The molecule having one unpaired electron is

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(A) According to Molecular Orbital Theory $(MOT)$,the electronic configurations are as follows:$A].$ For $NO$ ($15$ electrons): $\sigma 1s^2, \sigma^*

Vedclass · Accepted Answer

$NO$

Vedclass · Answer

(A) According to Molecular Orbital Theory $(MOT)$,the electronic configurations are as follows:$A].$ For $NO$ ($15$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1$. It has $1$ unpaired electron.$B].$ For $CO$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired.$C].$ For $CN^{-}$ ($14$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2$. All electrons are paired.$D].$ For $O_2$ ($16$ electrons): $\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \sigma 2p_z^2, \pi 2p_x^2, \pi 2p_y^2, \pi^* 2p_x^1, \pi^* 2p_y^1$. It has $2$ unpaired electrons.Thus,$NO$ is the correct molecule.

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