In which of the following compounds nitrogen exhibits highest oxidation state

Options

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

Explanation

To determine in which of the given compounds nitrogen exhibits the highest oxidation state, we need to analyze the oxidation states of nitrogen in each compound.

### Step-by-Step Solution:

1. **Identify the Compounds:**
The compounds given are:
- \( NH_3 \)
- \( N_3H \)
- \( N_2OH \)
- \( N_2H_4 \)

2. **Calculate the Oxidation State of Nitrogen in Each Compound:**

- **For \( NH_3 \):**
- Hydrogen has an oxidation state of +1.
- Let the oxidation state of nitrogen be \( x \).
- The equation is: \( x + 3(+1) = 0 \)
- Solving gives: \( x + 3 = 0 \) → \( x = -3 \)

- **For \( N_3H \):**
- Hydrogen has an oxidation state of +1.
- Let the oxidation state of nitrogen be \( x \).
- The equation is: \( 3x + 1 = 0 \)
- Solving gives: \( 3x = -1 \) → \( x = -\frac{1}{3} \)

- **For \( N_2OH \):**
- Hydrogen has an oxidation state of +1 and oxygen has -2.
- Let the oxidation state of nitrogen be \( x \).
- The equation is: \( 2x + (-2) + 1 = 0 \)
- Solving gives: \( 2x - 1 = 0 \) → \( 2x = 1 \) → \( x = \frac{1}{2} \)

- **For \( N_2H_4 \):**
- Hydrogen has an oxidation state of +1.
- Let the oxidation state of nitrogen be \( x \).
- The equation is: \( 2x + 4(+1) = 0 \)
- Solving gives: \( 2x + 4 = 0 \) → \( 2x = -4 \) → \( x = -2 \)

3. **Summarize the Oxidation States:**
- \( NH_3 \): -3
- \( N_3H \): -1/3
- \( N_2OH \): +1/2
- \( N_2H_4 \): -2

4. **Determine the Highest Oxidation State:**
Among the calculated oxidation states, the highest oxidation state of nitrogen is in \( N_2OH \), which is +1/2.

### Conclusion:
The compound in which nitrogen exhibits the highest oxidation state is \( N_2OH \).