Ebook chemistry practical application part 1

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 globalmedia@dkpd com Table of Contents 1 Theoretic[.]

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: globalmedia@dkpd.com

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

where `h' is the order of the group l i is the dimension of the ith representation l j is the dimension of the

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 A 3 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 SAF s by y(a), y(e) and y¢(e)

From the projection operator theorem

The E representation is doubly degenerate

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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 1 A g ® 1 B 3u 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 D 2h group table

Since y0 É A 1g and y1 É B 3u

é (x01) = A 1g é(x) B 3u = A 1g B 3u B 3u = A 1g

é (y01) = A 1g é(y) B 3u = A 1g B 2u B 3u = B 1g

é (z01) = A 1g é(z) B 3u = A 1g B 1u B 3u = B 2g

\ 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., d x2 _ y2 , d z2 , s, p x , p y , p z so the

formation The D, splitting between T 2g & e g* depends upon the strength of the ligands

Chemistry : Practical Application

The molecular orbital diagram for [FeF 6 ] 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 — — —

e*

g - -

t 2g - - -

e g xx xx

t 1u 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 y i 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 C 2v point group :

The C 2v 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 C 2v 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,

C 2v 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

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The total can be written as

C 2v 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 BF 3 cannot have degenerate orbitals :

BF3 belongs to D 3h 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 D 3h 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 = sy =

The multiplication table is—

The matrices do not form a group since the product 1sz , is y , is x 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

H 2 O molecule to

Theoretical Representations

set up 6 × 6 matrices and then confirm by explicit matrix multiplication the group multiplication (i)

C 2 T V = T¢ v and (ii) s v s¢ v = C 2

Places orbitals h1 and h2 on the H atoms and s, p x , p y and p z on the O atom The z-axis is the C2 axis, X

lies perpendicular to s¢v , y lies perpendicular to s v 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

C2 : (h2, h1, s, _ p x , _p y , p z ) ¬ (h1, h2, s, p x , p y , p z) is reproduced by

sv = (h2, h1, s, p x , _ p y , p z ) ¬ (h1, h2, s, p x , p y , p z ) is reproducedby

Chemistry : Practical Application

s¢v = (h1, h2, s, _ p x , p y , p z ) ¬ (h1, h2, s, p x , p y , p z ) is reproducedby

(i) To confirm the correct representation of C2sv = s¢v , we write D (C2) D(s v)

=

(ii) Similarly, to confirm the correct representation of s v s¢v = C2, we write

Theoretical Representations

The irreducible components of representations generated by a set of a type atomic orbitals in XY 3 molecules of C 3v and D 3h symmetry

C 3v:

In case methane molecule is distorted to (a) C 3v , point group due to bond lengthening and (b) to C 2v point group due to scissor action of molecular vibration More d-orbitals would become available for bonding

(a) In C 3v 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 C 3v , the d

orbitals span A1 + E + E Therefore, all five d orbitals may contribute to the bonding

(b) In C 2v 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

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 (d xy ) may participate in bonding

The ground state of NO 2 (C 2v ) is A 1 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) [C 2v 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

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 D 6h , where m spans E 1u (x,y) and A 2u (z) and the group term is A 1g Then using A 2u × A 1g = A 2u , E 1u × A 1g = E 1u , A 2u × A 2u = A 1g and E 1u × E 1u = A 1g + A 2g + E 2g we conclude that the upper term is either E 1u or A 2u

The symmetry elements listed and named the point groups to which following molecule belongs :

(i) Staggered CH 3 CH 3 , (ii) Chair and boat cyclohexane, (iii) B 2 H 6 , (iv) [Co(en) 3 ] 3 + , (v) S 8 which of them molecules can be polar and chiral

(i) Staggered CH3CH3 : E, C3, C2, 3sd ; D3d

(ii) Chair C6H12 : E, C3, C2, 3sd ; D 3d

Boat C6H12: E, C2, sv, s¢v , C 2v

(iii) B2H6 : E, C2, 2 C¢2; sn ; D 2h

(iv) [Co(en)3]3+ : E, 2C3, 3C2; D3

(v) Crown S8 : E, C4, C2,4C¢2, 4sd , 2S8; D 4d

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 BF 3 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 D 3h, 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 A 1 ® A 2 is forbidden for electric dipole transmission is C 3v 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

Since A1 is not present in any product, the transition dipole moment must be zero

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Raman Spectra

The selection rule for Raman spectra :

For Raman scattering it is necessary that atleast one of the integrals of the type be

non-zero 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 C 3v group is used to obtain the following irreducible representations for the quadratic and binary cartesian coordinates

Quadratic and Binary Cartesian Coordinates Representation

Z2 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

a i =

a i ® number of irreducible representations (symmetry types) of one kind i in T r , 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