`S-O` bond length is maximum in:

Options

A. `SOB r_(2)`
B. `SOCl_(2)`
C. `SOF_(2)`
D. `SO(CH_(3))_(2)`

Answer

To determine which compound has the maximum S-O bond length among the given options (SOBr2, SOCl2, SOF2, and SO(CH3)2), we can analyze the influence of the surrounding atoms on the bond length.

### Step-by-Step Solution:

1. **Understanding Bond Length**: The bond length between sulfur (S) and oxygen (O) can be affected by the electronegativity of the atoms bonded to sulfur. Higher electronegativity of surrounding atoms can lead to a decrease in bond length due to increased s-character in the hybridization of the sulfur atom.

2. **Analyzing the Given Compounds**:
- **SOBr2**: Bromine (Br) is less electronegative than chlorine and fluorine, which means it will have a lesser effect on increasing the s-character of the S-O bond.
- **SOCl2**: Chlorine (Cl) is more electronegative than bromine but less than fluorine, so it will also increase the s-character but not as much as fluorine.
- **SOF2**: Fluorine (F) is the most electronegative element, which will significantly increase the s-character of the S-O bond, resulting in a shorter bond length.
- **SO(CH3)2**: Methyl groups (CH3) are not electronegative and do not pull electron density away from sulfur. This means that the s-character will not increase significantly, leading to a longer S-O bond.

3. **Comparing the Effects**:
- The presence of highly electronegative atoms (F, Cl, Br) in SOF2, SOCl2, and SOBr2 will lead to shorter S-O bonds due to increased s-character.
- In SO(CH3)2, the absence of highly electronegative atoms means that the S-O bond will not be significantly shortened, resulting in a longer bond length.

4. **Conclusion**: Among the given options, the S-O bond length is maximum in **SO(CH3)2** because the methyl groups do not increase the s-character significantly, allowing for a longer bond length compared to the other compounds with more electronegative atoms.

### Final Answer:
The S-O bond length is maximum in **SO(CH3)2**.