Which of the following has the least polarity in the bond?

Options

A. `H-F`
B. `H-Cl`
C. `H-O`
D. `H-S`

Answer

To determine which of the given compounds has the least polarity in the bond, we need to analyze the electronegativity of the elements involved in each bond. The compounds given are HF (hydrogen fluoride), HCl (hydrogen chloride), HO (hydroxyl or water), and HS (hydrogen sulfide).

### Step 1: Identify the electronegativities of the elements involved
- **Fluorine (F)**: 4.0
- **Oxygen (O)**: 3.5
- **Chlorine (Cl)**: 3.0
- **Sulfur (S)**: 2.5
- **Hydrogen (H)**: 2.1 (used as a reference)

### Step 2: Calculate the electronegativity differences for each bond
1. **HF**:
- Electronegativity difference = |4.0 - 2.1| = 1.9
2. **HCl**:
- Electronegativity difference = |3.0 - 2.1| = 0.9
3. **HO**:
- Electronegativity difference = |3.5 - 2.1| = 1.4
4. **HS**:
- Electronegativity difference = |2.5 - 2.1| = 0.4

### Step 3: Compare the electronegativity differences
- HF: 1.9 (most polar)
- HO: 1.4
- HCl: 0.9
- HS: 0.4 (least polar)

### Conclusion
The bond with the least polarity is in **HS (hydrogen sulfide)**, as it has the smallest electronegativity difference of 0.4.

### Final Answer
**HS** has the least polarity in the bond.

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