Acidified potassium dichromate oxidises
I. iodides to iodine
II.sulphides to sulphur
III. Tin (IV)to tin (II)
The appropriate option with correct choices are

Options

A. Only I
B. II and III
C. I and III
D. I and II

Answer

To solve the question regarding the oxidation reactions involving acidified potassium dichromate, we will analyze each statement provided in the question step by step.

### Step 1: Understand the role of acidified potassium dichromate
Acidified potassium dichromate (K2Cr2O7 in acidic medium) is a strong oxidizing agent. It can oxidize certain substances, and we need to identify which of the given substances are oxidized by it.

### Step 2: Analyze the first statement: "Iodides to iodine"
Iodides (I⁻) can be oxidized to iodine (I2) by potassium dichromate. The oxidation reaction can be represented as:
\[ 2 I^- \rightarrow I_2 + 2 e^- \]
This statement is **correct**.

### Step 3: Analyze the second statement: "Sulphides to sulphur"
Sulphides (like H2S) can also be oxidized by potassium dichromate to form elemental sulfur (S). The reaction can be represented as:
\[ H_2S \rightarrow S + 2 H^+ + 2 e^- \]
This statement is also **correct**.

### Step 4: Analyze the third statement: "Tin (IV) to tin (II)"
Tin (IV) (Sn⁴⁺) being reduced to tin (II) (Sn²⁺) indicates a reduction process, not oxidation. The reaction can be represented as:
\[ Sn^{4+} + 2 e^- \rightarrow Sn^{2+} \]
Since this is a reduction process, this statement is **incorrect**.

### Step 5: Conclusion
Based on the analysis:
- Statement I (iodides to iodine) is correct.
- Statement II (sulphides to sulphur) is correct.
- Statement III (tin (IV) to tin (II)) is incorrect.

Thus, the appropriate option with the correct choices is the one that includes the first two statements but not the third.

### Final Answer
The correct answer is **Option 4** (I and II are correct).

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