Question

The calculated bond order in $O_2^− $ ion is

• A. 1
• B. 1.5
• C. 2
• D. 2.5

Answer 1

Answer: (B)

First write the molecular orbital configuration of ion and then find its bond order as

Bond order $=\frac{N_{b}-N_{a}}{2}$

where, $N_{b}=$ number of bonding electrons,

$N_{a}=$ number of electrons in antibonding orbitals.

The $MO$ electronic configuration of

$O _{2}^{-}(8+8+1=17)=\sigma 1 s^{2}, \overset{*}{\sigma} 1 s^{2}, \sigma 2 s^{2}, \overset{*}{\sigma} 2 s^{2} \sigma 2 p_{z}^{2}, \pi 2 p_{x}^{2} \approx \pi 2 p_{y}^{2}, \overset{}{\pi} 2 p_{x}^{2} \approx \overset{*}{\pi} 2 p_{y}^{1}$

$B. O .=\frac{10-7}{2}=1.5$