To draw the Lewis dot structure for boron tetrafluoride (BF4), we start by determining the number of valence electrons available, the hybridization, electron geometry, molecular geometry, and polarity of the molecule.
a. Number of Valence Electrons
The total number of valence electrons can be calculated by adding the valence electrons from boron and fluorine atoms. Boron has 3 valence electrons and each fluorine has 7. Since there are four fluorine atoms, the calculation is as follows:
- Boron: 3 electrons
- Fluorine: 4 x 7 = 28 electrons
Total = 3 + 28 = 31 valence electrons.
b. Hybridization
The hybridization of BF4 can be determined by looking at the steric number, which is the number of bonding pairs and lone pairs around the central atom. In BF4, boron is bonded to four fluorine atoms and has no lone pairs. Thus, the steric number is 4, indicating that the hybridization is sp3.
c. Electron Geometry
With a steric number of 4 and no lone pairs, the electron geometry of BF4 is Tetrahedral.
d. Molecular Geometry
Since all positions around boron are occupied by bonding pairs (the four fluorine atoms), the molecular geometry is also Tetrahedral.
e. Polarity
BF4 has a tetrahedral geometry, and although the bonds between boron and fluorine are polar due to the significant electronegativity difference, the symmetrical shape results in the dipoles canceling each other out. Therefore, BF4 is non-polar.
In summary, the Lewis dot structure shows a boron atom in the center with four fluorine atoms bonded to it, and all aspects of its bonding and geometry contribute to these properties.