Question

The pair of species with the same bond order is

• A. $NO, CO$
• B. $N_2, O_2$
• C. $O_2^{2-} , B_2$
• D. $O_2^+ , NO^+$

Answer 1

The Correct Option is C

To determine which pair of species have the same bond order, we first need to calculate the bond order for each given species. The bond order can be found using the formula:

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

Let's analyze each option:

1. NO, CO
   - For nitric oxide (NO):
   It has a total of 11 valence electrons.
   Using molecular orbital theory, the 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}^1.
   Bond order = \frac{8-3}{2} = 2.5.
   - For carbon monoxide (CO):
   It has a total of 10 valence electrons.
   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.
   Bond order = \frac{8-2}{2} = 3.
2. N_2, O_2
   - For nitrogen molecule N_2: It has a total of 10 valence electrons.
   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.
   Bond order = \frac{8-2}{2} = 3.
   - For oxygen molecule O_2: It has a total of 12 valence electrons.
   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 \ \pi^*_{2p_x}^1 \ \pi^*_{2p_y}^1.
   Bond order = \frac{8-4}{2} = 2.
3. O_2^{2-}, B_2
   - For peroxide ion O_2^{2-}: It has a total of 14 valence electrons.
   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 \ \pi^*_{2p_x}^2 \ \pi^*_{2p_y}^2.
   Bond order = \frac{8-6}{2} = 1.
   - For boron molecule B_2: It has a total of 6 valence electrons.
   Configuration: \sigma_{1s}^2 \ \sigma^*_{1s}^2 \ \sigma_{2s}^2 \ \sigma^*_{2s}^2 \ \pi_{2p_x}^1 \ \pi_{2p_y}^1.
   Bond order = \frac{4-2}{2} = 1.
4. O_2^+ , NO^+
   - For superoxide ion O_2^+: It has a total of 11 valence electrons.
   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 \ \pi^*_{2p_x}^1.
   Bond order = \frac{8-3}{2} = 2.5.
   - For nitrosonium ion NO^+: It has a total of 10 valence electrons.
   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.
   Bond order = \frac{8-2}{2} = 3.

Based on the bond order calculations, the pair O_2^{2-} and B_2 have the same bond order of 1. Therefore, the correct option is O_2^{2-} , B_2.