Question

Using MOT, predict which of the following species has the shortest bond length ?

• A. $O_2^{+}$
• B. $O_2^{-}$
• C. $O_2^{2-}$
• D. $O_2^{2+}$

Answer 1

The correct option is D $O_2^{2+}$

Molecular orbital configuration of $O_2^{+}$ :

${\sigma }_{1s^2}$ $>$ ${\sigma }_{1s^2}^{\ast }$ $>$ ${\sigma }_{2s^2}$ $>$ ${\sigma }_{2s^2}^{\ast }$ $>$ ${\sigma }_{2p_x{}^2}$ $>$ ${\pi }_{2p_y{}^2}$ = ${\pi }_{2p_z{}^2}$ $>$ ${\pi }_{2p_y{}^1}^{\ast }$ = ${\pi }_{2p_z{}^0}^{\ast }$
Bond order of $O_2^{+}$ = $\frac{N_b-N_a}{2}$ = $\frac{10-5}{2}$ = 2.5
Molecular orbital configuration of $O_2^{-}$ :

${\sigma }_{1s^2}$ $>$ ${\sigma }_{1s^2}^{\ast }$ $>$ ${\sigma }_{2s^2}$ $>$ ${\sigma }_{2s^2}^{\ast }$ $>$ ${\sigma }_{2p_x{}^2}$ $>$ ${\pi }_{2p_y{}^2}$ = ${\pi }_{2p_z{}^2}$ $>$ ${\pi }_{2p_y{}^2}^{\ast }$ = ${\pi }_{2p_z{}^1}^{\ast }$
Bond order of $O_2^{-}$ = $\frac{10-7}{2}$ = 1.5
Molecular orbital configuration of $O_2^{2-}$ :

${\sigma }_{1s^2}$ $>$ ${\sigma }_{1s^2}^{\ast }$ $>$ ${\sigma }_{2s^2}$ $>$ ${\sigma }_{2s^2}^{\ast }$ $>$ ${\sigma }_{2p_x{}^2}$ $>$ ${\pi }_{2p_y{}^2}$ = ${\pi }_{2p_z{}^2}$ $>$ ${\pi }_{2p_y{}^2}^{\ast }$ = ${\pi }_{2p_z{}^2}^{\ast }$
Bond order of $O_2^{2-}$ = $\frac{10-8}{2}$ = 1
Molecular orbital configuration of $O_2^{2+}$ :

${\sigma }_{1s^2}$ $>$ ${\sigma }_{1s^2}^{\ast }$ $>$ ${\sigma }_{2s^2}$ $>$ ${\sigma }_{2s^2}^{\ast }$ $>$ ${\sigma }_{2p_x{}^2}$ $>$ ${\pi }_{2p_y{}^2}$ = ${\pi }_{2p_z{}^2}$ $>$ ${\pi }_{2p_y{}^0}^{\ast }$ = ${\pi }_{2p_z{}^0}^{\ast }$
Bond order of $O_2^{2+}$ = $\frac{10-4}{2}$ = 3

As Bond order $\alpha$ $\frac{1}{Bondlength}$

So, $O_2^{2+}$ has the shortest bond length.