Question

The correct order of $O-O$ bond length is:

• A. $O_2^{+}<O_2<O_2^{-}$
• B. $O_2^{-}<O_2<O_2^{+}$
• C. $O_2^{+}<O_2^{+}<O_2$
• D. none of these

Answer 1

The correct option is A $O_2^{+}<O_2<O_2^{-}$

Bond length is inversely proportional to bond order

Bond length$\propto \frac{1}{Bondorder}$

Bond order can be foundd on apply molecular orbital theory in the given examples B.O$=\frac{BMOelectron-ABMOelectron}{2}$

$O_2^{+}$ (Total electron$=15$)

$\Rightarrow {\sigma }_{1s^2},{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2},{\sigma }_{2s^2}^{\ast },{\sigma }_{2p_z^2},\left({\pi }_{2p_x^2}{\pi }_{2p_y^2}\right),\left({\pi }_{2p_x^1}^{\ast }{\pi }_{2p_x}^{\ast }\right)$

B.O$=\frac{6-1}{2}=2.5$

$O_2$ (Total electron$=16$)

$\Rightarrow {\sigma }_{1s^2},{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2},{\sigma }_{2s^2}^{\ast },{\sigma }_{2p_z^2},\left({\pi }_{2p_x^2}{\pi }_{2p_y^2}\right),\left({\pi }_{2p_x^1}^{\ast }{\pi }_{2p_y^1}^{\ast }\right)$

B.O$=\frac{6-2}{2}=2$

$O_2^{-}$ (Total electron$=17$)

$\Rightarrow {\sigma }_{1s^2},{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2},{\sigma }_{2s^2}^{\ast },{\sigma }_{2p_z^2},\left({\pi }_{2p_x^2}{\pi }_{2p_y^2}\right),\left({\pi }_{2p_x^1}^{\ast }{\pi }_{2p_y^1}^{\ast }\right)$

B.O$=\frac{6-3}{2}=1.5$

$\therefore$ Order of bond order$=O_2^{-}<O_2<O_2^{+}$

Order of Bond length$=O_2^{+}<O_2<O_2^{-}$