`NH_(2)OH` reacts with ferric sulphate as follows `:`
`2NH_(2)OH + 4Fe^(3+) rarr N_(2) O + H_(2) O + 4Fe^(2+)+4H^(+)` . The eq. wt. of `NH_(2) OH` in this reaction is `:`

Options

A. ( mol. wt. ) `//` 1
B. ( mol. wt. ) `//`2
C. ( mol. wt.) `//` 3
D. ( mol. wt. ) `//` 4

Answer

To find the equivalent weight of `NH2OH` (hydroxylamine) in the given reaction with ferric sulfate, we will follow these steps:

### Step 1: Determine the oxidation states of nitrogen in `NH2OH` and `N2O`
1. In `NH2OH`, the oxidation state of nitrogen can be calculated as follows:
- Let the oxidation state of nitrogen be \( x \).
- The two hydrogen atoms contribute \( +2 \) (since each hydrogen has an oxidation state of +1).
- The oxygen contributes \( -2 \).
- The overall charge of the molecule is 0.
- Therefore, we can set up the equation:
\[ x + 2 - 2 = 0 \implies x = -1 \]
- Thus, the oxidation state of nitrogen in `NH2OH` is \(-1\).

2. In `N2O`, the oxidation state of nitrogen can be calculated as follows:
- Let the oxidation state of nitrogen be \( y \).
- There are two nitrogen atoms, so their contribution is \( 2y \).
- The oxygen contributes \( -2 \).
- The overall charge is 0:
\[ 2y - 2 = 0 \implies 2y = 2 \implies y = +1 \]
- Thus, the oxidation state of nitrogen in `N2O` is \( +1 \).

### Step 2: Calculate the change in oxidation state
- The change in oxidation state for nitrogen in the reaction from `NH2OH` to `N2O` is:
\[ \text{Change} = +1 - (-1) = 2 \]

### Step 3: Determine the number of moles of electrons transferred
- In the reaction, each `NH2OH` molecule loses 2 electrons (as the oxidation state changes from \(-1\) to \(+1\)).
- Since there are 2 moles of `NH2OH` in the balanced equation, the total number of electrons lost is:
\[ 2 \text{ moles of } NH2OH \times 2 \text{ electrons/mole} = 4 \text{ electrons} \]

### Step 4: Calculate the n-factor
- The n-factor is defined as the number of moles of electrons lost or gained per mole of the substance.
- For `NH2OH`, the n-factor is:
\[ n = \frac{\text{Total electrons lost}}{\text{Moles of } NH2OH} = \frac{4}{2} = 2 \]

### Step 5: Calculate the molecular weight of `NH2OH`
- The molecular weight of `NH2OH` can be calculated as follows:
- Nitrogen (N): 14 g/mol
- Hydrogen (H): 1 g/mol (2 atoms contribute 2 g/mol)
- Oxygen (O): 16 g/mol
- Therefore, the molecular weight of `NH2OH` is:
\[ 14 + 2 + 16 = 32 \text{ g/mol} \]

### Step 6: Calculate the equivalent weight
- The equivalent weight of a substance is given by the formula:
\[ \text{Equivalent weight} = \frac{\text{Molecular weight}}{n \text{-factor}} \]
- Substituting the values we have:
\[ \text{Equivalent weight of } NH2OH = \frac{32 \text{ g/mol}}{2} = 16 \text{ g/equiv} \]

### Final Answer
The equivalent weight of `NH2OH` in this reaction is **16 g/equiv**.

---