JEE Main 2019 — 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.

(a)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(b)   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.

(c)   $$O_2$$ 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, so it is paramagnetic.

(d)   NO has 15 electrons.

Moleculer orbital configuration of NO 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 is 1 unpaired electron, So it is Paramagnetic.