Which of the following pair has same bond order ?

Question

Accepted Answer

To determine which pair of molecules or ions has the same bond order, we need to calculate the bond order for each species. Bond order is a concept from molecular orbital theory and is calculated as:

$$ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} $$

Let's analyze each option:

### Option 1: $N_2, Cl_2$

- **$N_2$**: Nitrogen molecule has 10 valence electrons. The molecular orbital 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 $. The bond order is $ \frac{(10 - 4)}{2} = 3 $.
- **$Cl_2$**: Chlorine molecule has 14 valence electrons. The molecular orbital 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 \pi_{2p_x}^*^2 \pi_{2p_y}^*^2 \sigma_{2p_z}^*^2 $. The bond order is $ \frac{(8 - 6)}{2} = 1 $.

These bond orders are not the same, so Option 1 is incorrect.

### Option 2: $O_2^{2+}, F_2$

- **$O_2^{2+}$**: This ion has 14 valence electrons. The molecular orbital 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 $. The bond order is $ \frac{(10 - 4)}{2} = 3 $.
- **$F_2$**: Fluorine molecule has 14 valence electrons. The molecular orbital 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}^*^2 \pi_{2p_y}^*^2 \sigma_{2p_z}^*^2 $. The bond order is $ \frac{(8 - 6)}{2} = 1 $.

These bond orders are not the same, so Option 2 is incorrect.

### Option 3: $NO^+, CO$

- **$NO^+$**: This ion has 10 valence electrons. The molecular orbital configuration is similar to $ N_2 $: $ \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 bond order is $ \frac{(10 - 4)}{2} = 3 $.
- **$CO$**: Carbon monoxide has 10 valence electrons. The molecular orbital 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 $. The bond order is $ \frac{(10 - 4)}{2} = 3 $.

Both $NO^+$ and $CO$ have the same bond order of 3, making Option 3 correct.

### Option 4: $O_2, N_2$

- **$O_2$**: Oxygen molecule has 12 valence electrons. The molecular orbital 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 $. The bond order is $ \frac{(8 - 4)}{2} = 2 $.
- **$N_2$**: As calculated earlier, the bond order is 3.

These bond orders are not the same, so Option 4 is incorrect.

In conclusion, Option 3 is the correct answer because both $NO^+$ and $CO$ have the same bond order of 3.