Triatomics and Beyond

1) Complex, so we deal with simple symmetrical molecules

2) Same principles apply to orbital combinations as with Diatomics:

i) Compatible symmetry ii) Compatible energy (within 1 Rydberg, 1

Ry)

3) The number of valence AO’s must equal the number of Mo’s

4) MO’s must conform to the symmetry of the molecule.

5) Orbitals of the same energy and the same number of nodes mix.

BeH2

BeH2 is the simplest triatomic molecule.

Linear the gas phase.

The relative energies for the AO’sof Be and H are:

1s (Be) = -9.38 Ry 1s (H) = -0.99 Ry2s (Be) = -0.61 Ry2p (Be) = +0.14 Ry

Which atomic orbitals will combine to make σ MOs?

Which will combine to make p MOs?

Which will not combine remaining σ or p nonbonding MOs?

2pz

2pxy

Be H

2s

1s

1s

Along bond axis

BeH2 MO Diagram

Lewis Structure?Electron Configuration?BO?HOMO?LUMO?Lewis Acid?

2pxy

2pz

Be H * 2

Along bond axis

CO2

Lewis Structure?Shape Family?Valence atomic orbitals on C and O: 2s and 3 x 2p

Consider s and p MO’s formed separately.6 s and 6 p MO’s will be formed (12 possile for each)

Order of energies:

C O * 2

2pz

2px

y

2pxy

2pz

Along bond axis Along bond axis

2s (O) + 2s(C) small2s (O) + 2pz(C) smallest2pz(O) + 2s(C) large2pz(O) + 2pz(C) largest

Valence MO Diagram for CO2

Free atomFree atom

2s (O) + 2s(C) small1s, 2s, 3 *, s 2s

2s (O) + 2pz(C) smallest3 , s 2s, 3 , s 4 *s

2pz(O) + 2s(C) large4s, 5 *s , 4s, 3 s

2pz(O) + 2pz(C) largest6s*, 5s, 4s, 5s,

2px(O) + 2px(C) largest1p , 2p, 3 *, p 2p

2py(O) + 2py(C) largest1p , 2p, 3 *, p 2p 1s

2s

3 *s4s

5 *s

6 *s

BH3

What orbital combinations are possible now?

2pz

2pxy

B H * 3Lewis structure?

Shape Family?

Along Bonding Plane

BH3 MO Diagram

CH4 - The third dimension…

B H * 3

2pz

2pxy

Along Bonding Plane

Frontier MO Theory

BH3H-

BH3 + H- —> BH4-

Reactions take place during collisions.

Bonds are formed and/or broken.

That must mean that there is some kind of orbital interaction.

Which orbitals are most likely interact in forming the new bond?

In general, reactions take place via the interaction of the HOMO of one component with the LUMO of the other because these are the closest in energy.

These orbitals are known as the “frontier orbitals”.

Free atom

Free atom

Electron delocalization (Resonance)In resonance structures, the only electrons that move are:

Delocalized electrons are always found in orbitals.

As orbitals are usually found at higher energy than the orbitals, the

HOMO and LUMO of molecules with multiple bonds are usually orbitals.

As a result of this, we often look only at the orbitals and construct

MO diagrams.

Ethylene

1s

2s

3s4s5s1p

2 *p

*s ’s

C: 2*(2s + 3*(2p)) => 8 AO’s

H: 4*(1s ) => 4 AO’s

=> 12 AO’s

=> 12 MO’s

-MO diagram of Ethylene

Nodes…Pi-bond order…Sigma bond order

When Ethylene reacts…

Ethyne?

Ozone

Nodes…Sigma & Pi-bond order…Total bond orderLewis BOFormal Charge:

Total bond order = bond order + bond order

-7.3eV

-11

1.2

-9.5

0.2

2

-12

Ethylene Butadiene

Nodes…

The importance of the HOMO/LUMO gap. Note: this is not two isolated double bonds but a single -system spread out over four carbons.

HOMO

LUMO

HOMO

LUMO12.2 ev

9.7ev

Benzene

The polygon method for determining -MOs of monocyclic unsaturated molecules:

Works for any monocyclic molecule with contiguous atomic p orbitals.

Benzene can’t be considered to have “three double bonds and three single bonds”. It has three p bonds with bond order _____.

Accordingly, all six C-C bonds in benzene are 140 pm

(whereas pure C-C bonds are 154 pm and pure C=C bonds are 134 pm).

The -MOs of Benzene

How many pi-electrons?

Nodes…(Cuts?)

Aromatic Stabilization(1,3,5-hexatriene)

0

1

2

3