The correct order of dipole moments of HX bond in the given compounds is :

Options

A. `HF lt H_2S lt H_2O`
B. `HF lt H_2Sgt H_2O`
C. `HF gt H_2O gt H_2 S`
D. `HF gt H_2O lt H_2S`

Answer

To determine the correct order of dipole moments of the HX bond in the compounds HF, H2O, and H2S, we need to analyze the polarity of each bond based on electronegativity and bond length. Here’s a step-by-step solution:

### Step 1: Understand Dipole Moment
The dipole moment (μ) is a measure of the polarity of a bond and is given by the formula:
\[ \mu = q \times d \]
where:
- \( q \) is the charge difference (related to electronegativity),
- \( d \) is the distance between the charges (bond length).

### Step 2: Analyze Electronegativity
The electronegativity values for the relevant elements are:
- Fluorine (F) - highest electronegativity,
- Oxygen (O) - second highest,
- Sulfur (S) - lower than O.

This means that the bond in HF will have the highest charge separation due to the high electronegativity of F, followed by H2O (O), and then H2S (S).

### Step 3: Determine Bond Lengths
The bond lengths generally follow this trend:
- HF has the shortest bond length,
- H2O has a longer bond length than HF but shorter than H2S,
- H2S has the longest bond length.

### Step 4: Compare Charge and Bond Length
Since the dipole moment depends on both charge and bond length, we need to consider:
- In HF, the charge is very high due to the high electronegativity of F, and the bond length is short.
- In H2O, the charge is also significant (due to O), but the bond length is longer than HF.
- In H2S, the charge is the least significant (due to S), and the bond length is the longest.

### Step 5: Order of Dipole Moments
Based on the analysis:
- HF will have the highest dipole moment due to high charge and short bond length.
- H2O will have a moderate dipole moment due to significant charge but longer bond length than HF.
- H2S will have the lowest dipole moment due to the least charge and longest bond length.

Thus, the order of dipole moments is:
\[ \mu_{HF} > \mu_{H2O} > \mu_{H2S} \]

### Final Answer
The correct order of dipole moments of the HX bond in the given compounds is:
\[ \text{HF} > \text{H2O} > \text{H2S} \]