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The number of electron(s) in the $\sigma 2p$ molecular orbital in $N_2^{2+}$ is/are

$\begin{array}{1 1}(a)\;2&(b)\;3\\(c)\;0&(d)\;1\end{array}$

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Electronic configuration of $N_2^+:(\sigma 1s)^2 < (\sigma ^*1s)^2 < (\sigma 2s)^2 < (\sigma^* 2s)^2 < (\pi 2p_x^2=\pi 2p_y^2) < \sigma 2p_z^0$
$\therefore$ The no of electrons in $\sigma 2p$ is zero.
Hence (c) is the correct answer.
answered Mar 5, 2014 by sreemathi.v
 

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