The correct order of bond dissociation energy among $\mathrm{N}_2, \mathrm{O}_2, \mathrm{O}_2^{-}$is shown…
The correct order of bond dissociation energy among $\mathrm{N}_2, \mathrm{O}_2, \mathrm{O}_2^{-}$is shown in which of the following arrangements?
$\mathrm{N}_2>\mathrm{O}_2^{-}>\mathrm{O}_2$
$\mathrm{O}_2^{-}>\mathrm{O}_2>\mathrm{N}_2$
$\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}$
$\mathrm{O}_2>\mathrm{O}_2^{-}>\mathrm{N}_2$
Solution
The bond order of $\mathrm{N}_2, \mathrm{O}_2$, and $\mathrm{O}_2^{-}$are respectively 3,2 and $1.5$
Since higher bond order implies higher bond dissociation energy hence the correct order will be
$
\mathrm{N}_2>\mathrm{O}_2>\mathrm{O}_2^{-}
$