The correct order of equatorial FSF bond angle in the folllowing compound.
(I)`SF_(4)` (II) `OSF_(4)` (III) `H_(2)CSF_(4)`

Options

A. `IIIgtIIgtI`
B. `IgtIIIgtII`
C. `IgtIIgtIII`
D. `IIgtIIIgtI`

Answer

To determine the correct order of the equatorial F-S-F bond angle in the compounds \( SF_4 \), \( OSF_4 \), and \( H_2CSF_4 \), we will analyze the molecular structures and the effects of lone pairs on bond angles.

### Step 1: Draw the Lewis Structures
1. **For \( SF_4 \)**:
- Sulfur has 6 valence electrons and forms 4 bonds with fluorine, using 4 electrons for bonding and leaving 2 electrons as a lone pair.
- The structure is a seesaw shape due to the presence of one lone pair.

2. **For \( OSF_4 \)**:
- Oxygen has 6 valence electrons and forms a bond with sulfur, which has 4 fluorine atoms bonded to it.
- The structure will have a bent shape due to the lone pairs on oxygen and the lone pair on sulfur.

3. **For \( H_2CSF_4 \)**:
- Carbon forms a double bond with sulfur and has 4 fluorine atoms bonded to sulfur.
- There are no lone pairs on sulfur in this case, leading to a tetrahedral arrangement.

### Step 2: Analyze the Bond Angles
- **In \( SF_4 \)**:
- The presence of a lone pair on sulfur causes greater lone pair-bond pair repulsion, which decreases the F-S-F bond angle. This angle is expected to be the smallest among the three compounds.

- **In \( OSF_4 \)**:
- The bond angle will be larger than in \( SF_4 \) due to the presence of an oxygen atom, which has lone pairs that can influence the bond angles but is less repulsive than the lone pair in \( SF_4 \).

- **In \( H_2CSF_4 \)**:
- The absence of lone pairs on sulfur leads to the highest bond angle among the three compounds, as bond pair-bond pair repulsions are minimized.

### Step 3: Compare the Bond Angles
- From the analysis:
- \( SF_4 \) has the smallest F-S-F bond angle due to lone pair repulsion.
- \( OSF_4 \) has a larger F-S-F angle than \( SF_4 \) but smaller than \( H_2CSF_4 \).
- \( H_2CSF_4 \) has the largest F-S-F bond angle due to the absence of lone pairs.

### Conclusion
The correct order of equatorial F-S-F bond angles is:
\[ F-S-F \text{ angle in } H_2CSF_4 > F-S-F \text{ angle in } OSF_4 > F-S-F \text{ angle in } SF_4 \]

Thus, the final order is:
\[ H_2CSF_4 > OSF_4 > SF_4 \]