The oxidation number of .N. in `NH_2 OH` is

Options

A. `1//3`
B. 0
C. `-1`
D. 1

Explanation

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

### Step 1: Assign Variables
Let the oxidation number of nitrogen be represented as \( x \).

### Step 2: Identify Oxidation States of Other Atoms
In the compound NH₂OH:
- Each hydrogen (H) has an oxidation number of +1. Since there are 2 hydrogen atoms attached to nitrogen and 1 hydrogen atom attached to oxygen, the total contribution from hydrogen is:
\[ 2 \times (+1) + 1 \times (+1) = +2 + +1 = +3 \]
- The oxygen (O) has an oxidation number of -2.

### Step 3: Set Up the Equation
Since the compound is neutral (no overall charge), 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 values we have:
\[ x + 2 - 2 = 0 \]
This simplifies to:
\[ x + 1 = 0 \]

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

### Conclusion
The oxidation number of nitrogen in hydroxylamine (NH₂OH) is \(-1\).

### Final Answer
The oxidation number of nitrogen in NH₂OH is \(-1\).

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