen atoms are present in COC bridges ; the graphite layers thus lose their unsaturated character and buckle. Graphite fluoride (poly carbon monofluoride) is obtained by direct fluorination of graphite at a temperature of ~ 600°. The product (CF)n is non-stoichiometric. The layer spacing is ~ 8Å and the layers are most likely buckled. In electrically conducting lamellar compounds, various atoms, molecules or ions are inserted between the carbon sheets. A large number of compounds are formed spontaneously when graphite and the reactants are brought into contact. Alkali metals, halogens, FeCl3, UC14, FeS2 and MoO3 form lamellar compounds spontaneously. The manner in which the invading reactant species increase the conductivity of graphite is not definite, but apparently they do so by either adding electrons to or remaining electrons from the conduction level of graphite. Spring Reaction : It is used for the preparation of sodium thiosulphate. It consists in treating a mixture of sodium sulphide and sodium sulphite with calculated quantity of iodine. Na2S + Na2SO3 + I2 ¾¾® Na2S2O3 + 2NaI Lonic Mobility : Lonic mobility or conductivity measurements in aqueous solution give results in the order Cs+ > Rb+ > K+ > Na + >Li +
Chemistry : Practical Application The reason for this anomaly is that the ions are hydrated in solutions. Since Li+ is very small, it is heavily hydrated. This makes the radius of hydrated ion large, and hence it moves very slowly. Cs+ is least hydrated and radius of hydrated Cs+ ion is smaller.
Sulphur is reduced by a solution of sodium in liquid NH3 but not by a solution of sodium in water. This is because alkali metals are stronger reducing agents than hydrogen and so will react with water and liberate hydrogen. The metals can exist for some time in liquid ammonia. S + Na/NH3 ¾¾® Na2S S is reduced from oxidation state O to _II. In the crown ether complex dibenzo-18-crown-6 there are two benzene rings and 18 atoms makes up a crown-shaped ring and six of the ring atoms are oxygen. These six oxygen atoms may complex with a metal ion with larger metal ions of group I. Similarly benzo-12-crown-4 has a ring of 12 atoms, four of which are oxygen. The structure of these compounds are as :
Non-transition Elements Structure of Basic Beryllium Acetate : Basic beryllium acetate [Be4O(CH3COO)6] is formed if Be(OH)2 is evaporated with acetic acid. The structure comprises a central oxygen atom surrounded by four beryllium atoms located at the corners of a tetrahedron, with six acetate groups (Ac) arranged along the six edges of the tetrahedron. Similar structure is there for a series of stable covalent molecules of formula [Be4O(R6)] where R may be NO3_, HCOO_, CH3COO_, C2H5COO_ C6H5COO_ etc.
Structure of Beryllium Oxalate : In beryllium oxalate [Be(OX)2]2_, Be2+ ion is tetrahedrally co-ordinated. Similar structures are formed with b-diketones (acetylacetone and catechol). Boric Acid : Boric acid (H3BO3) is written as B(OH)3. ft behaves as a weak monobasic acid. It does not donate protons like most acids, but rather it accepts OH~. B(OH)3 + 2H2O
[B(OH)4]_
or, [H3BO3] Beryllium and Aluminium borohydrides : These are covalent and volatile. In these the [BH4]_ groups acts as a ligand. One or more H atoms in a [BH4]_ act as a bridge and bond to the metal, forming a three centre bond with two electrons. The structures are as
Chemistry : Practical Application
Boron nitride is a white slippery solid. One B atom and one N atom together have the same number of
valency electrons as two C atoms. Thus boron nitride has almost the same structure as graphite, with sheets made up of hexagonal rings of alternate B and N atoms joined together.
Borazine (B3N3H6) is sometimes called `inorganic benzene' because its structure shows some formal similarity with benzene, with delocalized electrons and aromatic character.
Non-transition Elements Structure of Oxides of Halogens Oxides of Halogen : Oxygen is less electronegative than F, so binary compounds of F and O are fluorides of oxygen. The other halogens are less electronegative than oxygen and thus form oxides. There is only a small difference in electronegativity between halogens and oxygen, so the bonds are largely covalent, except I2O4 and I4O9 which are stable and ionic. OF2, C12O, Br2O: 2F2 + 2OH ¾¾® 2F_ + F2O + H2O 2HgO + 2Cl2 ¾¾® HgCl2 · HgO + C12O
The structures of OF2, C12O and Br2O are all related to a tetrahedron with two positions occupied by alone pair of electrons.
Repulsion between the bond pairs reduces the bond angle in F2O from the tetrahedral angle of 109°28' to 103°. In C12O (and presumably Br2O) the bond angle is increased because of steric crowding of larger halogen atoms. In other way in OF2, the bonding electrons are nearer of F atom because of higher electronegativity, so the repulsion between the lone pairs exceeds that of bonding pairs, resulting in decrease of bond angle. In C12O bonding electrons are near to O atoms because of its greater electronegativity. Hence repulsion between the bonding pairs of electrons exceeds that between the lone pairs. Hence the bond angle is greater than 109°28'. O2F2 : Its structure is similar to that of H2O2. The bond angle (FOO) is 109.5°. OO bond distance is shorter (1.22Å) than H2O2 (1.48Å)
Chemistry : Practical Application
C1O2 : It is formed by the following reaction : 3C1O3_ + 2H+ ¾¾® 2C1O2 + ClO4_ + H2O It is paramagnetic and contains an odd number of electrons and is, therefore, highly reactive. C1O2 molecule is analogous to that of NO2 in that it contains a three electron bond and has an angular shape but C1O2 does not dimerise probably because the odd electron is delocalized. The bond lengths are both
1.47Å and are shorter than for single bond and resonance is thought to occur between two structures. The ClOCl bond angle is 118°.
C12O7 : The molecule is polar and electron diffraction analysis reveals that two C1O4 tetrahedra are linked through oxgen atom in its structure.
I2O5 : I2O5 solid is a three dimensional network with strong intermolecular I...O interactions linking molecules together. Non-transition Elements
Bond Angle in C12O is Less than C1O2 : This is because the repulsion for the bonded pairs is less in C1O2 than that in C12O due to non-bonding electrons on the central atom in C1O2. The shorter bond length in C1O2 results from resonance, with the unpaired electron on the chlorine or the oxygen atom. Preparation of Compounds : (a) 3C12+ 6NaOH ¾¾® 5NaCl + NaC1O3 + 3H2O 4NaClO3 NaCl + 3NaC1O4
4NaClO4 + H2SO4 ¾¾® NaHSO4+ HC1O4 (b) I2 + 6OH_ ¾¾® 5I_ + IO3_ + 3H2O IO3_ + H+ ¾¾® HIO3 2HIO3
I2O5 + H2O
(c) 2C12_ + 2HgO ¾¾® HgCl2.HgO + C12O (d) C12 + 6OH_ ¾¾® 5Cl_ + ClO3_ + 3H2O 3C1O3 + 2H+ ¾¾® 2ClO2 + ClO4_ + H2O (e) 3Br2 + 6KOH ¾¾® 5KBr + KBrO3 (f) 2F2 + 2OH_ ¾¾® 2F_ + F2O + H2O
(g)
Br2 + O3
BrO3
(h) 2Br2 + HgO ¾¾® Br2O + HgO Reaction of Cyanogen : Cyanogen (C2N2) disproportionates upon treatment with strong base, as do Cl2, Br2 and I2 (C2N2 is a pseudo-halogen). C2N2 + 2OH_ ¾¾® CN_ + CNO_ + H2O Structure of Silicates Silicates are the compounds in which the anions present are either discrete SiO4_ tetrahedra or a number of such units Chemistry : Practical Application joined together through corners. These discrete tetrahedra or a number of tetrahedra linked together are
present as the anions in the silicates. However, when they link together, they do so by corners and never by edges. Thus the anions present in the silicates may be SiO44_ (for discrete tetrahedra), Si2O76_ (for two tetrahedra linked through one corner) and so on. Classification of Silicates : On the basis of corners (0; 1, 2, 3 or 4) of the SiO4 tetrahedra shared with other neo-silicate. Orthosilicates [SiO4]4_ : They contain discrete (SiO4)4_ tetrahedra, i.e., they share no corners(Fig.)
General Formula : M2III [SiO4], where M may be Be, Mg, Fe, Mn or Zn or MIV [SiO4] e.g., ZrSiO4. Different structures are formed depending on the coordination number adopted by metal. For example, in willemite Zn2[SiO4] and phenacite Be2 [SiO4], the Zn and Be atoms have a coordination number of 4 and occupy tetrahedral holes. In forsterite Mg2[SiO4], the Mg has a coordination number of 6 and occupies octahedral holes. When octahedral sites are occupied, it is quite common to get isomorphous replacement of one divalentmetal ion by another of similar size, without changing the structure Zircon ZrSiO4 has a coordination number S. The structure is not close packed. Pyrosilicates (Si2O7)6_ - soro-silicates, Disilicates : Two tetrahedral units are joined by sharing the O at one corner, thus giving (Si2O7)6_. The structure possessed by them are called island silicates structure(Fig) Non-transition Elements
Example. Therteveitite Sc2(Si2O7), in this Sc3+ ions are octahedrally coordinated. Hemimorphite Zn4(OH)2 (Si2O7). H2O. Cyclic or Ring Silicates : If two oxygen atoms per tetrahedron are shared to form closed rings such that the structures with the general formula
are obtained, the silicates containing these
anion are called cyclic silicates. The typical examples of such anions are structures are shown in Fig.
Their
Examples : Wollartnite Ca3 (Si3O9), Benitoite BaTi [Si3O9], Beryl Be3Al2 [Si6OI8]. Chain Silicates, Linearpolyanion, Metasilicates : Simple chain silicate or pyroxenes are formed by the sharing of the O atoms on two corners of each tertahedron with other tetrahedra. This gives the formula Fig.
Chemistry : Practical Application
Examples : Spodumene LiAl [(SiO3)2], Diopsite CaMg[(SiO3)2], Wollastonite Ca3[(SiO3)3]
Double chains can be formed when two simple chains are joined together by shared oxygen.These minerals are called amphiboles. There are several ways of forming double chains giving formulae like
and others.
is shown in Fig.
A typical amphibole is termolite Ca2Mg5(Si4O11 )2 (OH)2. Asbestos minerals are all amphiboles. Sheet Silicates : The sharing of three corners (i.e., three oxygen of each tetrahedron results in an infinite two dimensional sheet structure of the formula Fig. There are strong bonds within the SiO sheet, but much weaker forces hold each sheet to the next one. Thus these minerals Non-transition Elements tend.to cleave into thin sheets. These include clay minerals, mica.
Three Dimensional Silicates : If all the four corners (i.e., all the four oxygen atoms) are shared with other tetrahedra, three dimensional network of formula SiO2 is obtained. For example quartz, tridymite, cristobalite etc. Interhalogen Compounds Covalent compounds formed whentwo different halogen react are termed interhalogen compounds. Interhalogen compoundsare morer reactive than halogens because A—B bond is weaker than B—B. General Properties Reactivity : Interhalogen compounds are more reactive than the halogen (except F2). This is because the A—X bond in interhalogen is weaker than the X—X bond in the halogens. The order of reactivity of some interhalogen compounds has been found to be ClF3 > BrF5 > IF7 > CIF > BrF3 > IF5 > BrF > IF3 > IF
Chemistry : Practical Application Hydrolysis : Hydrolysis gives halide and oxohalide ions. Oxohalide ion is always formed from the larger halogen present. Oxidation state of the larger halogen does not change during hydrolysis. ICl
Cl_ + OI_
2ICl3 + 3H2O ¾¾® 5HCl + HIO3 + ICl BrF5 + 3H2O ¾¾® 5HF + HBrO3 (F_) (BrO3_) IF7
+ 7F-
Halogenation : Interhalogen compounds are good halogenatig agent. ClF + CsF ¾¾® Cs+ + ClF2_ 2ClF + AsF5 ¾¾® FCl2 + AsF6_ 6ClF + 2Al ¾¾® 2AlF3 + 3Cl2 6ClF + U ¾¾® UF6 + 3Cl2 6ClF + S ¾¾® SF6 + 3Cl2 4ClF3 + 6MgO ¾¾® 6MgF2 + 2Cl2 + 3O2 4ClF3 + 2Al2O3 ¾¾® 2AlF3+ 2Cl2 + 3O2 2ClF3 + 2Al2O3 ¾¾® 2AgF2 + Cl2 + 2ClF 2ClF3 + 2NH3 ¾¾® 6HF + Cl2 + 2ClF 2ClF3 + 2NH3 ¾¾® 6HF + Cl2 + N2 ClF3 + BF3 ¾¾® [ClF2]+ [BF4]_ ClF3 + SbF5 ¾¾® [ClF2] + [SbF6]_ ClF3 + PtF5 ¾¾® [ClF2] + [PtF6]_ ClF3 + UF4 ¾¾® UF6 + ClF
4ClF3 + 3Pu ¾¾® 3PuF4 + 2Cl2 4ClF3 + 3N2H4 ¾¾® 12HF + 3N2 + 2Cl2 4BrF3 + SiO2 ¾¾® 3SiF4 + 2Br2 + 3O2 Non-transition Elements (B2O3, As2O5, I2O5, CuO, TiO2 react similarly) 2CiF + AsF5 ¾¾® FCl2 + AsF6_ ClF3 + AsF5 ¾¾® ClF2 + AsF6_ ClF3 + CsF ¾¾® Cs2 + ClF4_ 2BrF5 + SiO2 ¾¾® SiF4 + 2BrF3 + O2 BrF5 + CsF ¾¾® Cs + [BrF6]_ IF5 + KI ¾¾® K+ + [_IF6]_ 2IF7 + SiO2 ¾¾® 2IOF5 + SiF4 IF7 + CsF ¾¾® Cs+ + [IF8]_ Self Ionisation : In liquid state many interhalogen have an appreciable electrical conductivity due to self ionisation. 2ICl
2BrF3 2IF5
I+ + [ICl2]_ [BrF2]+ + [BrF4]_ [IF4]+ + [IF6]_
The Structure of following Compounds :
C1F3, I2C16, IF5, IF7. The structure of these compounds can be represented as: CIF3 : The molecule is T-shaped (structure 3) with bond angle of 87°40'. The distortion from 90° is because of repulsion^between the lone pairs. Equatorial bonds (those in triangle) are different from apical bonds (those pointing ups and down)
Chemistry : Practical Application The Structure of BrF3 is also T-shapes :
I2Cl6 I2Cl6 : In the solid state, two Tshaped IC13 molecules join together, forming a planar dimeric molecule (IC13)2 or I2C16. IC13 gas decomposes into IC1 and- Cl2, so its structure is not known. The liquid undergoes self ionisation as
2ICI3
IC12+ + IC14_
IF5 : The IF5 (or AX5) compound has structure based on a square pyramid, that is octahedral with one position occupiedby lone pair. The central atom is displaced slightly below the plane.
