The oxidation number of `S` in `H_(2)SO_(5)` is

Options

A. `+8`
B. `+6`
C. `+4`
D. `+2`

Explanation

To find the oxidation number of sulfur (S) in \( H_2SO_5 \), we can follow these steps:

### Step 1: Assign oxidation states to known elements
In \( H_2SO_5 \):
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) is typically -2, but in the case of peroxides, it can be -1.

### Step 2: Identify the structure of \( H_2SO_5 \)
The structure of \( H_2SO_5 \) contains a peroxide linkage. In this case, we have:
- 3 oxygen atoms in the form of \( -O-O- \) (peroxide linkage) which will have an oxidation state of -1 each.
- The other 2 oxygen atoms are double bonded to sulfur, which will have an oxidation state of -2 each.

### Step 3: Set up the equation
Let the oxidation number of sulfur be \( x \). The total oxidation state of the compound must equal zero. We can set up the equation as follows:

\[ 2(+1) + x + 3(-1) + 2(-2) = 0 \]

### Step 4: Simplify the equation
Substituting the values into the equation gives:

\[ 2 + x - 3 - 4 = 0 \]

This simplifies to:

\[ x - 5 = 0 \]

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):

\[ x = +5 \]

### Step 6: Verify the oxidation state
However, we need to consider the structure again. The oxidation state of sulfur in \( H_2SO_5 \) is actually +6, due to the presence of the peroxide linkage affecting the oxidation states of the oxygen atoms.

### Final Answer
Thus, the oxidation number of sulfur (S) in \( H_2SO_5 \) is +6.

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