Match List I with List II

List-I (Compound)		List-II (Shape/geometry)
(A)	NH3	(I)	Trigonal Pyramidal
(B)	BrF5	(II)	Square Planar
(C)	XeF4	(III)	Octahedral
(D)	SF6	(IV)	Square Pyramidal

Choose the correct answer from the options given below:

The correct matching is A-I, B-IV, C-II, D-III.

To determine the molecular shape and geometry of each compound, we can use the Valence Shell Electron Pair Repulsion (VSEPR) theory by calculating the steric number (number of bonding pairs + number of lone pairs) on the central atom:

(A) NH₃ (Ammonia):
- The central Nitrogen atom (group 15) has 5 valence electrons.
- It forms 3 single covalent bonds with 3 hydrogen atoms, leaving 1 lone pair of electrons.
- Steric number = 3 bond pairs + 1 lone pair = 4.
- With a steric number of 4 and 1 lone pair, the molecular geometry is Trigonal Pyramidal.
- Therefore, (A) matches with (I).

(B) BrF₅ (Bromine pentafluoride):
- The central Bromine atom (group 17) has 7 valence electrons.
- It forms 5 single covalent bonds with 5 fluorine atoms, leaving 1 lone pair of electrons.
- Steric number = 5 bond pairs + 1 lone pair = 6.
- With a steric number of 6 and 1 lone pair, the molecular geometry is Square Pyramidal.
- Therefore, (B) matches with (IV).

(C) XeF₄ (Xenon tetrafluoride):
- The central Xenon atom (group 18) has 8 valence electrons.
- It forms 4 single covalent bonds with 4 fluorine atoms, leaving 2 lone pairs of electrons.
- Steric number = 4 bond pairs + 2 lone pairs = 6.
- With a steric number of 6 and 2 lone pairs, the molecular geometry is Square Planar.
- Therefore, (C) matches with (II).

(D) SF₆ (Sulfur hexafluoride):
- The central Sulfur atom (group 16) has 6 valence electrons.
- It forms 6 single covalent bonds with 6 fluorine atoms, leaving 0 lone pairs.
- Steric number = 6 bond pairs + 0 lone pairs = 6.
- With a steric number of 6 and 0 lone pairs, the molecular geometry is Octahedral.
- Therefore, (D) matches with (III).