To determine the order of dipole moments for the compounds BF3, NF3, and NH3, we need to analyze the molecular geometry and the electronegativity of the atoms involved in each compound.
### Step-by-Step Solution:
1. **Analyze BF3 (Boron Trifluoride)**:
- Boron (B) has 3 valence electrons, and each Fluorine (F) has 7 valence electrons.
- In BF3, Boron forms three bonds with three Fluorine atoms, using all three of its valence electrons.
- The molecular geometry is trigonal planar with bond angles of 120°.
- The dipole moments of the B-F bonds point towards the Fluorine atoms. However, due to the symmetry of the trigonal planar shape, the dipole moments cancel each other out.
- **Dipole Moment of BF3 = 0**.
2. **Analyze NF3 (Nitrogen Trifluoride)**:
- Nitrogen (N) has 5 valence electrons, and each Fluorine has 7 valence electrons.
- In NF3, Nitrogen forms three bonds with three Fluorine atoms, using three of its valence electrons, leaving one lone pair on Nitrogen.
- The molecular geometry is trigonal pyramidal due to the presence of the lone pair.
- The dipole moments of the N-F bonds point towards the Fluorine atoms, and the lone pair contributes to the net dipole moment.
- The dipole moment of NF3 is approximately **0.4 D** (Debye).
3. **Analyze NH3 (Ammonia)**:
- Nitrogen has 5 valence electrons, and each Hydrogen (H) has 1 valence electron.
- In NH3, Nitrogen forms three bonds with three Hydrogen atoms, using three of its valence electrons, leaving one lone pair on Nitrogen.
- The molecular geometry is also trigonal pyramidal.
- The dipole moments of the N-H bonds point towards the Nitrogen atom, and the lone pair also contributes to the net dipole moment.
- The dipole moment of NH3 is approximately **1.47 D**.
### Conclusion:
Now that we have the dipole moments for each compound:
- BF3: 0 D
- NF3: 0.4 D
- NH3: 1.47 D
Thus, the correct order of dipole moments is:
**NH3 > NF3 > BF3**.
