Question

What is the number of unpaired electron(s) in the highest occupied molecular orbital of the following species : $N_2;N_2^{+};O_2;O_2^{+}$?

• A. $0,1,2,1$
• B. $2,1,2,1$
• C. $0,1,0,1$
• D. $2,1,0,1$

Answer 1

Answer: (A)

$N_2$
$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\frac{\sigma 2p_z^2}{HOMO}$
$N_2^{+}-\sigma 1s^2{\sigma }^{\ast }1s^2{\sigma }^{\ast }2s^2{\sigma }^{\ast }2s^2\pi 2p_x^2=\pi 2p_y^2\frac{\sigma 2p_z^1}{HOMO}$
$O_2-\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2$
$\pi 2p_x^2=\pi 2p_y^2$
${\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^1(HOMO)$
$O_2^{+}-\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2=\pi 2p_y^2$
${\pi }^{\ast }2p_x^1={\pi }^{\ast }2p_y^0(HOMO)$
$N_2\Rightarrow 0$ unpaired $e^{-}\text{in HOMO }$
$N_2^{+}\Rightarrow 1$ unpaired $e^{-}\text{in HOMO }$
$O_2\Rightarrow 2$ unpaired $e^{-}\text{in HOMO }$
$O_2^{+}\Rightarrow 1$ unpaired $e^{-}\text{in HOMO }$