Acidified potassium dichromate is treated with hydrogen sulphide. In the reaction the oxidation number of chromium

Options

A. increases from +3 to +6
B. decreases from +6 to +3
C. remains unchanged
D. decreases from +6 to +2

Answer

To solve the question regarding the change in the oxidation number of chromium when acidified potassium dichromate is treated with hydrogen sulfide, we can follow these steps:

### Step 1: Identify the Reactants and Products
- The reactant is acidified potassium dichromate, which is represented as \( K_2Cr_2O_7 \) (potassium dichromate) in the presence of sulfuric acid (\( H_2SO_4 \)).
- The other reactant is hydrogen sulfide, \( H_2S \).
- The products of the reaction are potassium sulfate (\( K_2SO_4 \)), chromium sulfate (\( Cr_2(SO_4)_3 \)), water (\( H_2O \)), and sulfur (\( S \)).

### Step 2: Write the Balanced Chemical Equation
The balanced chemical reaction can be represented as:
\[ K_2Cr_2O_7 + 3H_2S + 4H_2SO_4 \rightarrow Cr_2(SO_4)_3 + 3K_2SO_4 + 7H_2O + 3S \]

### Step 3: Determine the Oxidation State of Chromium in Reactants
- In \( K_2Cr_2O_7 \):
- Potassium (\( K \)) has an oxidation state of +1.
- Oxygen (\( O \)) has an oxidation state of -2.
- Let the oxidation state of chromium be \( x \).

The equation for the oxidation state in \( K_2Cr_2O_7 \) is:
\[ 2(+1) + 2x + 7(-2) = 0 \]
Simplifying this:
\[ 2 + 2x - 14 = 0 \implies 2x - 12 = 0 \implies 2x = 12 \implies x = +6 \]
Thus, the oxidation state of chromium in \( K_2Cr_2O_7 \) is +6.

### Step 4: Determine the Oxidation State of Chromium in Products
- In \( Cr_2(SO_4)_3 \):
- The sulfate ion \( SO_4^{2-} \) has an oxidation state of -2.
- Let the oxidation state of chromium be \( y \).

The equation for the oxidation state in \( Cr_2(SO_4)_3 \) is:
\[ 2y + 3(-2) = 0 \]
Simplifying this:
\[ 2y - 6 = 0 \implies 2y = 6 \implies y = +3 \]
Thus, the oxidation state of chromium in \( Cr_2(SO_4)_3 \) is +3.

### Step 5: Analyze the Change in Oxidation State
- The oxidation state of chromium changes from +6 in \( K_2Cr_2O_7 \) to +3 in \( Cr_2(SO_4)_3 \).
- Therefore, the oxidation number of chromium decreases from +6 to +3.

### Conclusion
The correct answer is that the oxidation number of chromium decreases from +6 to +3.

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