When the molecules ${\mathrm{N}}_2$ , ${\mathrm{N}}_2\mathrm{O}$ and ${\mathrm{N}}_2{\mathrm{O}}_4$ are arranged in order of decreasing $\mathrm{N}-\mathrm{N}$ bond length, which order is correct?

Find the bond order in each compound to compare the bond length of the $\mathrm{N}-\mathrm{N}$ bond between the supplied values. The bond order of the ${\mathrm{N}}_2$ molecule is $3$. In this instance, nitrogen is so triple bound. Bond order $2$is present in the molecule ${\mathrm{N}}_2\mathrm{O}$. In this instance, the nitrogen atoms are doubly linked. The bond order of${\mathrm{N}}_2{\mathrm{O}}_4$ is $1$. In this instance, the nitrogen atoms are solely connected. Bond length is higher in the molecule with a lower amount of bond order. Therefore, molecules are grouped as follows in order of increasing $\mathrm{N}-\mathrm{N}$ bond length:

${\mathrm{N}}_2{\mathrm{O}}_{4 }, {\mathrm{N}}_2\mathrm{O}, {\mathrm{N}}_2$