Which one of the following pairs of species have the same bond order?

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$O_{2}, {NO}^{+}$

${CN}^{-}, CO$

$N_{2}, O_{2}$

$CO, NO$

answer is B.

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Detailed Solution

Bond order $= \frac{e_{B M O}^{-} - e_{A B M O}^{-}}{2}$

Pair	BMO e-s	ABMO e- s	Bond order
CN^-	10	4	3
CO	10	4	3
O₂	10	6	2
NO⁺	10	4	3
N₂	10	4	3
O₂	10	6	2
CO	10	4	3
NO	10	5	2.5

Electronic Configuration
CN^-: $σ_{1 s}^{2} σ_{1 s}^{2} σ_{2 s}^{2} σ_{2 s}^{2} π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}$
CO: ${σ^{2}}_{ls} σ_{1 s}^{2} σ_{2 s}^{2} σ_{2 s}^{2} π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}$
O2^–: $σ_{1 s}^{2} σ_{1 s^{2}}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{z}}^{2} π_{2 p_{x}}^{2}$ $= π_{2 p_{y}}^{2} π_{2 p_{x}}^{* 2} = π_{2 p_{y}}^{* 1}$
NO⁺: $σ_{2}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} σ_{2 p_{x}}^{2} π_{2 p_{y}}^{2} = π_{2 p_{z}}^{2}$
N₂: $σ^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{2} = π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}$