Find the valency factor (n) for `NH_(2)OH` in given reaction :
`Fe^(3+)+NH_(2)OHrarrFe^(2+)+N_(2)O+H^(+)+H_(2)O`

Explanation

To find the valency factor (n) for `NH2OH` in the reaction:

`Fe^(3+) + NH2OH → Fe^(2+) + N2O + H^+ + H2O`

we need to determine the change in oxidation state of the nitrogen in `NH2OH` as it converts to `N2O`.

### Step-by-Step Solution:

1. **Identify the oxidation states:**
- In `NH2OH`, we need to find the oxidation state of nitrogen (N).
- The formula for calculating oxidation states is:
\[ \text{Sum of oxidation states} = 0 \text{ (for neutral compounds)} \]
- For `NH2OH`, let the oxidation state of nitrogen be \( x \):
- The two hydrogens contribute \( +1 \times 2 = +2 \)
- The hydroxyl group (OH) contributes \( -1 \)
- Therefore, we have:
\[ x + 2 - 1 = 0 \]
Simplifying gives:
\[ x + 1 = 0 \implies x = -1 \]

2. **Determine the oxidation state in the product (N2O):**
- In `N2O`, there are two nitrogen atoms. Let the oxidation state of nitrogen be \( y \):
- The oxygen contributes \( -2 \)
- Thus, we have:
\[ 2y - 2 = 0 \]
Simplifying gives:
\[ 2y = 2 \implies y = +1 \]

3. **Calculate the change in oxidation state:**
- The change in oxidation state for nitrogen from `NH2OH` to `N2O` is:
\[ \text{Change} = \text{Final state} - \text{Initial state} = +1 - (-1) = +1 + 1 = +2 \]

4. **Determine the valency factor (n):**
- The valency factor (n) is equal to the total change in oxidation state per molecule of the reducing agent (in this case, `NH2OH`).
- Since the change is +2, the valency factor \( n \) for `NH2OH` is:
\[ n = 2 \]

### Final Answer:
The valency factor (n) for `NH2OH` in the given reaction is **2**.