Arrange the following in decreasing order of dipole moment <

Options

A. `BF_(3) lt H_(2)S lt H_(2)O`
B. `H_(2)S lt BF_(3)ltH_(2)O`
C. `H_(2)O lt H_(2)S lt BF_(3)`
D. `BF_(3) lt H_(2)O lt H_(2)S`

Explanation

To arrange the compounds BF3, H2S, and H2O in decreasing order of dipole moment, we will follow these steps:

### Step 1: Analyze the structure of BF3
- **Structure**: BF3 has a trigonal planar geometry due to sp2 hybridization of boron. It has three fluorine atoms symmetrically arranged around the boron atom.
- **Dipole Moment**: The individual bond dipoles (B-F) cancel each other out due to symmetry, resulting in a net dipole moment of zero.

**Conclusion**: Dipole moment of BF3 = 0.

### Step 2: Analyze the structure of H2S
- **Structure**: H2S has a bent shape due to the presence of two lone pairs on the sulfur atom, which leads to sp3 hybridization.
- **Dipole Moment**: The bond dipoles (H-S) do not cancel out completely because of the bent shape, resulting in a net dipole moment.

**Conclusion**: Dipole moment of H2S > 0.

### Step 3: Analyze the structure of H2O
- **Structure**: H2O also has a bent shape due to two lone pairs on the oxygen atom, leading to sp3 hybridization.
- **Dipole Moment**: The bond dipoles (H-O) are directed towards the oxygen atom, and since oxygen is more electronegative than sulfur, the dipole moment is stronger than that of H2S.

**Conclusion**: Dipole moment of H2O > Dipole moment of H2S.

### Step 4: Compare the dipole moments
- We have established that:
- Dipole moment of BF3 = 0
- Dipole moment of H2S > 0
- Dipole moment of H2O > Dipole moment of H2S

### Final Arrangement
Thus, the decreasing order of dipole moments is:
**H2O > H2S > BF3**