Question

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

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

Answer 1

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

1. $O_2^{2+}$:

Valence electrons $=6+6-2=10$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\pi 2p\right)}^4{\left(\sigma 2p\right)}^2$

BO $=\frac{1}{2}\times (6-0)=3$;

2. $O_2^{+}$:

Valence electrons $=6+6-1=11$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^1$

BO $=\frac{1}{2}\times (6-1)=2.5$;

3. $O_2^{-}$:

Valence electrons $=6+6+1=13$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^3$

BO $=\frac{1}{2}\times (6-3)=1.5$;

4. $O_2^{2-}$:

Valence electrons $=6+6+2=14$

${\left(\sigma 2s\right)}^2{(\sigma \ast 2s)}^2{\left(\sigma 2p\right)}^2{\left(\pi 2p\right)}^4{\left({\pi }^{\ast }2p\right)}^4$

BO $=\frac{1}{2}\times (6-4)=1$;

Bond length $\propto \frac{1}{\text{bond order}}$

$\therefore O_2^{2+}$ has the highest bond order and smallest bond length.

Hence, the correct option is $\text{A}.$