Which of the following is paramagnetic

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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.

Similar Questions

In the conversion of $N_2$ into $N_2^+$ the electron will be lost from which of the following molecular orbital ?

Which of the following molecular orbitals has maximum number of nodal planes ?

Molecular orbital theory was developed mainly by

Select the $CORRECT$ statement if internuclear axis is $y-$ axis

Number of molecules having bond order $2$ from the following molecule is. . . . . . . $\mathrm{C}_2, \mathrm{O}_2, \mathrm{Be}_2, \mathrm{Li}_2, \mathrm{Ne}_2, \mathrm{~N}_2, \mathrm{He}_2$