The answer to each of the following is a non-negative integer
Find the number lone pairs in `XeO_2F_2`

Answer

To find the number of lone pairs in \( XeO_2F_2 \), we can follow these steps:

### Step 1: Determine the Valence Electrons
First, we need to find the total number of valence electrons for \( XeO_2F_2 \):
- Xenon (Xe) has 8 valence electrons.
- Oxygen (O) has 6 valence electrons, and there are 2 oxygen atoms, contributing \( 2 \times 6 = 12 \) valence electrons.
- Fluorine (F) has 7 valence electrons, and there are 2 fluorine atoms, contributing \( 2 \times 7 = 14 \) valence electrons.

Adding these together:
\[ 8 (Xe) + 12 (O) + 14 (F) = 34 \text{ valence electrons} \]

### Step 2: Draw the Lewis Structure
Next, we will draw the Lewis structure for \( XeO_2F_2 \):
- Place the central atom \( Xe \) in the center.
- Attach the two oxygen atoms and two fluorine atoms to the xenon.
- Since oxygen is more electronegative than xenon, it will occupy the axial positions, while fluorine will occupy the equatorial positions.

### Step 3: Count the Bonds
Now, we need to account for the bonds formed:
- Each \( O \) atom forms a double bond with \( Xe \) (2 bonds per oxygen).
- Each \( F \) atom forms a single bond with \( Xe \).

This gives us:
- 2 double bonds from the oxygen atoms (4 electrons).
- 2 single bonds from the fluorine atoms (2 electrons).

Total bonding electrons:
\[ 4 (from \, O) + 2 (from \, F) = 6 \text{ bonding electrons} \]

### Step 4: Calculate Remaining Electrons
Now, we subtract the bonding electrons from the total valence electrons:
\[ 34 \text{ total valence electrons} - 6 \text{ bonding electrons} = 28 \text{ remaining electrons} \]

### Step 5: Distribute Remaining Electrons as Lone Pairs
Next, we distribute the remaining electrons as lone pairs:
- Each fluorine atom will have 6 remaining electrons (3 lone pairs each).
- Each oxygen atom will have 4 remaining electrons (2 lone pairs each).
- The xenon atom will have the remaining electrons as lone pairs.

### Step 6: Count the Lone Pairs
Now, we can count the lone pairs:
- Each fluorine contributes 3 lone pairs (2 fluorine atoms contribute \( 3 \times 2 = 6 \) lone pairs).
- Each oxygen contributes 2 lone pairs (2 oxygen atoms contribute \( 2 \times 2 = 4 \) lone pairs).
- The xenon will have 1 lone pair.

Total lone pairs:
\[ 6 (from \, F) + 4 (from \, O) + 1 (from \, Xe) = 11 \text{ lone pairs} \]

### Final Answer
The total number of lone pairs in \( XeO_2F_2 \) is **11**.

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