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 Question:
The question asks to find which of the given molecular transformations result in both an increase in bond order and no change in magnetic behavior (remaining either diamagnetic or paramagnetic).

Step 2 : Key Formulas and Approach:
We will use Molecular Orbital (MO) Theory to determine the bond order and magnetic properties of each species.
The formula for Bond Order (BO) is:
\[ \text{BO} = \frac{N_b - N_a}{2} \]
where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
If a species contains unpaired electrons, it is paramagnetic; otherwise, it is diamagnetic.

Step 3 : Detailed Explanation:

• Process (i): $\text{N}_2 \rightarrow \text{N}_2^+ + \text{e}^-$
$\text{N}_2$ has 14 electrons. Its MO configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2 \sigma_{2p_z}^2$.
Bond Order of $\text{N}_2 = \frac{10 - 4}{2} = 3$. All electrons are paired, so it is diamagnetic.
$\text{N}_2^+$ has 13 electrons. One electron is removed from the bonding orbital ($\sigma_{2p_z}$).
Bond Order of $\text{N}_2^+ = \frac{9 - 4}{2} = 2.5$. It has one unpaired electron, so it is paramagnetic.
Here, the bond order decreases, and the magnetic behavior changes from diamagnetic to paramagnetic.

• Process (ii): $\text{O}_2 \rightarrow \text{O}_2^+ + \text{e}^-$
$\text{O}_2$ has 16 electrons. Its MO configuration is: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$.
Bond Order of $\text{O}_2 = \frac{10 - 6}{2} = 2$. It has two unpaired electrons, so it is paramagnetic.
$\text{O}_2^+$ has 15 electrons. One electron is removed from an antibonding orbital ($\pi_{2p}^*$).
Bond Order of $\text{O}_2^+ = \frac{10 - 5}{2} = 2.5$. It has one unpaired electron, so it is still paramagnetic.
Here, the bond order increases ($2 \rightarrow 2.5$), and there is no change in magnetic behavior (both are paramagnetic).

• Process (iii): $\text{O}_2 + \text{e}^- \rightarrow \text{O}_2^-$
$\text{O}_2^-$ has 17 electrons. The extra electron is added to an antibonding orbital ($\pi_{2p}^*$).
Bond Order of $\text{O}_2^- = \frac{10 - 7}{2} = 1.5$.
Here, the bond order decreases from $2$ to $1.5$.

Step 4 : Final Answer:
Only process (ii) is associated with an increase in bond order and no change in magnetic behavior.
This corresponds to Option (A).