Polyatomic Molecular Orbital Theory – Notes | CHE 3340, Study notes of Inorganic Chemistry

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Polyatomic

Molecular Orbital

Diagrams

Chapter 2

Simple Polyatomics

 H

3

+ and H

3

 Simplest polyatomics

 H

3

+ Transitory gas phase ion

 (^) Detected by spectroscopy

 What controls the SHAPE?

 Is it Linear or Triangular?

 Look at MO theory for answer.

 Only valence AO’s are 1s  3 total

  3 possible LCAO’s

 ie., 3 MO’s

Linear H 3 H A —H B —H

 Now draw MO C

diagram

 Central Atom at

left

 Usually less

electronegative

 Terminal atoms (or

ligand) at right

 Usually more

electronegative

1 s B 1 s A,^1 s C Central H Terminal H's

Linear H 3 H A —H B —H  (^) Now put in MO's C

 Start at bottom with

completely bonding

MO

 (^) No nodes

 Next MO with 1

node

 Last MO with 2

nodes

 (^) E of MO's with  # nodes between neighbor atoms

1 s B 1 s

A,^1 s C

g

u

g

Central

H

Terminal

H's

Triangular H 3

 Point group =

 MO’s take same

form, but now each 

has different

symmetry than

above

H H H A B C

a ' 1 s 1 s 1 s

1                A B C A C s s s s s e 1 1 1

 2      a 1 ' e' D 3h

Triangular H 3

 Now 2 & 3 are degenerate—same E

Why? Symmetry!!!

 Can make argument based on # nodes

 Net # bonding & antibonding interactions

 In triangular molecule

 2  & 3 both have one node

 In 2

 (^) Now A & C are next to each other  (^) Have opposite sign  (^) Antibonding interactions  (^) Higher in E than in linear molecule

How to Determine Symmetry Labels for MO’s?

1. Determine Point group for molecule

2. Draw LGO’s (ligand group orbitals)

3. See How transform under symmetry

operations

4. Assign label from character table

(Appendix 3) based on above.

Or

1. Determine Point group for molecule

2. Draw LGO’s (ligand group orbitals)

3. Use Appendix 4 to assign symmetry label

Triangular H 3

 Create MO

diagram

 First put in atomic

orbitals

 Central on Left

 Terminal on Right

Note: both central

& terminal atoms

same

 Energy same

1 s B 1 s A, 1 s C Central H Terminal H's

H H

H

Triangular H 3  (^) E of MO's with  # nodes between neighbor atoms 1 s B 1 s A, 1 s C a 1 ' Central H Terminal H's e' H H H

Triangular H 3

 E  with  # nodes

between neighbor

atoms

 In H

3

 2 e  in bonding

MO

 1 e ^ in

Antibonding MO

Bond Order 

1 s B 1 s A, 1 s C a 1 ' Central H Terminal H's e'

H H

H

6 1

Bonds 3

Bond Order Net Bond Order 2 1    14

Walsh Diagram  (^) Correlates 2 structures so that can see which is more favorable 180° linear 60° cyclic H—H—H

H H

H

e’ a 1

g

u

g

(nb)

Which is Preferred Structure? Net bond order

 For H

 (^) Linear is favored as BO = 1  (^) Net = ½  (^) Linear, 3 rd^ e in non-bonding MO  (^) Triangular, 3rd^ e in anti-bonding MO

 For H

3

 (^) Triangular preferred as BO = 1  (^) Net BO =1/  (^) Now lower E of bonding MO in triangular dominates

Linear vs. Cyclic MO’s– N = 4

Linear vs. Cyclic MO’s– N = 5