Question

Molecular orbital electronic configuration for $' X'$ anion is : $K K^{*}(\sigma 2 s)^{2}\left(\sigma^{*} 2 s\right)^{2}\left(\pi 2 p_{x}\right)^{2}\left(\pi 2 p_{y}\right)^{2}\left(\sigma 2 p_{z}\right)^{2}$ $\left(\pi^{*} 2 p_{x}\right)^{1}$. The anion ' $X$' is :

• A. $ N_{2}^{-} $
• B. $ O_{2}^{-} $
• C. $ N_{2}^{2-} $
• D. $ O_{2}^{2-} $

Answer 1

Answer: (A)

Given, electronic configuration of anion $X$ is :
$\sigma 1 s^{2}, \sigma^{*} 1 s^{2}, \sigma 2 s^{2}, \sigma^{*} 2 s^{2}, \sigma 2 p_{x}^{2}, \pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2} \pi^{*} 2 p_{x}^{1}$
$\therefore $ total number of electrons of anion $X=15$
Hence, the anion $X$ is $N _{2}^{-}$