Solved: CHEMISTRY Question for NEET

NEET CHEMISTRYMedium

The pair of species with the same bond order is:

O₂²⁻, B₂

O₂⁺, NO⁺

NO, CO

N₂, O₂

Formula: $\text{Bond Order} = \frac{1}{2} (N_b - N_a)$ , where $N_b$ is the number of bonding electrons and $N_a$ is the number of antibonding electrons.
Analyze Option A ( $O_2^{2-}, B_2$ ):

$O_2^{2-}$ (Peroxide Ion): Total electrons = $16 + 2 = 18$ . Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} \sigma_{2p_z}^2 (\pi_{2p_x}^2 = \pi_{2p_y}^2) (\pi_{2p_x}^{*2} = \pi_{2p_y}^{*2})$ . $N_b = 10, N_a = 8$ . $\text{Bond Order} = 0.5(10 - 8) = 1$ .
$B_2$ (Boron Molecule): Total electrons = $10$ . Configuration: $\sigma_{1s}^2 \sigma_{1s}^{*2} \sigma_{2s}^2 \sigma_{2s}^{*2} (\pi_{2p_x}^1 = \pi_{2p_y}^1)$ . $N_b = 6, N_a = 4$ . $\text{Bond Order} = 0.5(6 - 4) = 1$ .
Result: Both have a Bond Order of 1.

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