IF7 : The structure of IF7 is usual a pentagonal bypyramid. It is probably the only known example of a non-transition element using three d orbitals for bonding.
Non-transition Elements The Polyhalides Ions, Prepared :
Halide ions often react with molecules of halogens or interhalogens and form polyhalide ions. Iodine is only slightly soluble in water. Its solubility is greatly increased if same iodide ions are present in the solution. The increase in solubility is due to the forma'tion of polyhalide ion, in this case triiodide ion I3 ~. This is stable both in aqueous solution and in ionic crystals. I2 + I_ ¾¾® I3_ Br3_ and Cl3_ ions are less stable. No F3_ compounds are known presumably because fluorine has no available d-orbitals and, therefore, cannot expand its octet. Cl_ + Cl2 ¾¾® Cl3_ Many polyhalides are known which contain two or three different halogens.
IC1 + KC1 ¾¾® K + [IC13]_ IC13 + KC1 ¾¾® K+[IC14]_ IF5 + CsF ¾¾® Cs + [IF6]_ 2C1F + AsF5 ¾¾® [FC12]_ [AsF6]BrF3 +C1F3 ¾¾® [ClF2] + [BrF4]_ Structure and Shape of I3_ Ion : I3_ ion has linear shape as shown in Fig.
Chemistry : Practical Application In some cases I—I—I bond angle in I3_ ion is not equal to 180° and the two I—I bond lengths are unequal (unsymmetrical structure) as shown below in some ionic salts.
Structure of I5_ Ion : The structure of pentaiodide in (CH3)4NI5 is quite different from those of other polyhalides containing five halogens. It is planar and V-shaped.
Structure and Shape of IC12_ Ion : This ion has symmetrical linear shape which results from sp3d hybridisation of I-atom (central atom). Due to the presence of one unit of negative charge on ICI2_ ion, Iatom (central atom) can be regarded as having eight electrons (instead of seven) in its valence-shell. The Lewis structure of IC12_ ion,
shows that I-atom is surrounded by two s-bps and the
remaining six electrons remain as three lps on I-atom. Thus, since s-bps + Ips = 2 + 3 = 5, I-atom is sp3d hybridised in IC12_ ion. Now the two axial hybrid orbitals, each of which has one electron, overlaps with the 3px orbital (singlyfilled) of two Cl-atoms to form two ICl s-bonds. Due to the presence of Ips, IC12_ ion assumes linear (symmetrical) shape Non-transition Elements
Five electrons pairstrigonal bipramid with thre positions occupied by lone pairs The three lone pairs of electrons (Ips) occupy the basal positions of the trigonal bipyramid.
Structure and Shape of IC14_ Ion : In this ion I-atom is the central atom. Due to the presence of one unit of negative charge on IC14_ ion, I-atom (central atom) can be regarded as having eight electrons (instead
shows that I-atom is of seven) in its valence-shell. The Lewis structure of this ion, surrounded by four s-bps and two Ips. Thus since s -bps + Ips =4 + 2 = 6, I-atom is assumed to be sp3d2 hybridised. Out of the six hybrid orbitals two axial hybrid orbitals contain two Ips, because in order to have minimum lone pair repulsion, the lone pairs must be as far from each other as possible. Other four hybrid orbitals (equatorial hybrid orbitals) which are singly-filled overlap with the singly-filled 2px orbitals on four Cl-atoms and form four ICl a-bonds. Although the spatial arrangement of six electron pairs round I-atom is octahedral, due to the presence of two lone pairs of electrons in the axial hybrid orbitals, the shape of IC14_ ion gets distorted and becomes square planar as shown in Fig.
Chemistry : Practical Application
(BrF4_ also has similar structure and same explanation) Structure and Shape of C1F2+ : In this ion Cl-atom is the central atom. Due to the presence of one unit of positive charge on C1F2+ ion, Cl-atom (central atom) can be regarded as having six electrons (instead
of seven) in its valence-shell. The Lewis structure of C1F2+ ion,
shows that Cl-atom has s-
bps =2 and Ips = 2, and, since s-bps + Ips = 4, Cl-atom is supposed to be sp3 hybridised. Two sp3 hybrid orbitals have lps while each of the remaining two sp3 hybrid orbitals has one electron. These two singly-filled sp3 hybrid orbitals make a head-to-head (linear) overlap with singly-filled 2px orbitals on each of the two F-atoms and form two sp3 (Cl)2px (F) s-bonds. Although the spatial arrangement of four electron pairs round Cl-atom is tetrahedral, due to the presence of two lone pairs of electrons, the shape of C1F2 + ion gets distorted and becomes angular or V-shape as shown in Fig.
The ClF bond length and ClFF bond angle found in C1F2+ ion depends on the nature of the anion attached to this cation. (The IC12+ ion also has similar structure and shape.) Non-transition Elements Important Reactions of Xenon Compounds : 1. XeF2 + H2 ¾¾® 2HF + Xe XeF4 + 2H2 ¾¾® 4HF + Xe XeF6 + 3H2 ¾¾® 6HF + Xe 2. XeF2 + 2HC1 ¾¾® 2HF + Xe + Cl2 XeF4 + 4KI ¾¾® 4KF + Xe + 2I2
SO42_ + XeF2 + Ce2III (SO4)3 ¾¾® 2CeIV (SO4)2 + Xe + F2 (In above reactions xenon fluoride oxidises Cl_ to Cl2, I_ to I2 and Ce (III) to Ce (IV). 3. XeF4 + 2SF4 ¾¾® Xe + 2SF6 XeF4 + Pt ¾¾® Xe + PtF6 XeF4 + C6H6 ¾¾® Xe + C6H5F + HF (XeF4 is used as fluorinating agent.) 4. 2XeF2 + 2H2O ¾¾® 2Xe + 4HF + O2
3XeF4 + 6H2O ¾¾® 2Xe + XeO3 + I O2 XeF6 + 2H2O ¾¾® XeOF4 + 2HF 2XeF6 + 6H2O ¾¾® XeO3 + 12HF (XeO3 is a explosive solid.) 5. 2XeF6 + SiO2 ¾¾® XeOF4+SiF4 6. XeO3 + 2XeF6 ¾¾® 3XeOF4 XeO3 + XeOF4 ¾¾® 2XeO2F2 7. XeO3 + NaOH ¾¾® Na + [HXeO4)_ Sodium xenate 2[HXeO4]_ + 2OH ¾¾® [XeO6]4_ + Xe + O2 + 2H2O 8. Xenon fluorides act as a fluoride donor and forms complexes with covalent pentafluorides.
XeF2 forms complexes with PF5, AsF5, SbF6, NbF5, TaF5, RuF5, OsF5, RhF5, IrF5 and PtF5. These are thought to have the structure
Chemistry : Practical Application XeF2 · MF5 [XeF]+ [MF6]_ XeF2.2MF5 [XeF] + [M2F11]_ 2XeF2. MF5 [Xe2F3] + [MF6]_ XeF4 forms only a few complexes (withPF5, As F5 and SbF5 only) XeF6 forms complexes such as XeF6. BF3; XeF6.GeF4; XeF6. AsF5; XeF6- SbF5. 9. XeF6 may also act as fluoride acceptor. With RbF and CsF it reacts as XeF6 + RbF ¾¾® Rb + [XeF7]_ On heating XeF7_ ion decomposes 2Rb+ [XeF7]_
XeF6 + Rb2 [XeF8]
Starting from xenon and fluorine and other appropriate reagents, is prepared potassium perxenate. Using VSEPR theory, the probable structure of perxenate ion, predicted : From xenon and fluorine, potassium perxenate can be prepared as follows : Xe +3F2 ¾¾® XeF6 1 : 20 XeF6 + 3H2O ¾¾® XeO3 + 16HF XeO3 is soluble in water, but does not ionise. However, in alkaline solution above pH 10-5, it forms the xenate ion XeO3 + KOH ¾¾® K + [HXeO4]_
Potassium Xenate + 2H2O Xenates contain Xe (+ VI) and they slowly disproportionate in solution to perxenates (which contain Xe (+ VIII) and Xe K + [HXeO4]_ + 2OH- ¾¾® K4 + [XeO6]4_ + Xe + O2 Potassium perxenate + 2H2O K4XeO6 For the structure of perxenate ion please refer to page number 27. Non-transition Elements A brief VSEPR explanation of structures of xenon fluorine compounds : The structure of xenon and fluorine compounds can be summarised as follows : Compound Structure No. of No. of VSEPR explanations electron lone pairs pairs XeF2 Linear 5 3 5 electron pairs forms trigonal bypyramid with three lone pairs at equitorial positions XeF4 Square planar 6 2 6 electron pairs form an octahedron with two position occupied by lone pairs XeF6 Distorted octahedron 7 1 Pentagonal bipyramid with one lone pair XeO3 Pyramid 7 1 Three n bonds so the remaining H electron pairs form a tetrahedron with one lone pair XeO2F2 Trigonal bipyramid 7 1 Two p bonds so the remaining 5 electron pairs form trigonal bipyramid with one lone pair at equitoria) position XeOF4 Square pyramidal 7 1 One p bond so the remaining 6 electron pairs form an octahedron with one position occupied by lone pair
XeO4 Tetrahedral 8 0 Four p bonds so remaining four electron pairs form a tetrahedron XeO3F2 Trigonal bipyramid 8 0 Three p bonds so remaining five electron pairs form a trigonal bipyramid [XeO6]4_ Octahedral 8 0 Two p bonds so the remaining 6 electron pairs form an octahedron
Chemistry : Practical Application The Structures of Compounds Drawn : (i) XeF2 (ii) XeF4 (iii) XeF6 (iv) XeO3 (v) XeOF4 (vi) XeO2F2 (vii) XeO4 (viii) [XeO6]4_ XeF2 : There are 10 valence electrons (8 from valence shell of xenon 5s2 5p6 and two form the two bonded fluorine atoms) to be filled in orbitals. These require five orbitals which can be formed by the hybridisation of one 5s, three 5p and one 5d orbitals (sp3d hybridization). These are directed towards the five corners of a trigonal bipyramid. Two of these contain shared electrons, the other three lone pairs. For greatest stability, the shared pairs are as far apart as possible, so Xe - F bonds are at 180° to each other and the structure is linear. Xenon Atom
Formation of XeF2 molecule by sp3d hybridization. Electrons supplied by fluorine atoms are represented by dotted arrows. XeF4 : In a similar manner we have 8 + 4 or 12 valence electrons to accommodate (8 being electrons of the central atom and 4 being electrons from 4 fluorine atoms). These require 6 orbitals, hence sp^d2 hybridization. The structure is square planar and the four fluorine atoms are in the same plane along with
xenon atom. The six orbitals are directed towards the six corners of an octahedron. The two lone pairs are one above one below the plane containing Xe and 4 fluorine atoms. Non-transition Elements Xenon Atom
XeF6 : Here we have 8 + 6 or 14 electrons t.o accommodate. These need 7 orbitals. Hence sp3d3 hybridization resulting in distorted octahedron. Xenon Atom
The shapes of oxygen containing compounds of Xe are correctly, predicted by the valence bond method. (Electrons in p bonds (double bonds) must be subtracted before counting the number of electron pairs which determine the primary shape of the molecule. XeC3 : Oxygen shares three of the four lone pairs of xenon. Hence only 8 valence electrons are to be accommodated and hybridization around xenon is sp3. As three orbitals contain bond pairs and one lone pair, so the geometry is described as trigonal pyramid. The OXeO bond angle is found to be 103° and is reasonably close to the expected » 109°, taking into account the greater repulsion between lone pair and bond pairs as compared to bond pair-bond pair repulsion.
Chemistry : Practical Application Xenon Atom
XeOF4 : Here oxygen atom does not contribute any electrons but shares one of the lone pairs of xenon. Hence the structure is like that of XeF4 in which one of the orbitals containing lone
pairs around xenon is replaced by XeO bond. The structure is described as square pyramid. The six orbitals, as before, are arranged towards the six corners of a regular octahedron. The structure of XeO2F2, [XeO6]4_ and XeO4 can be described on similar lines. The structures are shown below :
XeF2 : The outer electronic configuration of xenon and fluorine atoms are
Non-transition Elements
Assume that bonding involves the 5pz orbital of Xe and 2pz orbitals of two F atoms. For bonding to occur, orbitals with the same symmetry must overlap. These three atomic orbitals combine resulting in the formation of three molecular orbitals, one bonding, one non-bonding and one antibonding as represented below
The original three atomic orbitals contain four electrons (two in the Xe 5pz and one in each of the F 2pz). These occupy the molecular orbitals of the lowest energy. Two of these occupy bonding MO and two occupy non-bonding MO, their energies being in the order. bonding MO < non-bonding MO < antibonding MO. The bonding may be described as three centre, four electrons s -bonding. A linear combination of the atoms gives the best overlap of the orbitals.
Chemistry : Practical Application
Geometry Around Xenon : Xenon difluoride can act as a fluoride ion donor, and as a result, forms a number of addition compounds with fluoride acceptor such as ASF5 and SbF5. For example, 2XeF2 + SbF5 ¾¾® [Xe2F3] + and [SbF6]_ The Xe2F3 + cation has the planar structure as given below:
XeF4 also form addition compounds with PF5, AsF5 and SbF5. XeF4 + 3SbF5 ¾¾® [XeF3]+ + [Sb3F16]_ The structure of XeF3+ cation can be represnted as follows: Due to the presence of one unit positive charge on XeF2+ ion, Xe atom can be regarded as having seven electrons (instead of eight)
Non-transition Elements (a) Given the geometry of phophorus and iron in the ioning compound formed by mixing equimolar ratio of PCl5 and FeCl3. (b) When boron oxide (B2O3) is heated in a stream of ammonia at 1200°C, a solid is formed which gives the analysis: B = 43.5 and N = 56.5%, comments, (c) Give (i) the products of the reaction between trimethyl-phosphine oxide and sulphur tetrafluoride and (ii) their structures. (a) PCl5 + FeCl3
[PCl4]+ + [FeCl4]
In [PCl4]+ the central atom P is sp3 hybridised and the ionis tetrahedral; the complex [FeCl4]_ is also tetrahedral. (b) When boron oxide (B2O3) is heated in a stream of ammonia at 1200°C boron nitride is formed.
