Chem 373- Lecture 28: Heteronuclear Diatomic Molecules

Lecture 28: Heteronuclear Diatomic Molecules The material in this lecture covers the following in Atkins.

14.7 Heteronuclear diatomic molecules (a) Polar bonds (b)Electronegativity (c) The variation principle (d) Two simple cases

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Handout for this lecture

Heteronuclear diatomic molecules

A heteronuclear diatomic molecule is a diatomic molecule formed from atoms of two different elements CO HCl

The electron distribution in a covalentbond between the atoms is not evenly shared

Because it is energetically favorable for the electron pair to be closer to one atom than the other

The imbalance results in a polar covalent bond in which the electron pair is hared unequally by the atoms

The bond in HF, for instance, is polar , with the electron pair closer to F.

+ −−

δ δH F

H F

Heteronuclear diatomic molecules Polar BondIn homonuclear diatomics the atomic orbitals from the two nuclei contribute equally to the molecularorbitals :

1 of Hg 2σ : 11 1

2 1σg

A Bs sS

= ++( )

σg 2 of F :12 2

2 1σ σ σ

gA Bp p

S=

++( )

πu of N2 : 12 2

2 1π π π

uA Bp p

S=

++( )

In heteronuclear diatomics the atomic orbitals from the two nuclei do not contribute equally to the molecular orbitals :

1 of HAuσ :=+

+C s C s

SH H Au Au1 6

2 1( )

σ of FI : = ++

C p C pS

F I2 52 1σ σ( )

π of NP : =++

C p C p

SN N P P2 3

2 1π π

( )

The ability of an element to attract electrons is a measure of its : Electronegativity

Electronegativity according to Pauling :

χA = Electronegativity of A χB = Electronegativity of B

|

{

χ χA =

0.102 D(A -B) -12

D(A - A) -12

D(B -B)

−

×

B

AB bond energy AA bond energy

BB bond energy


Electronegativity


Electronegativity

Electronegativity according to Mulliken :

χA = Electronegativity of A

χA I EA ea= +12

( )

Ionization potential of A

Electron affinity of A

L earg value indicatesthat electron is difficult to remove poor electron donor

L earg value indicatesthat element is a good electron acceptor

Heteronuclear diatomic molecules Electronegativity

We have for the total energy of the trial wavefunction

= C + C A Bψ ψ ψA B

E =H dv

dv

ψ ψψ ψ

*

*∫∫

This energy is optimal if E

C and

EC

A B

δδ

δδ

= =0 0

We now express theenergy in terms of thecoefficients C and CA B

ψ ψ

ψ ψ ψ ψ

*

( )( )

dv

C C C C dvA A B B A A B B

=∫

+∫ +

= +∫ +

= + ∫∫

+ ∫

( )

(

C C C C dv

C dv C dv

C C dv

A A B B A B A B

A A B B

A B A B

2 2 2 2

2 2 2 2

2

2

ψ ψ ψ ψ

ψ ψ

ψ ψ

= + +C C C C SA B A B2 2 2

1 1 Overlap S

Linear variationHeteronuclear diatomic molecules

Heteronuclear diatomic moleculesψ ψ

ψ ψ ψ ψ

*

( ) ( )

H dv

C C H C C dvA A B B A A B B

=∫

+∫ +

= + ∫∫

+ ∫

C H dv C H dv

C C H dv

A A A B B B

A B A B

2 2

2

ψ ψ ψ ψ

ψ ψ )

