MO theory for polyatomics

Orbitals are treated as objects that can be labeled by their irreducible representations

This can be done within either the VBT or LCAO-MO approaches

VBT

Find the irreducible reps of the bond pairs; this allows us to label the bonds into different energy groups but gives no indication of the relative energy of each type of irred. rep.

CH₄

These transform as a₁ + t₂

a₁ is also an s orbital; t₂ is also p_x, p_y, p_z orbitals which implies an sp³ hybrid

SF₄

find irred reps of bonds; find hybrid

C2v	E	C2	σv(xz)	σv'(yz)
a1	1	1	1	1	z	x2, y2, z2
a2	1	1	-1	-1	Rz	xy
b1	1	-1	1	-1	x, Ry	xz
b2	1	-1	-1	1	y, Rx	yz
bonds	4	0	2	2
lp	1	1	1	1

bonds 2a₁ + b₁ + b₂ sp³

hybrid?

lp transforms as a₁ therefore dsp³ hybrid!

LCAO MO

attempt to use symmetry to convert polyatomic problems into something like a diatomic problem.

1. Break the molecule up into two parts: the center atom and all the atoms bonded to it.

2. Find the irreducible representations of the basis orbitals on the center atom, generally by inspection of the character table.

3. Create "Group Orbitals" from the basis orbitals on the peripheral atoms using symmetry operations.

4. Estimate overlaps remembering that only orbitals of the same irreducible rep can overlap with each other.

H₂O

Basis set: H, 1s; O 2s, 2p

Irreducible reps of the O basis orbitals are read directly from the character table:

2s – a₁

2p_z – a₁

2p_x – b₁

2p_y – b₂

The irreducible reps of the H orbitals must be done as a group since they are not located at the "center" of the molecule.

Reducible rep of H 1s orbitals:

This total rep transforms as a₁ + b₂

What do the a₁ and b₂ group orbitals look like?

Use projection operators:

Note that the sum is over all rotations, not just classes!

Use the rotation operator on the H s orbitals:

	E	C2	σv(xz)	σv'(yz)
	s1	s2	s2	s1

Since s₁ and s₂ are connected by symmetry, we do not need to use the rotation operator on s₂.

Now, use the projection operator for each group orbital:

The wavefunctions are the result of the projection operator normalized:

Ψ(a₁) = 2N(s₁+s₂), where N is the normalization constant.

Normalized wavefunctions are found by squaring the function and setting it equal to 1:

Ψ² = 1 = 4N²(s₁+s₂)²

s₁² = s₂² = 1; s₁s₂ = 0, so

Likewise

Group Theoretical assessment of the energies of π orbitals

Consider all trans butadiene. Find the relative orbital energies of the π orbitals and label them with their irreducible representations.

Point Group: C_2h

Find the transformation properties of the p orbitals that form the π orbitals:

This gives 2b_g + 2a_u

Now use projection operators to find the composition of each molecular orbital:

P(b_g-1) = [(1)p₁ + (–1)p₄ + (1)(–p₄) + (–1)(–p₁)] = 2(p₁ – p₄)

Ψ(b_g-1) = 2^–½(p₁–p₄)

P(b_g-2) = [(1)p₂ + (–1)p₃ + (1)(–p₃) + (–1)(–p₂)] = 2(p₂ – p₃)

Ψ(b_g-2) = 2^–½(p₂ – p₃)

P(a_u-1) = [(1)p₁ + (1)p₄ + (–1)(–p₄) + (–1)(–p₁)] = 2(p₁ + p₄)

Ψ(a_u-1) = 2^–½(p₁ + p₄)

P(a_u-2) = [(1)p₂ + (1)p₃ + (–1)(–p₃) + (–1)(–p₂)] = 2(p₂ + p₃)

Ψ(a_u-2) = 2^–½(p₂ + p₃)

Now draw the MO diagram