Question

The molecular orbital theory supports paramagnetic behaviour of

• A. $B{e}_2$
• B. $C_2$
• C. $N_2$
• D. $O_2$

Answer 1

Answer: (D)

When all the electrons in a molecule are paired it is diamagnetic and when unpaired electrons are present it is paramagnetic. Considering the filling of electrons in molecular orbitals the e.c. of molecules is

$B{e}_2(8)=(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2$

$C_2(12)=(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,\left({\sigma }^{\ast }2s^2\right)$

${\left(\pi 2p_y\right)}^2,{\left(\pi 2p_z\right)}^2$

$N_2(14)=(\sigma 1s{)}^2,{\left({\sigma }^{\ast }ls\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2$

${\left(\pi 2p_y\right)}^2,{\left(\pi 2p_z\right)}^2,{\left(\sigma 2p_x\right)}^2$

$O_2(16)=(\sigma 1s{)}^2,{\left({\sigma }^{\ast }1s\right)}^2,(\sigma 2s{)}^2,{\left({\sigma }^{\ast }2s\right)}^2$

${\left(\pi 2p_y\right)}^2,{\left(\sigma 2p_x\right)}^2$

${\left(\pi 2p_z\right)}^1{\left(\pi 2p_z\right)}^1$

Unpaired electron is present only in $O_2$, so it is paramagnetic.