Identify the molecule $(X)$ with maximum number of lone pairs of electrons (obtained using Lewis dot structure) among $\mathrm{HNO}_3, \mathrm{H}_2 \mathrm{SO}_4, \mathrm{NF}_3$ and $\mathrm{O}_3$. Choose the correct bond angle made by the central atom of the molecule $(X)$.

Among $\mathrm{H}_2 \mathrm{~S}, \mathrm{H}_2 \mathrm{O}, \mathrm{NF}_3, \mathrm{NH}_3$ and $\mathrm{CHCl}_3$, identify the molecule $(\mathrm{X})$ with lowest dipole moment value. The number of lone pairs of electrons present on the central atom of the molecule $(X)$ is :

Two p-block elements X and Y form fluorides of the type $\mathrm{EF}_3$. The fluoride compound $\mathrm{XF}_3$ is a Lewis acid and $\mathrm{YF}_3$ is a Lewis base. The hybridizations of the central atoms of $\mathrm{XF}_3$ and $\mathrm{YF}_3$ respectively are

The formal charges on the atoms marked as (1) to (4) in the Lewis representation of $\mathrm{HNO}_3$ molecule respectively are