Question

$N_2$ and $O_2$ are converted into monocations, $N_2^{+}$ and $O_2^{+}$ respectively. which of the following statements is wrong ?

• A. In $N_2$ the $N-N$ bond weakens
• B. $N_2^{+}$ becomes diamagnetic
• C. In $O_2,$ paramagnetism decreases
• D. In $O_2,$ the $O-O$ bond order increases

Answer 1

The correct option is B $N_2^{+}$ becomes diamagnetic

For $N_2$ : $\sigma 1s^2<{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\pi 2p_y^2=2p_z^2<\sigma 2p_x^2<{\sigma }^{\ast }2p_x<{\pi }^{\ast }2p_y={\pi }^{\ast }2p_z$ $\text{For}{O}_2:$ $\sigma 1s^2{\sigma }^{\ast }1s^2<\sigma 2s^2<{\sigma }^{\ast }2s^2<\sigma 2p_x^2<\pi 2p_y^2=\pi 2p_z^2<{\pi }^{\ast }2p_y^1={\pi }^{\ast }2p_z^1<{\pi }^{\ast }2p_y={\pi }^{\ast }2p_z<{\sigma }^{\ast }2p_x$

a) In $N^{2+}$, the electron is removed from bonding molecular orbital. Hence bond order decreases and $N-N$ bond weakens.

b) In $N^{2+}$, there is one unpaired electron, it is paramagnetic. This statement is wrong.

c) In $O^{2+}$ the no of unpaired electrons decreases and hence paramagnetism decreases.

d) In $O^{2+}$, the electron is removed from anti bonding molecular orbital. Hence $O-O$ bond order increases.