Question

Regarding the molecular orbital (MO) energy levels for homonuclear diatomic molecules, the INCORRECT statement(s) is (are):

• A. Bond order of \( \text{Ne}_2 \) is zero
• B. The highest occupied molecular orbital (HOMO) of \( \text{F}_2 \) is \( \sigma \)-type
• C. Bond energy of \( \text{O}_2^+ \) is smaller than the bond energy of \( \text{O}_2 \)
• D. Bond length of \( \text{Li}_2 \) is larger than the bond length of \( \text{B}_2 \)

Hint:

In molecular orbital theory: - Bond order helps predict stability and bond strength. - Removing electrons from antibonding orbitals increases bond order and strength. - The HOMO type ($\pi$ or $\sigma$) must be determined based on proper MO filling order.

Answer 1

The Correct Option is B, C

(A) Bond order of \( \text{Ne}_2 \) is zero

Electronic configuration of \( \text{Ne}_2 \): 10 bonding and 10 antibonding electrons.

\[ \text{Bond order} = \frac{10 - 10}{2} = 0 \quad \Rightarrow \text{Correct} \]

(B) HOMO of \( \text{F}_2 \) is \( \sigma \)-type

Electronic configuration of \( \text{F}_2 \) based on MO theory:

\[\sigma_{1s}^2 \, \sigma_{1s}^{*2} \, \sigma_{2s}^2 \, \sigma_{2s}^{*2} \, \sigma_{2p_z}^2 \, \pi_{2p_x}^2 = \pi_{2p_y}^2 \, \pi_{2p_x}^{*1} = \pi_{2p_y}^{*1}\]
 

So, the HOMO is the \( \pi^* \) orbital — not a \( \sigma \)-type orbital.

\[ \Rightarrow \text{Incorrect} \]

(C) Bond energy of \( \text{O}_2^+ \) is smaller than \( \text{O}_2 \)

Bond order of \( \text{O}_2 \) is 2; for \( \text{O}_2^+ \), one electron is removed from an antibonding orbital, increasing bond order to 2.5.

Since bond energy \( \propto \) bond order:

\[ \text{O}_2^+ \text{ has higher bond energy than } \text{O}_2 \Rightarrow \text{Incorrect} \]

(D) Bond length of \( \text{Li}_2 \) is larger than \( \text{B}_2 \)

Bond length generally increases with the size of the atom. Lithium is a larger atom than boron. So:

\[ \text{Bond length of Li}_2 > \text{Bond length of B}_2 \Rightarrow \text{Correct} \]

Final Answer: \( \boxed{\text{(B), (C)}} \)