STATEMENT-1 : `[Fe(CN)_(6)]^(4-) to Fe^(3+)+CO_(2)+NO_(3)^(-)`, the equivalent mass of reactant is 3.74.
STATEMENT-2 : "Equivalent mass of reactant" = ("Mol.mass")/(61)`.

Options

A. If both the statement are TRUE and STATEMENT -2 is the correct explanation of STATEMENT-1
B. If both the statement are TRUE but STATEMENT-2 is NOT the correct explanation of STATEMENT-1
C. If STATEMENT- is 1 TRUE and STATEMENT-2 is FALSE
D. If STATEMENT-1 is FALSE and STATEMENT-2 is TRUE

Answer

To solve the question, we will analyze both statements step by step.

### Step 1: Understanding the Reaction
We have the reaction:
\[ [Fe(CN)_6]^{4-} \rightarrow Fe^{3+} + CO_2 + NO_3^{-} \]

### Step 2: Determine the Oxidation States
1. **Iron (Fe)** in \([Fe(CN)_6]^{4-}\):
- The oxidation state of Fe in \([Fe(CN)_6]^{4-}\) is +2.
- It changes to +3 in \(Fe^{3+}\).
- This indicates that Fe undergoes oxidation by losing 1 electron.

2. **Carbon (C)** in cyanide (\(CN^{-}\)):
- The oxidation state of nitrogen (N) is -3.
- Let the oxidation state of carbon be \(x\).
- The total charge of the cyanide ion is -1:
\[ x - 3 = -1 \implies x = +2 \]
- Carbon goes from +2 to +4 in \(CO_2\), indicating oxidation.
- There are 6 carbon atoms, and each undergoes a change of +2:
\[ \text{Total change for carbon} = 6 \times 2 = 12 \]

3. **Nitrogen (N)** in cyanide:
- The oxidation state of nitrogen in cyanide is -3 and changes to +5 in \(NO_3^{-}\).
- The change in oxidation state for nitrogen is:
\[ -3 \rightarrow +5 \implies \text{Change} = 8 \]
- There are 6 nitrogen atoms, so:
\[ \text{Total change for nitrogen} = 6 \times 8 = 48 \]

### Step 3: Calculate the Total Change in Oxidation State (N factor)
- Total change in oxidation state (N factor):
\[ N = 1 \text{ (for Fe)} + 12 \text{ (for C)} + 48 \text{ (for N)} = 61 \]

### Step 4: Calculate the Molar Mass of \([Fe(CN)_6]^{4-}\)
- Molar mass calculation:
- Atomic mass of Fe = 56 g/mol
- Atomic mass of C = 12 g/mol (6 atoms)
- Atomic mass of N = 14 g/mol (6 atoms)

\[ \text{Molar mass} = 56 + (6 \times 12) + (6 \times 14) = 56 + 72 + 84 = 212 \text{ g/mol} \]

### Step 5: Calculate the Equivalent Mass
- Equivalent mass formula:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{N \text{ factor}} = \frac{212}{61} \approx 3.475 \text{ g/equiv} \]

### Step 6: Evaluate the Statements
- **Statement 1**: The equivalent mass of the reactant is 3.74. (This is incorrect as we calculated it to be approximately 3.475).
- **Statement 2**: Equivalent mass of the reactant = Molar mass / 61. (This is correct).

### Conclusion
- Statement 1 is false.
- Statement 2 is true.