JEE Main 2022 — Chemical Bonding And Molecular Structure Question with Solution

Those species which have unpaired electrons are called paramagnetic species.

And those species which have no unpaired electrons are called diamagnetic species.

B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

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

Here two unpaired electrons present. So it is paramagnetic.

$$O_2^{2-}$$ has 18 electrons.

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

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

Here is no unpaired electron so it is diamagnetic.

$$O_2^{+}$$ 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, so it is paramagnetic.

$$C_2$$ has 12 electrons.

Moleculer orbital configuration of $$C_2$$

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

Here no unpaired electron present, so it is diamagnetic.

$$C_2^{-}$$ has 13 electrons.

Moleculer orbital configuration of $$C_2^{-}$$ is

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

Here 1 unpaired electron present, so it is paramagnetic.

Li₂ has 6 electrons.

Li₂ = $${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }$$ $${\sigma }_{2s^2}$$

Here no unpaired electron present, so it is diamagnetic.

Configuration of $$H{e}_2^{+}$$ (3 electrons) is = $${\sigma }_{1s^2}$$ $${\sigma }_{1s^1}^{\ast }$$

Here 1 unpaired electron present, so it is paramagnetic.