Question

Which one of the following molecules shows an increase in bond order after loss of an electron from the highest occupied molecular orbital?

• A. F$_2$
• B. N$_2$
• C. C$_2$
• D. B$_2$

Hint:

Lose from Bonding orbital $\rightarrow$ Bond order decreases (weakens the bond).
Lose from Antibonding orbital $\rightarrow$ Bond order increases (strengthens the bond).
Among the options, only oxygen ($\text{O}_2$) and fluorine ($\text{F}_2$) have antibonding molecular orbitals as their HOMO.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Molecular Orbital (MO) Theory:
- Loss of a bonding electron decreases bond order.
- Loss of an anti-bonding electron increases bond order.

Step 2: Detailed Explanation:
$\bullet$ (A) $F_{2$ (18e):} Configuration $\sigma_{1s}^{2} \sigma_{1s}^{*2} \dots \pi_{2p}^{4} \pi_{2p}^{*4}$. The HOMO is $\pi_{2p}^{*}$, which is an anti-bonding orbital. Removing an electron ($F_{2} \rightarrow F_{2}^{+}$) increases bond order from 1 to 1.5.
$\bullet$ (B) $N_{2$ (14e):} HOMO is $\sigma_{2p_{z}}$ (bonding). Loss $\rightarrow$ BO decreases ($3 \rightarrow 2.5$).
$\bullet$ (C) $C_{2$ (12e):} HOMO is $\pi_{2p}$ (bonding). Loss $\rightarrow$ BO decreases ($2 \rightarrow 1.5$).
$\bullet$ (D) $B_{2$ (10e):} HOMO is $\pi_{2p}$ (bonding). Loss $\rightarrow$ BO decreases ($1 \rightarrow 0.5$).

Step 3: Final Answer:
Only $F_{2}$ has its HOMO as an anti-bonding orbital.
This corresponds to option (A).