The pair of species with the same bond order is:

Question

Accepted Answer

1. **Formula:** $	ext{Bond Order} = \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.
2. **Analyze Option A ($O_2^{2-}, B_2$):**
   - **$O_2^{2-}$ (Peroxide Ion):** Total electrons = $16 + 2 = 18$. Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$.
     $N_b = 10, N_a = 8$. $	ext{Bond Order} = 0.5(10 - 8) = 1$.
   - **$B_2$ (Boron Molecule):** Total electrons = $10$. Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^1 = \pi_{2p_y}^1)$.
     $N_b = 6, N_a = 4$. $	ext{Bond Order} = 0.5(6 - 4) = 1$.
   - **Result:** Both have a Bond Order of 1.
3. **Analyze other options:**
   - $O_2^+$ (BO = 2.5) vs $NO^+$ (BO = 3).
   - $NO$ (BO = 2.5) vs $CO$ (BO = 3).
   - $N_2$ (BO = 3) vs $O_2$ (BO = 2).

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