Question

The correct order of bond dissociation energy among $N_2, O_2, O_2^-$ is shown in which of the following arrangements ?

• A. $N_{2}> O_{2}^{-}>O_{2}$
• B. $O^{-}_{2}> O_{2}>N_{2}$
• C. $N_{2}> O_{2}>O_{2}^{-}$
• D. $O_{2}> O_{2}^{-}>N_{2}$

Answer 1

The Correct Option is C

Bond energy $\propto$ Bond order bondorder :-

$N_{2} =Nb=10, Na=4$
$B.O.=\left(N_{2}\right)=\frac{10-4}{2}=3$
$O_{2}=Nb=10, Na=6$
$B.O_{\left(o_2\right)}=\frac{10-6}{2}=2$
$O_{2}^{-}=Nb=10, Na=7$
$B.O._{\left(o_2\right)}=\frac{10-7}{2}=\frac{3}{2}=1.5$

Hence the order of B.O.

$N_{2}>\, O_{2} >\, O_{2}^{-}$