Which of the following is paramagnetic

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Question

(a) Paramagnetic property arise through unpaired electron.

${B_2} 	o $$\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\pi

Vedclass · Accepted Answer

${B_2}$

Vedclass · Answer

(a) Paramagnetic property arise through unpaired electron.

${B_2} 	o $$\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\pi 2{p_x}^1 = \pi 2{p_y}^1}\limits_{{m{(2 unpaired electron)}}} $

${C_2} 	o $ $\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\pi 2{p_x}^2.\pi 2{p_y}^2}\limits_{{m{(No unpaired electron)}}} $

${N_2} 	o $ $\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2\pi 2{p_z}^2}\limits_{{m{(No unpaired electron)}}} $

${F_2} 	o $$\mathop {\sigma {s^2},{\sigma ^*}1{s^2},\sigma 2{s^2},{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2,\pi 2{p_z}^2,}\limits_{{m{(No\,\, unpaired\,\, electron)}}} $${\pi ^*}2{p_y}^2,

{\pi ^*}2{p_z}^2$So only ${B_2}$

exist unpaired electron and show the paramagnetism.

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