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In the molecular orbital diagram for the molecular ion, $\mathrm{N}_2^{+}$, the number of electrons in the $\sigma_{2 p}$ molecular orbital is :
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The correct answer is:
$1$
$1$
Total electrons in $\mathrm{N}_2^{+}=(7 \times 2)-1=13$
$\mathrm{N}_2^{+} \rightarrow \sigma_{1 s^2}, \sigma_{1 s^2}^*, \sigma_{2 s^2}, \sigma_{2 s^2}^*,\left[\pi_{2p_x}^2=\pi_{2 p_y}^2\right] \sigma_{2 p_z}^1$
Number of electron in $\sigma_{2 p_z}$ is $1$.
$\mathrm{N}_2^{+} \rightarrow \sigma_{1 s^2}, \sigma_{1 s^2}^*, \sigma_{2 s^2}, \sigma_{2 s^2}^*,\left[\pi_{2p_x}^2=\pi_{2 p_y}^2\right] \sigma_{2 p_z}^1$
Number of electron in $\sigma_{2 p_z}$ is $1$.
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