Bond order of 1.5 is shown by- | vedclass

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Question

Bond order $=\frac{	ext { bonding M.O. } {-} 	ext {antibonding M.O. }}{2}$

Bond order of $\mathrm{O}_{2}^{+}=\frac{10-5}{2}=2.5$

Bond order of

Vedclass · Accepted Answer

$O_{2}^-$

Vedclass · Answer

Bond order $=\frac{	ext { bonding M.O. } {-} 	ext {antibonding M.O. }}{2}$

Bond order of $\mathrm{O}_{2}^{+}=\frac{10-5}{2}=2.5$

Bond order of $\mathrm{O}_{2}^{-}=\frac{10-7}{2}=1.5$

Bond order of $\mathrm{O}_{2}^{2-}=\frac{10-8}{2}=1.0$

Bond order of $\mathrm{O}_{2}=\frac{10-6}{2}=2$

Bond order of $1.5$ is shown by-

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