The number of paramagnetic species among the following is B2

Question

The number of paramagnetic species among the following is ___________. B_2, Li_2, C_2, C_2^-, O_2^2-, O_2^+ and He_2^+

DivineJEE · Accepted Answer

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 _{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.

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 is no unpaired electron so it is diamagnetic.

O_2^{+}

has 15 electrons.

Moleculer orbital configuration of

O_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^1}^ * = \,\,\pi _{2p_y^0}^ *

Here 1 unpaired electron present, so it is paramagnetic.

C_2

has 12 electrons.

Moleculer orbital configuration of

C_2

=

{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\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 _{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^1}}

Here 1 unpaired electron present, so it is paramagnetic.

Li₂ has 6 electrons.

Li₂ =

{\sigma _{1{s^2}}}\,\,\sigma _{1{s^2}}^ * \,

{\sigma _{2{s^2}}} \,

Here no unpaired electron present, so it is diamagnetic.

Configuration of

He_2^ +

(3 electrons) is =

{\sigma _{1{s^2}}}

\sigma _{1{s^1}}^ *

Here 1 unpaired electron present, so it is paramagnetic.

Explanation

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