In which of the following compounds,nitrogen exhibits highest oxidation state?

Options

A. `N_(3)H`
B. `NH_(2)OH`
C. `N_(2)H_(4)`
D. `NH_(3)`

Explanation

To determine the compound in which nitrogen exhibits the highest oxidation state, we will calculate the oxidation state of nitrogen in each of the given compounds: N3H, NH2OH, N2H4, and NH3.

### Step 1: Calculate the oxidation state of nitrogen in N3H
Let the oxidation state of nitrogen be \( X \).

The compound N3H consists of:
- 3 nitrogen atoms (3X)
- 1 hydrogen atom (H has an oxidation state of +1)

Setting up the equation for the neutral compound:
\[ 3X + 1 = 0 \]

Solving for \( X \):
\[ 3X = -1 \]
\[ X = -\frac{1}{3} \]

### Step 2: Calculate the oxidation state of nitrogen in NH2OH
Let the oxidation state of nitrogen be \( X \).

The compound NH2OH consists of:
- 1 nitrogen atom (X)
- 2 hydrogen atoms (2 × +1)
- 1 oxygen atom (-2)

Setting up the equation:
\[ X + 2 + (-2) = 0 \]

Solving for \( X \):
\[ X + 0 = 0 \]
\[ X = -1 \]

### Step 3: Calculate the oxidation state of nitrogen in N2H4
Let the oxidation state of nitrogen be \( X \).

The compound N2H4 consists of:
- 2 nitrogen atoms (2X)
- 4 hydrogen atoms (4 × +1)

Setting up the equation:
\[ 2X + 4 = 0 \]

Solving for \( X \):
\[ 2X = -4 \]
\[ X = -2 \]

### Step 4: Calculate the oxidation state of nitrogen in NH3
Let the oxidation state of nitrogen be \( X \).

The compound NH3 consists of:
- 1 nitrogen atom (X)
- 3 hydrogen atoms (3 × +1)

Setting up the equation:
\[ X + 3 = 0 \]

Solving for \( X \):
\[ X = -3 \]

### Summary of Oxidation States
- In N3H: \( -\frac{1}{3} \) (approximately -0.33)
- In NH2OH: \( -1 \)
- In N2H4: \( -2 \)
- In NH3: \( -3 \)

### Conclusion
The highest oxidation state of nitrogen among the given compounds is in **N3H**, where nitrogen has an oxidation state of \( -\frac{1}{3} \).