Question

Ψ1, Ψ2, Ψ3 and Ψ4 are four Hückel molecular orbitals of benzene with orbital energies E1, E2, E3, and E4, respectively.
Ψ1 = $\frac{1}{2}$(ФB + ФC - ФE - ФF)
Ψ2 = $6^{-\frac{1}{2}}$(ФA - ФB + ФC - ФD + ФE - ФF)
Ψ3 = $6^{-\frac{1}{2}}$(ФA + ФB + ФC + ФD + ФE - ФF)
Ψ4 = $12^{-\frac{1}{2}}$(2ФA + ФB - ФC - 2ФD - ФE + ФF)
The correct order of the orbital energies is
(The six carbon atoms of benzene are denoted by A to F and ФJ, is the 2pz, orbital of Jth carbon of benzene.)

• A. E1 < E2 = E3 < E4
• B. E4 < E1 = E3 < E2
• C. E3 < E1 = E4 < E2
• D. E3 < E2 < E1 = E4

Answer 1

Answer: (C)

E3 < E1 = E4 < E2

Answer 2

The correct option is (C) : E3 < E1 = E4 < E2.