Question

which of the following has the largest bond order?
o n₂
o n₂⁻
o n₂²⁻
o n₂⁺
o n₂²⁺

Explanation:

Step1: Recall Bond Order Formula

The formula for bond order (BO) in a diatomic molecule/ion is \( BO = \frac{1}{2}(N_b - N_a) \), where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of antibonding electrons.

Step2: Determine Electron Configurations

• For \( \text{N}_2 \): The electron configuration is \( (\sigma_{1s})^2 (\sigma_{1s}^)^2 (\sigma_{2s})^2 (\sigma_{2s}^)^2 (\pi_{2p})^4 (\sigma_{2p})^2 \). So \( N_b = 10 \), \( N_a = 4 \). \( BO = \frac{1}{2}(10 - 4) = 3 \).
• For \( \text{N}_2^- \): Add 1 electron (antibonding, since it goes to \( \pi_{2p}^* \)). \( N_b = 10 \), \( N_a = 5 \). \( BO = \frac{1}{2}(10 - 5) = 2.5 \).
• For \( \text{N}_2^{2-} \): Add 2 electrons (both to \( \pi_{2p}^* \)). \( N_b = 10 \), \( N_a = 6 \). \( BO = \frac{1}{2}(10 - 6) = 2 \).
• For \( \text{N}_2^+ \): Remove 1 electron (from \( \sigma_{2p} \), bonding). \( N_b = 9 \), \( N_a = 4 \). \( BO = \frac{1}{2}(9 - 4) = 2.5 \).
• For \( \text{N}_2^{2+} \): Remove 2 electrons (from \( \sigma_{2p} \), bonding). \( N_b = 8 \), \( N_a = 4 \). \( BO = \frac{1}{2}(8 - 4) = 2 \).

Step3: Compare Bond Orders

Comparing the bond orders: \( \text{N}_2 \) (3) > \( \text{N}_2^- \) (2.5) = \( \text{N}_2^+ \) (2.5) > \( \text{N}_2^{2-} \) (2) = \( \text{N}_2^{2+} \) (2). So \( \text{N}_2 \) has the largest bond order.

Answer:

\( \text{N}_2 \) (the first selected option, \( \text{N}_2 \))