In H2S2O8 the oxidation states of S are

Options

A. +2
B. +4
C. +6
D. -4

Answer

To determine the oxidation states of sulfur (S) in H₂S₂O₈, we can follow these steps:

### Step 1: Write the formula and identify the elements
The compound is H₂S₂O₈, which contains hydrogen (H), sulfur (S), and oxygen (O).

### Step 2: Assign known oxidation states
- The oxidation state of hydrogen (H) is +1.
- The oxidation state of oxygen (O) in peroxides is -1, and in other cases, it is -2. In H₂S₂O₈, there are two oxygen atoms in a peroxide linkage and six oxygen atoms with an oxidation state of -2.

### Step 3: Set up the equation
Let the oxidation state of sulfur be represented as X. The overall charge of the molecule is neutral (0). Therefore, we can set up the equation based on the contributions of each element:
\[ 2X + 2(+1) + 6(-2) + 2(-1) = 0 \]

### Step 4: Simplify the equation
Now, simplify the equation:
- The contribution from hydrogen: \( 2(+1) = +2 \)
- The contribution from six oxygen atoms: \( 6(-2) = -12 \)
- The contribution from two peroxide oxygen atoms: \( 2(-1) = -2 \)

Putting it all together:
\[ 2X + 2 - 12 - 2 = 0 \]
\[ 2X - 12 = 0 \]

### Step 5: Solve for X
Now, solve for X:
\[ 2X = 12 \]
\[ X = 6 \]

### Step 6: Conclusion
The oxidation state of sulfur (S) in H₂S₂O₈ is +6.

### Summary of oxidation states:
- In H₂S₂O₈, the oxidation states of sulfur are +6.