The correct order of decreasing bond angles in `H_(2)S, NH_(3), BF_(3)` and `SiH_(4)` is

Options

A. `H_(2)S ltSiH_(4) ltNH_(3) ltBF_(3)`
B. `NH_(3) ltH_(2)S ltSiH_(4) ltBF_(3)`
C. `H_(2)S ltNH_(3) ltSiH_(4) ltBF_(3)`
D. `H_(2)S ltNH_(3) ltBF_(3) ltSiH_(4).`

Explanation

To determine the correct order of decreasing bond angles in \( H_2S, NH_3, BF_3, \) and \( SiH_4 \), we will analyze the molecular geometry and hybridization of each compound.

### Step 1: Analyze \( SiH_4 \)
- **Molecular Geometry**: Tetrahedral
- **Bond Pairs**: 4 bond pairs, 0 lone pairs
- **Bond Angle**: The ideal bond angle for a tetrahedral geometry is \( 109.5^\circ \).
- **Hybridization**: \( sp^3 \)

### Step 2: Analyze \( BF_3 \)
- **Molecular Geometry**: Trigonal planar
- **Bond Pairs**: 3 bond pairs, 0 lone pairs
- **Bond Angle**: The bond angle in a trigonal planar geometry is \( 120^\circ \).
- **Hybridization**: \( sp^2 \)

### Step 3: Analyze \( NH_3 \)
- **Molecular Geometry**: Trigonal pyramidal
- **Bond Pairs**: 3 bond pairs, 1 lone pair
- **Bond Angle**: The presence of a lone pair decreases the bond angle from the ideal \( 109.5^\circ \) to approximately \( 107^\circ \).
- **Hybridization**: \( sp^3 \)

### Step 4: Analyze \( H_2S \)
- **Molecular Geometry**: Bent
- **Bond Pairs**: 2 bond pairs, 2 lone pairs
- **Bond Angle**: The bond angle is further decreased due to the presence of two lone pairs, resulting in a bond angle of about \( 92^\circ \).
- **Hybridization**: \( sp^3 \)

### Step 5: Order the Bond Angles
Now that we have the bond angles:
- \( BF_3 \): \( 120^\circ \)
- \( SiH_4 \): \( 109.5^\circ \)
- \( NH_3 \): \( 107^\circ \)
- \( H_2S \): \( 92^\circ \)

### Final Order of Decreasing Bond Angles
The correct order of decreasing bond angles is:
\[ BF_3 > SiH_4 > NH_3 > H_2S \]