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)

$N_{2}^{-}$

Answer 2

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 $N _{2}^{-}$