Q.

The N is having sp hybridized with the two bound pairs and the cation of the ${\mathrm{N}}_2{\mathrm{O}}_5$ is ${\mathrm{NO}}_2^{+}$. Similar to this, the ${\mathrm{N}}_2{\mathrm{O}}_5$anion is $N{O}_3$, with the N having hybridized with the three bound pairs.
Thus, the ${\mathrm{N}}_2{\mathrm{O}}_5$ cation is ${\mathrm{NO}}_2^{+}$ and the ${\mathrm{N}}_2{\mathrm{O}}_5$ anion is $N{O}_3$, and the N atom in the cation has hybridised due to the two bound pairs in the sp state.

${\mathrm{N}}_2{\mathrm{O}}_5\to \underset{\left({180}^{\circ }\right)}{{\mathrm{NO}}_2^{+}}+\underset{\left({120}^{\circ }\right)}{{\mathrm{NO}}_3^{-}}$