JEE MAIN - Chemistry (2025 - 2nd April Morning Shift - No. 14)
Among $\mathrm{SO}_2, \mathrm{NF}_3, \mathrm{NH}_3, \mathrm{XeF}_2, \mathrm{ClF}_3$ and $\mathrm{SF}_4$, the hybridization of the molecule with nonzero dipole moment and highest number of lone-pairs of electrons on the central atom is
$\mathrm{sp}^3$
$\mathrm{dsp}^2$
$\mathrm{sp}^3 \mathrm{~d}^2$
$\mathrm{sp}^3 \mathrm{~d}$
Explanation
| Molecule | Hybridisation | Dipole Moment | Lone pair on the central atom |
|---|---|---|---|
| $\mathrm{SO}_2$ | $\mathrm{sp}^2$ | Non-zero | 1 |
| $\mathrm{NF}_3$ | $\mathrm{sp}^3$ | Non-zero | 1 |
| $\mathrm{NH}_3$ | $\mathrm{sp}^3$ | Non-zero | 1 |
| $\mathrm{XeF}_2$ | $\mathrm{sp}^3$ | Non-zero | 3 |
| $\mathrm{ClF}_3$ | $\mathrm{sp}^3d$ | Non-zero | 2 |
| $\mathrm{SF}_4$ | $\mathrm{sp}^3d$ | Non-zero | 1 |
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