B2O3
(BN)x
Boron nitride is a white slippery solid. One B atom and one N atom together have the same number of valency electrons as two C atoms. Thus boron nitride has almost the same structure as graphite, with sheets made up of hexagonal rings of alternate B and N atoms joined together. The sheets are staked one on top of theother, giving a layer structure. The formula BN is in completely agreement with the result of analysis, B = 43.5% and N = 56.4. (c) In the reaction between trimethyl phosphine oxide and sulphur-tetrafluoride, the SF4 is a selective fluorinaing agent, which converts P = O group to PF2.
2(CH3)3 P = O + SF4 ¾¾®
Chemistry : Practical Application The entries of Column I matched with appropriate entries of Column II. Column I Column II 1. Nido compounds A. Aluminium alkyl derivative 2. Spring reactions B. Boranes 3. Stock method C. Poly phosphates 4. Fulminating metal D. Cui 5. Amphiboles E. Cross linked polymers 6. Ziegler catalyst F. Nitrides of Ag, Au, Hg 7. RSiCl3 G. Hypo 8. Electrides H. Carboranes 9. Rochow Process I. SmOu6" 10. Grahm's salt J. Crown ether complexes (l)_(H); (2)_(G); (3)_(B); (4)_(F); (5)_(I); (6)_(A); (7)_(E); (8)_(J); (9)_(D); (10)_(C).
Lanthanides and Actinides 6 Lanthanides and Actinides The electronic configurations for following atoms or ions : Ce3+ (Z = 58), Eu2+ (Z = 63), Gd (Z = 64), Lu3+ (Z = 71) The electronic configurations are as follows Ce3+ (Z = 58) — [Xe] 4f1 5d0 6s0 Eu2+ (Z = 63) — [Xe] 4f1 6s0 Gd (Z = 64) — [Xe] 4f1 5d1 6s2 Lu3+ (Z = 71) — [Xe] 4f14 5d0 6s0 The 4f electrons in lanthanide elements do not affect their chemistry and crystal field stabilization : The 4f electrons in the antipenultimate shell are very effectively screened from their chemical environment outside the atom by 5s and 5p electrons. Consequently the 4f electrons do not take part in bonding. They are neither removed to produce ions nor do they take any significant part in crystal field stabilization of complexes. The octahedral splitting of f-orbitals D0 is only about 1 kJ mol_1. Whether f-orbitals are filled or empty has little effect on the normal chemical properties. However, it does effect their spectra and their magnetic properties.
Chemistry : Practical Application The important oxidation states of lanthanides, + 3 most stable M+ and M2+ are less stable than M3+ in lanthanide series. Reason : Whatever the electronic configuration of lanthanides in the ground state, all of them form ions in oxidation state +3. This is not directly evident from the electronic configuration. It is actually due to the fact that the magnitude of energy required to remove an electron from the gaseous ion in its lower oxidation state (i.e., ionisation energy) and that of released when two gaseous ions combine with water to form aquated species (i.e., hydration energy) are such that all the + 4 species (except Ce4+) and all the + 2 species (except Eu2+) revert to the tripositive state. This leads to the conclusion that ionisation and hydration energies are such that + 3 state is more stable than +2 or +4 states in aqueous solution.
In the solid state too the combination of ionisation and lattice energies is more negative than for + 2 and + 4 state. Therefore, + 3 state are also the most common in solid compounds. This is reflected from their reduction potential values of the couple : Ln3+(aq) + 3e_ ® Ln The + 2 and + 4 oxidation states are shown by those elements which by doing so attain stable f0, f7 and f14 configuration, e.g., Ce4+ : [Xe] 4f0; Tb4+ : [Xe] 4f7 Eu2+ : [Xe] 4f7 ; Yb2+ : [Xe] 4f14 The +1V oxidation state of cerium : Ce (IV) is the only tetrapositive lanthanide species sufficiently stable to exist in aqueous solution as well as in the solid compounds. It is rare to find 4 + ions in solution, because high charge on the ion leads to it being heavily hydrated, and except in strongly acidic solutions the hydrated Ce4+ is hydrolysed, giving polymeric species and H+. Ce(IV) in solution is obtained by treatment of Ce(III) Lanthanides and Actinides solutions with very powerful oxidising agents such as ammonium peroxodisulphate (NH4)2S2O8 . Ce(IV) solutions are widely used as oxidising agent in volumetric analysis instead of KMnO4 and K2Cr2O7. Burettes containing Ce4+ion must be washed with acid, since washing with water gives the hydrated ion [Ce(H2O)]4+. Ce (IV) is also used in organic reactions, like oxidation of alcohols, aldehydes and ketones at cc-carbon atoms. Sm2+, Eu2+ and Yb2+ ions in solutions are good reducing agents but an aqueous solution of Ce4+ is a good oxidising agent : The most stable oxidation state of lanthanides is + 3. Hence ions in + 2 state tend to change to + 3 state by loss of electron and those in + 4 state tend to change + 3 state by gain of electron. The lanthanide contraction and its cause : On moving along the lanthanide series from La to Lu there is a continuous decrease in the size of
lanthanide ion with increase in atomic number, this steady decrease in size is known as Lanthanide contraction. Cause of Lanthanide Contraction : As we move through the lanthanide series, 4f electrons are being added, one at each step. The mutual shielding effect of f-electrons is very little, being even smaller than that of d-electrons. This is due to the shape of the f-orbitals. The nuclear charge, however, increases by one at each step. Hence the inward pull experienced by the 4f electron increases. This causes a reduction in the size of the 4fn shell. The sum of successive reduction gives the total lanthanide contraction. In going from La3+ to Lu3+, the ionic radius shrinks from 106 pm to 85 pm i.e., accounting to 21 pm. Lanthanide Contraction Lanthanide contraction effects the properties of post lanthanide elements (elements of third transition series). For example,
Chemistry : Practical Application Atomic and Ionic Radii : Due to lanthanide contraction atomic radii of elements just following (i.e., Hf and onwards) are close to the elements just above them in their respective groups. Hence these elements are very much similar in properties and often very difficult to separate, e.g., Zr and Hf, Nb and Ta, Mo and W, Ru, Rh, Pd and Os, Ir, Pt, etc. This is why second and third row transition elements are quite close in properties while properties of first and second row are not. Densities : Because of lanthanide contraction, the atomic size of Hf and succeeding elements becomes very small and their densities are high. That is why density of third row transition elements are almost double to those of elements of second transition series. Basicity of Ions : Due to lanthanide contraction from La3+ to Lu3+ covalent character increases between Lanthanide(III) ions and OH_ ions. Thus La(OH)3 is most basic while Lu(OH)3 is least basic. According to Fajan's'rule (Basicity decreases as ion size decreases). These differences in basicity are reflected in (a) thermal decomposition of oxy salts (more basic oxy salt decompose less easily); (b) hydrolysis (more basic ions hydrolyse less readily); (c) oxidation potential for the couple Ln ¾® Ln3+ + 3e_ regularly goes on decreasing;
(d) formation of complexes; (e) basic strength of lanthanide oxides M2O3 decreases with rise in atomic number. The applications of lanthanide metals and their compounds: Important applications are as follows 1. In the formation of important alloys e.g., Misch Metal : 50% Ce, 40% La, 7% Fe, 3% other metals. Lanthanides and Actinides This is added to steel to improve its strength and workability. Small amount of misch metal is also used as lighter flints. Pyrophoric Alloys : 40% Ce, 44% La + Nd, 5% Fe and rest is Al, Ca, C and Si. It is used in the preparation of ignition devices such as tracer bullets and shells. 2. La2O3 is used in Crooke's lenses, which gives protection from UV light by absorbing it. 3. CeO2 is used in polish glass. 4. Gas mantles are treated with a mixture of 1 % CeO2 and 99% ThO2 to increase the amount of light emitted by coal gas flames. 5. Lanthanide oxides are also used as phosphorus in colour TV tubes. 6. Lanthanide elements are present in warm super-conductors such as La(2 _ x) Bax CuO(4 _ y) and YBa2Cu2O7 and others. 7. 3-diketone complexes of Eu3 + and Pr3+ dissolved in organic solvents are used as lanthanide shift reagents in NMR spectroscopy. The type of electronic transitions, responsible for colour in lanthanides : All but one of the lanthanide ions show absorption in the Visible or near UV region. The exception is Lu3+ which has a full f shell. In the spectra of lanthanides spin orbit coupling is more important than crystal field splitting. The colours are due to Laporte-forbidden f-f transition i.e., transitions between the J states of 4n configuration since the change in the subsidiary quantum number is zero. The forbitals are
deep inside the atom. Thus they are well shielded from environmental factors such as nature and number of ligands which form the complexes, and from vibration of the ligands. Thus the position of absorption band (i.e., colour) does not change with different ligands. Vibration of the ligands changes the external field, Chemistry : Practical Application however, this only splits the various spectroscopic states by about 100 cm_1. Therefore, the absorption bands are weak but sharp Lanthanides are used for wavelength calibration of instruments because of their sharp absorption bands. Strong and broader peaks in spectra of Ce3+, Tb3+, Sm2+, Eu2+ and Yb2+ due to the transitions of 4f n ® 5dl type Laporte-permitted bands. These bands are considerably influenced by chemical environment. Charge transfer spectra are possible due to the transfer of an electron from the ligand to metal. This is more probable if the metal is in high oxidation state or ligand has reducing properties. Yellow colour of Ce4+ or blood red colour of Sm2+ arise from charge transfer spectra rather than f-f spectra. The d-d spectra is different from f-f spectra : For an f-electrons the subsidiary quantum number l = 3, so m1 may have values 3, 2, 1, 0, -1, -2, -3. Thus a large number of transitions are usually possible. This is an marked contrast to the transition elements where d-d spectra give absorption bands whose position changes from ligand to ligand and the width of the peak is greatly brodened because of the vibration of the ligands. It is also possible to get transitions from 4f to the 5d level. Such transitions give broader peaks and their position is affected by the nature of the ligands. Ce3+ and Tb3+ are colourless but show strong absorption in UV region, reason : Ce3+ and Tb3+ are colourless because they do not absorb in the visible region. However, they show exceptionally strong absorption in UV region, because of transitions from 4f to 5d. Absorption is very strong due to two reasons. Since Dl = 1 this is an allowed transition and so gives stronger absorption than forbidden f-f transition. Furthermore, promotion of electrons in these ions is easier than for other ions. The electronic configuration of Ce3+ is f1 and Tb3+ is f8. Loss of one electron gives the extra stability of an empty or half full shell, f-d peaks are broad in contrast to the narrow f-f peaks. Lanthanides and Actinides The magnetic properties of lanthanides are different than transition elements : For the first row of transition elements magnetic moment must be calculated using simple spin only formula
For the second and third row of transition elements magnetic
moment from the unpaired electron spin and that from orbital motion is taken into account. This is the case with lanthanides also, but here a fundamental difference is that spin contribution S and orbital contribution L couple together to five a new quantum number J. J = L - S when the shell is less than half J = L + S when the shell is more than half The magnetic moment (m) in Bohr magneton (BM) is given by
m= where g = Lande splitting factor and is given by
g= Sm3+ and Eu3+ are exception because the spin orbit coupling constant is only about 300cm_1 i.e., the difference in energy between the ground state and the next state is small. Thus the energy of thermal motion is sufficient to promote some electrons and partially populate the higher state. Because of this the magnetic moments are not solely determined by the ground state configuration. Measuring the magnetic moment at a low temperature prevents the population of higher energy levels. Few organometallic groups form complexes with lanthanides : This is because 4f orbitals are well shielded and are inside the atom, thus they cannot take part in n back bonding.
Chemistry : Practical Application Shift Reagents The shift reagents are ions in the rare earth (lanthanide series) coordinated to organic ligands, generally bdiketone complexes of Eu3+ and Pr3 +. More recent shift reagents are Eu (dpm)3 and Eu (fod)3. Their name came from tris (dipivalomethanato) europium and tris-1, 1, 1, 2, 2, 3, 3-heptafluoro-7, 7-dimethyl-3, 5-octanedionatoeuropium. Shift reagents provide a method for spreading out NMR absorption patterns without increasing the
strength of the applied magnetic field. Addition of shift reagents to appropriately functionalized samples results in substantial magnifications of the chemical shift differences of non-equivalent protons. There are two major applications of the shift reagent to structure determination (i) To simplify the spectrum. (ii) To assign the protons from the data on the response curve. Effectiveness of the shift reagent analysis may be increased by the use of low temperature NMR. Actinides show higher oxidation states than lanthanides; reason : Actinides show a greater multiplicity of oxidation states. The 5f orbitals extend into space beyond the 6s and 6p orbitals and participate in bonding, while 4f orbitals are totally shielded by outer orbitals and thus unable to take part in bonding. The participation of the 5f orbitals explains the higher oxidation states shown by the earlier actinide elements. The greater extension of 5f orbitals compared-with 4f is shown by the difference in electron resonance spectra. Since in the first half of the actinide series (i.e., lower actinides), the energy required for the conversion of 5f ® 6d is less than that required for the conversion of 4f ® 5d, the lower actinides should, on a comparative basis, show more higher oxidation states such as + 4, + 5, + 6 and + 7. Lanthanides and Actinides Correspondingly, since in the second half of the actinide series (i.e., higher actinides) the energy required for the conversion of 4f ® 5d should show more lower oxidation states such as + 2. Like lanthanides + 3 oxidation state is more common in actinides. Actinides form oxocations but lanthanide don't; reason : Actinides show higher oxidation states but lanthanides do not. When the oxidation state of an actinide increases to + 6 the ion is no longer a simple ion. There is a high charge density on the ion which leads to the formation of oxygenated ion e.g., UO22+, such oxycations are stable in aqueous solution. Reason for the observation that a large number of oxides of actinide elements are non-stoichiometric but the same is not true for the lanthanide elements : The compounds of indefinite structure and composition are known as non-stoichiometric compounds. They are formed because of the ability of the metal to exist in more than one oxidation states. Lanthanides forms non-stoichiometric compounds in lesser number as compared to actinides because greater variation of oxidation states compared to lanthanides or rare earth is to be expected among actinide metals because the atomic radii are larger and the energy levels of the valence electrons are closer together. Several stable oxidation states are encountered for most actinide metals.
