$\mathrm{SiH}_4$ :
Si belongs to group 14 , has 4 valence electrons, undergoes $\mathrm{sp}^3$ - hybridization to form a tetrahedral geometry. $\mathrm{Be}$ in $\mathrm{BeCl}_2$ is in sp-hybridization and forms linear shape.
$\mathrm{S}$ in $\mathrm{SF}_4$ is in $\mathrm{dsp}^3$ - hybridization and forms a squarepyramidal geometry with one lone pair.
$\mathrm{SnCl}_2$ has $\mathrm{Sn}$ in $\mathrm{sp}^2$ - hybridization and forms an angular geometry with one lone pair.
Thus, only I is correct.
