## CHM 501

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

The point group for pyridine is C2v and the axes are labeled as:

Find the irreducible representations of the p orbitals used for the pi basis set in the C2v point group:

 C2v E C2 σv(xz) σv'(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 σv(xz) σv'(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:

Ψ(a2(1)) = (p2 – p6)/ 2½

Ψ(a2(2)) = (p3 – p5)/ 2½

Ψ(b2(1)) = p1

Ψ(b2(2)) = (p2 + p6)/ 2½

Ψ(b2(3)) = (p3 + p5)/ 2½

Ψ(b2(1)) = p4

Now the MO diagram can be constructed:

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