Arrange $\mathrm{O}_2, \mathrm{O}_2\left[A s \mathrm{~F}_6\right], \mathrm{KO}_2$ in the increasing order of…

Arrange $\mathrm{O}_2, \mathrm{O}_2\left[A s \mathrm{~F}_6\right], \mathrm{KO}_2$ in the increasing order of bond length of $\mathrm{O}-\mathrm{O}$ bond.
  1. $\mathrm{O}_2 < \mathrm{KO}_2 < \mathrm{O}_2\left[\mathrm{AsF}_6\right]$
  2. $\mathrm{KO}_2 < \mathrm{O}_2 < \mathrm{O}_2\left[\mathrm{AsF}_6\right]$
  3. $\mathrm{O}_2\left[\mathrm{AsF}_6\right] < \mathrm{KO}_2 < \mathrm{O}_2$
  4. $\mathrm{O}_2\left[\mathrm{AsF}_6\right] < \mathrm{O}_2 < \mathrm{KO}_2$

Solution

Bond length $\propto \frac{1}{\text { Bond order }}$ In $\left.\mathrm{K}_2 \mathrm{O}_2\right)$ or $\mathrm{O}_2^{-}$bond order is 1.5 . In $\mathrm{O}_2$ bond order is 2 . In $\mathrm{O}_2\left(\mathrm{AsF}_6\right)$ or $\mathrm{O}_2^{+}$bond order is 2.5 . Hence, the correct order is $\mathrm{O}_2\left(\mathrm{AsF}_6\right) < \mathrm{O}_2 < \mathrm{KO}_2$.

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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