JEE Advanced 2023 Paper 2 Q01 Chemistry Chemical Bonding Molecular Orbital Theory Medium

JEE Advanced 2023 Paper 2 · Q01 · Molecular Orbital Theory

The correct molecular orbital diagram for $F_2$ molecule in the ground state is one in which the molecular orbitals for the $2p$ shell are arranged in the order $\sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}$. Choose the option that depicts this ordering correctly.

  1. A. Order: $\pi^* < \sigma^* < \pi < \sigma$
  2. B. Order: $\sigma^* < \pi^* < \pi < \sigma$
  3. C. Order: $\sigma^* < \pi^* < \sigma < \pi$ (energy increases upward; $\sigma_{2p_z}$ below the $\pi$ pair, $\pi^*$ above $\pi$, and $\sigma^*$ at the top)
  4. D. Order: $\pi^* < \pi < \sigma^* < \sigma$
Reveal answer + step-by-step solution

Correct answer:C

Solution

For $F_2$ ($Z=9$, $14$ electrons), the MO order in increasing energy is: $\sigma_{1s} < \sigma^*_{1s} < \sigma_{2s} < \sigma^*_{2s} < \sigma_{2p_z} < \pi_{2p_x} = \pi_{2p_y} < \pi^*_{2p_x} = \pi^*_{2p_y} < \sigma^*_{2p_z}$. The standard '$\sigma$ below $\pi$' ordering for $O_2, F_2$ is the relevant choice — option (C).

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