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GATE 2024 CY – Question 44

Physical Chemistry · Molecular Structure and Chemical Bonding: Born-Oppenheimer approximation; Valence bond theory treatment for hydrogen molecule and linear combination of atomic orbitals – molecular orbitals (LCAO-MO) theory. Concept of hybridization and hybrid orbitals. Applications of LCAO-MO theory to H2+, H2; Molecular orbital theory (MOT) of homo- and heteronuclear diatomic molecules. Hückel theory, applications to simple molecules, small conjugated π-electron systems. · 2 marks · Multiple choice

$\psi_1,\psi_2,\psi_3$, and $\psi_4$ are four Hückel molecular orbitals of benzene with orbital energies $E_1,E_2,E_3$, and $E_4$, respectively.

$$\psi_1=\tfrac12(\phi_B+\phi_C-\phi_E-\phi_F)$$
$$\psi_2=6^{-1/2}(\phi_A-\phi_B+\phi_C-\phi_D+\phi_E-\phi_F)$$
$$\psi_3=6^{-1/2}(\phi_A+\phi_B+\phi_C+\phi_D+\phi_E+\phi_F)$$
$$\psi_4=12^{-1/2}(2\phi_A+\phi_B-\phi_C-2\phi_D-\phi_E+\phi_F)$$
\nThe correct order of the orbital energies is (The six carbon atoms of benzene are denoted by A to F and $\phi_J$ is the $2p_z$ orbital of $J$th carbon of benzene.)

  1. $E_1<E_2=E_3<E_4$
  2. $E_4<E_1=E_3<E_2$
  3. $E_3<E_1=E_4<E_2$
  4. $E_3<E_2<E_1=E_4$

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Correct answer: (C) $E_3<E_1=E_4<E_2$

Explanation

The six Hückel molecular orbitals of benzene have the energies (with $\beta<0$):

OrbitalNodesEnergy
all in phase0$\alpha+2\beta$ (lowest)
one nodal plane (a degenerate pair)1$\alpha+\beta$
two nodal planes (a degenerate pair)2$\alpha-\beta$
alternating signs3$\alpha-2\beta$ (highest)

**Identify the given orbitals.**
- $\psi_3$ has all coefficients equal: the bonding orbital, $E_3=\alpha+2\beta$ (lowest).
- $\psi_1$ and $\psi_4$ each have one nodal plane (the two degenerate members of the pair), so $E_1=E_4=\alpha+\beta$.
- $\psi_2$ alternates in sign around the ring: $E_2=\alpha-2\beta$ (highest).

Order: $E_3<E_1=E_4<E_2$ (option **C**).