Which of the following are correct about the reaction,
`FeS_(2)+O_(2)rarrFe_(2)O_(3)+SO_(2)`

Options

A. Eq.wt.of `FeS_(2)` is `M//11`
B. Eq.wt.of `SO_(2)` is `M//5`
C. `S` has `-2` oxidation state in `FeS_(2)`
D. `1` mole of `FeS_(2)` requires `7//4` mole of `O_(2)`

Explanation

To determine which statements are correct regarding the reaction \( \text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 \), we will analyze the oxidation states, equivalent weights, and the stoichiometry of the reaction step by step.

### Step 1: Write the balanced reaction
The reaction given is:
\[ \text{FeS}_2 + \text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 + \text{SO}_2 \]

### Step 2: Determine oxidation states
- In \( \text{FeS}_2 \):
- Let the oxidation state of Fe be \( +2 \) (since there are two sulfur atoms).
- Let the oxidation state of sulfur (S) be \( x \).
- The equation becomes:
\[ 2 + 2x = 0 \]
\[ 2x = -2 \]
\[ x = -1 \]
So, the oxidation state of S in \( \text{FeS}_2 \) is \( -1 \).

- In \( \text{Fe}_2\text{O}_3 \):
- The oxidation state of Fe is \( +3 \) (since there are three oxygen atoms, each with an oxidation state of \( -2 \)).

- In \( \text{SO}_2 \):
- The oxidation state of sulfur is \( +4 \) (since \( 2 + 4 = 0 \)).

### Step 3: Identify changes in oxidation states
- For Fe:
- From \( +2 \) in \( \text{FeS}_2 \) to \( +3 \) in \( \text{Fe}_2\text{O}_3 \) (oxidation).

- For S:
- From \( -1 \) in \( \text{FeS}_2 \) to \( +4 \) in \( \text{SO}_2 \) (oxidation).

### Step 4: Calculate equivalent weights
1. **Equivalent weight of \( \text{FeS}_2 \)**:
- The molecular weight of \( \text{FeS}_2 \) is calculated as:
\[ \text{Fe} = 55.85 \, \text{g/mol}, \, \text{S} = 32.07 \, \text{g/mol} \]
\[ \text{Molecular weight of } \text{FeS}_2 = 55.85 + 2 \times 32.07 = 119.99 \, \text{g/mol} \]
- The valence factor (n) for \( \text{FeS}_2 \) is 11 (total electrons lost).
- Therefore, the equivalent weight of \( \text{FeS}_2 \) is:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n} = \frac{119.99}{11} \]

2. **Equivalent weight of \( \text{SO}_2 \)**:
- The molecular weight of \( \text{SO}_2 \) is:
\[ \text{Molecular weight of } \text{SO}_2 = 32.07 + 2 \times 16.00 = 64.07 \, \text{g/mol} \]
- The valence factor for \( \text{SO}_2 \) is 5 (total electrons gained).
- Therefore, the equivalent weight of \( \text{SO}_2 \) is:
\[ \text{Equivalent weight} = \frac{64.07}{5} \]

### Step 5: Calculate moles of \( \text{O}_2 \) required
From the balanced equation, we find that:
- 4 moles of \( \text{FeS}_2 \) require 11 moles of \( \text{O}_2 \).
- Therefore, 1 mole of \( \text{FeS}_2 \) requires:
\[ \frac{11}{4} \text{ moles of } \text{O}_2 \]

### Conclusion
Based on the above calculations:
- The equivalent weight of \( \text{FeS}_2 \) is \( \frac{m}{11} \) (Correct).
- The equivalent weight of \( \text{SO}_2 \) is \( \frac{m}{5} \) (Correct).
- The oxidation state of S in \( \text{FeS}_2 \) is \( -1 \) (Correct).
- 1 mole of \( \text{FeS}_2 \) requires \( \frac{11}{4} \) moles of \( \text{O}_2 \) (Correct).

Thus, the correct statements are:
- A: Equivalent weight of \( \text{FeS}_2 \) is \( \frac{m}{11} \) (Correct).
- B: Equivalent weight of \( \text{SO}_2 \) is \( \frac{m}{5} \) (Correct).
- C: S has oxidation state \( -1 \) in \( \text{FeS}_2 \) (Correct).
- D: 1 mole of \( \text{FeS}_2 \) requires \( \frac{11}{4} \) moles of \( \text{O}_2 \) (Correct).