JEE Main 2017 — Chemical Bonding And Molecular Structure Question with Solution

Those species which have unpaired electrons are called paramagnetic species.

(a)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

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

Here is no unpaired electron, So it is Diamagnetic.

(b)    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.

(c)   $$O_2^{2-}$$ 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 also no unpaired electron present, so it is diamagnetic.

(d)   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.