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The correct order of dipole moments for the given species is

  • A.\(\text{BF}_{3} = \text{NH}_{4}^{+} < \text{NF}_{3} < \text{NH}_{3}\)
  • B.\(\text{BF}_{3} < \text{NH}_{4}^{+} < \text{NF}_{3} < \text{NH}_{3}\)
  • C.\(\text{NH}_{4}^{+} < \text{BF}_{3} < \text{NH}_{3} < \text{NF}_{3}\)
  • D.\(\text{BF}_{3} < \text{NH}_{4}^{+} < \text{NH}_{3} < \text{NF}_{3}\)

Answer: A

The correct order is BF3 = NH4+ < NF3 < NH3.

BF3 is trigonal planar and NH4+ is a regular tetrahedron. In both, the bond dipoles cancel exactly by symmetry, so both have zero dipole moment and are equal, which rules out every option that puts one below the other.

NH3 and NF3 are both pyramidal with a lone pair on nitrogen. In NH3, nitrogen is more electronegative than hydrogen, so the N-H bond dipoles point towards N and add to the lone-pair moment, about 1.5 D in all. In NF3, fluorine is more electronegative than nitrogen, so the N-F bond dipoles point away from N and largely cancel the lone-pair moment, leaving only about 0.2 D. So NF3 < NH3, and the options placing NF3 last are wrong.

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