Which of the following `"molecules"//"species"` has the minimum number of lone pairs?

Options

A. `ICl_3`
B. `BF_4^(-)`
C. `SnCl_2`
D. `XeF_4`

Answer

To determine which of the given molecules or species has the minimum number of lone pairs, we will analyze each one step by step.

### Step 1: Analyze ICl3
- **Valence Electrons Calculation**: Iodine (I) has 7 valence electrons, and each chlorine (Cl) also has 7 valence electrons.
- **Bonding**: Iodine bonds with 3 chlorine atoms.
- **Lone Pairs Calculation**:
- Iodine: 7 (valence) - 3 (bonds) = 4 electrons left, which means 2 lone pairs.
- Each chlorine has 3 lone pairs (since it forms 1 bond).
- **Total Lone Pairs in ICl3**:
- Iodine: 2 lone pairs + (3 Cl × 3 lone pairs) = 2 + 9 = 11 lone pairs.

### Step 2: Analyze BF4^-
- **Valence Electrons Calculation**: Boron (B) has 3 valence electrons, and each fluorine (F) has 7 valence electrons.
- **Total Valence Electrons**: 3 (B) + 4 × 7 (F) + 1 (negative charge) = 3 + 28 + 1 = 32 electrons.
- **Bonding**: Boron forms 4 bonds with fluorine.
- **Lone Pairs Calculation**:
- Boron: 0 lone pairs (all electrons are used in bonding).
- Each fluorine has 3 lone pairs.
- **Total Lone Pairs in BF4^-**:
- 0 (B) + (4 F × 3 lone pairs) = 0 + 12 = 12 lone pairs.

### Step 3: Analyze SnCl2
- **Valence Electrons Calculation**: Tin (Sn) has 4 valence electrons, and each chlorine has 7 valence electrons.
- **Total Valence Electrons**: 4 (Sn) + 2 × 7 (Cl) = 4 + 14 = 18 electrons.
- **Bonding**: Tin forms 2 bonds with chlorine.
- **Lone Pairs Calculation**:
- Sn: 4 (valence) - 2 (bonds) = 2 electrons left, which means 1 lone pair.
- Each chlorine has 3 lone pairs.
- **Total Lone Pairs in SnCl2**:
- 1 (Sn) + (2 Cl × 3 lone pairs) = 1 + 6 = 7 lone pairs.

### Step 4: Analyze XeF4
- **Valence Electrons Calculation**: Xenon (Xe) has 8 valence electrons, and each fluorine has 7 valence electrons.
- **Total Valence Electrons**: 8 (Xe) + 4 × 7 (F) = 8 + 28 = 36 electrons.
- **Bonding**: Xenon forms 4 bonds with fluorine.
- **Lone Pairs Calculation**:
- Xe: 8 (valence) - 4 (bonds) = 4 electrons left, which means 2 lone pairs.
- Each fluorine has 3 lone pairs.
- **Total Lone Pairs in XeF4**:
- 2 (Xe) + (4 F × 3 lone pairs) = 2 + 12 = 14 lone pairs.

### Conclusion
Now that we have calculated the total number of lone pairs for each species:
- ICl3: 11 lone pairs
- BF4^-: 12 lone pairs
- SnCl2: 7 lone pairs
- XeF4: 14 lone pairs

**The species with the minimum number of lone pairs is SnCl2 with 7 lone pairs.**

### Final Answer
**SnCl2 has the minimum number of lone pairs.**

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