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The pair of species with the same bond order is
A
$N_2$, $O_2$
B
$O_2^{2-}$, $B_2$
C
$O_2^+$, $NO^+$
D
NO, CO
Detailed Solution
$N_2$ = 3, $O_2$ = 2
$O_2^{2-}$ (18 e⁻) = $\frac{1}{2}(10 - 8) = 1$; $B_2$ (10 e⁻) = $\frac{1}{2}(6 - 4) = 1$
$O_2^+$ = 2.5, $NO^+$ = 3
NO = 2.5, CO = 3
So $O_2^{2-}$ and $B_2$ have the same bond order, 1.
$O_2^{2-}$ (18 e⁻) = $\frac{1}{2}(10 - 8) = 1$; $B_2$ (10 e⁻) = $\frac{1}{2}(6 - 4) = 1$
$O_2^+$ = 2.5, $NO^+$ = 3
NO = 2.5, CO = 3
So $O_2^{2-}$ and $B_2$ have the same bond order, 1.