Search any question & find its solution
Question:
Answered & Verified by Expert
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.
Options:
Solution:
2129 Upvotes
Verified Answer
The correct answer is:
$\mathrm{O}_2\left[\mathrm{AsF}_6\right] < \mathrm{O}_2 < \mathrm{KO}_2$
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 $\mathrm{O}_2\left(\mathrm{AsF}_6\right) < \mathrm{O}_2 < \mathrm{KO}_2$.
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 $\mathrm{O}_2\left(\mathrm{AsF}_6\right) < \mathrm{O}_2 < \mathrm{KO}_2$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.