Question
Calculate lone pair AlF4-
Answers
GPT-4o mini
To calculate the number of lone pairs in the tetrafluoroaluminate ion (\( \text{AlF}_4^- \)), we can follow these steps:
1. **Determine the total number of valence electrons**:
- Aluminum (Al) is in group 13 and has 3 valence electrons.
- Each fluorine (F) atom is in group 17 and has 7 valence electrons. Since there are four fluorine atoms, they contribute \( 4 \times 7 = 28 \) valence electrons.
- The ion has an overall charge of -1, which means we need to add one additional electron.
\[
\text{Total valence electrons} = 3 + 28 + 1 = 32
\]
2. **Determine the number of bonding pairs**:
- The aluminum atom will form four single bonds with the four fluorine atoms. Each bond uses 2 electrons, so with 4 bonds, we use \( 4 \times 2 = 8 \) electrons.
3. **Calculate the remaining electrons**:
\[
\text{Remaining electrons} = 32 - 8 = 24
\]
These remaining electrons will typically be distributed as lone pairs on the fluorine atoms.
4. **Distributing remaining electrons as lone pairs**:
- Each fluorine atom needs 6 more electrons (3 lone pairs) to complete its octet (since each bond already contributes 1 electron).
- For 4 fluorine atoms, we need \( 4 \times 6 = 24 \) electrons, which is exactly what we have remaining.
Thus, in the tetrafluoroaluminate ion (\( \text{AlF}_4^- \)):
- Each fluorine atom has 3 lone pairs (making a total of \( 4 \times 3 = 12 \) lone pairs).
- The aluminum atom does not have any lone pairs because it is bonded to all four fluorine atoms.
In conclusion, the total lone pairs in \( \text{AlF}_4^- \) is **12**, which are all located on the fluorine atoms. The aluminum atom has zero lone pairs.
1. **Determine the total number of valence electrons**:
- Aluminum (Al) is in group 13 and has 3 valence electrons.
- Each fluorine (F) atom is in group 17 and has 7 valence electrons. Since there are four fluorine atoms, they contribute \( 4 \times 7 = 28 \) valence electrons.
- The ion has an overall charge of -1, which means we need to add one additional electron.
\[
\text{Total valence electrons} = 3 + 28 + 1 = 32
\]
2. **Determine the number of bonding pairs**:
- The aluminum atom will form four single bonds with the four fluorine atoms. Each bond uses 2 electrons, so with 4 bonds, we use \( 4 \times 2 = 8 \) electrons.
3. **Calculate the remaining electrons**:
\[
\text{Remaining electrons} = 32 - 8 = 24
\]
These remaining electrons will typically be distributed as lone pairs on the fluorine atoms.
4. **Distributing remaining electrons as lone pairs**:
- Each fluorine atom needs 6 more electrons (3 lone pairs) to complete its octet (since each bond already contributes 1 electron).
- For 4 fluorine atoms, we need \( 4 \times 6 = 24 \) electrons, which is exactly what we have remaining.
Thus, in the tetrafluoroaluminate ion (\( \text{AlF}_4^- \)):
- Each fluorine atom has 3 lone pairs (making a total of \( 4 \times 3 = 12 \) lone pairs).
- The aluminum atom does not have any lone pairs because it is bonded to all four fluorine atoms.
In conclusion, the total lone pairs in \( \text{AlF}_4^- \) is **12**, which are all located on the fluorine atoms. The aluminum atom has zero lone pairs.