Question

Question 28

Diamagnetic species are those which contain no unpaired electrons. Which among the following are diamagnetic?

(a) $N_2$

(b) $N_2^{2-}$

(c) $O_2$

(d) $O_2^{2-}$

Answer 1

Answers: (a) $N_2$ and (d) $O_2^{2-}$.

(a) Electronic configuration of $N_2=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2.$

It has no unpaired electron indicates diamagnetic species.

(b) Electronic configuration of $N_2^{2-}ion=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2p_x^2\approx \pi 2p_y^2,\sigma 2p_z^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$

It has no two unpaired electrons, paramagnetic in nature.

(c) Electronic configuration of $O_2=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^1\approx {\pi }^{\ast }2p_y^1$

The presence of two unpaired electrons shows its paramagnetic nature.

(d) Electronic configuration of $O_2^{2-}ion=\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,{\pi }^{\ast }2p_x^2\approx {\pi }^{\ast }2p_y^2$

It contains no unpaired electron, therefore, it is diamagnetic in nature.