Oxidation state of S in `H_(2)SO_(5) and H_(2)S_(2)O_(8)` respectively are

Options

A. `+6, +6`
B. `+6, +4`
C. `+8, +7`
D. `+4, +4`

Answer

To determine the oxidation states of sulfur (S) in the compounds \( H_2SO_5 \) and \( H_2S_2O_8 \), we will follow these steps:

### Step 1: Identify the oxidation states of hydrogen and oxygen
- In compounds, hydrogen (H) typically has an oxidation state of +1.
- Oxygen (O) typically has an oxidation state of -2.

### Step 2: Write the general formula for oxidation state calculation
For a neutral compound, the sum of the oxidation states must equal zero. We can express this as:
\[ \text{Sum of oxidation states} = 0 \]

### Step 3: Calculate the oxidation state of sulfur in \( H_2SO_5 \)
1. Write the equation based on the number of atoms:
- There are 2 hydrogen atoms, 1 sulfur atom, and 5 oxygen atoms.
- Let the oxidation state of sulfur be \( x \).

The equation becomes:
\[ 2(+1) + 1(x) + 5(-2) = 0 \]
Simplifying this gives:
\[ 2 + x - 10 = 0 \]
\[ x - 8 = 0 \]
\[ x = +8 \]

Thus, the oxidation state of sulfur in \( H_2SO_5 \) is **+8**.

### Step 4: Calculate the oxidation state of sulfur in \( H_2S_2O_8 \)
1. Write the equation based on the number of atoms:
- There are 2 hydrogen atoms, 2 sulfur atoms, and 8 oxygen atoms.
- Let the oxidation state of sulfur be \( y \).

The equation becomes:
\[ 2(+1) + 2(y) + 8(-2) = 0 \]
Simplifying this gives:
\[ 2 + 2y - 16 = 0 \]
\[ 2y - 14 = 0 \]
\[ 2y = 14 \]
\[ y = +7 \]

Thus, the oxidation state of sulfur in \( H_2S_2O_8 \) is **+7**.

### Final Answer:
- The oxidation states of sulfur in \( H_2SO_5 \) and \( H_2S_2O_8 \) are +8 and +7, respectively.

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