Question

The molecular orbital theory supports paramagnetic behaviour of

• A. $Be _{2}$
• 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

$Be _{2}(8)=(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2} $

$C _{2}(12)=(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s^{2}\right) $

$\left(\pi 2 p_{y}\right)^{2},\left(\pi 2 p_{z}\right)^{2} $

$N _{2}(14)=(\sigma 1 s)^{2},\left(\sigma^{*} l s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2} $

$\left(\pi 2 p_{y}\right)^{2},\left(\pi 2 p_{z}\right)^{2},\left(\sigma 2 p_{x}\right)^{2} $

$O _{2}(16)=(\sigma 1 s)^{2},\left(\sigma^{*} 1 s\right)^{2},(\sigma 2 s)^{2},\left(\sigma^{*} 2 s\right)^{2} $

$\left(\pi 2 p_{y}\right)^{2},\left(\sigma 2 p_{x}\right)^{2}$

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

Unpaired electron is present only in $O _{2}$, so it is paramagnetic.