The Lewis diagram for BF4- is: The electron-pair geometry around the B atom in BF4- is fill in the blank 1. There are lone pair(s) around the central atom, so the geometry of BF4- is fill in the blank 3.
The Correct Answer and Explanation is:
Correct Answers:
- Blank 1: tetrahedral
- Blank 2: 0
- Blank 3: tetrahedral
Explanation
The Lewis structure of the tetrafluoroborate ion, BF₄⁻, can be determined by first calculating the total number of valence electrons. Boron (B) has 3 valence electrons, and each fluorine (F) atom has 7. Since there are four fluorine atoms and one additional electron due to the negative charge, the total number of valence electrons is:
3 (from B) + 4 × 7 (from F) + 1 (extra electron) = 32 valence electrons
To draw the Lewis structure:
- Place boron in the center as it is the least electronegative element.
- Connect each fluorine to the boron using single bonds. This uses up 4 × 2 = 8 electrons.
- Distribute the remaining 24 electrons (32 – 8 = 24) as lone pairs around the fluorine atoms so that each fluorine achieves an octet. Each F receives 6 electrons (3 lone pairs), which accounts for the remaining 24 electrons.
Each fluorine now has 8 electrons around it (6 from lone pairs and 2 from bonding), and boron has 8 electrons through four single bonds, which is unusual because neutral boron often only forms three bonds. However, in BF₄⁻, boron accepts one more electron to form four bonds due to the negative charge.
The electron-pair geometry considers both bonding and lone electron pairs around the central atom. Since boron in BF₄⁻ has four bonding pairs and no lone pairs, the geometry is tetrahedral. This arrangement minimizes electron pair repulsion according to VSEPR (Valence Shell Electron Pair Repulsion) theory.
Because there are no lone pairs on the central atom, the molecular geometry is the same as the electron-pair geometry, which is also tetrahedral. This symmetry leads to a highly stable and nonpolar ion despite the presence of polar B–F bonds.
