The correct order of bond orders of C_{2}^{2- | vedclass

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(B) To determine the bond order,we use the Molecular Orbital Theory $(MOT)$ configuration: $1$. For $C_{2}^{2-}$ ($14$ electrons): $\sigma 1s^2, \sigm

Vedclass · Accepted Answer

$O_{2}^{2-} < N_{2}^{2-} < C_{2}^{2-}$

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(B) To determine the bond order,we use the Molecular Orbital Theory $(MOT)$ configuration: $1$. For $C_{2}^{2-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2$. Bond order = $(10-4)/2 = 3$. $2$. For $N_{2}^{2-}$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2, \pi^* 2p_x^1 = \pi^* 2p_y^1$. Bond order = $(10-6)/2 = 2$. $3$. For $O_{2}^{2-}$ ($18$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2 = \pi 2p_y^2, \pi^* 2p_x^2 = \pi^* 2p_y^2$. Bond order = $(10-8)/2 = 1$. Thus,the order of bond orders is $O_{2}^{2-} (1) < N_{2}^{2-} (2) < C_{2}^{2-} (3)$.

The correct order of bond orders of $C_{2}^{2-}$,$N_{2}^{2-}$,and $O_{2}^{2-}$ is,respectively.

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