Oxidation number of nitrogen is highest in :

Options

A. `N_3H`
B. `N_2O_4`
C. `NH_4OH`
D. `NH_3`

Explanation

To determine the oxidation number of nitrogen in the given compounds, we will analyze each compound step by step.

### Step 1: Analyze the first compound, N3H
1. Let the oxidation number of nitrogen be \( X \).
2. The total oxidation state of the molecule must equal zero since it is neutral.
3. The equation can be set up as:
\[ 3X + 1 = 0 \]
4. Solving for \( X \):
\[ 3X = -1 \implies X = -\frac{1}{3} \]
5. Therefore, the oxidation number of nitrogen in N3H is \(-\frac{1}{3}\).

### Step 2: Analyze the second compound, N2O4
1. Let the oxidation number of nitrogen be \( X \).
2. The total oxidation state of the molecule must equal zero.
3. The equation can be set up as:
\[ 2X + 4(-2) = 0 \]
4. Solving for \( X \):
\[ 2X - 8 = 0 \implies 2X = 8 \implies X = 4 \]
5. Therefore, the oxidation number of nitrogen in N2O4 is \( +4 \).

### Step 3: Analyze the third compound, NH4OH
1. Let the oxidation number of nitrogen be \( X \).
2. The total oxidation state of the molecule must equal zero.
3. The equation can be set up as:
\[ X + 4(+1) + 1 + 4(-2) = 0 \]
4. Simplifying gives:
\[ X + 4 - 8 + 1 = 0 \implies X - 3 = 0 \implies X = -3 \]
5. Therefore, the oxidation number of nitrogen in NH4OH is \( -3 \).

### Step 4: Analyze the fourth compound, NH3
1. Let the oxidation number of nitrogen be \( X \).
2. The total oxidation state of the molecule must equal zero.
3. The equation can be set up as:
\[ X + 3(+1) = 0 \]
4. Solving for \( X \):
\[ X + 3 = 0 \implies X = -3 \]
5. Therefore, the oxidation number of nitrogen in NH3 is \( -3 \).

### Conclusion
After analyzing all four compounds, we find:
- N3H: Oxidation number of nitrogen = \(-\frac{1}{3}\)
- N2O4: Oxidation number of nitrogen = \( +4\)
- NH4OH: Oxidation number of nitrogen = \(-3\)
- NH3: Oxidation number of nitrogen = \(-3\)

The highest oxidation number of nitrogen is found in **N2O4**, which is \( +4 \).

### Final Answer
The compound with the highest oxidation number of nitrogen is **N2O4**.

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