Match List - I with List - II

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

Choose the correct answer from the options given below 

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

To match each chemical species in List-I with its correct geometry in List-II, we can use the Valence Shell Electron Pair Repulsion (VSEPR) theory. The geometry of a molecule depends on the number of bonding pairs and lone pairs of electrons around the central atom.

(a) ${\mathrm{PCl}}_5$ (Phosphorus pentachloride):
The central phosphorus (P) atom belongs to Group 15 and has 5 valence electrons. It forms 5 single covalent bonds with 5 chlorine atoms, leaving no lone pairs on the phosphorus atom. Thus, the steric number is:
$\mathrm{Steric\; Number}=5\text{bond pairs}+0\text{lone pairs}=5$
A steric number of 5 corresponds to a trigonal bipyramidal geometry.
Therefore, (a) matches with (iv).

(b) ${\mathrm{SF}}_6$ (Sulfur hexafluoride):
The central sulfur (S) atom belongs to Group 16 and has 6 valence electrons. It forms 6 covalent bonds with 6 fluorine atoms, leaving no lone pairs on the sulfur atom. The steric number is:
$\mathrm{Steric\; Number}=6\text{bond pairs}+0\text{lone pairs}=6$
A steric number of 6 corresponds to an octahedral geometry.
Therefore, (b) matches with (iii).

(c) ${\mathrm{BrF}}_5$ (Bromine pentafluoride):
The central bromine (Br) atom belongs to Group 17 and has 7 valence electrons. It forms 5 covalent bonds with 5 fluorine atoms, which uses 5 valence electrons, leaving 2 valence electrons as 1 lone pair on bromine. The steric number is:
$\mathrm{Steric\; Number}=5\text{bond pairs}+1\text{lone pair}=6$
For a steric number of 6 with 5 bonding pairs and 1 lone pair, the shape of the molecule is square pyramidal.
Therefore, (c) matches with (i).

(d) ${\mathrm{BF}}_3$ (Boron trifluoride):
The central boron (B) atom belongs to Group 13 and has 3 valence electrons. It forms 3 covalent bonds with 3 fluorine atoms, leaving no lone pairs on the boron atom. The steric number is:
$\mathrm{Steric\; Number}=3\text{bond pairs}+0\text{lone pairs}=3$
A steric number of 3 corresponds to a trigonal planar geometry.
Therefore, (d) matches with (ii).

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