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Which of the following is paramagnetic
${B_2}$
${C_2}$
${N_2}$
${F_2}$
Solution
(a) Paramagnetic property arise through unpaired electron.
${B_2} \to $$\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\pi 2{p_x}^1 = \pi 2{p_y}^1}\limits_{{\rm{(2 unpaired electron)}}} $
${C_2} \to $ $\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\pi 2{p_x}^2.\pi 2{p_y}^2}\limits_{{\rm{(No unpaired electron)}}} $
${N_2} \to $ $\mathop {\sigma 1{s^2}{\sigma ^*}1{s^2},\sigma 2{s^2}{\sigma ^*}2{s^2},\sigma 2{p_x}^2,\pi 2{p_y}^2\pi 2{p_z}^2}\limits_{{\rm{(No unpaired electron)}}} $
${F_2} \to $$\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_{{\rm{(No\,\, unpaired\,\, electron)}}} $${\pi ^*}2{p_y}^2,
{\pi ^*}2{p_z}^2$So only ${B_2}$
exist unpaired electron and show the paramagnetism.