Question

From elementary molecular orbital theory we can deduce the electronic configuration of the singly positive nitrogen molecular ion as

• A. $\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{2} \sigma 2 p_{z}^{1}$
• B. $\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2}=\pi * 2 p_{y}^{1}$
• C. $\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{1} \pi 2 p_{x}^{2}=\pi 2 p_{y}^{1}$
• D. $\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \sigma 2 p_{z}^{2} \pi 2 p_{x}^{2}$

Answer 1

Answer: (A)

$N _{2}^{+}$(13 electrons ): $\sigma 1 s^{2} \sigma^{*} 1 s^{2} \sigma 2 s^{2} \sigma^{*} 2 s^{2} \pi 2 p_{x}^{2} =\pi 2 p_{y}^{2} \sigma 2 p_{z}^{1}$