Question

Which of the following species is not paramagnetic?

• A. $CO$
• B. $O_2$
• C. $B_2$
• D. $NO$

Answer 1

The correct option is A $CO$

$CO$ – diamagnetic
The molecular orbital electronic configuration of $CO$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^2=\pi 2p_z^2\sigma 2p_x^2{\pi }^{\ast }2p{y}^0={\pi }^{\ast }2p_z^0{\sigma }^{\ast }2p_x^0$
Since all the electrons are paired up, $CO$ is diamagnetic

$O_2$
The molecular orbital electronic configuration of $O_2$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_x^2\pi 2p_y^2=\pi 2p_z^2{\pi }^{\ast }2p{y}^1={\pi }^{\ast }2p_z^1{\sigma }^{\ast }2p_x^0$
Since there are unpaired electrons, $O_2$ is paramagnetic

$B_2$
The molecular orbital electronic configuration of $B_2$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_y^1=\pi 2p_z^1\sigma 2p_x^0{\sigma }^{\ast }2p_x^0{\pi }^{\ast }2p{y}^0={\pi }^{\ast }2p_z^0$
Since there are unpaired electrons, $B_2$ is paramagnetic

$NO$
The molecular orbital electronic configuration of $NO$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_x^2\pi 2p_y^2=\pi 2p_z^2{\pi }^{\ast }2p{y}^1={\pi }^{\ast }2p_z^0{\sigma }^{\ast }2p_x^0$
Since there are unpaired electrons, $NO$ is paramagnetic