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 Question:
This question asks us to identify which of the given diatomic molecules experienced an increase in its bond order after losing one electron from its Highest Occupied Molecular Orbital (HOMO).

Step 2: Key Formula or Approach:
According to Molecular Orbital (MO) Theory, the bond order (B.O.) of a molecule is given by:
\[ \text{B.O.} = \frac{N_b - N_a}{2} \] Where:
- $N_b$ is the number of electrons in bonding molecular orbitals.
- $N_a$ is the number of electrons in antibonding molecular orbitals.
- Removing an electron from a bonding orbital decreases $N_b$, which decreases the bond order.
- Removing an electron from an antibonding orbital decreases $N_a$, which increases the bond order.
Thus, the molecule must have its HOMO as an antibonding molecular orbital.

Step 3: Detailed Explanation:
Let's analyze the molecular orbital configurations for each molecule:

• (A) F$_2$ (18 electrons):
- MO configuration:
\[ \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*2} \] - The HOMO is the $\pi_{2p}^*$ orbital, which is an antibonding orbital.
- For $\text{F}_2$: $N_b = 10$, $N_a = 8 \implies \text{B.O.} = \frac{10 - 8}{2} = 1$.
- For $\text{F}_2^+$ (after losing 1 electron from $\pi_{2p}^*$): $N_b = 10$, $N_a = 7 \implies \text{B.O.} = \frac{10 - 7}{2} = 1.5$.
- The bond order increases from 1 to 1.5.

• (B) N$_2$ (14 electrons):
- MO configuration:
\[ \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2 \] - The HOMO is the $\sigma_{2p_z}$ orbital, which is a bonding orbital.
- Removing an electron decreases the bond order from 3 to 2.5 ($\text{N}_2^+$).

• (C) C$_2$ (12 electrons):
- MO configuration:
\[ \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \] - The HOMO is the $\pi_{2p}$ orbital, which is a bonding orbital.
- Removing an electron decreases the bond order from 2 to 1.5 ($\text{C}_2^+$).

• (D) B$_2$ (10 electrons):
- MO configuration:
\[ \sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^1 \pi_{2p_y}^1 \] - The HOMO is the $\pi_{2p}$ orbital, which is a bonding orbital.
- Removing an electron decreases the bond order from 1 to 0.5 ($\text{B}_2^+$).

Step 4: Final Answer:
Only $\text{F}_2$ experiences an increase in bond order because its HOMO is an antibonding orbital ($\pi_{2p}^*$). This corresponds to option (A).