Introducing

H dvA A Aα ψ ψ= ∫ α ψ ψB B BH dv= ∫ b β ψ ψ= ∫ A BH dv

Linear variation theory

ψ ψ α α β*H C C C CA A B B A B∫ = + +2 2 2

Heteronuclear diatomic molecules Linear variation theory

ψ ψ α α β*H C C C CA A B B A B∫ = + +2 2 2 E =

H dv

dv

ψ ψψ ψ

*

*∫∫

ψ ψ*∫ = + +C C C C SA B A B2 2 2

EnergyKinetic

of electron in orbital energy + attraction energy

from both nuclei + repulsion from all other electrons

Aψ : Always negative stabilizing

EnergyKinetic

of electron in orbital energy + attraction energy

from both nuclei+ repulsion fromall other electrons

Bψ : Always negative stabilizing


ψ ψ α α β*H C C C CA A B B A B∫ = + +2 2 2 E =

H dv

dv

ψ ψψ ψ

*

*∫∫

ψ ψ*∫ = + +C C C C SA B A B2 2 2

Overlap between and A Bψ ψ

Resonance integral represents energy of overlap density S

β


ψ ψ α α β*H C C C CA A B B A B∫ = + +2 2 2 E CA(

*

*,C ) =H dv

dvB

ψ ψψ ψ

∫∫

ψ ψ*∫ = + +C C C C SA B A B2 2 2

hus

E C C C C C C S

C C C

A B A B A B

A A B B A B

( , )[ ]2 2

2 2

2

2

+ + =

+ +α α β

This energy is optimal if E

C and

EC

A B

δδ

δδ

= =0 0


We shall now differentiate respect to Cand make use of that for the optimal C

AA

dE C C dC oA B A( , ) / =hus

dE C CdC

C C C C S E C C S

C C

A B

AA B A B A B

A A B

( , )[ ] [ ]2 2 2 2

2 2

+ + + +

+

2

α β

Thus

E C C C C C C S

C C C CA B A B A B

A A B B A B

( , )[ ]2 2

2 22

2

+ + =+ +α α β

o

rE C ES CA A B

:( ) ( )α β− + − = 0


We shall now differentiate respect to Cand make use of that for the optimal C

BB

dE C C dC oA B B( , ) / =hus

dE C CdC

C C C C S E C C S

C C

A B

BA B A B B A

B B A

( , )[ ] [ ]2 2 2 2

2 2

+ + + +

+

2

α β

hus

E C C C C C C S

C C C

A B A B A B

A A B B A B

( , )[ ]2 2

2 2

2

2

+ + =

+ +α α β

o

rE C ES CB B A

:( ) ( )α β− + − = 0

Heteronuclear diatomic molecules Linear variation theory We have derived :( ) ( )( ) ( )α ββ α

A A BA B B

E C ES CES C E C− + − =

− + − =00

This is a homogeneous linear equation in the unknown C and CA B

Homogeneousx A y

A x A y

eq :A11 + =

+ =1222

021 0

Has only solutions ifthe secular determinants :

AA

1121

0 A A

1222

=

Our secular equationreads

α ββ α

AB

-E -ES-ES -E

= 0

This gives us aquadratic equation inE from which we can determine the two roots E and E+ -

Here E corresponds tothe orbital of lower energyand E to the orbital of higher energy

+

-


A substitution of E = E intothe secular equation

+

( ) ( )( ) ( )α ββ α

A A BA B B


− + − =00

affords the coefficients C and CA

+B+

We thus obtain the orbital

energy E

+

+

ψ ψ ψ= ++ +C C

with

A A B B

A substitution of E = E into the secular equation

-

( ) ( )( ) ( )α ββ α

A A BA B B


− + − =00

affords the coefficients C and CA

-B-

We thus obtain the orbital

energy E

-

-

ψ ψ ψ= +− −C C

with

A A B B


αA

αB

E+

E−

ψ ψ ψ+ = ++ +C CA A B B

ψ ψ ψ- = +− −C CA A B B

This is an in - phasebonding orbital withthe largest contributionfrom of lowest energyAψ

This is an out - of-phase anti - bonding orbital withthe largest contribution from of highest energy

