Question

Which of the $O^{-}_{2}, O^{2-}_{2}$, $BN$ and $NN$ is/are paramagnetic?

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

Answer 1

Answer: (D)

(a) Total number of electrons in $N _{2}$ molecule $=14 .$

$\therefore EC$ of $N _{2}= KK \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \pi 2 p_{x}^{2}$

$\approx \pi 2 p_{y}^{2}, \sigma 2 p_{z}^{2}$

Presence of no unpaired electrons indicates that $N _{2} $ molecule is diamagnetic.

(b) Total number of electrons in $O _{2}^{2-}$ ion $=18$

$\therefore EC$ of $O _{2}^{2-}= KK \sigma 2 s ^{2},\overset{*}{\sigma} 2 s ^{2}, \sigma 2 p_{z}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{2} \approx \overset{*}{\pi} 2 p_{y}^{2}$

Hence, it is also diamagnetic.


(c) Total number of electrons in $BN$ (boron nitride) molecule $=5$ (from B) $+7$ (from $N )=12$

$\therefore EC$ of $BN = KK \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \pi 2 p_{x}^{2}$

$\approx \pi 2 p_{y}^{2}$

Hence, it is also diamagnetic.

(d) Total number of electrons in $O _{2}^{-}=17$

$EC$ of $O _{2}^{-}= KK \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2}, \sigma 2 p_{z}^{2}$

$\pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \overset{*}{\pi} 2 p_{x}^{2},\overset{*}{\pi} 2 p_{y}^{1 }$

One unpaired electron in anti-bonding $\overset{*}{\pi} 2 p_{y}$ orbital indicates that $O _{2}^{-}$ ion is paramagnetic.