Match List - I with List - II and choose the correct answer.

List - I		List - II
(a)	PCl5	(i)	Square pyramidal
(b)	SF6	(ii)	Priangular planar
(c)	BrF5	(iii)	Octahedral
(d)	BF3	(iv)	Trigonal bipyramidal

The correct matching option is (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii).

Let us determine the molecular geometry of each compound step-by-step using the Valence Shell Electron Pair Repulsion (VSEPR) theory:

(a) PCl₅ (Phosphorus pentachloride):
Phosphorus (P) is in Group 15 and has 5 valence electrons. It forms 5 single bonds with five chlorine (Cl) atoms, using all 5 of its valence electrons. This gives a steric number of 5 (5 bond pairs and 0 lone pairs). According to VSEPR theory, a steric number of 5 with no lone pairs results in a trigonal bipyramidal geometry.
Therefore, (a) matches with (iv).

(b) SF₆ (Sulfur hexafluoride):
Sulfur (S) is in Group 16 and has 6 valence electrons. It forms 6 single bonds with six fluorine (F) atoms. This results in a steric number of 6 (6 bond pairs and 0 lone pairs). A steric number of 6 with no lone pairs gives an octahedral geometry.
Therefore, (b) matches with (iii).

(c) BrF₅ (Bromine pentafluoride):
Bromine (Br) is in Group 17 and has 7 valence electrons. It forms 5 single bonds with five fluorine (F) atoms, leaving 2 non-bonding valence electrons, which form 1 lone pair. This gives a steric number of 6 (5 bond pairs and 1 lone pair). The electron geometry is octahedral, but the presence of one lone pair reduces the molecular symmetry to a square pyramidal shape.
Therefore, (c) matches with (i).

(d) BF₃ (Boron trifluoride):
Boron (B) is in Group 13 and has 3 valence electrons. It forms 3 single bonds with three fluorine (F) atoms. This results in a steric number of 3 (3 bond pairs and 0 lone pairs). A steric number of 3 with no lone pairs gives a trigonal planar geometry (spelt as "Priangular planar" in the question's list).
Therefore, (d) matches with (ii).

Combining these matches, we get:
(a) → (iv)
(b) → (iii)
(c) → (i)
(d) → (ii)