Question
The correct order of $$N - O$$ bond lengths in $$NO,NO_2^ - ,NO_3^ - $$ and $${N_2}{O_4}$$ is
A.
$${N_2}{O_4} > NO_2^ - > NO_3^ - \, > NO$$
B.
$$NO > NO_3^ - \, > {N_2}{O_4} > NO_2^ - $$
C.
$$NO_3^ - \, > NO_2^ - > {N_2}{O_4} > NO$$
D.
$$NO > {N_2}{O_4} > NO_2^ - > NO_3^ - $$
Answer :
$$NO_3^ - \, > NO_2^ - > {N_2}{O_4} > NO$$
Solution :
As the bond order increases, bond length decreases and bond order is highest for $$NO,$$ i.e. 2.5 and least for $$NO_3^ - ,\,$$ i.e. 1.33. So, the order of bond length is $$\mathop {NO_3^ - }\limits_{1.33} > \mathop {NO_2^ - }\limits_{1.5} > \mathop {{N_2}{O_4}}\limits_{1.5} > \mathop {NO}\limits_{2.5} $$