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

$O_2^{-}(8+8+1=17)=\sigma 1s^2,\stackrel{\ast }{\sigma }1s^2,\sigma 2s^2,\stackrel{\ast }{\sigma }2s^2\sigma 2p_z^2,\pi 2p_x^2\approx \pi 2p_y^2,\pi 2p_x^2\approx \stackrel{\ast }{\pi }2p_y^1$

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