Actinides have a greater tendency to form complexes than lanthanides; reason : Actinides have a greater tendency to form complexes because of higher nuclear charge, small size of their atoms and greater multiplicity of oxidation states as compared to lanthanides. The account for the fact that magnetic moment of Eu3+ and Sm3+ complexes are higher than those calculated by spin-only values and also by the total angular momentum J values of the ground state.
Chemistry : Practical Application For La3+ (f°) and two of the lanthanides Gd3+ (f1) and Lu3+ (f14), which have S term symbol L = 0, i.e., no orbit effect. Their magnetic moment can be calculated by the relationship.
ms = The other lanthanides ions do not obey this simple relationship. The 4f electrons are well shielded from the enternal field by the overlying 5s and 5p electrons. Thus, the magnetic effect of the motion of the electron in its orbital is not quenched out. Thus, the magnetic moment must be calculated taking into account both the magnetic moment from the unpaired electron spins and that from the orbital motion. In lanthanides the spin contribution S and orbital contribution L couple together to give a new quantum number J. J = L _ S when the shell is less than half full. and J = L + S when the shell is more than half full The magnetic moment m is calculated in Bohr magnetons (BM) by
m=
Where g = The agreement of calculated magnetic moment for Eu3+ is poor, and that for Sm3+ is not very good. The reason is that with Eu3+ the spin orbit coupling constant is only about 300 cm-1. This mean that the difference in energy between the ground state and the next state is small. Thus, the energy of thermal motion is sufficient to promote same electrons and partially populate the higher state. Because of this the magnetic moment are not solely determined by the ground state configuration. Measuring the magnetic
moment at a low temperature prevents the population of higher energy levels. Lanthanides and Actinides Since these excited states have higher J values than the ground state, the actual magnetic moments are higher than those of calculated by considering ground states only. The magnetic moment of trivalent gadolinium (Z = 64) complexes can be obtained by spin only formula but not for trivalent terbium complexes; reason : Gd3+ had 4f7 configuration i.e., half-filled f-orbitals. So L or ML = mi = 0 (- 3, - 2, - 1, 0, + 1, + 2, + 3). Therefore, the orbital contribution to the magnetic moment is zero and we have only spin contribution. For 4f0, 4f7 and S term symbol
= 0 i.e. no orbit effect.
So the relationship ms =
works with Gd3+ (f7) ms = = But Tb3+ has 4f8 configuration and ML ¹ 0. For it, therefore, there is both orbital and spin contribution of magnetic moment and resultant quantum number J used for calculating magnetic moment is obtained by Russell-Saunders coupling of L and S quantum numbers. The magnetic moment of Sm3+ (f5) utilising the equation meff = g {J (J + 1)}1/2, at room temperature, but it is lower than the actual experimental magnetic moment of 1.54 BM at room temperature; reason : Effective magnetic moment of lanthanide ions are given by the following equation provided that the ground state symbols of lanthanide(III) ions are known meff = g{J(J+1)1/2 BM
or, meff =
where g = Lande splitting factor and is given by
Chemistry : Practical Application
g= For Sm3 + i.e., f5system number of unpaired electrons, n = 5 Term symbol = 6H5/2 S = n/2 = 5/2 L=5 J = 5/2 By putting the values of S, L and J in the above equation for calculating g, we get the value of g = 217. Then, we can calculate meff from the given equation which comes out to be 0.84 BM, while the mexp = 1.54. The explanation of lower value of meff than mexp is given in question no. 20. The Explanation : (i) Which is more basic Nd2O3 or Dy2O3 ? (ii) Which is thermally more stable : Sm(NO3)3 or Tm (NO3)3 ? (iii) La3+, Lu3+, Yb2+, Ce4+ are diamagnetic while Sm3+ has low paramagnetism. (i) Nd2O3 is more basic than Dy2O3.Nd3+ has greater ionic radius than Dy3+. Thus Dy2O3 will have greater covalent character (Dy3+ being smaller) and will thus be less basic than Nd2O3. (ii) A crystal with smaller ion has always more stable lattice having higher m.p. Tm3+ being smaller than Sm3+, Tm (NO3)3 should form a more stable lattice and thus be more thermally stable.
(iii) Sm3+ has 4f5 configuration having unpaired electrons whereas La3+, Lu3+, Yb2+ and Ce4+ have empty or complete 4f orbitals. Lanthanides and Actinides The separation of lanthanides is a difficult task; reason : All the lanthanides have similar outer electronic configuration and display mainly + 3 oxidation state in their compounds, therefore, lanthanides have exceedingly similar chemical properties. Their similarity is much closer than that of ordinary transition elements because lanthanides differ mainly in the number of 4f electrons which are buried deep in the atoms of lanthanides and thus don't influence their properties. Moreover, due to lanthanide contraction there is a very small difference in the size of all the fifteen trivalent lanthanide ions. Thus, for all practical purposes, the size of these ions is almost identical which results in similar chemical properties of these elements. The principle of separation of lanthanides by ion exchange method : This is the most rapid and effective method for the separation and purification of the lanthanides. This method is based on the following principles : (i) Lanthanide ions on contact with synthetic resins containing COOH or SO3H group undergo proton exchange. Ln3+ + 3H-resin ® Ln (resin)3 + 3H+ (ii) The bonding of the lanthanide ion to the resin depends on its size, i.e., the smaller the size of the lanthanide ion, the more firmly it is bound to the resin and vice-versa. Since lanthanide ions are hydrated, therefore, size of the hydrated ions should be considered for binding purpose. Hydration of the ions depends upon size i.e., smaller the size of the ion greater will be the hydration. Therefore in case of lanthanide ions, the smallest lanthanide ion, namely Lu3+ will be the most heavily hydrated. Thus it will have the maximum size and, therefore, the least firmly bound to the resin while reverse will be the case with La3+ which will be the most firmly bound to the resin.
Statistical Thermodynamics 7 Statistical Thermodynamics Important Features of Maxwell-Boltzmann (MB) M.-B. Statistics B.-E. Statistics F.-D. Statistics 1. The laws of classical The laws of quantum In this statistics also, laws mechanics are applied mechanics are applied of quantum mechanics are according to which according to which applied. individual atoms or individual atoms or molecules have definite molecules have only positions and momenta. quantized values of energy. 2. It treats the different It treats the different It also treats the particles as particles as particles as indistin- indistinguishable. distinguishable. guishable. 3. Any number of particles Any number of particles Only one particle is may occupy the same may occupy the same supported to occupy a energy level. energy level. particular energy level. 4. Particles obeying M.-B. It is applied to those It is applied to those Statistics are called particles which have particles which have a Maxwellons or symmetric wave functions symmetric wave functions Boltzmannons. like photons or atoms and like electrons, protons or
molecules having even no. atoms and molecules of particles in the nucleus. having odd no. of Such particles are called particles in the nucleus. Bosons. Such particles are called Fermions.
Chemistry : Practical Application The Ensemble A collection of a large number of systems which are identical with the system under consideration in a number of aspects such as total volume, total number of molecules etc. is called ensemble of system. Microcanonical Ensemble : Each system in this ensemble is like the isolated system in the thermodynamic sense. Such an ensemble of systems in which each member has the same values of N, V and E is called a microcanonical ensemble. Canonical Ensemble : If all the members (systems) of an ensemble have the same value of N, V and T then it is called a canonical ensemble. Each system in this is like a closed system, in the thermodynamic sense. It is the most common used ensemble in statistical mechanics. Grand canonical ensemble : In this, each system is an open system. Hence matter can flow between the systems and the composition of each one may fluctuate. Consequently the number of molecules in different systems is not same. However V, T and m for each component is same for each member of the ensemble. In a sample of atomic hydrogen at 25°C, proportion of atoms in the first excited electronic state if it lies 1000 kj mol_1 above the ground state. The proportion of atoms at 107 K also, calculated: At 25° C (298.16 K), RT = 8.314 × 298.16 Jmol_1 = 2.5 kJ mol_1 DEij = E1 _ Eground = 1000 kJ mol_1
Hence
= exp [ _ DEij/RT] = e_1000 kJ mol_1/2.5 kJ mol_1 = e_400 = 10_174 Statistical Thermodynamics It means, essentially whole sample is in its electronic ground state at room temperature.
At 107 K
= e_1000/8.314 × 107 = 0.99
i.e., about 1% of the total population will be in the ground state. The possible number of ways of distribution of 2 particles among 4 energy states, when, calculated : (1) particles are distinguishable and there is no restriction on the occupancy of the energy state (boltzons), (2) particles are indistinguishable (corrected boltzons), (3) particles are Fermions, (4) particles are bosons. (1) Boltzons : No. of energy states gi = 4 No. of particles = 2 (a) When both particles are in the same energy state
W1 =
= 1, but particles are indistinguishable hence there will be a total number of waysof
=4×1=4
(b) When one particle occupies one state
W2 =
= 2,
and total number of ways of this 2 × 6 = 12. Hence Wtotal = 4 + 12 = 16 (2) Corrected Boltzons : Particles are indistinguishable, hence number of ways of distribution of two particles in various energy states = 2 !
Chemistry : Practical Application Total number of ways, Wtotol =
(3) Fermions : WFermion =
=8
=
=6
(4) Bosons : WBoson =
=
=
= 10
The ratio of iodine molecules in the ground, and second excited vibrational states at room temperature. The vibrational energy levels are separated by 214.6 cm_1. The partition function is defined as
Z= For a molecule having constant energy separation between energy states it can be written as (ground energy state is taken as zero)
Z = 1 + exp (E/KT) + exp (_2E/KT) + exp (_3E/KT) + . . . This expansion series is geometrical progression which can be summed up as Z = [1 _ exp (_ E/KT)]_1 It follows then that the proportion of molecules in the energy state with energy Ei will be Pi = exp (_ Ei/KT)/Z = exp (_ Ei/KT) (1 _ exp (_ E/KT) The fraction of molecules in nth energy state, having energy n·E, will then be Pn = [1 _ exp (_ E/KT)] . exp (_ nE/KT)] Statistical Thermodynamics The vibrational energy levels have a constant separation between neighbours and E = 214.6 cm_1 At T = 298 K,
KT/hc = = 207.2 cm_1 Therefore,
E/KT =
= 1.036
The relative population in various vibrational levels will then be
P0 = (1 _ e_1.036).e° = 0.645 P1 = 0.645 × e_1.036 = 0.229 P2 = 0.645 × e_2 × 1.036 = 0.081 P3 = 0.645 5 e_3 × 1.036 = 0.029 Molar entropy at 0 K of CO would be R In 2, proved : If at 0 K, CO in crystalline form assumes any of the two configurations namely CO and OC and both these have slightly different energy, then 50% molecules may have configuration CO and remaining OC. If there are N molecules, N/2 molecules will have configuration CO and N/2 molecules will have OC configuration. \ Total number of configurations possible will be
W=
Chemistry : Practical Application By taking logarithms and using Stirlings approximation for large numbers, we have
In W = N In N _ N _ 2
= N In 2
\ The entropy (S), using Boltzmann equation S = K In W, for CO at 0 K will be S = NK In 2 For 1 mole of CO, N is Avogadro number and N · K = R Hence S = R In 2
Partition Function According to Boltzmann distribution law, the fraction of molecules which in the most probable state at temperature T possesses the energy Ei is given by
= where ni is the number of molecules having energy Ei at temperature T, N is the total number of molecules and gi is the degeneracy (or statistical weight) of energy level Ei. The denominator of the above equation which gives the sum of terms gi e_Ei/KT for all the energy levels is called the partition function. It is usually represented by Q. Thus Partition function Q = Sgi e_Ei/KT The partition function is so called because it indicates how the particles are distributed among the various energy states. The nuclear partition function of ortho H2 and ortho D2 molecules, calculated : The nuclear partition function of any molecule or atom is the degeneracy of the ground state nuclear energy level. The degeneracy of the ortho state is given as Statistical Thermodynamics = (I + 1) (2I + 1) where I is the spin quantum number of the nucleus. For hydrogen atom I = 1/2
=
=
=3
=3
For deuterium atom I = 1
= (1 + 1) (2 × 1 + 1) = 6
=
=6
The vibrational frequencies of the three modes of H2O are 1654 cm_1, 3825 cm_l and 3935cm_l. The vibrational partition function at 300 K, calculated : The hc v_i/KbT terms for each vibrational frequencies are _
= = 7.929 e_7.929 = 3.60 × 10_4
= = 18.336 e_18.336 = 1.088 × 10_8
= = 18.863 e_18.863 = 6.425 × 10_9
Chemistry : Practical Application
The total vibrational partion function
=
=
»1
For a condensed phase entropy is given by the relation S = NKb, In q + E/T, where q is the partition function. Using this relation an expression for the entropy of an ideal gas, derived: The difference between the ideal gases and the condensed phase is that in the condensed phase the particles can be considered as independently localised (distinguishable particles) whereas in the former case the particles are independently non-localised. The thermodynamic properties will differ in two cases by the difference in the number of complexions in two cases i.e.,
WD (localised) =
WD (non-localised) = p ® sign of product. Now Entropy is given as S = Kb In WD
=
=
(For condensed phase)
(For ideal phase)
Now
=
... (1)
Statistical Thermodynamics
S=
\S= \ The entropy of non-distinguishable particles (ideal gas) can be obtained from the equation (1) if q is multiplied by the factor e/N. The relative numbers of distinguishable states in ice and in water at 273 K, calculated : DHfus = 6.0 kj mol_1 at 273 K, K = l.38 × 10_23JK_1 The number of distinguishable states in a molecular system is given by the thermodynamic probability W. This is related to entropy as S = K In W For ice, let S1 = K In W1 For water, let S2 = K In W2 Therefore, DS = S2 _ S1 = K In W2 _ K In W1 = 2.303 K log (W2/W1) DS = DH/T = 6000 J mol_1/273K
= 21.978 J K_1 mol_1 log (W2/W) = 21.978 ¸ (2.303 × 1.38 × 10_23) = 6.9 × 1023 \ W2/W1 = 106.9 × 1023 This calculation suggests that there are roughly 1024 times as many ways to arrange a mole of water molecules in liquid as in ice. If we assume that the state of each molecule could be described independently of all other molecules then the entropy change for 1 molecule of ice and 1 molecule of liquid will be
Chemistry : Practical Application DSm/NA = 2l.978 ¸ 6.023 × 1023 = 3.649 × 10_23 and hence log W2/W1 = 1.149 or, W2/W1 = 14 There are 14 as many ways to arrange an individual molecule in the liquid compared to that in the crystal of ice. The molar residual entropy of a crystal in which the molecules can adopt 6 orientations of equal energy at 0 K, calculated : The residual entropy is given as S = K In WN = N K In W W = Number of orientation = 6 N = Number of particles = NA = 1 mol \ NAK = R = gas constant S = R In 6 = 8.314 J K_1 mol_1 × 2.303 log 6 = 14.89 JK_1 mol_1
The translational partition function of a molecule of oxygen gas at 1 atm and 298 K moving in a vessel of volume 24.4 dm3, calculated : We know that translational partition function
qt = V = 24.4 × 10_3 m3, p = 3.1416, T = 298 K, K = 1.38 × 10_23 JK_1, h = 6.62 × 10_34 J s, m = M/NA = 5.313 × 10_26 kg for O2
\ qt = = 4.28 × 1030 Statistical Thermodynamics That is, about 1030 quantum levels are thermally accessible even at room temperature for O2 molecule. The total entropy of a molecule is the sum of the contributions due to translation vibration and rotation, which particular one of these contributions is mainly responsible for the difference in entropy of S° (SbH3, g) = 232.7 J K_1 mol_1 and S° (NF3, g) = 260 JK_1 mol_1 although the molar mass of SbH3 is higher than that of NF3. Usually with increasing molar mass the entropies of gases increases. Thus molar entropy of SbH3 should
be greater than NF3 but S° (NF3) > S° (SbH3). This is so because the reduced mass hence moment of inertia of SbH3 is small therefore the rotational entropy will be small. The lighter hydrogen atoms also lead to higher vibrational frequency and a lower vibrational entropy.
