Oxidation number of S in `(CH_(3))_(2)S` is

Options

A. 0
B. `+1`
C. `-2`
D. `+3`

Explanation

To determine the oxidation number of sulfur (S) in the compound (CH₃)₂S, we can follow these steps:

### Step 1: Identify the structure of the compound
The compound (CH₃)₂S consists of two methyl groups (CH₃) attached to a sulfur atom.

### Step 2: Assign oxidation numbers to hydrogen and carbon
- The oxidation number of hydrogen (H) is +1.
- The oxidation number of carbon (C) in the methyl group (CH₃) is -3, since each carbon is bonded to three hydrogens.

### Step 3: Calculate the total contribution from the methyl groups
Each methyl group (CH₃) contributes:
- 1 carbon (C) with an oxidation number of -3
- 3 hydrogens (H) with an oxidation number of +1 each

For one methyl group:
- Contribution from C: -3
- Contribution from H: +3 (3 × +1)

Thus, the total contribution from one methyl group is:
\[ -3 + 3 = 0 \]
Since there are two methyl groups, the total contribution from both is:
\[ 0 + 0 = 0 \]

### Step 4: Set up the equation for the oxidation number of sulfur
Let the oxidation number of sulfur be \( x \). The overall charge of the molecule is neutral (0), so we can set up the equation:
\[ \text{Total oxidation number} = 0 \]
This gives us:
\[ x + 2(-3) + 6(+1) = 0 \]
Where:
- \( x \) is the oxidation number of sulfur
- \( 2(-3) \) is the contribution from the two carbons
- \( 6(+1) \) is the contribution from the six hydrogens (3 from each methyl group)

### Step 5: Simplify the equation
Now, simplify the equation:
\[ x - 6 + 6 = 0 \]
This simplifies to:
\[ x = 0 \]

### Step 6: Conclusion
The oxidation number of sulfur in (CH₃)₂S is -2.

### Summary
The oxidation number of sulfur (S) in the compound (CH₃)₂S is -2.

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