When acidified `K_(2)Cr_(2)O_(7)` solution is added to `Sn^(2+)` salts then `Sn^(2+)` changes to

Options

A. Sn
B. `Sn^(3+)`
C. `Sn^(4+)`
D. `Sn^(+)`

Explanation

To solve the question regarding the reaction of acidified \( K_2Cr_2O_7 \) with \( Sn^{2+} \) salts, we will follow these steps:

### Step 1: Identify the Reactants
We have two reactants:
1. Acidified \( K_2Cr_2O_7 \) (potassium dichromate in acidic solution)
2. \( Sn^{2+} \) salts (tin(II) ions)

### Step 2: Understand the Reaction
In an acidic medium, \( K_2Cr_2O_7 \) acts as an oxidizing agent. It will oxidize \( Sn^{2+} \) ions to a higher oxidation state.

### Step 3: Write the Chemical Equation
The balanced chemical equation for the reaction can be written as:
\[ Cr_2O_7^{2-} + 4H^+ + 3Sn^{2+} \rightarrow 2Cr^{3+} + 3Sn^{4+} + 7H_2O \]

### Step 4: Determine the Changes in Oxidation States
- For chromium in \( Cr_2O_7^{2-} \):
- Let the oxidation state of chromium be \( x \).
- The equation for the oxidation state is \( 2x + 7(-2) = -2 \).
- Solving gives \( x = +6 \) for chromium in \( Cr_2O_7^{2-} \).

- For chromium in \( Cr^{3+} \):
- The oxidation state is +3.

Thus, chromium is reduced from +6 to +3.

- For tin:
- \( Sn^{2+} \) is oxidized to \( Sn^{4+} \).

### Step 5: Conclusion
The \( Sn^{2+} \) ions are oxidized to \( Sn^{4+} \) ions when reacted with acidified \( K_2Cr_2O_7 \).

### Final Answer
When acidified \( K_2Cr_2O_7 \) solution is added to \( Sn^{2+} \) salts, \( Sn^{2+} \) changes to \( Sn^{4+} \).

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