JEE Main 2020 — Chemical Bonding And Molecular Structure Question with Solution

Magnetic moment = 1.73 BM

$$\therefore$$ Unpaired electron = 1

(1) $$O_2$$ has 16 electrons.

Moleculer orbital configuration of $$O_2$$ is

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

Here 2 unpaired electron present.

(2)   $$O_2^{+}$$ has 15 electrons.

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

$${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\sigma }_{2p_z^2}{\pi }_{2p_x^2}={\pi }_{2p_y^2}{\pi }_{2p_x^1}^{\ast }={\pi }_{2p_y^0}^{\ast }$$

Here 1 unpaired electron present.

(3)   $$O_2^{-}$$ has 17 electrons.

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

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

Here 1 unpaired electron present.

Hence $$O_2^{-}$$ & $$O_2^{+}$$ have one unpaired electron.