Examples of p electron charge and p electron bond order calculations

Ethene

1 (doubly) occupied MO: y₁ = 1/Ö2 (f₁ +f₂)

p electron charge on atom 1 = q₁= 2((1/Ö2)² = 1

p bond order between atoms 1 and 2 = p₁₂ = 2((1Ö2)(1Ö2)) = 1

(i.e. p bond order for double bond = 1)

1, 3-Butadiene

2 occupied MOs (both doubly occupied),

y₁ = 0.37f₁ + 0.60f₂ + 0.60f₃ + 0.37 f₄ E= a + 1.62b

y₂ = 0.60f₁ + 0.37f₂ – 0.37f₃ – 0.60f₄ E= a + 0.62b

p electron charge on atom 1 = q₁ = 2 (0.37² + 0.60²) @ 1

p electron charge on atom 2 = q₂ = 2 (0.60² + 0.37²) @ 1

Similarly, q₃ = , q₄ = 1

p electron bond order between atoms 1 and 2 = p₁₂

= 2 ((0.37 x 0.60) + (0.60 x 0.37)) = 0.89 (similarly, p₃₄=0.89)

p electron bond order between atoms 1 and 2 = p₁₂

= 2 ((0.37 x –0.37) + (0.60 x 0.60)) = 0.45

Summary: for 1,3-butadiene we obtain:

So according to Hückel theory, the central bond in 1,3-butadiene has less double bond character than the outer bonds, as in a simple drawing

Next: Cycloaddition Reactions