Chemistry · Chemical Bonding

JEE Main 2025 — 2 April, Morning Shift — Question 6

Among $\mathrm{SO}_{2}, \mathrm{NF}_{3}, \mathrm{NH}_{3}, \mathrm{XeF}_{2}, \mathrm{CIF}_{3}$ and $\mathrm{SF}_{4}$ , the hybridization of the molecule with non-zero dipole

moment and highest number of lone-pairs of electrons on the centre atom is :

Option A:
$s p^{3}$
Option B:
$d s p^{2}$
Option C:
$s p^{3} d^{2}$
Option D:
$s p^{3} d$
Correct

Answer: D

Step-by-step solution

$\mathrm{ClF}_{3}$ has highest no. of lone pairs on central atom which is $s p^{3} d$ hybrid and has non-zero dipole moment

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Bonding
Topic: Hybridization and VSEPR

A molecule with the formula A X 4 Y has all its elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and…

Answer: D

Step-by-step solution

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