The oxidation state of sulphur in `H_(2)SO_(5)` and chromium in `K_(2)Cr_(2)O_(7)` respectively is :-

Options

A. 8, 6
B. 4, 6
C. 8, 8
D. 6, 6

Answer

To find the oxidation states of sulfur in \( H_2SO_5 \) and chromium in \( K_2Cr_2O_7 \), we will follow these steps:

### Step 1: Determine the oxidation state of sulfur in \( H_2SO_5 \)

1. **Assign oxidation states to known elements**:
- Hydrogen (H) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation**:
Let the oxidation state of sulfur (S) be \( x \).
The formula for \( H_2SO_5 \) consists of:
- 2 hydrogen atoms: \( 2 \times (+1) = +2 \)
- 1 sulfur atom: \( 1 \times x = x \)
- 5 oxygen atoms: \( 5 \times (-2) = -10 \)

The total charge of the molecule is 0, so we can write the equation:
\[ 2 + x - 10 = 0 \]

3. **Solve for \( x \)**:
Rearranging the equation gives:
\[ x - 8 = 0 \implies x = +8 \]

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

### Step 2: Determine the oxidation state of chromium in \( K_2Cr_2O_7 \)

1. **Assign oxidation states to known elements**:
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.

2. **Set up the equation**:
Let the oxidation state of chromium (Cr) be \( y \).
The formula for \( K_2Cr_2O_7 \) consists of:
- 2 potassium atoms: \( 2 \times (+1) = +2 \)
- 2 chromium atoms: \( 2 \times y = 2y \)
- 7 oxygen atoms: \( 7 \times (-2) = -14 \)

The total charge of the molecule is 0, so we can write the equation:
\[ 2 + 2y - 14 = 0 \]

3. **Solve for \( y \)**:
Rearranging the equation gives:
\[ 2y - 12 = 0 \implies 2y = 12 \implies y = +6 \]

Thus, the oxidation state of chromium in \( K_2Cr_2O_7 \) is **+6**.

### Final Answer:
The oxidation states are:
- Sulfur in \( H_2SO_5 \): **+8**
- Chromium in \( K_2Cr_2O_7 \): **+6**