True and False Statements : (1) Statistical mechanics bridges the quantum mechanics and thermodynamics. (2) Thermodynamic probabilty and general probability are the same. (3) The expression Q = qN/N ! is true for the system containing distinguishable particles. (4) Spontaneous process leads to the most probable state of the system. (5) The translational entropy is directly proportional to the molecular weight of the sample.] (1) True, (2) False, (3) False, (4) True, (5) True. The molecules of a gas which have two quantum states of energies 0 and e and degeneracies of g1 and g2 respectively considered. The contribution of these quantum states to the molar heat capacity of the gas at constant volume, calculated :
Chemistry : Practical Application This problem can be solved by evaluating the molecular partition function viz., q = g1 e_e1/KT + g2 e_e2/KT = g1 + g2 e_e/KT (... e1= 0 and e2 = e) From Boltzmann distribution we know that the relative population of the two quantum states is given by
=
= For 1 mole of the gas n1 + n2 = NA where NA is Avogadro's Number. Hence
=
or n1 =
\ n2 = NA _ n1 = \ The molar internal energy of the gas is given by
U = n1e1 + n2e2 = Hence molar heat capacity
CV =
=
Statistical Thermodynamics
= Matched thermodynamic functions in terms or the Molecular Partition function q = Sgi Exp. (_ ei /KT), Volume (V) and Temperature (T). Property Relation (1) P (a) (nR/T2) [¶2 In q/¶(T_2)]v (2) E (b) _ nRT In q _ RT (1 _ In NA) (3) H (c) nRT2 (¶ In q/¶T)v
(4) CV (d) nRT (¶ In q/¶T) + nR In q + nR (1 _ NA) (5) S (e) _nRT In q + nRT V (¶ In q/¶V)T _ nRT (1 _ In NA) (6) A (f) nRT (¶ In q/¶V)T (7) G (g) nRT2 (¶ In g/¶T)V + nRT V(¶ In q/¶V)T (1)_(f), (2)_(c), (3)_(g), (4)_(a), (5)_(d), (6)_(b), (7)_(e). Starting with the expressions for the translational partition function and other expression show that PV = nRT : We know that, A = _ KT In Q and Q = qN/N ! \ In Q = N In q _ N In N + N \ A = _ KTN In q + KTN In N _ KTN
and
= _ KNT (¶ In q/¶V)T, N
But qt = (2p mKT/h2)3/2 \ In qt = In (2 p mKT/h2)3/2 +
In V and (¶ In qt /¶V)T, N =
Chemistry : Practical Application Hence (¶A/¶V)T, N = _ KNT/V But (¶A/¶V)T, N = _ P
\ PV = NKT The total number of complexions in arranging four distinguishable systems among energy levels 0, 1e, 2e, 3e such that total energy is 3e : The possible ways of arranging the particles in the energy levels with total energy 3e are — (1) three particles in 0 and 1 in energy level 3e (2) to particles in energy 0, 1 in energy level e and 1 in 2e. (3) one particle in energy level 0 and three in energy levele. The number of complexion in each case
W1 =
=4
W2 =
W3 =
= 12
=4
\ Wtotal = 4 + 12 + 4 = 20 The entropy change of one mole of helium when it is heated from 300 K to 600 K at constant pressure, calculated :
S = R log = R log AT5/2
where A =
Now S600 = R log (A·6005/2) Statistical Thermodynamics S300 = R log (A·3005/2)
DS = R log
=
× 1.98 × 2.303 log 2
= 3.43 cal deg_1 mole_1 The heat capacity of diamond at 1080 K, qD being 1860 K, calculated :
Here, T >
, then
CV = 3R
xm =
=
= 1.7222
CV =
» » 3 × 1.987 (1 _ 0.1483 + 0.01571) cal K_1 mol_1 = 5.171 cal K_1 mol_1
The heat capacity of an element at a temperature equal to its characteristic temperature, calculated :
Since T = qD, T >
and xm =
=1
\ CV = 3R
Chemistry : Practical Application = 3R [1 _ 0.05 + 0.001786] = 3 × 1.987 × 0.9679 = 5.77cal K_1 mol_1 Using the principle of equipartition of energy, the translational, vibrational and rotational contribution to the heat capacity (CV) of the hydrogen molecule, indicated :
At about 300 K, CV of H2 is » R (experimental) corresponding to CV (trans) + CV (rot) for a diatomic molecule. On the basis of the classical principle of equipartition of energy, CV = 3/2 R and CV (rot) = R the vibrqtional and electronic contributions are absent since the corresponding excitations are absent the energy separation being large. At about 50 K, CV = 3/2 R (experimental value) corresponding CV (trans) only, the rotational, vibrational and electronic excitations being absent. In between these two extreme temperatures (i.e., 300 K and 50 K) the calculated values do not agree with the experiment value. Explanation: (1) Hydrogen is a molecule of very low mass and the moment of inertia is low, so separation between consecutive rotational energy levels is large and there is only partition rotational excitation among the hydrogen molecules and the extent of this excitation varies with temperature. (2) Nuclear spin contribution to rotational heat capacity should be taken into account. The CV(ortho) and CV(para) are obtained from the corresponding partition functions
CV(ortho) =
and
qortho = 3S (2J + 1) e_Bh J (J + 1)/KT, J = 1, 3, 5 .... Statistical Thermodynamics
where B =
and
CV(para) = The factors 3 and 1 are statistical weight factors. From these separate rotational heat capacities can be calculated and the experimental value will be CR = 1/4 Cpara + 3/4 Cortho The results obtained in this manner were found to be excellent agreement with experimental heat capacities. The rotational entropy of N2 gas at 25° C, I = 1.39 × 10_46 kg m2, evaluated :
Srot = R + R In
,
Here s = 2, T = 298.2 K
In
= In
= In 51.518 = 3.9419
\ Srot = 8.314 [1 + 3.9419] = 41.08 JK_1 mol_1 No has a doubly degenerate ground state and doubly degenerate electronically excited state 121.1 cm_1 above the ground state. The electronic partition function (Zelec) at 25° C, calculated: Electronic partition function Zel, is Zel = g0 + g1 exp (_DEelec/KT) = 2 + 2 e_121.1/207.2 Zel = 2.8
Chemistry : Practical Application The heat capacity of steam at 100° C, estimated : Vibrational modes of H2O lie in the region 1500 _ 3500 cm_1 which are larger than KT [KT = 260 cm_1 at 100° C] so they are all inactive. Translations and rotations are active hence CV = 1/2 R [3 + 3 + 0] = 3R or 24.9 J K_1 mole_1 Standard third law entropies are less than statistical entropies for CO and N2O, whereas it is reverse for (CH3)2Cd and CH3CC13, the reason : CO and N2O have third law entropies lower than the statistical values, because at T = 0° K, the crystals of these substances are not perfect. Hence the assumption that S° = 0 at 0 K is not correct. The explanation for higher third law entropies in case of (CH3)2Cd and CH3CC13 is that the rotational contribution towards entropy has been taken for the molecule as a whole whereas actually in such molecules groups of atoms can rotate as a unit about other group to which they are joined by single bonds e.g., CH3 around the bond linking it to Cd or CCl3 around the bond linking it to CH3.
Non-equilibrium Thermodynamics 8 Non-equilibrium Thermodynamics Onsager Reciprocity Relation in Mathematical Form : The Onsager reciprocal relationship is LiK = LKi, The above equation means that for two irreversible processes which couple, the cross coefficients are equal. It is possible for more than two forces to couple. There exists a criterion which allows one to deduce `a priori' the number of effective couplings. This is Curie's principle of symmetry. The principle states that a macroscopic phenomenon in the system never has more elements of symmetry than the cause that produces it. For example, the chemical affinity (which is a scalar quantity) can never cause a vectorial heat flux and the corresponding coupling coefficient disappears. A coupling is possible only between phenomenon which have the same tensor symmetry. Thus Onsager reciprocity relation is not valid for a situation when the fluxes have different tensorial character. Irreversible Thermodynamics The branch of science dealing with the study of thermodynamic properties of the systems which are not in Chemistry : Practical Application equilibrium and involve transport processes which are irreversible is termed as non-equilibrium or irreversible thermodynamics. Irreversible thermodynamics is applicable to those systems which are not too far from equilibrium. Assumption of Local Equilibrium Temperature (T), pressure (P), partial specific Gibbs function (G), and entropy (S) for a cell in a nonequilibrium state depend upon the internal energy (U) and the mass units (m,) of the molecular species `/' exactly in the same manner as in an equilibrium situation. In other words each cell of a system in nonequilibrium is treated exactly in the same manner (statistically as well as thermodynamically) as if equilibrium exists in each of these cells. Entropy Production Arising from Heat Interaction : Entropy changes occur in a system due to the interaction of the system with the environment and the irreversible processes that occur internally. These changes may be expressed as dS = dSext + dSint
where the subscripts ext and int refer to external and internal effects respectively. Consider two systems A and B of fixed composition. Let dqA and dqB represent the heat interaction between systems A and B with the environment during an interval of time dt. During the same interval of time, the heat interaction between the two systems is dqAB.
Heat interaction between two systems Non-equilibrium Thermodynamics If the temperature of each system remains unchanged the energy balance at system A is given by the expression dqA.net = dqA _ dqAB while at system B, the energy balance is given by dqB.net = dqAB _ dqB The entropy change of system A is given by
dSA = and of system B is given by
dSB = The change of entropy of the combined system is represented as dS = dSA + dSB
= Hence, the entropy,change due solely to heat interaction with the environment is given by
(dSAB)ext = while the entropy change due to internal effects resulting from heat interaction between systems A and B is given by
(dSAB)int =
=
>0
This entropy production is associated with irreversible Chemistry : Practical Application heat interaction between systems A and B. It should be noted that if TA > TB and dqAB > 0 then (dSAB)int > 0. On the other hand, if TA < TB, then dqAB < 0 and hence (dSAB)int is again >0. The rate of production of internal entropy is given by
=
> 0 ... (1)
If TA and TB differ by only a small amount then equation (1) becomes
=
... (2)
From equation (2), the rate of entropy production per unit volume, s, can be readily obtained.
Thus s =
=
=
... (3)
... (4)
where A.dx represents volume and JQ is the heat flux. The negative sign in equation (3) indicates that the heat flow takes place in the direction of negative temperature gradient. Entropy Production due to Flow of Matter
Isothermal flow through a semipermeable membrane A fluid consisting of several species flows steadily and isothermaiiy through a distance dx. A quantity of heat dq must Non-equilibrium Thermodynamics be transferred in order to maintain the isothermal flow. It is assumed that no phase changes occur in the fluid and that properties such as enthalpy and chemical potential vary infinitesimally across the distance dx. Thus the change in entropy occurs only due to heat transfer through internal irreversible process. This change is given by dS = dSint + dq/T ... (1) Rearranging equation (1) we have TdSint = T dS _ dq ... (2)
The first law of thermodynamics can be applied to this isothermal, steady state process, to give
=0
or, dq =
... (3)
Also, at constant temperature dH = dG + T dS ... (4) Since in an open system, the chemical potential mi of the ith species is given by mi = (¶G/¶ni) T, P, nj ... (5) hence we can write
dG =
\
... (6)
=
... (7)
Equation (3) can now be written as
... (8)
dq =
Chemistry : Practical Application Hence, T dSint = T dS _
=
... (10)
... (9)
The rate of entropy production per unit volume can thus be written as
s=
=-
... (11)
Since dni/A dt is the molar flux Jni, hence
... (12)
s=
=
> 0 ... (13)
From the above two cases it is clear that the rate of internal entropy production per unit volume is the sum of the products of the flux and the conjugate driving force. This may he expressed as
s=
>0
where Ji is either the energy flux or mass flux and Xi is a function of a potential gradient such as temperature gradient or chemical potential gradient. Linear Law This law holds only when system under study is close to equilibrium i.e., only removed slightly from equilibrium. Some examples of this law are Ohms law, Fourier's law. The causes which give rise to irreversible process (e.g., temperature Non-equilibrium Thermodynamics gradient, potential difference, etc.) are termed as driving forces (denoted by X) and the effects produced (e.g., flow of heat, flow of current, etc.) are termed as fluxes (denoted by J). Thus the linear relations near the equilibrium have been of the following type. In the above linear relations, the fluxes are proportional to the forces. Relations of this type are termed as phenomenological relations and the coefficients Lik are termed as phenomenological coefficients.
