The bond length of the species O_2,O_2^+ and | vedclass

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(A) The bond length is inversely proportional to the bond order.Bond order is calculated as: $Bond \; Order = \frac{N_b - N_a}{2}$.For $O_2$ $(16 \; e

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$O_2^- > O_2 > O_2^+$

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(A) The bond length is inversely proportional to the bond order.Bond order is calculated as: $Bond \; Order = \frac{N_b - N_a}{2}$.For $O_2$ $(16 \; e^-)$: $Bond \; Order = \frac{10 - 6}{2} = 2.0$.For $O_2^+$ $(15 \; e^-)$: $Bond \; Order = \frac{10 - 5}{2} = 2.5$.For $O_2^-$ $(17 \; e^-)$: $Bond \; Order = \frac{10 - 7}{2} = 1.5$.The bond order sequence is $O_2^+ (2.5) > O_2 (2.0) > O_2^- (1.5)$.Since bond length is inversely proportional to bond order,the bond length order is $O_2^- > O_2 > O_2^+$.

The bond length of the species $O_2$,$O_2^+$ and $O_2^-$ are in the order of

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