CHM 501

Homework 12

Due November 1, 2001

Use group theoretical techniques to construct an MO energy diagram for the pi orbitals in pyridine. What is the pi bond order?

Answer:

The point group for pyridine is C_2v and the axes are labeled as:

Find the irreducible representations of the p orbitals used for the pi basis set in the C_2v point group:

C2v	E	C2	v(xz)	v'(yz)
a1	1	1	1	1
a2	1	1	–1	–1
b1	1	–1	1	–1
b2	1	–1	–1	1
p orbitals	6	–2	–6	2

The six p orbitals transform as 2a₂ + 4b₂. Now use the rotation and projection operators to find out what each orbital looks like.

C2v	E	C2	v(xz)	v'(yz)
Rp1	p1	–p1	–p1	p1
Rp2	p2	–p6	–p2	p6
Rp3	p3	–p5	–p3	p5
Rp4	p4	–p4	–p4	p4

Now use the projection operators:

Pp₁(a₂) = p₁ + –p₁ + p₁ + –p₁ = 0

Pp₂(a₂) = p₂ + –p₆ + p₂ + –p₆ = 2(p₂ – p₆)

Pp₃(a₂) = p₃ + –p₅ + p₃ + –p₅ = 2(p₃ – p₅)

Pp₄(a₂) = p₄ + –p₄ + p₄ + –p₄ = 0

Pp₁(b₂) = p₁ + p₁ + p₁ + p₁ = 4p₁

Pp₂(b₂) = p₂ + p₆ + p₂ + p₆ = 2(p₂ + p₆)

Pp₃(b₂) = p₃ + p₅ + p₃ + p₅ = 2(p₃ + p₅)

Pp₄(b₂) = p₄ + p₄ + p₄ + p₄ = 4p₄

Find the wavefunctions by using the results of the projection operators and normalizing:

(a₂(1)) = (p₂ – p₆) / 2^½

(a₂(2)) = (p₃ – p₅) / 2^½

(b₂(1)) = p₁

(b₂(2)) = (p₂ + p₆) / 2^½

(b₂(3)) = (p₃ + p₅) / 2^½

(b₂(1)) = p₄

Now the MO diagram can be constructed:

The filled orbitals are all net bonding, so the pi bond order is 3.