Question

Choose the correct option

Molecule		Shape
A	\(BrF_5\)	i	T-shape
B	\(H_2O\)	ii	See-saw
C	\(ClF_3\)	iii	Bent
D	\(SF_4\)	iv	Square Pyramidal

• A. (A)- IV, (B)- III, (C)- I, (D)- II
• B. (A)- III, (B)- IV, (C)- I, (D)- II
• C. (A)- II, (B)- I, (C)- III, (D)- IV
• D. (A)- I, (B)- III, (C)- IV, (D)- II

Answer 1

The Correct Option is A

Analyze Each Molecule Based on VSEPR Theory:
BrF₅: The molecule has five bonded pairs and one lone pair around bromine, leading to a square pyramidal shape.
H₂O: Water has two bonded pairs and two lone pairs, giving it a bent shape.
ClF₃: Chlorine trifluoride has three bonded pairs and two lone pairs, resulting in a T-shape.
SF₄: Sulfur tetrafluoride has four bonded pairs and one lone pair, leading to a see-saw shape.

Match Each Molecule with the Correct Shape:
(A) BrF₅ - Square pyramidal
(B) H₂O - Bent
(C) ClF₃ - T-shape
(D) SF₄ - See-saw

Conclusion: Based on the shapes identified above, the correct answer is Option (1).

Answer 2

To determine the correct shape of the given molecules, we need to understand the molecular geometry based on their Lewis structures and VSEPR (Valence Shell Electron Pair Repulsion) theory.

1. BrF₅: This molecule consists of bromine surrounded by five fluorine atoms and one lone pair of electrons. According to VSEPR theory, BrF₅ adopts a square pyramidal shape due to five bond pairs and one lone pair around the central bromine atom.
2. H₂O: Water consists of an oxygen atom bonded to two hydrogen atoms, with two lone pairs of electrons on the oxygen. The electronic geometry is tetrahedral, but due to the presence of lone pairs, the molecular shape is bent (or angular).
3. ClF₃: In this molecule, chlorine is surrounded by three fluorine atoms and two lone pairs of electrons. Following the VSEPR model, ClF₃ is T-shaped as two lone pairs occupy equatorial positions in a trigonal bipyramidal electronic geometry.
4. SF₄: Sulfur is bonded to four fluorine atoms and has one lone pair. The structure of SF₄ is a see-saw shape due to the lone pair occupying one of the equatorial positions in a trigonal bipyramidal geometry.

Using the information above, we can match each molecule to its correct shape:

• (A) BrF₅ - IV Square Pyramidal
• (B) H₂O - III Bent
• (C) ClF₃ - I T-shape
• (D) SF₄ - II See-saw

Therefore, the correct option is:

(A)- IV, (B)- III, (C)- I, (D)- II