Question
Calculate lone pairs BF4-
Answers
GPT-4o mini
To determine the number of lone pairs in the tetrafluoroborate ion (\( \text{BF}_4^- \)), we first need to analyze its structure:
1. **Identify the central atom**: Boron (B) is the central atom.
2. **Count the valence electrons**:
- Boron has 3 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are four fluorine atoms: \( 4 \times 7 = 28 \) electrons.
- The overall charge of \( -1 \) adds one additional electron.
- Total valence electrons = \( 3 + 28 + 1 = 32 \).
3. **Determine the bonding**: Boron forms single bonds with each of the four fluorine atoms, which uses up 8 valence electrons (2 for each bond):
- \( 4 \text{B-F bonds} = 4 \times 2 = 8 \text{ electrons} \).
4. **Calculate remaining electrons**:
- Total bonds = 8 electrons (used)
- Total available = 32 electrons
- Remaining electrons = \( 32 - 8 = 24 \) electrons.
5. **Distribution of remaining electrons**: The remaining electrons will be placed as lone pairs on the fluorine atoms, which typically have three lone pairs each. Each fluorine needs a total of 8 electrons (octet rule), so with one bond to boron (which counts as 2 electrons), each fluorine will have:
- \( 8 \text{ electrons} - 2 \text{ (bond)} = 6 \text{ (lone pair electrons)} \).
Accordingly:
- Each of the 4 fluorine atoms will have 3 lone pairs (6 electrons).
- There are no lone pairs on the central boron atom because it is bonded to four fluorine atoms.
In summary, the tetrafluoroborate ion \( \text{BF}_4^- \) has:
- **Boron**: 0 lone pairs
- **Fluorine**: Each of the 4 fluorine atoms has 3 lone pairs.
Thus, \( \text{BF}_4^- \) has a total of \( 4\text{ F} \times 3 \text{ lone pairs} = 12 \text{ lone pairs} \).
So, to directly respond to your question about lone pairs in \( \text{BF}_4^- \): there are **12 lone pairs** in total (from the four fluorine atoms). The boron does not have any lone pairs.
1. **Identify the central atom**: Boron (B) is the central atom.
2. **Count the valence electrons**:
- Boron has 3 valence electrons.
- Each fluorine (F) has 7 valence electrons, and there are four fluorine atoms: \( 4 \times 7 = 28 \) electrons.
- The overall charge of \( -1 \) adds one additional electron.
- Total valence electrons = \( 3 + 28 + 1 = 32 \).
3. **Determine the bonding**: Boron forms single bonds with each of the four fluorine atoms, which uses up 8 valence electrons (2 for each bond):
- \( 4 \text{B-F bonds} = 4 \times 2 = 8 \text{ electrons} \).
4. **Calculate remaining electrons**:
- Total bonds = 8 electrons (used)
- Total available = 32 electrons
- Remaining electrons = \( 32 - 8 = 24 \) electrons.
5. **Distribution of remaining electrons**: The remaining electrons will be placed as lone pairs on the fluorine atoms, which typically have three lone pairs each. Each fluorine needs a total of 8 electrons (octet rule), so with one bond to boron (which counts as 2 electrons), each fluorine will have:
- \( 8 \text{ electrons} - 2 \text{ (bond)} = 6 \text{ (lone pair electrons)} \).
Accordingly:
- Each of the 4 fluorine atoms will have 3 lone pairs (6 electrons).
- There are no lone pairs on the central boron atom because it is bonded to four fluorine atoms.
In summary, the tetrafluoroborate ion \( \text{BF}_4^- \) has:
- **Boron**: 0 lone pairs
- **Fluorine**: Each of the 4 fluorine atoms has 3 lone pairs.
Thus, \( \text{BF}_4^- \) has a total of \( 4\text{ F} \times 3 \text{ lone pairs} = 12 \text{ lone pairs} \).
So, to directly respond to your question about lone pairs in \( \text{BF}_4^- \): there are **12 lone pairs** in total (from the four fluorine atoms). The boron does not have any lone pairs.