Match List I (Molecule/Ion) with List II (Shape) : List I List II A...

For $\mathrm{SF}_4$, the central atom $\mathrm{S}$ has $6$ valence electrons. It forms $4$ bond pairs and has $1$ lone pair. The hybridization is $sp^3d$ and the shape is see-saw.

For $\mathrm{XeF}_4$, the central atom $\mathrm{Xe}$ has $8$ valence electrons. It forms $4$ bond pairs and has $2$ lone pairs. The hybridization is $sp^3d^2$ and the shape is square planar.

For $\mathrm{I}_3^-$, the central atom $\mathrm{I}$ has $7$ valence electrons plus $1$ from the negative charge, giving $8$ electrons. It forms $2$ bond pairs and has $3$ lone pairs. The hybridization is $sp^3d$ and the shape is linear.

For $\mathrm{H}_2\mathrm{O}$, the central atom $\mathrm{O}$ has $6$ valence electrons. It forms $2$ bond pairs and has $2$ lone pairs. The hybridization is $sp^3$ and the shape is bent.

Thus, the correct matching is A-4, B-1, C-2, D-3.

Answer: A-4, B-1, C-2, D-3