Question

Match the following:

• A. A-II, B-IV, C-I, D-III
• B. A-III, B-I, C-IV, D-II
• C. A-II, B-III, C-I, D-IV
• D. A-III, B-I, C-II, D-IV

Hint:

A quick shortcut for 14-electron systems (like $\text{N}_2$ or $\text{O}_2^{2+}$): they always have a bond order of 3.0 and are diamagnetic.
Each addition or removal of an electron changes the bond order by 0.5.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We need to match each diatomic molecule/ion in List-I with its correct bond order and magnetic nature in List-II.

Step 2: Key Formula or Approach:
We use Molecular Orbital Theory (MOT) to determine the bond order and magnetic behavior:
\[ \text{Bond Order} = \frac{N_b - N_a}{2} \] If a species contains unpaired electrons in its molecular orbitals, it is paramagnetic; otherwise, it is diamagnetic.

Step 3: Detailed Explanation:

• Let us evaluate each species:

• A. $\text{C}_2$ (12 electrons):
Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \pi_{2p_x}^2 \pi_{2p_y}^2$.
Here, $N_b = 8$ and $N_a = 4$.
\[ \text{Bond Order} = \frac{8 - 4}{2} = 2 \] Since all electrons are paired, $\text{C}_2$ is diamagnetic.
This matches with III.

• B. $\text{O}_2^{2+}$ (14 electrons):
Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2$.
Here, $N_b = 10$ and $N_a = 4$.
\[ \text{Bond Order} = \frac{10 - 4}{2} = 3 \] Since all electrons are paired, it is diamagnetic.
This matches with I.

• C. $\text{O}_2$ (16 electrons):
Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*1} \pi_{2p_y}^{*1}$.
Here, $N_b = 10$ and $N_a = 6$.
\[ \text{Bond Order} = \frac{10 - 6}{2} = 2 \] It contains two unpaired electrons in the antibonding $\pi^*$ orbitals, so it is paramagnetic.
This matches with II.

• D. $\text{O}_2^-$ (17 electrons):
Electronic configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^{*2} \pi_{2p_y}^{*1}$.
Here, $N_b = 10$ and $N_a = 7$.
\[ \text{Bond Order} = \frac{10 - 7}{2} = 1.5 \] It has one unpaired electron, so it is paramagnetic.
This matches with IV.

• Thus, the correct match is A-III, B-I, C-II, D-IV.


Step 4: Final Answer:
The correct matching sequence is (D) A-III, B-I, C-II, D-IV.