Matched Phenomenological Laws to define the transport processes and relations. Process Relation Law
(1) Heat Transfer (a) J =
(i) Newton's
(2) Mass Transfer (b) J =
(ii) Fourier's
(3) Momentum Transfer (c) J =
(4) Flow of electricity (d) J =
(iii) Ohm's
(iv) Pick's
(1) (c) (ii), (2) (a) (iv), (3) (d) (i), (4) (b) (iii) Prigogines principle of minimum entropy production, proved : When a system consisting of n fluxes and n forces is allowed to age, it eventually attains a state of equilibrium. It is necessary, therefore to impose certain constraints on the system to achieve the steady state. Let us impose the constraints in such a manner that forces X1, X2, ... XK are fixed at constant values while the remaining forces XK + 1, Xk + 2, ... Xn are allowed to vary. The entropy production is given by
s=
... (1)
Chemistry : Practical Application Substituting the values of Ji i.e.,
Ji = inequation (1) we obtain
... (2)
s=
...(3)
Differentiating equation (3) with respect to unrestrained forces Xi and making use of Onsager reciprocal relations viz., Lij = Lji we get
=
=2Ji... (4)
If now we take all the flows corresponding to the unrestrained forces as zero, as a criterion of the steady state, then we can write
= 0 ... (5) The entropy production s, being a positive definite, equation (5) gives the condition of a minimum. It is a mathematical form of Prigogine's principle of minimum entropy production according to which at the steady state, all the flows corresponding to the unrestricted forces vanish. Gibbs Equation for an Open System : For an open system, we can write the following equation (3) by combining equations (1) and (2) which is a form of Gibbs equation. dE = TdS _ PdV _ dq¢ ... (1) dGT,P = _ dq¢ ...(2) TdS = dE + PdV _ dG ...(3) Substituting dG in terms of chemical potentials TdS = dE + PdV _ Sm dni Dividing by
TdS = (dE/T) + (PdV/T) _ Non-equilibrium Thermodynamics
Thus the total entropy change consists of the part supplied by the environment and of the entropy production with in the system due to chemical reaction. Donder's Concept of Extent of Reaction : For a chemical reaction in a closed system the law of conservation of mass is applicable. For the formation of ammonia N2 + 3H2
2 NH3
2 × 17 _ 1 × 28 _ 3 × 2 = 0 This can be written as =0 where V is the stoichiometric coefficient of the component c, the value being positive for products and negative for the reactants. The degree of progress or extent of the reaction (x) also called chemical variable, according to de Donder's is given by the equation dmC = VC MC dx where dmC is the change in the mass of the component c. The principle of the conservation of mass for a reaction in a closed system can also be stated as (SVCMC) dx = 0 Stationary State A system is said to be in a stationary state if its macroscopic properties such as temperature, pressure, composition and entropy do not change with time inspite of the possible occurrence of irreversible process. Of the macroscopic properties, the intensive ones, though unchanged in time, will generally still vary from point to point in the system. The stationary states may be classified into the following two classes— (1) Equilibrium stationary states i.e., the stationary states attained in a system in equilibrium.
Chemistry : Practical Application (2) Non-equilibrium stationary states i.e., the stationary states attained in a system not in equilibrium. The two differ mainly in the fact that in the former, the value of any intensive property not only remains
constant with time but also has the same value at every point whereas in the latter, the value of any intensive property though remains constant with time at any point but differs from point to point and thus a continuous inflow of energy or matter and energy is required from the surroundings to the system. Thus equilibrium stationary states can be attained only in an isolated system, the non equilibrium stationary states can be attained in open systems.
Molecular Orbital and Valance Bond 9 Molecular Orbital and Valance Bond Valence Bond Wave The valence bond wave function for the HF molecule (assuming that it is formed from Is orbital of H and 2pz orbital of F) in the following three cases : (a) HF is purely covalent (b) HF is purely ionic (c) HF is 70% covalent and 30% ionic. Let `H' atom be represented by A and `F' atom by B. Let the electron numbered 1 be the 1s electron of H atom and the electron numbered 2 be the pz electron of F atom. (a) For purely covalent structure of HF, the wave function is written as ycovalent = yA(1) yB(2) + yA(2) yB(1) (b) For purely ionic structure H+ F_, both the electrons are on the F atom so that yionic = yB (l) yB (2)
Chemistry : Practical Application (c) HF is a resonance hybrid shown as H _ F « H+ F_ \ y = C1 ycovalent + C2 yionic with C12 + C22 = 1 Since C12 = 0.70 and C22 = 0.30 (given)
hence C1 = = 0.83, C2 = = 0.54 y = 0.83 [yA(l) yB(2) + yA(2) yB (1)] + 0.54 [yB(1) yB(2)] Molecular Orbital Wave A molecular orbital wave function for the bond between H and Cl in HCl, constructed, assuming that the bond is formed from the 1s electron of H atom and a 3p electron of Cl atom: Let H atom be designated by A and Cl atom by B. Let the 1s electron of H atom be labelled as electron 1, and 3p electron of Cl atom be labelled as electron 2. yMO = y1y2 where y1 = C1 yA (1) + C2yA(2) and y2 = C1yA (2) + C2 yB(2) \ ymo = [C1yA(1) + C2yB(1)] [C1yA(2) + C2 yB(2)] = C12 [yA(l) yA(2)] + C1C2 {yA(l) yB(2) + {C1C2 yA(2) yB(l)] + C22 [yB(1) yB(2)] The first and the last term are ionic where as middle terms are covalent. The Huckel molecular orbital energies of the planar radical (i.e., 1 nonpaired electron) CH2CHCH2: The Secular determinantal equation is
=0
Molecular Orbital and Valance Bond
or
=0
with x = This yields x3 _ 2x = 0 with roots x = 0, _
,+
, therefore,
E1 = E2 = a E3 = The orbitals corresponding to E2 is called a non bonding orbital since its energy is unchanged from the atomic value a = Hii, which is the 2p atomic orbital energy in this system. The trial wave function for H2+ and H2 molecule. In H2 molecule, pointed out which one are ionic and which one are covalent : Both H2+ molecule ion and H2 molecule are made up of two 1s nuclei. Let the nuclei be a and b. So trial wave function for H2+ molecule ion is y = C1 fa (1) + C2 fb(1) IIIrd for H2 y = y1 y2
=
...(1)
where fi are the atomic wave functions. On expanding equation (1) we have y = C12 fa (1) fa(2) + C22 fb(1) fb(2) + C1C2[fa (1) fb(2) + fb(1) fa(2)] The first and second terms are ionic terms whereas third and fourth terms are covalent ones.
Chemistry : Practical Application If is the Hamiltonian and using the trial function for H2+ molecule ion, the energies of bonding and antibonding orbitals, calculated : The average energy of H2+ ion using the trial wave function y = C1fa (1) + C2 fb (1) can be written as
e_ =
= ... (1)
Let
= Haa;
= Hbb
= Hab;
= Hba
= Saa;
= Sbb
= Sab;
= Sba
Now, both fa and fb are normalised 1s atomic orbital wave functions, Haa = Hbb, Saa = Sbb = 1. It can also be shown that Hab = Hba. The equation (1) becomes
e_ =
...(2)
Using the variation theorem and minimising the energy with respect to C1 and C2. The equation obtained after rearranging and setting
= 0 and
= 0, we get
(Haa _ e) C1 + (Hab _ e Sab) C2 = 0 ... (3) (Hab _ e Sab) C1 + (Haa _ e ) C2 = 0 ... (3) Molecular Orbital and Valance Bond The solution of these equation other that C1 = C2 = 0, is if the following determinant is zero,i.e.,
=0 or (Haa _ e)2 _ (Hab _ e Sab)2 = 0 This is called `Secular determinant'. The solutions of the quadratic equation (5) are
e1 =
...(6)
e2 =
... (7)
Equation (6) is the energy for the bonding and equation (7) is the energy for the antibonding orbitals. From the above problem, the corresponding wave functions, found out :
Substituting the value of e1 in equation (3) of the above answer we get
=0
or
=0
or C1 = C2 = C IIIrd for e = e2, C1 = _ C2 = C Since the total probability is unity, with C1 = C2 and y, we get
1=
=
= = C2 (2 + S)
Chemistry : Practical Application or, C = Similarly for C1 = _ C2 = C, the value of C will be
C=
So the wave functions for bonding and antibonding orbitals are
ybonding =
yantibonding = The combination of the following atomic orbitals give molecular orbitals or not : (a) S and pz (b) px and Px (c) py and dyz Only px and px atomic orbitals will give molecular orbitals as the overlap of these will be finite. The overlap of S and pz and py and dyz will be zero, considering x axis as the molecular axis. Assuming that the ionic character in H_Br bond is 11% the fraction of the contribution of ionic character to the valence bond wave function, calculated : yAB = ycovalent + lyionic
The % ionic character is given by \ 11 = 100l2/ (1 + l2) or, l2 = 11/89
or, l =
= 0.35
Molecular Orbital and Valance Bond The molecular orbital wave function for H2_ molecule anion: H2_ molecule anion contains three electrons which may be labelled as 1, 2 and 3. It contains two nuclei which may be designated as A and B. If in general there are n electrons, than the configurational wave function of the system would be given by
y = y1 y2 y3 .... yn ... (1) Where y1y2 ... are one electron molecular orbital wave functions. \ according to equation (1) we have yMO = y1 y2 y3 where y1 = C1yA (1) + C2yB. ... (1) y2 = C3yA (2) + C4 yB ... (2) y3 = C5 yA (3) + C6 yB ... (3) The valence bond structure of a diatomic molecule, described: The ground state electron configuration of a Cl atom is [Ne] 3s2 3px2 3py2 3pz1 . This configuration suggests that a o bond can be formed between two atoms by spin pairing of the electrons in the 3pz orbitals. The remaining six valence electrons on each atom are not involved in the bonding. This description is consistent with the Lewis structure
The valence-bond wave function for the bonding pair is y = yCl 3pzA (1) yCl 3pzB (2) + yCl 3pzA (2) yCl 3pzB (1) The molecular orbital in equation y = N {y1s (A) + y1s (B)}, normatized : We need to find the factor N in the given equation that ensures that òy2 dt = 1
Chemistry : Practical Application
To proceed we substitute LCAO into this integral and make use of the fact that the atomic orbitals are individually normalized. The additional integral required is the overlap integral (S) S = òyA * yB dt = 0 When we substitute the wave function we find
= N2 {1 + 1 + 2s} = 1 \ The normalization factor is
N= Li2 and Be2 are likely to exist if only the valency s orbitals contribute to molecular orbitals: Each Li atom supplies one valence electron, which fills the bonding s orbitals, to give a bonding configuration. Each Be atom provides two valence electrons, which fill the bonding and antibonding combinations, resulting in no net bond. d orbitals can contribute to o and K orbitals in diatomic molecules : A dz2 orbital has cylindrical symmetry around z and so can contribute to a orbitals. The dzx and dyz orbitals have p symmetry with respect to the axis and so can contribute to p orbitals. The two lowest electron states of H2 can be understood in terms of the MO theory using spatial functions of the form,
y+= All possible two-electron functions which are antisymmetric with respect to electron exchange, constructed: There are six possible ways to arrange the electrons in keeping with the Pauli principles.
Molecular Orbital and Valance Bond - ½ - ½ -½ y_ — _ _ _ _ — ½¯½¯½¯ -½--½½ y+ — _ _ _ _ — ½¯½½¯¯ y1 y2 y3 y4 y5 Ms = 0 1 0 0 _ 1 0
\ y1 =
=
y2 =
=
y3 =
=
y4 =
y5 =
y6 =
=
=
=
Electron He_ H++ ion System
The one electron He_ H++ ion system, considered written the Hamiltonian (Born-Oppenheimer approximation) for the electronic energy in atomic units :
=In MO theory ionic bond is a limiting case of covalent bond, explained : For a molecule AB, yMO = CA fA + CBfB
Chemistry : Practical Application If CA >> CB yMO = fA, a nonbonding M.O. If now we place the two valence electrons in fA we have only the ions A_ and B+ which will be held together by coulombic attraction only. The expectation value of the Hamiltonian for H2+ calculated using the ground state wave function of H atom : Explained given, the integral,
= R being the internuclear distance.
EH2+ =
=
,
=
=
(in a. u.)
(a) When R ® ¥, EH2+ = EH =
a.u.
(b) When R is very small, e_2R » 1 _2R, then (l + R) e_2R » 1 _ R _ 2R2
and EH2+ » EH _ (1 + 2R) »
=
+1
a.u.
At an intermediate value of R, addition of the interauclear repulsion energy
gives the total,
Molecular Orbital and Valance Bond
EH2+ = Calculated energy for various values of R, when plotted against R, does not show a minimum indicating no stability, infact, it is never less than function.
a.u., the energy of an H atom. So 1s is not a good trial
On the basis of the MO theory the stability of HeH and LiH predicted. The nature of polarity in LiH, would be : HeH — From screening rules Z* = 2 _ 0.35 = 1.65
EHe (ls) =
= _ 1.36 a.u.
EH (1s) = _ 0.5 a.u. Difference = 0.86 a.u = 23eV, which is very large, therefore there, is no effective combination.
LiH : ELi (ls) = _
× 32 = _ 4.5 a.u. CLi i.e.; MO formulation puts more electron density in the H side than on the Li side of the molecule, therefore, d+ d_ LiH is Li—————— H.
The Heitler London wave function of H2 used to show that the electron density between the nuclei increases by an amount 2s_1sA 1sB — s2 (1sA2 + 1sB2) in the bonding state and decreases by the same amount in the non bonding state : The electron density distribution in the symmetric and antisymmetric states is given by the equation
y±2 =
[1sA (1)2 _ 1sB (2)2 + 1sA (2)2 1sB(l)2
± 2{1sA(l) 1sB(2) . 1sA(2) 1sB(l)}] ...(1) The density of one electron is obtained by integrating equation (1) over the coordinates of the other electron so that,
r (1) =
=
[1sA(l)2 + 1sB(l)2 ± 2s 1sA(l)1sB(l)]
r (2) =
=
[1sA (2)2 + 1sB (2)2 ± 2s 1sA (2) 1sB (2)]
The total electron density
r = r (1) + r (2) = Change in electron density between the nuclei
= Molecular Orbital and Valance Bond
= = ± 2s . 1sA 1sB _ s2 [1sA2 + 1sB2] assuming s2 < 1 The carbon-carbon bond distance in cyclobutadiene as 1.4 Å, given and shown that it is a paramagnetic molecule by considering its p-electrons as freely moving particles in a two dimensional box : Cyclobutadiene is electrons . The single bond distance of 1.4 Å indicates delocalization of p electron \ The molecule can be treated as a two dimensional square box. The energy of the particle inside the box is
E=
.
