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 FSF bond angles in the compounds \( SF_4 \), \( OSF_4 \), and \( H_2CSF_4 \), we will analyze the molecular geometry and the influence of lone pairs and hybridization on the bond angles.

### Step 1: Analyze the structure of \( SF_4 \)

1. **Molecular Geometry**: The structure of \( SF_4 \) is trigonal bipyramidal with one lone pair of electrons occupying an equatorial position.
2. **Bond Angle**: The FSF bond angle in \( SF_4 \) is approximately \( 101^\circ \).

### Step 2: Analyze the structure of \( OSF_4 \)

1. **Molecular Geometry**: The structure of \( OSF_4 \) involves an oxygen atom bonded to sulfur, which introduces a double bond (S=O). This increases the p-character of the bonds.
2. **Bond Angle**: The presence of the double bond increases the bond angle due to the increased p-character, leading to a bond angle greater than that in \( SF_4 \).

### Step 3: Analyze the structure of \( H_2CSF_4 \)

1. **Molecular Geometry**: The structure of \( H_2CSF_4 \) is also trigonal bipyramidal, but there are no lone pairs on the central carbon atom.
2. **Bond Angle**: The FSF bond angle in \( H_2CSF_4 \) is approximately \( 97^\circ \), which is smaller than that in \( SF_4 \) due to the lack of lone pair repulsion.

### Step 4: Compare the bond angles

Now we can compare the bond angles:
- \( OSF_4 \) has the largest bond angle due to the double bond increasing p-character.
- \( SF_4 \) has a bond angle of \( 101^\circ \).
- \( H_2CSF_4 \) has the smallest bond angle of \( 97^\circ \).

### Conclusion: Correct Order of Bond Angles

Thus, the correct order of equatorial FSF bond angles is:
\[ OSF_4 > SF_4 > H_2CSF_4 \]

### Final Answer
The correct order of equatorial FSF bond angles is:
\[ OSF_4 > SF_4 > H_2CSF_4 \]

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