Question
How to calculate bond pairs XeF4
Answers
GPT-4o mini
To calculate the number of bond pairs in xenon tetrafluoride (XeF4), we need to follow these steps:
1. **Determine the Lewis structure**:
- Xenon (Xe) is the central atom and is surrounded by four fluorine (F) atoms.
- Xenon has 8 valence electrons, while each fluorine atom has 7 valence electrons, contributing a total of 28 valence electrons (4 F atoms × 7 electrons).
- In total, there are \(8 + 28 = 36\) valence electrons available for bonding.
2. **Form bond pairs**:
- Each Xe-F bond uses 2 valence electrons. Since there are 4 fluorine atoms, indicating 4 bonds, this accounts for \(4 \times 2 = 8\) electrons.
- There are 36 - 8 = 28 valence electrons left after forming the bond pairs.
3. **Distribute remaining electrons**:
- The remaining 28 electrons can be used to satisfy the octet rule for the fluorine atoms, with each fluorine requiring 6 additional electrons (3 lone pairs) to complete their octet.
- Therefore, 4 fluorine atoms will use \(4 \times 6 = 24\) electrons.
- This leaves us with 4 electrons.
4. **Assign lone pairs to the central atom**:
- The xenon atom can also hold additional lone pairs because it has access to more than 8 electrons (it can expand its octet).
- The 4 remaining electrons will remain as 2 lone pairs on the xenon atom.
5. **Count the bond pairs**:
- In XeF4, there are 4 Xe-F bond pairs. Each bond consisting of a shared pair of electrons between xenon and each of the four fluorine atoms.
In conclusion, **XeF4 has a total of 4 bond pairs**.
1. **Determine the Lewis structure**:
- Xenon (Xe) is the central atom and is surrounded by four fluorine (F) atoms.
- Xenon has 8 valence electrons, while each fluorine atom has 7 valence electrons, contributing a total of 28 valence electrons (4 F atoms × 7 electrons).
- In total, there are \(8 + 28 = 36\) valence electrons available for bonding.
2. **Form bond pairs**:
- Each Xe-F bond uses 2 valence electrons. Since there are 4 fluorine atoms, indicating 4 bonds, this accounts for \(4 \times 2 = 8\) electrons.
- There are 36 - 8 = 28 valence electrons left after forming the bond pairs.
3. **Distribute remaining electrons**:
- The remaining 28 electrons can be used to satisfy the octet rule for the fluorine atoms, with each fluorine requiring 6 additional electrons (3 lone pairs) to complete their octet.
- Therefore, 4 fluorine atoms will use \(4 \times 6 = 24\) electrons.
- This leaves us with 4 electrons.
4. **Assign lone pairs to the central atom**:
- The xenon atom can also hold additional lone pairs because it has access to more than 8 electrons (it can expand its octet).
- The 4 remaining electrons will remain as 2 lone pairs on the xenon atom.
5. **Count the bond pairs**:
- In XeF4, there are 4 Xe-F bond pairs. Each bond consisting of a shared pair of electrons between xenon and each of the four fluorine atoms.
In conclusion, **XeF4 has a total of 4 bond pairs**.