Since lx = ly = a, the side of the square;
E= The lowest energy level e1 corresponds to nx = ny = 1 and it is not degenerate. The next higher level with energy E2, is doubly degenerate, since the value of the quantum numbers nx and ny may be nx = 1, ny = 2 or nx = 2, ny = 1. The arrangement of the four p electrons is as follows.
E-
0a By Hund's principle, in level E2, there is one electron in each of the two states, so that the number of unpaired electrons
Chemistry : Practical Application is maximum. The presence of unpaired electrons means paramagnetic behaviour. In fact molecule has been experimentally shown to be a biradical. Molecular Orbital Energy Level The n molecular orbital energy level diagram sketched, for the allyl system under Huckel approximation and computed the delocalization energy of allyl cation : The HMO of the allyl system (C1 _ C2 _ C3) is y = a1 p1 + a2 p2 + a3 p3, where p1, p2 and p3 are the 2pz orbitals of the carbon atoms. The Huckel's secular determinant is
=0 and the corresponding polynomial is x3 _ 2x = 0. Here x=
where a and b are negative energy quantities.
a = H11 = H22 = ......... known as coulomb integral and b = H12 = H23 = ......... known as resonance integral. The solutions for x3 _ 2x = 0 are x = 0, + E1 = a +
b; E2 = a; E3 = a _
b
and _
, so that the allowed p energy levelsare
HMO energy level diagram : E E3 = a _
b———
½ E2 = a — - -¯ ½ ½ E1 = a +
b -¯ -¯ -¯
C3H5 Å C3H5· C3H5Q Molecular Orbital and Valance Bond The allyl cation, allyl radical and allyl anion have 2, 3 and 4p electrons respectively. The p bond energies for the three allyl system are
C3H5 Å : Ep =
C3H5 · : Ep =
C3H5 Q : Ep =
=
=
=
(p bond energy = Total p energy _ The energy of the electrons in the isolated carbon 2p orbitals). Since HMO encompasses all the three carbon atoms it may be assumed that the p electrons are delocalized and the HMO's are spread over the entire molecular frame work for all the three allyl systems. (C— C— C)Å (C— C— C— )· (C— C— C)Q The p-bond energies (equal to
) are due to delocalized structures. By taking only one ethylenic
double bond (i.e., localized p bond) the energy would be 2b. Therefore, the difference _ 2b = 0.828b (which is negative) is known as the delocalization energy which is the same for the three allyl systems. The wave function for the BMO, written down for a heteronuclear diatomic molecule AB assuming that the electron on an average spends 90% of its time on nucleus A and 10% of its tune on nucleus B : In the LCAO — MO scheme, yMO = CA yA + CB yB where the coefficients CA and CB are such that [CA]2 and [CB]2 determine the probability of finding the electron in the AOs yA and yB respectively. Then [CA]2 = 90% = 0.9 and [CB]2 = 10% = 0.1
Chemistry : Practical Application Hence CA = and CB =
= ± 0.95
= ± 0.32
Thus yMO = 0.95 yA + 0.32 yB The wave functions for the sp hybrid orbitals, constructed: The wave functions for the two sp hybrid orbitals are y1 = a1 fs + b1 fp ... (1) y2 = a2 fs + b2 fp ... (2) Since the s orbital is equally distributed between the two hybrid orbitals, we have a12 = a22 = 1/2 ... (3)
\ a1 = a2 =
... (4)
Since y is normalized we have òy12 dt = ò (a1fs + b1fp)2 dt ... (5) = a12 ò fs2 dt + b12 ò fp2 + 2a1 b1 ò fs fp dt = 1 ... (6) Since AO wave functions are orthogonal and individually normalized we have ò y12 dt = al2(l) + b12(l) + 2al bl (0) = al2 + b12 = 1 ... (7)
Since a1 =
hence b1 =
Again since y1 and y2 are orthogonal \ a1a2 + b1b2 = 0 ... (8) Substituting for a1, a2 and b1 we have
=0 hence b2 =
... (9)
Molecular Orbital and Valance Bond Thus the two sp hybrid orbital wave functions from equation (1) and (2) are
y1 =
y2 = The energy levels for benzene molecule calculated, using HMO theory and also the frequency required to have the lowest energy transition : Benzene is a cyclic molecule and has six atoms. The Huckel Secular determinant is
=0
If
then we have
=0 This 6 × 6 secular determinant leads to a sixth degree polynomial for E. The six roots of the polynomial are
E1 =
= a + 2b
Chemistry : Practical Application E2 =
=a+b
E3 =
=a_b
E4 =
= a _ 2b
E5 =
=
E5 =
=
=a_b
=a+b
Thus E2 and E6 and E3 and E5 are degenerate. The Huckel energy level diagram of benzeneis - — a _ 2b ½ ½——a_b E ½ ..... ..... ..... ...a ½ ½ -¯ - ¯ a + b ½ ½ - ¯ a + 2b The energy required to excite the electron from E2 to E3 i.e.,
v=
=
The simple Huckel molecular treatment of ethylene molecule (CH2 = CH2) :
This is a simple system containing two p orbitals. The determinant form of the secular equation is
=0 Molecular Orbital and Valance Bond
=0 The expansion of this determinant gives an equation x2 _ 1 = 0 x = ± 1, In other words E1 = a + b (bonding) E2 = a _ b (antibonding) The two p electrons in ethylene occupy the orbital of lowest energy in the ground state. Since b is negative, the lowest energy (E) is a + b. An energy level diagram showing the ground state of ethylene is - ¾¾¾¾ a _ b ½ E ½ ................ a ½ ½ ¾¾¾¾¾-¯ a + b The p electronic energy of ethylene is Ea = 2a + 2b
Because a is used to specify the zero of energy, the two energies, E = a ± b must correspond to binding and antibonding orbitals. \ We can write C1 (a _ E) + C2 b = 0 C1b + C2 (a _ E) = 0
Since
=x
C1x + C2 = 0. ü ý ... (1) C1 = C2 x = 0 þ If x = 1, C1 = _ C2 ... (2)
Chemistry : Practical Application Since every molecular orbital is normalised f = C1f1 + C2f2 \ Above equation gives rise to an equation C12 + C22 + 2C1C2 = 1 C12 + C22 = 1 where the overlap integral (S12 = 0) is neglected. On combining equation (2) and (3) we get
C1 =
and C2 = _
Hence the MO which corresponds to the energy level E1 = a + b is given by
y1 = By a similar procedure the molecular orbital whose energy is E2 = a _ b is given by
y2 = The two p electrons occupy the MO y.
(The bonding and antibonding orbitals in the Simple Huckel molecular orbital treatment of ethylene) In SCF method are explained by integral and : The integral describes the electron repulsion between two electrons in the field of the same core and between the two electrons in the neighbouring cores fi and fj. Molecular Orbital and Valance Bond The following structures with their p energy levels : (1) ¾¾¾ (a) Ep+ = 2a + 4b, Ep = 3a + 3b, Ep_ = 4a + 2b (2) (b) Ep = 4a + 5.123 b (3) (c) Ep = 4a + 4 b (4) (d) Ep = 4a + 4.962 b (5) (e) Ep = 6a + 6.988 b (1)—(d), (2)—(c), (3)—(a), (4)—(b), (5)—(e) The wave functions of the molecular orbitals of diborane in terms of its orbitals fB1, fB2 and fH, given
: The molecule has two three centre-two electron bonds. The exact state of hybridization of boron is possibly indecisive. Let us take the view that each boron is sp3 hybridized. Two of the hybrid orbitals of one boron atom overlap with the terminal hydrogen orbitals. The remaining two hybrid orbitals have only one electron and these overlap with two similar orbitals of the other boron atom, through the Is orbitals of the bridging hydrogens. Thus two B—H—B bridges result. These are referred to as the banana bonds. The wave functions of the MO's for the delocalized three centre bond may be written as H HH BB HH H
Chemistry : Practical Application The AO of the two boron atoms, f (B1) and f (B2) which are approximately sp3 and the orbital of hydrogen atom f (H) which is an `s' orbital have similar energy and the appreciable spatial overlap, but it is only the combination f (B1) + f (B2) which has the correct symmetry to combine linearly with f (H). The three normalized and orthogonal MO's have the approximate form.
Bonding : y1 =
Non bonding : y2 =
Antibonding : y3 =
The Solutions 10 The Solutions Ideal and Non-ideal Solutions Ideal Solutions : A solution of two components is said to be ideal if each component of the solution obeys Raoult's law at all temperatures and concentrations. Further, it is observed that the solution formed on mixing the two liquids is ideal if there is no volume change on mixing i.e., DVmix = 0 and there is no enthalpy change on mixing i.e., DHmix = 0 example is mixture of n-hexane and n-heptane. Non-ideal Solution : A solution formed by mixing two liquids is said to be non-ideal if it does not obey Raoults law or the interactions of A and B molecules in the solution are not similar to those of pure A and pure B or DVmix ¹ 0 and DHmix ¹ 0. Positive and Negative Deviations Positive Deviation Solutions : (1) Water + ethanol (2) Ethanol + Chloroform
Chemistry : Practical Application (3) Benzene + toluene. Negative Deviation Solutions : (1) Water + Nitric acid (2) Water + Sulphuric acid (3) Acetone + Chloroform. Konowaloffs Rule Konowaloffs rule states that in case of ideal solutions or solutions showing small positive or negative
deviations, at any fixed temperature, the vapour phase is always richer in the more volatile component as compared to the solution phase. In other words, the mole fraction of the more volatile, component is always greater in the vapour phase than in the solution phase. The rule is very useful in the separation of binary liquid solutions by distillation. A mixture of water and aniline boils at a temperature of 98.5° C at pressure of 760 mm. The vapour pressure of water at this temperature is 717mm. Composition of the Distillate. Here, pw° = 717mm pl° = 760 _ 717 = 43 mm Further Mw = 18, Ml = 93
\
=
=
= 0.31
i.e., in the distillate, weight of aniline : Weight of water = 0.31 : 1
Percentage by weight of aniline =
= 23.66
Percentage by weight of water = 100 _ 23.66 = 76.34 Through their gill systems, fish breathes the dissolved air present in water. Though the concentration of air in water is The Solutions not great, show that the air which a fish breathes is richer in oxygen percentage wise than ours. The partial pressure of oxygen in air is 0.20 atm and that of nitrogen is 0.80 atm at 20°C. The Henry's law constant for oxygen is 4.1 × 104 atm. for nitrogen it is 8.3 × 104 atm. The solubility of O2 and N2 in water in terms of mole fraction can be calculated as follows XO2 = PO2/KO 2, H2O XO2 = 0.20 atm/4.l × 104 atm
XO2 = 4.9 × 10_6 and XN2 = PN2/KN 2, H2O = 0.80 atm/8.30 × 104 atm = 9.6 × 10_6 Thus, the dissolved air a fish breath is thus 1/3 oxygen, which is greater than the 1/5 oxygen present in the air we breathe. Vapour pressure of a pure liquid is 100 torr at 25°C while its vapour pressure from 95 mole per cent solution at the same temperature is only 90 torr. The activity of the liquid in the solution. The activity coefficient for it. In this case vapour pressures also represent the fugacities of the liquid in the pure and solution states, the activities can thus be calculated. Pure liquid state will be the standard state of the liquid.
