Based upon VSEPR theory, match the shape (geometry) of the molecules in List-I with the molecules in List-II

List-I (Shape)	List-II (Molecules)
A T-shaped	I. $$XeF_4$$
B Trigonal planar	II. $$SF_4$$
C Square planar	III. $$ClF_3$$
D See-saw	IV. $$BF_3$$

We use VSEPR theory to determine the molecular geometry of each molecule in List-II:

XeF$$_4$$ (I): Xe has 8 valence electrons. Four are used for bonding with F atoms, leaving 2 lone pairs. Total electron pairs = 6 (octahedral arrangement). With 4 bonding pairs and 2 lone pairs (in trans positions), the shape is square planar.

SF$$_4$$ (II): S has 6 valence electrons. Four are used for bonding with F atoms, leaving 1 lone pair. Total electron pairs = 5 (trigonal bipyramidal arrangement). With 4 bonding pairs and 1 lone pair, the shape is see-saw (distorted tetrahedron).

ClF$$_3$$ (III): Cl has 7 valence electrons. Three are used for bonding with F atoms, leaving 2 lone pairs. Total electron pairs = 5 (trigonal bipyramidal arrangement). With 3 bonding pairs and 2 lone pairs (both in equatorial positions), the shape is T-shaped.

BF$$_3$$ (IV): B has 3 valence electrons. All three are used for bonding with F atoms, leaving 0 lone pairs. Total electron pairs = 3. The shape is trigonal planar.

Matching: A (T-shaped) → III (ClF$$_3$$), B (Trigonal planar) → IV (BF$$_3$$), C (Square planar) → I (XeF$$_4$$), D (See-saw) → II (SF$$_4$$).

The answer is Option D: A-(III), B-(IV), C-(I), D-(II).