Which of the following xenon compound has the same number of lone pairs as in `I_(3)^(-)` ?

Options

A. `XeO_(4)`
B. `XeF_(4)`
C. `XeF_(2)`
D. `XeO_(3)`

Answer

To determine which xenon compound has the same number of lone pairs as in \( I_3^- \), we will follow these steps:

### Step 1: Determine the number of lone pairs in \( I_3^- \)

1. **Identify the valence electrons of iodine**: Each iodine atom has 7 valence electrons.
2. **Calculate the total valence electrons for \( I_3^- \)**:
- There are 3 iodine atoms, contributing \( 3 \times 7 = 21 \) electrons.
- The negative charge adds 1 more electron, giving a total of \( 21 + 1 = 22 \) electrons.
3. **Draw the Lewis structure**:
- One iodine atom forms bonds with the other two iodine atoms. The central iodine atom forms 2 bond pairs (one with each of the other two iodine atoms).
- The two outer iodine atoms each have 3 lone pairs (6 electrons) and the central iodine has 1 bond pair and 3 lone pairs.
4. **Count the lone pairs**:
- The total lone pairs in \( I_3^- \) is \( 3 (from outer I) + 3 (from other outer I) + 3 (from central I) = 9 \) lone pairs.

### Step 2: Analyze the xenon compounds

Now we will analyze the given xenon compounds to find which one has 9 lone pairs.

#### Option 1: \( XeO_4 \)

1. **Valence electrons**: Xenon has 8 valence electrons.
2. **Bonding**: Xenon forms 4 double bonds with 4 oxygen atoms.
3. **Lone pairs**: Since all 8 electrons are used in bonding, there are **0 lone pairs** in \( XeO_4 \).

#### Option 2: \( XeF_4 \)

1. **Valence electrons**: Xenon has 8 valence electrons.
2. **Bonding**: Xenon forms 4 single bonds with 4 fluorine atoms.
3. **Lone pairs**: After forming 4 bonds, there are \( 8 - 4 = 4 \) electrons left, which means there are **2 lone pairs** on xenon.
4. **Fluorine lone pairs**: Each fluorine has 3 lone pairs, contributing \( 4 \times 3 = 12 \) lone pairs from fluorine.
5. **Total lone pairs**: \( 2 (from Xe) + 12 (from F) = 14 \) lone pairs.

#### Option 3: \( XeF_2 \)

1. **Valence electrons**: Xenon has 8 valence electrons.
2. **Bonding**: Xenon forms 2 single bonds with 2 fluorine atoms.
3. **Lone pairs**: After forming 2 bonds, there are \( 8 - 2 = 6 \) electrons left, which means there are **3 lone pairs** on xenon.
4. **Fluorine lone pairs**: Each fluorine has 3 lone pairs, contributing \( 2 \times 3 = 6 \) lone pairs from fluorine.
5. **Total lone pairs**: \( 3 (from Xe) + 6 (from F) = 9 \) lone pairs.

### Conclusion

The xenon compound that has the same number of lone pairs as \( I_3^- \) is \( XeF_2 \), which has a total of 9 lone pairs.