Ψ 1 , Ψ 2 , Ψ 3 and Ψ 4 are four Hückel molecular orbitals of benzene with orbital energies E 1 , E 2 , E 3 , and E 4 , respectively. Ψ 1 = $\frac{1}{2}$ (Ф B + Ф C - Ф E - Ф F ) Ψ 2 = $6^{-\frac{1}{2}}$ (Ф A - Ф B + Ф C - Ф D + Ф E - Ф F ) Ψ 3 = $6^{-\frac{1}{2}}$ (Ф A + Ф B + Ф C + Ф D + Ф E - Ф F ) Ψ 4 = $12^{-\frac{1}{2}}$ (2Ф A + Ф B - Ф C - 2Ф D - Ф E + Ф F ) The correct order of the orbital energies is (The six carbon atoms of benzene are denoted by A to F and Ф J , is the 2p z , orbital of J th carbon of benzene.)

Question

Ψ 1 , Ψ 2 , Ψ 3 and Ψ 4 are four Hückel molecular orbitals of benzene with orbital energies E 1 , E 2 , E 3 , and E 4 , respectively. Ψ 1 = $\frac{1}{2}$ (Ф B + Ф C - Ф E - Ф F ) Ψ 2 =  $6^{-\frac{1}{2}}$ (Ф A - Ф B + Ф C - Ф D + Ф E - Ф F ) Ψ 3 =  $6^{-\frac{1}{2}}$ (Ф A + Ф B + Ф C + Ф D + Ф E - Ф F ) Ψ 4 =  $12^{-\frac{1}{2}}$ (2Ф A + Ф B - Ф C - 2Ф D - Ф E + Ф F ) The correct order of the orbital energies is (The six carbon atoms of benzene are denoted by A to F and Ф J , is the 2p z , orbital of J th carbon of benzene.)

cdquestions Admin · Accepted Answer

The correct option is (C) : E 3 < E 1 = E 4 < E 2 .

Answer

E₁ < E₂ = E₃ < E₄

Answer

E₄ < E₁ = E₃ < E₂

Answer

E₃ < E₁ = E₄ < E₂

Answer

E₃ < E₂ < E₁ = E₄

The Correct Option is C

Solution and Explanation

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