Arrange O_2,O_2[AsF_6],and KO_2 in the increa | vedclass

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(D) The bond length is inversely proportional to the bond order: $	ext{Bond length} \propto \frac{1}{	ext{Bond order}}$.In $KO_2$,the species is $O_

Vedclass · Accepted Answer

$O_2[AsF_6] < O_2 < KO_2$

Vedclass · Answer

(D) The bond length is inversely proportional to the bond order: $ ext{Bond length} \propto \frac{1}{ ext{Bond order}}$.In $KO_2$,the species is $O_2^-$,which has a bond order of $1.5$.In $O_2$,the bond order is $2.0$.In $O_2[AsF_6]$,the species is $O_2^+$,which has a bond order of $2.5$.Since the bond order follows the order $O_2^+ (2.5) > O_2 (2.0) > O_2^- (1.5)$,the bond length follows the reverse order: $O_2[AsF_6] < O_2 < KO_2$.

Arrange $O_2$,$O_2[AsF_6]$,and $KO_2$ in the increasing order of bond length of the $O-O$ bond.

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