Total number of electrons present in (pi^*) m | vedclass

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Question

(A) The molecular orbital configuration for $O_2$ $(16 \ e^-)$ is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\

Vedclass · Accepted Answer

$6$

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(A) The molecular orbital configuration for $O_2$ $(16 \ e^-)$ is $(\sigma_{1s})^2(\sigma_{1s}^*)^2(\sigma_{2s})^2(\sigma_{2s}^*)^2(\sigma_{2p_z})^2(\pi_{2p_x})^2(\pi_{2p_y})^2(\pi_{2p_x}^*)^1(\pi_{2p_y}^*)^1$. Number of electrons in $(\pi^*)$ of $O_2 = 2$. For $O_2^{+}$ $(15 \ e^-)$,one electron is removed from the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{+} = 1$. For $O_2^{-}$ $(17 \ e^-)$,one electron is added to the $(\pi^*)$ orbital,so number of electrons in $(\pi^*)$ of $O_2^{-} = 3$. Total number of electrons in $(\pi^*)$ orbitals $= 2 + 1 + 3 = 6$.

Total number of electrons present in $(\pi^*)$ molecular orbitals of $O_2$,$O_2^{+}$,and $O_2^{-}$ is ............

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