JEE Chemistry 2024 Question with Solution

Given: Match the molecules $\mathrm{BrF_5}$, $\mathrm{H_2O}$, $\mathrm{ClF_3}$ and $\mathrm{SF_4}$ with their molecular shapes.

Find: The correct matching option using VSEPR theory.

Analyze each molecule based on VSEPR theory:

• $\mathrm{BrF_5}$ has five bonded pairs and one lone pair around bromine, so its shape is square pyramidal.
• $\mathrm{H_2O}$ has two bonded pairs and two lone pairs, so its shape is bent.
• $\mathrm{ClF_3}$ has three bonded pairs and two lone pairs, so its shape is T-shape.
• $\mathrm{SF_4}$ has four bonded pairs and one lone pair, so its shape is see-saw.

Thus the correct matching is:

• (A) $\mathrm{BrF_5}$ - IV
• (B) $\mathrm{H_2O}$ - III
• (C) $\mathrm{ClF_3}$ - I
• (D) $\mathrm{SF_4}$ - II

the solution contains a discrepancy: one section says "Option (1)", but the detailed matching shown there corresponds to Option (4).

Therefore, the correct option is D.

Given: Four molecules and four possible shapes.

Find: Which option gives the correct one-to-one matching.

Use the electron-pair arrangement around the central atom:

1. $\mathrm{BrF_5}$

• Central atom bromine has $5$ bond pairs and $1$ lone pair.
• Electron geometry is octahedral.
• With one lone pair, the molecular shape becomes square pyramidal.

2. $\mathrm{H_2O}$

• Oxygen has $2$ bond pairs and $2$ lone pairs.
• Electron geometry is tetrahedral.
• Due to two lone pairs, the molecular shape is bent.

3. $\mathrm{ClF_3}$

• Chlorine has $3$ bond pairs and $2$ lone pairs.
• Electron geometry is trigonal bipyramidal.
• Two equatorial lone pairs leave a T-shaped molecule.

4. $\mathrm{SF_4}$

• Sulfur has $4$ bond pairs and $1$ lone pair.
• Electron geometry is trigonal bipyramidal.
• One equatorial lone pair gives a see-saw shape.

So the final matching is: (A) - IV, (B) - III, (C) - I, (D) - II

Hence, the correct option is D.