Bψ


ES

A+ = +

+α β1

αB

αA

αB −αA

Ε+

Ε−

ES

B− = −

−α β1

The solutions

1. or << 1A BA Bα α α α

β≈ −| |

| |

ψ ψ ψ+ = +12

12A B

ψ ψ ψ- = +12

12A B

αB

αA

αB −αA

Ε+

Ε−


ES

AA

A BA

A B+ = + −

−≈ +

−α β α

α αα β

α α( )2 2

The solutions

2. or << 1A BA B

α α βα α

<<−

| || |

ψ ψ β αα α

ψ+ = + −−A

A

A BB

S( )

ES

BB

A BB

A B− = − −

−≈ −

−α β α

α αα β

α α( )2 2

ψ ψ β αα α

ψ− = + −−B

B

A BA

S( )

αB

αA

αB −αA

Ε+

Ε−


E B+ = +α β ζcot

The general solution

ψ ψ ζ ψ ζ+ = +A Bcos sin

E A− = −α ζcot

ψ ψ ζ ψ ζ− = − +B Asin cos

ζ βα α

=−

12

2arctan

| |

B A


Molecular orbitals og HF

Molecular Orbital Theory Diatomics Term symbols

HD+

Molecule Configuration Term symbol

( )1 1σ 2Σ+

2 1ST +

SYM Lz( )

Spin multiplicity

Σ Π ∆

LTz : 0 1 2Reflection


HD


( )1 2σ Σ+

2 1ST +

SYM Lz( )

Spin multiplicity

Σ Π ∆


HD− ( ) ( * )1 22 1σ σ 2Σ+

2 3He He ( ) ( *)1 22 2σ σ Σ+

LiNa ( ) ( *) ( )1 2 32 2 2σ σ σ Σ+ 2σ *

BeMg ( ) ( *) ( ) ( *)1 2 3 42 2 2 2σ σ σ σ Σ+σ1s s+

1s s−

3σ2 3s s+

4σ *2 3s s−


2 1ST +

SYM Lz( )

Spin multiplicity

Σ Π ∆



BAl ( )1 2π

CSi ( )1 4π

3 Σ− 1∆+ 1Σ+

NP+ ( ) ( )5 11 4σ π

NP ( ) ( )5 12 4σ π 1Σ+

1Σ+

2 Σ+

1π

5σ


2 1ST +

SYM Lz( )

Spin multiplicity

Σ Π ∆



5σ

π

6σ *

2πNO ( ) ( ) ( *)5 1 22 4 1σ π π 2Π− 2Π+

SO ( ) ( ) ( )5 1 22 4 2σ π π3Σ− ∆+ Σ−

FCl ( ) ( ) ( )5 1 22 4 4σ π π Σ−


Metal - main - groupdiatomics

1σ1πx 1πy

δ1 δ2

2πx 2πy

2σ

Main groupMetal

Heteronuclear diatomic molecules Metal - Metal

diatomics

1σg

1πxu1πyu

1δg1 1δg2

1δu1 1δu2

1πxg1πyg

1σu

What you should learn from this lecture

1. You should be able to construct in qualitative terms the compositionsand energies of molecular orbitals for heteronuclear diatomicmolecules

2.You

C C C C

changes

A A B B A A B B

should be able to account for how the coefficients changes in and + +

as A becomes more electronegative than B

ψ ψ ψ ψ ψ ψ= + = ++ + + +

3. You should know qualitatively howE E behaves when A and B havethe same electronegativity and whenA is much more electronegative than B

+ - and

What you should learn from this lecture 4. You should be able to figure out the electron configuration for maingroupdiatomics as well as the term symbol(in simple cases)

5. It would be really cool if you also could account for the bonding in diatomics containing transition metals. But it is not required

6

00

.

( ) ( )( ) ( )

You are required to be able to derive the equations

you will not be askedto solve for E or and

general case

α ββ α

A A BA B B

A B

E C ES CES C E C

HoweverC C in

the

− + − =− + − =

Chem 373- Lecture 28: Heteronuclear Diatomic Molecules

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