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(C) The bond order is calculated as $	ext{B.O.} = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the

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$O_2^{2-}$ and $B_2$

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(C) The bond order is calculated as $	ext{B.O.} = \frac{1}{2}(N_b - N_a)$,where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number of electrons in antibonding orbitals.$(A)$ $CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. $	ext{B.O.} = \frac{1}{2}(10 - 4) = 3$.$O_2^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^2$. $	ext{B.O.} = \frac{1}{2}(10 - 8) = 1$.$(B)$ $O_2^{-}$ ($17$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^2, \pi^* 2p_y^1$. $	ext{B.O.} = \frac{1}{2}(10 - 7) = 1.5$.$CO$ has $	ext{B.O.} = 3$.$(C)$ $O_2^{2-}$ has $	ext{B.O.} = 1$.$B_2$ ($10$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^1, \pi 2p_y^1$. $	ext{B.O.} = \frac{1}{2}(6 - 4) = 1$.$(D)$ $CO$ has $	ext{B.O.} = 3$.$N_2^{+}$ ($13$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2, \pi 2p_y^2, \sigma 2p_z^1$. $	ext{B.O.} = \frac{1}{2}(9 - 4) = 2.5$.Since $O_2^{2-}$ and $B_2$ both have a bond order of $1$,option $(C)$ is correct.

Among the following,the species with identical bond order are

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