When `SO_(2)` is passed through acidified `K_(2)Cr_(2)O_(7)` solution, `Cr_(2)(SO_(4))_(3)` is formed. The change in oxidation state of Cr is -

Options

A. `+4` to `+2`
B. `+5` to +3
C. `+6` to `+3`
D. `+7` to `+1`

Answer

To determine the change in oxidation state of chromium when \( SO_2 \) is passed through acidified \( K_2Cr_2O_7 \) solution, we can follow these steps:

### Step 1: Identify the oxidation states in \( K_2Cr_2O_7 \)

In potassium dichromate \( K_2Cr_2O_7 \):
- Potassium (K) has an oxidation state of +1.
- Oxygen (O) has an oxidation state of -2.
- Let the oxidation state of chromium (Cr) be \( x \).

The formula can be represented as:
\[ 2(+1) + 2(x) + 7(-2) = 0 \]
This simplifies to:
\[ 2 + 2x - 14 = 0 \]
\[ 2x - 12 = 0 \]
\[ 2x = 12 \quad \Rightarrow \quad x = +6 \]

### Step 2: Identify the oxidation state in \( Cr_2(SO_4)_3 \)

In \( Cr_2(SO_4)_3 \):
- The sulfate ion \( SO_4^{2-} \) has a charge of -2.
- Therefore, the oxidation state of chromium in \( Cr_2(SO_4)_3 \) can be calculated as follows:
\[ 2(x) + 3(-2) = 0 \]
This simplifies to:
\[ 2x - 6 = 0 \]
\[ 2x = 6 \quad \Rightarrow \quad x = +3 \]

### Step 3: Calculate the change in oxidation state

Now, we can determine the change in oxidation state of chromium:
- Initial oxidation state of Cr in \( K_2Cr_2O_7 \) is +6.
- Final oxidation state of Cr in \( Cr_2(SO_4)_3 \) is +3.

The change in oxidation state is:
\[ +6 \rightarrow +3 \]
This indicates a reduction, as the oxidation state decreases.

### Step 4: Determine the number of electrons gained

The change in oxidation state from +6 to +3 means that chromium has gained 3 electrons during the reaction.

### Final Answer

The change in oxidation state of chromium is from +6 to +3, indicating a reduction of 3.

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