=
\f=
f° =
atm
atm
a = f / f° =
a1 = Y1 X1 where X1 =
Chemistry : Practical Application Y1 = If the vapour pressure of two pure liquids are 400 and 910 mm Hg at 80°C. (i)The composition of the mixture that boils at 80°C and (ii) The composition of the vapour would be:
Taking 80°C as the normal boiling point of the mixture, applying Raoults law, XA × 400 + (1 _ XA) 910 = 760 torr 510 XA = 150, \ XA = 0.2941 Hence XB = 0.7059 The composition of the liquid mixture is thus 29.41% A and 70.59% B (mole per cent). In the Vapour phase, the molar ratio is 400 XA : 910 XB = 117.6 : 642.4 The composition of the vapour mixture is
= = 15.47 % of A
= = 84.53% of B The solubility of Ag2CrO4 in water is 8 × 10_5 mol kg_1 at 25°C and its solubility in 0.04 mol kg_l NaNO3 solution is 8.84 × 10_5 mol kg _1. The mean ionic activity coefficient of Ag2CrO4 in 0.04 mol kg_1 NaNO3: Ag2CrO4
2Ag+ + CrO42_
(s) (Solution) The concentration solubility product
KC = C2Ag + CCrO42_ = (16 × 10_5)2 × (8 × 10_5) (in pure water) The Solutions The activity solubility product, where (Y±)0 is the mean activity coefficient of Ag2CrO4 in the absence of added salt. In the presence of salt solution (NaNO3) KC = (17.68 × 10_5)2 (8.84 × 10_5) The activity solubility product, Ka = (17.68 × 10_5)2 (8.84 × 10_5) (Y+3) where Y+ is the mean activity coefficient of Ag2CrO4 in NaNO3 Solution. The two Ka values are same. (16 × 10_5)2 (8 × 10_5)(Y+)3 0 = (17.68 × 10_5)2 (8.84 × 10_5)(Y±)3
i.e.,
=
= 1.3492
Taking log, 3[log (Y±)0 _ log (Y±)] = log 1.3492 = 0.130076 log (Y+)0 _ log (Y+) = 0.043359 ... (1) By the Debye Huckel limiting law, log Y+ = _A|Z+ Z_| For
Ag2CrO4, |Z+ Z_| = 2, when Ag2CrO4 is dissolved in pure water
[16 × 10_5 × 12 + 8 × 10_5 × 22j
I=
= 24 × 10_5 = 2.4 × 10_4; = 1.5492 × 10_2 \ log (Y+)0 = _ A × 2 × 1.5492 × 10_2 = _ A × 3.0984 × 10_2
Chemistry : Practical Application Similarly when Ag2CrO4 is dissolved in 0.04m NaNO3, the total ionic strength = 0.04 + 2.652 × 10_4 = 0.0402652 and
= 0.2007
\ log (Y+) = _ A × 2 × 0.2007 = _ A × 0.4014 Substituting in equation (1), (_ 3.0984 × 10_2) A + (0.4014 A) = 0.043359 A (0.4014 _ 0.030984) = A × 0.37042 = 0.043359
\A=
= 0.11705
(Theoretical value of the Debye Huckel constant, A is 0.509 at 25°C. So, the value obtained above from solubility measurements is very much below the theoretical value). With this value of A = 0.11705, we can calculate Y+, thus
log Y+ = _ A |Z+ Z_| = _ 0.11705 × 2 × 0.2007 = _ 0.046984 Y+ = 0.89746 The expressions for the activities of NaCl, CaCl 2 and LaCl3 in terms of their molalities and mean ionic activity coefficients: aNaCl = m2Y2± aCaCl2 = 4m3 Y3± aLaCl3 = 27m4 Y4± The partial pressure of bromine over a bromine-carbon tetrachloride solution containing mole fraction of bromine equal to 0.025 is 10.27 torr. If the vapour pressure of pure bromine at the same temperature is 213 torr. The activity coefficient of bromine in the given solution : The solution is non-ideal, for which Raoults law is pi = aipi0 The Solutions For bromine designed by subscript A pA = aAPA0 aA = pA/pA0 = 10.27 torr/213 torr = 0.0482 From equation Yi = ai/xi YA = aA/xA = 0.0482/0.025=1.93 One mole of component A and two moles of component B are mixed at 27°C to form an ideal binary solution. Calculation of DVmix, DHmix, DGmix and DSmix. (R = 8.314 JK_1 mol_1)
For an ideal binary solution DVmix = DHmix = 0 T = 27°C = 300K nA = 4 moles, nB = 2 moles
Mole fractions, xA =
= 0.334,
\ xB = 0.666 DGmix = RT Sni In xi = RT (nA In xA + nB In xB) = 8.314 × 300 (0.334 In 334 + 0.666 In 0.66) = _ 1585.38 J DSmix = _ R Sni In xi = _ R (nA In xi + nB In xb) = 5.28 JK_1 Another way to find it is DSmix = _DGmix/T
=
= 5.28 JK_1
Chemistry : Practical Application For an electrolyte (strong) MX, the mean activity coefficient (Y±) and mean activity (a±) is defined this way. Strong electrolyte MX ionizes completely MX ® M+ (aq) + X_ (aq) The chemical potential of the solution is given by m (M+) + m (X_ ) = m0 (M+) + m0(X_) + RT In a(M+) a(X_) ...(1) where the activities are denoted by a molality and activity are related by the expression
a=
...(2)
where Y is the activity coefficient and m0 is the standard concentration (m0 = 1 mol kg_1 of solvent) when this is substituted in equation (1) we get m (M+) + m(X_) = m° (M+) + m° (X_)
+ RT In
... (3)
where Y(M+) and Y(X_) are the activity coefficients for the cation and anion respectively. We cannot determine these two activity coefficients separately since the non-ideality arises mainly because of the interaction between the two ions. Hence mean activity coefficient Y+ is given by the equation
Y+ = Y(M+) = Y(X_) = With this definition the chemical potential of cation and anion is given by
m(M+) = m°(M+) + RT In a (M+) = m° (M+) + RT In m(M+) + RT In Y+ The Solutions The mean activity of the electrolyte with a positive and a negative ;on is therefore given by
= For a multivalent electrolyte MpXq. The expression for equilibrium ions mean activity and mean activity coefficient: Mean activity of 8.25 × 10_4 m Al2(SO4)3 at 25 ° C given that the mean activity coefficient at this temperature is 0.9913 and the compound dissociates completely into ions. Calculated: Mp Xq = pMq + (aq) + qXp_ (aq) If dissociation of the saturated solution is complete the equilibrium constant K is given by K = [a (Mq+)]p [a(Xp_) q] Mean activity of the ions is given by a± = {[a (Mq+ )]p [a (Xp_)q ]}1/(p + q) and mean activity coefficient by Y± = {[Y (Mq+)]p [(Xp_)q ]}1/(p + q) Al2(SO4)3 = 2 Al3+ + 3SO42_ \ The mean molality m± is given by m± = [(2m)2 (3m)3]1/5
= (108)1/5m = 2.55 m where m = 8.25 × 10_5 mol kg_1 of solvent The mean activity is therefore given by a+ = (0.9913) (2.55) (m/m0) = 2.09 × 10_4 Supposed the excess free energy of two ionic species is given by GE = KXi Xj (ni, + nj) zizje2 where X, n and z denote the mole fraction, moles, and the charge of the species respectively and K is a constant. The expression for the activity coefficient of the ith species. Derived :
Chemistry : Practical Application If we combine all the concentration independent terms into one constant, we get
GE = K¢ Xi Xj (ni + nj) = From equations and m(excess) = RT In Y
and m(excess) = It follows that
RT In Y =
=
= K¢ Xj2
Thus, the activity coefficient of one component depends on the square of the mole fraction of the other component. This type of behaviour is observed in some nonionic compounds The factors on which the activity coefficient of a given electrolyte depends when it is in a solution
containing many electrolytes are : It depends on the following factors— (1) Total concentration of the electrolyte. (2) Individual characters of the different ions present in the mixture. (3) When the mixture contains strong electrolytes of the same ion type (say all univalent or all bivalent) then the activity coefficient of a given electrolyte is very dilute solutions is the same in all solutions of the same total normality. In more concentrated solutions, the specific-character of the ions influences the activity coefficient. The Solutions The influence of an ion in a mixture on the activity coefficient of a given electrolyte is : Lewis and Randall introduced a term called ionic strength, denoted by (i in order to represent the variation of activity coefficient of an electrolyte with concentration, in presence of other ions in a fixture. The ionic strength is the measure of the effective influence of all the ions in the mixture. It is obtained by multiplying the molality of each ion present in the solution with the square of its valency and dividing the sum of these products by two. Hence mathematically
m=
or, m = where m1, m2, m3,... are the molalities and z1, z2, z3 are the valencies of all the ions in the mixture. Thus the influence ofan ion in the mixture on the activity coefficient of a given electrolyte is proportional to the square of its valency. Thus bivalent ion will influence four times in comparison of univalent ion. The ionic strength of (1) 0.1 molal KCl and (2) 0.2 molal BaCl2. Calculated: (1) The molality of each ion in 0.1 molal KCl is 0.1, i.e., m+ = 0.1and m_ = 0.1 Also z+ = 1 and z_ = 1,
\m=
= 0.1
Note : For a single univalent electrolyte the ionic strength is always equal to the molality of the solution. (2) In 0.2 molal BaCl2 solution, m+ = 0.2, z+ = 2, m_ = 2 × 0.2 = 0.4 and z_ = 1, so that
m=
(0.2 × 22 + 0.4 × 12) = 0.6
Chemistry : Practical Application The value of the coefficient A in the Debye Huckel equation for aqueous solutions at 298.15K. The relative permittivity er of water at this temperature is 78.54. Calculated : In dilute solutions the activity coefficient Yi of an ion species with a charge number of zi is given by log Yi = _ Azi2I1/2 where I is the ionic strength and
A= where msolv is the mass of solvent in volume V and er is the relative permittivity.
\A=
= 0.509 kg1/2 mol_1/2 Debye Huckel theory used to calculate Y+, Y_ and Y+ and aNaCl for 0.001 molal sodium chloride in water at 25° C : log Yi = _ Azi2I1/2 = _ (0.509) (0.001)1/2 Y+ = Y_ = 0.964 log Y+ = Az + z _ I1/2 = (0.509) (1)(_1)(0.001)1/2 Y± = (Y +Y_)1/2 = 0.964 Excess thermodynamic properties or functions, described: An excess thermodynamic property ME is equal to the difference between the actual property M and the property for an ideal solution at the same T, P and x. Here M represents extensive properties V, U, H, Cp, S, A and G. ME is defined by the relation. The Solutions ME = M _ Mid, where M is the actual property of the system and Mid the calculated value for an ideal solution. The property change of mixing DM_ is equal to M_ _ Sxi M°i and the property change of mixing to form an ideal solution DMid is equal to Mid _ Sxi Mi° and so ME = DM _ DMid If a solution is ideal, its property Mid is equal to the sum of the properties of the pure components, i.e., Mid = SXi M° where M° is the property in pure, state. \ DMid = 0, this is the case for the properties volume, E, H and Cp. But it is not so for the properties entropy (S) and free energy. For excess entropy of the solution SE = DS _ DSid = DS _ (_R Sxi In xi)
= DS + R Sxi In xi where DS is the change in entropy, when the pure components are mixed together to form the solution. Similarly GE = DG _ Rt Sxi In xi In terms of the partial molal free energy G_i of the ith component we have G = Sxi G_i GE = DGE = DG _ DGid But DG = G _ SXi Gi° and DGid = RT Sxi In xi \ GE = G _ Sxi Gi° _ RT Sxi In xi =
= But G_i = Gi0 + RT In ai, \ GE = Sxi RT In ai _ RT Sxi In xi
=
= RT Sxi In Yi,
Chemistry : Practical Application Since Yi =
(rational activity coefficient)
Thus GE = RT Sxi In Yi
Since GE is a thermodynamic property which is additive with respect to RT In Yi, it follows that RT In Yi may be considered as a partial molal property. The excess free energy GE is very useful since it gives the most compact way of storing information about a real solution. GE may be expressed as a function of T, P and x. For instance, the Margules equation is GE = RT (A21x1 + A12x2)x 1x2, where A21 and A12 are parameters to be obtained from experimental data. From this equation it is possible to express the activity coefficients of the two components as In Y1 = x22[A12 + 2(A21 _ A12) x1] and In Y2 = x12 [A21 + 2(A12 _ A21) x2] For a solution containing acetone (component 1) and chloroform (component 2), the formation of the solution being exothermic the graphical representation of some excess functions are :
The Solutions Matched :
(1) HE (a)
(2) SE (b) (3) VE (c) HE _ TSE
(4) GE (d) (1) (d), (2) (a), (3) (b), (4) (c). For 0.005 m K2SO4 solution at 25° C. Calculated mean activity coefficient of electrolyte using DebyeHuckel limiting law (Debye Huckel constant A at 25° C is 0.509 kg1/2 mol_1/2): The ionic strength of K2SO4 solution
=
(2 × 0.005 × 12 + 0.005 × 22)
= 0.015 According to Debye-Huckel limiting law,
log Y± = = = _ 0.1247 = 1_.8753 \ Y+ = 0.750 Gibb's Duhem Equation, proved equation. If the temperature and pressure are kept constant the chemical potentials of the components of the solution do not vary independently with the change in composition of the solution. They are related by the equation Sni dmi = 0.
Proof : At constant temperature and pressure the free energy of the mixture is given by the equation
Chemistry : Practical Application G = Sni mi ... (1) Differentiating equation (1)
dG =
... (2)
The fundamental equation under these conditions is
dG =
... (3)
Subtracting (3) from (2) we get
=0 If C1 and C2 are the concentrations (mol dm_3) of the solute x in two immiscible solvents and solute
exist as a polymer in solvent II, then
= constant.
The Nernst distribution law is applicable only to the species which have the same form in both the solvents. Now let the equilibrium in solvent II be nx
xn ... (1)
If aa is the degree of association, the concentration of association form will be aaC2/n and that of monomer is (1_ aa)C2. The association constant is defined as
Ka =
or, (1 _ aa) C2 =
... (2)
So the Nerns't distribution law becomes
=
= K ... (4)
The Solutions If the compound x is completely present as polymer in solvent II, aa will be unity and the equation (4) becomes
= constant because n, Ka and K are all constants. The ionic strength in terms of their molality with the given electrolytes, Matched : Electrolytes Ionic strength (1) KNO3 (a) 3m (2) CuSO4 (b) 3m (3) Na2SO4 (c) m (4) CaCl2 (d) 6m (5) Na3PO4 (e) 4m (1)_(c), (2)_(e), (3)_(b), (4)_(a), (5)_(d). [Note : (3) and (4) can have either (a) or (b) as their answers] Mean activity (a+), mean molality (m±) and mean activity coefficient (Y+) of CaCI2 in a 0.01 molal solution. Y+ and Y_ may be assumed to be 0.5 and 0.8 respectively. Calculated : Here x = l, y = 2
m+ = 0.01, m_ = 0.02 a+ = m+ Y+ = 0.01 × 0.5 = 0.005 a_ = m_Y_ = 0.02 × 0.8 = 0.016 a+ = (ax + ay_)1/(x + y) = (0.005 × 0.0162)1/3 = 0.01086 m+ = (m1 + m2_)1/(x + y) = (0.01 × 0.022)1/3 = 0.01587 Y+ = (Yx+· Yy_)1/( x + y) = (0.5 × 0.82)1/3 = 0.684.
Chemistry : Practical Application Hydration number of an ion : The hydration number (Nw) of an ion is defined as the number of water molecules that have lost their translational degrees of freedom because of their association with the ion. Smaller ions bind more water than larger ions and cations somewhat more water than anions. For example mean hydration number of Li+ is 4, Na+ is 3, Rb+ is 1, F_ is 3, Cl_ is 2, I_ is 0.7. Many important physiological mechanisms are based on the hydration of ions and the consequent effects on ionic mobilities. The volume of aqueous solutions of NaCl containing 1000 g of water is given by V (mL) = 1001.38 + 16.6253 n2 + 1.7738 n23/2 + 0.1194 n22 where n2 is the molarity of NaCl. The volumes were measured at 25°C. (a) Calculate the partial molar volume of NaCl when n2 = 0.5000, (b) calculate the partial molar volume of water at this concentration.
v2 =
= 16.6253 + 1.5 × 1.7738 n20.5 + 2 × 0.1194 n2 Substituting n2 = 0.5000, we get v2 (n2 = 0.5000) = 18.626 mL The number of moles of water is
= 55.506 The total volume n2 = 0.5000 is given by 1010.35 mL. \ V = 1010.35 = n1v1 + n2v2 = 55.506 v1 + 9.313 v1 = 18.035 ml.
