Question

Among the following compounds the one that is paramagnetic is:

• A. ${{\mathrm{O}}_2}^{2-}$
• B. ${\mathrm{O}}_3$
• C. ${\mathrm{N}}_2\mathrm{O}$
• D. ${{\mathrm{O}}_2}^{-}$

Answer 1

The correct option is D

${{\mathrm{O}}_2}^{-}$

Explanation for correct option:

D. ${{\mathrm{O}}_2}^{-}$

Paramagnetic compounds:

• "Paramagnetic compounds/ ions are the ones that have at least one unpaired electron.
• In ${{\mathrm{O}}_2}^{-}$, the total number of valence electrons are $17$.
• The electronic configuration can be written as: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2{p_z}^2,\pi 2{p_x}^2=\pi 2{p_y}^2\pi *2{p_x}^2=\pi *2{p_y}^1$ .
• As ${{\mathrm{O}}_2}^{-}$ has one unpaired electron. Thus it is paramagnetic in nature.

Explanation for incorrect options:

A. ${{\mathrm{O}}_2}^{2-}$

• The total number of electrons present in ${O_2}^{2-}$are $18$.
• The electronic configuration can be written as: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\pi 2{p_x}^2=\pi 2{p_y}^2,\sigma 2{p_z}^2,{\pi }^{\ast }2{p_x}^2={\pi }^{\ast }2{p_y}^2$
• Hence it is not paramagnetic as it does not have any unpaired electron.

B. ${\mathrm{O}}_3$

• The total number of valence electrons is ${\mathrm{O}}_3$ are $24$.
• It is not paramagnetic in nature.

C. ${\mathrm{N}}_2\mathrm{O}$

• The total number of valence electrons in ${\mathrm{N}}_2\mathrm{O}$ is $18$.
• The electronic configuration can be written as: $\sigma 1s^2,{\sigma }^{\ast }1s^2,\sigma 2s^2,{\sigma }^{\ast }2s^2,\sigma 2{p_z}^2,\pi 2{p_x}^2=\pi 2{p_y}^2,{\pi }^{\ast }2{p_x}^2={\pi }^{\ast }2{p_y}^2$
• Hence, it is diamagnetic in nature.

Therefore, the correct option is D i.e. ${{\mathrm{O}}_2}^{-}$.