B.sc. III Chemistry

Paper C

Submited by :- Dr. Sangeeta Mehtani

Associate Professor Deptt. Of Chemistry

PGGCG, sec11 Chd

Quantum Mechanics

PART 2 B

Rules for linear combination

1. Atomic orbitals must be roughly of the same energy.

2. The orbital must overlap one another as much as

possible- atoms must be close enough for effective

overlap.

3. In order to produce bonding and antibonding MOs,

either the symmetry of two atomic orbital must remain

unchanged when rotated about the internuclear line or

both atomic orbitals must change symmetry in identical

manner.

Linear combination of atomic orbitals

Rules for the use of MOs

* When two AOs mix, two MOs will be produced

* Each orbital can have a total of two electrons

(Pauli principle)

* Lowest energy orbitals are filled first (Aufbau

principle)

* Unpaired electrons have parallel spin (Hund’s rule)

Bond order = ½ (bonding electrons – antibonding

electrons)

A B

A B

AB = N(cA A + cBB)

Linear Combination of Atomic Orbitals (LCAO)

2AB = (cA

2 A

2 + 2cAcB A B + cB2 B

2)

Overlap integral

The wave function for the molecular orbitals can be

approximated by taking linear combinations of atomic

orbitals.

Probability density

c – extent to which each AO

contributes to the MO

cA = cB = 1

+. +. . .+

bonding g

Amplitudes of wave

functions added

g = N [A + B]

Constructive interference

2AB = (cA

2 A

2 + 2cAcB A B + cB2 B

2)

electron density on original atoms,

density between atoms

The accumulation of electron density between the nuclei put the

electron in a position where it interacts strongly with both nuclei.

The energy of the molecule is lower

Nuclei are shielded from each other

Amplitudes of wave

functions

subtracted.

Destructive interference

Nodal plane perpendicular to the

H-H bond axis (en density = 0)

Energy of the en in this orbital is

higher.

+. -. ..

node

antibonding u = N [A - B]

cA = +1, cB = -1 u

+ -

A-B

The electron is excluded from internuclear region destabilizing

Antibonding

When 2 atomic When 2 atomic orbitalsorbitals combine there are 2 combine there are 2

resultant resultant orbitalsorbitals..

low energy bonding orbitallow energy bonding orbital

high energy high energy antibondingantibonding orbital orbital1sb 1sa

s1s

s*

E1s

MolecularMolecular

orbitalsorbitals

EgEg. s . s orbitalsorbitals

Molecular potential energy curve shows the variation

of the molecular energy with internuclear separation.

Looking at the Energy Profile

Bonding orbital

called 1s orbital

s electron

The energy of 1s orbital

decreases as R decreases

However at small separation, repulsion becomes large

There is a minimum in potential energy curve

11.4 eV

109 nm

H2

Location of

Bonding orbital

4.5 eV

LCAO of n A.O n M.O.

The overlap integral

dS BA

*

The extent to which two atomic orbitals on different atom overlaps : the overlap integral

S > 0 Bonding S < 0 anti

S = 0 nonbonding Bond strength depends on the

degree of overlap

Homonuclear Diatomics

• MOs may be classified according to:

(i) Their symmetry around the molecular axis.

(ii) Their bonding and antibonding character.

• s1s s1s* s2s s2s* s2p y(2p) = z(2p)

y*(2p) z*(2p)s2p*.

dx2-dy2 and dxy

Cl4Re ReCl4

2-

A

B g- identical

under inversion

u- not identical

Place labels g or u in this diagram

sg

*g

s*u

u

First period diatomic molecules

s1s2 H E

nerg

y

H H2

1s 1s

sg

su*

Bond order =

½ (bonding electrons – antibonding electrons)

Bond order: 1

s1s2, s*1s2 He E

nerg

y

He He2

1s 1s

sg

su*

Molecular Orbital theory is powerful because it allows us to predict whether

molecules should exist or not and it gives us a clear picture of the of the

electronic structure of any hypothetical molecule that we can imagine.

Diatomic molecules: The bonding in He2

Bond order: 0

Three Conditions for Overlap/Combining of

Atomic Orbitals

Symmetry of the orbitals must be such that regions with the same sign of overlap.

Overlap of s atomic orbital with p atomic orbitals

Energies of the overlapping orbitals must be similar.

The distance between overlapping orbitals must be short to be effective.

Hybridization

An orthonormal set of hybrid orbitals is created by applying a transformation on the orthonormal hydrogenic orbitals. The sp3, sp2 or sp hybrid orbitals are linear combinations of the AO’s, they appear as the resulting interference between s and p orbitals

QUANTUM MECHANICAL

PRINCIPLES OF HYBRIDIZATION

The hybrid orbitals are formed by linear combination of atomic orbitals belonging to the same atom.

