Question

The calculated bond order in $H_2^{-}$ ion is:

• A. $0$
• B. $1/2$
• C. $-1/2$
• D. $1$

Answer 1

Answer: (B)

$1/2$

$H_2^{-}$: valence electrons=$2+1=3$

${\left(\sigma 1s\right)}^2{\left({\sigma }^{\ast }1s\right)}^1$

Bond order=$\frac{1}{2}(N_B-N_{AB})$

$\frac{1}{2}(2-1)=0.5$