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

<ol> <li>Molecules with non-zero dipole moment </li> </ol> <ul> <li><p>SO₂ (bent) </p></li> <li><p>NF₃ (pyramidal) </p></li> <li><p>NH₃ (pyramidal) </p></li> <li><p>ClF₃ (T-shaped) </p></li> <li><p>SF₄ (seesaw) </p> <p>(XeF₂ is linear → zero dipole)</p></li> </ul> <ol> <li>Lone-pair count on central atom </li> </ol> <ul> <li><p>SO₂: 1 </p></li> <li><p>NF₃: 1 </p></li> <li><p>NH₃: 1 </p></li> <li><p>SF₄: 1 </p></li> <li><p>ClF₃: 2 </p> <p>→ ClF₃ has the highest number (2) of lone pairs among those with nonzero dipole.</p></li> </ul> <ol> <li><p>Hybridization of Cl in ClF₃ </p> <p>• Total electron domains = 3 bonding pairs + 2 lone pairs = 5 </p> <p>• For 5 domains → $sp^3d$ hybridization </p></li> </ol> <p>Answer: Option D ($sp^3d$).</p>