QJEEM17Ch1090: JEE Main 2017 Chemistry - Class 11 JEEM17Ch1090 Solution: Which of the following is paramagnetic ?

Those species which have unpaired electrons are called paramagnetic species.

(a)   NO⁺ has 14 electrons.

Moleculer orbital configuration of NO⁺ is

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

Here is no unpaired electron, So it is Diamagnetic.

(b)    CO has 14 electrons.

Moleculer orbital configuration of CO is

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

Here is no unpaired electron so it is diamagnetic.

(c)   $O_2^{2 - }$ has 18 electrons.

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

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

Here also no unpaired electron present, so it is diamagnetic.

(d)   B₂ has 10 electrons.

Molecular orbital configuration of B₂ is

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

Here two unpaired electrons present.

So it is paramagnetic.