Question

Removal of electron results in increase in bond order of the bond in :-

• A. $CN$
• B. $O_{2}$
• C. $C_{2}$
• D. $N_{2}$

Answer 1

Answer: (B)

According to MOT, electronic configuration of $O_{2}$ molecules is following-
$\sigma 1 s ^2<\sigma^* 1 s ^2<\sigma 2 s ^2<\sigma * 2 s ^2<$ $\sigma 2 p_z^2 < \pi 2 p _{ x }^2=\pi 2 p _{ y }^2 < \pi^* 2 p _{ x }^1=\pi^* 2 p _{ y }^1$
Bond order $=\frac{\text{Bonding electrons - Anti Bonding electrons}}{2}$
Removal of electron will take place from highest energy orbital.
In $O_{2}$ will remove electron from anti-bonding orbital, if number of anti-bonding electron decrease, bond order increase.
In rest of the options, electron will remove from bonding orbital hence it decrease their bond order.
So, option 2 is correct.