The oxidation number of nitrogen in `NH_(2)OH` is :

Options

A. Zero
B. `+1`
C. `-1`
D. `-2`

Answer

To find the oxidation number of nitrogen in the compound NH₂OH (hydroxylamine), we can follow these steps:

### Step 1: Identify the oxidation states of known elements
- The oxidation number of hydrogen (H) is +1.
- The oxidation number of oxygen (O) is -2.

### Step 2: Set up the equation
Let the oxidation number of nitrogen (N) be represented as \( x \). In the molecule NH₂OH, there are:
- 2 hydrogen atoms contributing \( 2 \times (+1) = +2 \)
- 1 oxygen atom contributing \( -2 \)

### Step 3: Write the equation based on the overall charge
Since NH₂OH is a neutral molecule, the sum of the oxidation numbers must equal 0. Therefore, we can write the equation:
\[ x + 2 (+1) + (-2) = 0 \]

### Step 4: Simplify the equation
Substituting the known values into the equation gives:
\[ x + 2 - 2 = 0 \]
This simplifies to:
\[ x = 0 \]

### Step 5: Solve for \( x \)
From the equation, we find:
\[ x = -1 \]

### Conclusion
The oxidation number of nitrogen in NH₂OH is -1.