Number of `sigma` bonds , `pi` bonds and lone pair on Xe form of `XeOF_(4)`

Options

A. Option 1. `4,1,1`
B. Option 2. `4 , 2, 1`
C. Option 3. `5,1,1`
D. Option 4. `6, 2, 0`

Answer

To determine the number of sigma bonds, pi bonds, and lone pairs on the xenon atom in the molecule \( \text{XeOF}_4 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** has 8 valence electrons.
- **Oxygen (O)** has 6 valence electrons.
- **Fluorine (F)** has 7 valence electrons, and since there are 4 fluorine atoms, they contribute \( 4 \times 7 = 28 \) valence electrons.

**Total valence electrons** = \( 8 + 6 + 28 = 42 \) valence electrons.

### Step 2: Draw the Lewis Structure
- Start by placing the xenon atom in the center.
- Connect the four fluorine atoms to xenon with single bonds. Each bond uses 2 electrons.
- Connect the oxygen atom to xenon with a double bond. This uses 4 electrons (2 for the sigma bond and 2 for the pi bond).
- After forming these bonds, count the electrons used:
- 4 single bonds (4 F atoms) = 8 electrons.
- 1 double bond (O) = 4 electrons.

**Total used so far** = \( 8 + 4 = 12 \) electrons.

### Step 3: Calculate Remaining Electrons
- **Remaining electrons** = \( 42 - 12 = 30 \) electrons.
- These remaining electrons will be placed as lone pairs on the xenon atom.

### Step 4: Assign Lone Pairs
- Xenon will have 1 lone pair of electrons left after forming the bonds (since 30 electrons are left, and we already used 12).

### Step 5: Count Sigma and Pi Bonds
- Each single bond (4 F atoms) contributes 1 sigma bond:
- Total from F = 4 sigma bonds.
- The double bond with O contributes:
- 1 sigma bond and 1 pi bond.

**Total sigma bonds** = \( 4 + 1 = 5 \) sigma bonds.
**Total pi bonds** = \( 1 \) pi bond.

### Final Summary
- **Number of Sigma Bonds**: 5
- **Number of Pi Bonds**: 1
- **Number of Lone Pairs on Xenon**: 1