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The number of electrons in the $\sigma$ 2p molecular orbital in $N_2^+$ is

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

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The molecular orbital configuration of $N_2^+$ is $ (\;\sigma\; 2s)^2\;(\sigma\; 2s^*\;)^2\;(\pi \;2p)^4\;(\sigma \;2p)^1$
The number of electrons in the $\sigma$ 2p molecular orbital in $N_2^+$ is 1.
Hence (b) is the correct answer.
answered Dec 6, 2013 by sreemathi.v
edited Apr 3, 2014 by mosymeow_1
 

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