Which of the following would have a permanent dipole moment ?

Options

A. `BF_(3)`
B. `SiF_(4)`
C. `SF_(4)`
D. `XeF_(4)`

Answer

To determine which of the given compounds has a permanent dipole moment, we need to analyze the molecular geometry and the electronegativity of the atoms involved in each compound. Let's evaluate the compounds step by step.

### Step 1: Analyze BF3 (Boron Trifluoride)
- **Valence Electrons**: Boron has 3 valence electrons, and each fluorine has 1 valence electron. Thus, BF3 has a total of 3 + (3 × 1) = 6 valence electrons.
- **Bonding**: Boron forms three single bonds with three fluorine atoms, using all its valence electrons.
- **Lone Pairs**: There are no lone pairs on boron.
- **Hybridization**: The hybridization is sp².
- **Geometry**: The geometry is trigonal planar.
- **Electronegativity**: Fluorine is more electronegative than boron (4.0 vs 2.0). The dipole moments of the three B-F bonds point towards the fluorine atoms.
- **Dipole Moment**: The dipole moments cancel each other out due to the symmetrical shape, resulting in a net dipole moment of zero. Therefore, BF3 is non-polar.

### Step 2: Analyze SiF4 (Silicon Tetrafluoride)
- **Valence Electrons**: Silicon has 4 valence electrons, and each fluorine has 1 valence electron. Thus, SiF4 has a total of 4 + (4 × 1) = 8 valence electrons.
- **Bonding**: Silicon forms four single bonds with four fluorine atoms.
- **Lone Pairs**: There are no lone pairs on silicon.
- **Hybridization**: The hybridization is sp³.
- **Geometry**: The geometry is tetrahedral.
- **Electronegativity**: Again, fluorine is more electronegative than silicon (4.0 vs 1.9). The dipole moments of the four Si-F bonds point towards the fluorine atoms.
- **Dipole Moment**: The dipole moments do not cancel out due to the tetrahedral shape, resulting in a net dipole moment. Therefore, SiF4 is polar and has a permanent dipole moment.

### Conclusion
Among the compounds analyzed, **SiF4** has a permanent dipole moment, while **BF3** does not.

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