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 $N_2=KK\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2,\pi 2p_x^2$

$\approx \pi 2p_y^2,\sigma 2p_z^2$

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 $O_2^{2-}=KK\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2,\sigma 2p_z^2$

$\pi 2p_x^2\approx \pi 2p_y^2,\stackrel{\ast }{\pi }2p_x^2\approx \stackrel{\ast }{\pi }2p_y^2$

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 $BN=KK\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2,\pi 2p_x^2$

$\approx \pi 2p_y^2$

Hence, it is also diamagnetic.

(d) Total number of electrons in $O_2^{-}=17$

$EC$ of $O_2^{-}=KK\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2,\sigma 2p_z^2$

$\pi 2p_x^2\approx \pi 2p_y^2,\stackrel{\ast }{\pi }2p_x^2,\stackrel{\ast }{\pi }2p_y^1$

One unpaired electron in anti-bonding $\stackrel{\ast }{\pi }2p_y$ orbital indicates that $O_2^{-}$ ion is paramagnetic.