QJEEM21Ch3119: JEE Main 2021 Chemistry - Class 11 JEEM21Ch3119 Solution: Number of paramagnetic oxides among the following given oxides…

Li₂O $\Rightarrow$ 2Li⁺ O^{2 $-$}

CaO $\Rightarrow$ Ca²⁺ O^{2 $-$}

Na₂O₂ $\Rightarrow$ 2Na⁺ O $_2^{2 - }$

KO₂ $\Rightarrow$ O $_2^{- }$

MgO $\Rightarrow$ Mg²⁺ O^{2 $-$}

K₂O $\Rightarrow$ 2K⁺ O^{2 $-$}

O^{2 $-$} $\Rightarrow$ Complete octet, diamagnetic

$O_2^{2−}$ has 18 electrons.

Moleculer orbital configuration of $O_2^{2−}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^2}^ *$

Here is no unpaired electron so it is diamagnetic.

$O_2^{−}$ has 17 electrons.

Moleculer orbital configuration of $O_2^{−}$ is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ *$

Here is 1 unpaired electron so it is paramagnetic.