Question

The pair of molecules having same number of lone pair of electrons on central atom is

• A. $SF_{4}$, $XeF_{4}$
• B. $ClF_{3}$, $BrF_{5}$
• C. $ClF_{3}$, $XeF_{4}$
• D. $SF_{4}$, $ClF_{3}$

Hint:

Calculate total electron pairs = (valence central + attached)/2. Subtract bonds for lone pairs. Use VSEPR geometries: ClF3 T-shaped (2 lone), XeF4 square planar (2 lone).

Answer 1

The Correct Option is C

1. Lone pairs on central atom are determined by valence electrons minus bonding electrons, divided by 2.
2. For SF$_4$: S (6 valence e$^-$), 4 F (8 bonds), lone pairs = (6 + 4 - 8)/2 = 1.
3. For XeF$_4$: Xe (8 valence e$^-$), 4 F (8 bonds), lone pairs = (8 + 4 - 8)/2 = 2. Wait, no: each F brings 1 for bond, but standard VSEPR: XeF4 has 12 e- pairs: 4 bonds, 2 lone.
4. For ClF$_3$: Cl (7 valence), 3 F (6 bonds), lone pairs = (7 - 3)/2 wait, total e- pairs = (7 +3)/2 =5, 3 bonds, 2 lone.
5. BrF$_5$: Br (7), 5 F, 6 pairs: 5 bonds, 1 lone.
6. Thus, ClF$_3$ (2 lone) and XeF$_4$ (2 lone) match, so (3).