For the reaction: `I^(-)+ClO_(3)^(-)+H_(2)SO_(4)+Cl^(-)+HSO_(4)^(-)+I_(2)` The correct statement(s) in the balanced equation is/are
(1) Stoichiometric coefficient of `HSO_(4)^(-)` is 6
(2) Iodide is oxidized.
(3) Oxidation number of chlorine changes by 5 units
(4) `H_(2)O` is one of the products

Options

A. 1, 2
B. 1,4
C. 1, 2, 3
D. 1, 2, 4

Answer

To solve the question, we need to balance the given chemical reaction and analyze the statements provided. Let's break it down step by step.

### Step 1: Write the Unbalanced Reaction
The unbalanced reaction is:
\[ I^- + ClO_3^- + H_2SO_4 + Cl^- \rightarrow HSO_4^- + I_2 + H_2O \]

### Step 2: Identify Oxidation States
- For \( I^- \), the oxidation state is -1.
- For \( ClO_3^- \), let the oxidation state of Cl be \( x \):
\[ x + 3(-2) = -1 \implies x - 6 = -1 \implies x = +5 \]
- For \( Cl^- \), the oxidation state is -1.
- In \( HSO_4^- \), the oxidation state of S is +6 (not directly needed for this question).
- For \( I_2 \), the oxidation state is 0.

### Step 3: Determine Changes in Oxidation States
- Iodine changes from -1 (in \( I^- \)) to 0 (in \( I_2 \)), indicating oxidation.
- Chlorine changes from +5 (in \( ClO_3^- \)) to -1 (in \( Cl^- \)), indicating reduction.

### Step 4: Balance the Reaction
To balance the reaction, we can use the half-reaction method or trial and error. The balanced equation is:
\[ 6 I^- + ClO_3^- + 6 H_2SO_4 \rightarrow Cl^- + 6 HSO_4^- + 3 I_2 + 3 H_2O \]

### Step 5: Analyze Each Statement
1. **Stoichiometric coefficient of \( HSO_4^- \) is 6**:
- From the balanced equation, there are indeed 6 \( HSO_4^- \) on the product side. **This statement is correct.**

2. **Iodide is oxidized**:
- Iodine goes from -1 to 0, which is an increase in oxidation state, indicating oxidation. **This statement is correct.**

3. **Oxidation number of chlorine changes by 5 units**:
- Chlorine changes from +5 to -1, which is a change of 6 units (not 5). **This statement is incorrect.**

4. **\( H_2O \) is one of the products**:
- The balanced equation shows that 3 \( H_2O \) is produced. **This statement is correct.**

### Conclusion
The correct statements are:
- (1) Stoichiometric coefficient of \( HSO_4^- \) is 6 (Correct)
- (2) Iodide is oxidized (Correct)
- (3) Oxidation number of chlorine changes by 5 units (Incorrect)
- (4) \( H_2O \) is one of the products (Correct)

### Final Answer
The correct statements are 1, 2, and 4.

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