Question

Which among the following processes is/are associated with increasing bond order but no change in diamagnetic/paramagnetic behaviour?
(i) $\text{N}_2 \rightarrow \text{N}_2^+ + \text{e}^-$
(ii) $\text{O}_2 \rightarrow \text{O}_2^+ + \text{e}^-$
(iii) $\text{O}_2 + \text{e}^- \rightarrow \text{O}_2^-$

• A. (ii) only
• B. (i) and (ii)
• C. (ii) and (iii)
• D. (iii) only

Hint:

Removing an electron from an antibonding orbital ($\pi^*$ or $\sigma^*$) always increases the bond order.
Since $\text{O}_2$ has unpaired electrons in its antibonding $\pi^*$ orbitals, removing one electron to form $\text{O}_2^+$ increases the bond order from 2 to 2.5 while keeping it paramagnetic.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
According to Molecular Orbital (MO) Theory, bond order and magnetic behavior depend on the distribution of electrons in bonding molecular orbitals ($BMO$) and antibonding molecular orbitals ($ABMO$).
Step 2: Key Formula or Approach:
The formulas and properties are evaluated using:
- Bond Order = \( \frac{1}{2} (N_b - N_a) \), where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
- A molecule is paramagnetic if it contains one or more unpaired electrons, and diamagnetic if all electrons are completely paired.
Step 3: Detailed Explanation:
Let us check each process using molecular orbital configurations:
Process (i): \( \text{N}_2 \rightarrow \text{N}_2^+ \)
- \( \text{N}_2 \) ($14$ electrons): Bond order = \( \frac{14-4}{2} = 3 \). It has no unpaired electrons, so it is diamagnetic.
- \( \text{N}_2^+ \) ($13$ electrons): An electron is removed from a bonding orbital ($\sigma_{2p_z}$). Bond order decreases to $2.5$. It contains $1$ unpaired electron, making it paramagnetic.
- Result: Bond order decreases, magnetic property changes.
Process (ii): \( \text{O}_2 \rightarrow \text{O}_2^+ \)
- \( \text{O}_2 \) ($16$ electrons): Bond order = \( \frac{10-6}{2} = 2 \). It contains $2$ unpaired electrons in its antibonding orbitals ($\pi^_{2p_x}$ and $\pi^_{2p_y}$), so it is paramagnetic.
- \( \text{O}_2^+ \) ($15$ electrons): An electron is removed from an antibonding orbital ($\pi^$). Bond order increases to \( \frac{10-5}{2} = 2.5 \). It still contains $1$ unpaired electron, so it remains paramagnetic.
- Result: Bond order increases, and it remains paramagnetic (no change in magnetic behavior). This matches the criteria.
Process (iii): \( \text{O}_2 \rightarrow \text{O}_2^- \)
- \( \text{O}_2 \) ($16$ electrons): Bond order = $2$, paramagnetic.
- \( \text{O}_2^- \) ($17$ electrons): An electron is added to an antibonding orbital ($\pi^$). Bond order decreases to \( \frac{10-7}{2} = 1.5 \). It contains $1$ unpaired electron, remaining paramagnetic.
- Result: Bond order decreases.
Therefore, only process (ii) satisfies both conditions.
Step 4: Final Answer:
The process associated with increasing bond order and no change in magnetic behavior is (ii) only.