As the s-orbital is spherically symmetrical its charge density will be equally distributed among n possible hybrid orbitals.

Each wave function is normalized.

The wave functions of the hybrid orbitalls as well as atomic orbitals are orthogonal to each other.

QUANTUM MECHANICAL

PRINCIPLES OF HYBRIDIZATION

In the formation of sp2 hybrid orbitals as well as atomic orbitals the first hybrid orbital may be considered to have maximum charge density along X-axis. Then py and pz will not contribute towards this hybrid orbital

In the formation of sp3 hybrid orbitals the second orbital may be considered to be in a plane say xz then py will not contribute towards this hybrid orbital

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

The four sp3 hybrid orbitals are as follows: Ψ 1 = a 1 ф s + b 1 ф py +d 1 ф pz Ψ 2 = a 2 фs + b 2 ф py +d 2 ф pz

Ψ 3 = a 3 ф s + b 3 ф py +d 3 ф pz

Ψ 4 = a 4 ф s + b 4 ф py +d 4 ф pz

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

Since the s-orbital is equally distributed among the four hybrid orbital, we will get the following: a 2 1 = a 2 2 = a 2 3 = a 2 4 = 1/4 Hence, a 1 = a 2 = a 3 = a 4 = 1/4 1/2 = 1/2

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

Let us suppose that ф 1 is developed along the x-axis, that is this hybrid orbital will get contribution from s and px orbitals only. Because of this the contribution from ф py and ф pz will get vanished. Hence we can write: c 1 = d 1 = 0 As Ψ1 is in normal form, we will get: a 2 1 + b 2 1 + c 2 1 + d 2 1 = 1 Or we can rewrite it as: a 2 1 + b 2 1 = 1

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

(Because c 1 = d 1 = 0) Therefore b 1 = (1 – a 2 1)

1/2 = (1 – 1/4) 1/2 = (3) 1/2 / 2 The requirement of orthogonality condition for Ψ1 and Ψ2 , Ψ1 and Ψ3, ф1 and Ψ4, it will give: a 1 a 2 + b 1 b 2 = 0 a 1 a 3 + b 1 b 3 = 0 a 1 a 4 + b 1 b 4 = 0

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

Hence, b 2 = b 3 = b 4 = – a 1 a 2 / b 1 = – a1 a 3 / b1 = – a 1 a 4 / b 1

= – [ (1/2) (1/2) ]/[ (3)1/2 / 2]

= – 1/ (2 (3) 1/2)

Suppose that Ψ2 lies in the XZ plane, the hybrid orbital will have contributions form s , px and pz orbitals only. The contribution of py to ф2 would be equal to zero. Hence,

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

c2 = 0 The normalization requirements for Ψ2 will be: a 2 2 + b 2 2 + c 2 2 + d 2 2 = 1 Or we can rewrite it as: a2 2 + b 2 2 + d 2 2 = 1 (Because c 2 = 0)

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

Because d 2 2 = 1 – (a 2 2 + b 2 2) = 1 – (1/4 + 1/12) = 2/3 Therefore d2 = (2/3)1/2 The requirement of orthogonality condition for and Ψ2 , Ψ3 and Ψ 2 , Ψ 4 will give: a 2 a 3 + b 2 b3 + d2 d3 = 0 a 2 a 4 + b 2 b4 + d2 d4 = 0 Hence, d3 = d4

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

= [ - a2 a3 - b2 b3 ]/ d2

Or [- a2 a4 + b2 b4] / d2

= – [(1/4 + 1/12)]/[ (2/3) 1/2]

= – 1/ (6) 1/2

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

The normalization requirement for Ψ3 will be given as: a 2 3 + b 2 3 + c 2 3 + d 2 3 = 1 c 2 3 = 1- (a 2 3 + b 2 3 + d 2 3) = 1 – (1/4 + 1/12 + 1/6) = 1/2 Therefore c3 = + 1/2

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

The normalization requirement for Ψ4 and orthogonality condition between Ψ3 and Ψ4 gives:

c4 = – 1/ (2) 1/2

CALCULATION OF THE COEFFICIENTS OF

ATOMIC ORBITAL IN SP3 HYBRID ORBITALS

Hence, the four sp3 hybrid orbital’s wave functions are:

Ψ 1 = 1/2 ф s + {(3) 1/2} / 2 ф px Ψ 2 = 1/2 ф s – {1/2(3) 1/2} ψ px + (2/3) 1/2 ф pz

Ψ 3 = 1/2 ф s – {1/2(3) 1/2} ф px + 1/(2)sup>1/2 ф py – 1/(6) sup>1/2 ф pz

Ψ 4 = 1/2 ф s – {1/2(3) 1/2} ф px + 1/(2)sup>1/2 ф py – 1/(6) sup>1/2 ф pz