MHTCET 15 Apr 2026 Shift 1 Paper

Concept:Dipole moment depends on molecular shape and the resultant of bond dipoles and lone pair effects.Explanation:In $\mathrm{BF}_{3}$, the molecule is trigonal planar and perfectly symmetrical.The three B–F bond dipoles are equal and arranged at $120^\circ$, so they cancel each other.Hence, $\mu\left(\mathrm{BF}_{3}\right) = 0$.Both $\mathrm{NH}_{3}$ and $\mathrm{NF}_{3}$ are pyramidal because nitrogen has one lone pair.In $\mathrm{NH}_{3}$, nitrogen is more electronegative than hydrogen, so N–H bond dipoles point towards nitrogen.The lone pair dipole acts in the same direction, so the effects add up, giving a high dipole moment.In $\mathrm{NF}_{3}$, fluorine is more electronegative than nitrogen, so N–F bond dipoles point towards fluorine.But the lone pair dipole on nitrogen points opposite to these bond dipoles, causing partial cancellation.Therefore, $\mu\left(\mathrm{NF}_{3}\right)$ is smaller than $\mu\left(\mathrm{NH}_{3}\right)$.Thus, the correct order is:$$\mu\left(\mathrm{NH}_{3}\right) > \mu\left(\mathrm{NF}_{3}\right) > \mu\left(\mathrm{BF}_{3}\right)$$Answer:Option D: $\mathrm{NH}_{3} > \mathrm{NF}_{3} > \mathrm{BF}_{3}$