JEE Chemistry 2025 Question with Solution

Given: The molecules are $\mathrm{SO_3}$, $\mathrm{NF_3}$, $\mathrm{NH_3}$, $\mathrm{XeF_2}$, $\mathrm{ClF}$, and $\mathrm{SF_6}$.

Find: The hybridization of the molecule that has non-zero dipole moment and one or more lone pairs on the central atom.

Check each molecule using shape, lone pairs, and dipole moment:

1. $\mathrm{SO_3}$: trigonal planar, no lone pair on central atom, so it does not satisfy the condition.
2. $\mathrm{NF_3}$: central atom has one lone pair, shape is trigonal pyramidal, and dipole moment is non-zero. Its hybridization is $sp^3$.
3. $\mathrm{NH_3}$: central atom has one lone pair, shape is trigonal pyramidal, and dipole moment is non-zero. Its hybridization is $sp^3$.
4. $\mathrm{XeF_2}$: central atom has lone pairs, but the molecule is linear and symmetric, so dipole moment is zero. Its hybridization is $sp^3d$.
5. $\mathrm{ClF}$: the solution discusses this entry inconsistently and later analyzes $\mathrm{ClF_3}$ instead. Based on the explicit list in the question, this does not determine the required answer.
6. $\mathrm{SF_6}$: no lone pair on central atom and dipole moment is zero.

Thus, from the molecules explicitly listed, the ones satisfying the condition are $\mathrm{NF_3}$ and $\mathrm{NH_3}$, and both have hybridization $sp^3$.

Therefore, the correct option is A.

Note: The solution marks option D, but its own detailed analysis states that the listed molecules satisfying the condition have hybridization $sp^3$. Hence the working supports A, not D.

Given: We must identify a molecule with at least one lone pair on the central atom and $\mu \ne 0$.

Find: Its hybridization.

Use steric number:

• For $\mathrm{NF_3}$: $\text{SN} = 3 + 1 = 4 \Rightarrow sp^3$
• For $\mathrm{NH_3}$: $\text{SN} = 3 + 1 = 4 \Rightarrow sp^3$
• For $\mathrm{XeF_2}$: $\text{SN} = 2 + 3 = 5 \Rightarrow sp^3d$, but linear symmetry gives $\mu = 0$

So the only listed molecules meeting both conditions give hybridization $sp^3$.

Therefore, the correct option is A.