In `NH_(4)NO_(2)`, the oxidation number of nitrogen will be

Options

A. `+3`
B. `+5`
C. `-3` and +3
D. `+3` and +5

Explanation

To determine the oxidation number of nitrogen in \( NH_4NO_2 \), we can break the compound into its constituent ions and analyze each part separately.

### Step 1: Identify the ions in \( NH_4NO_2 \)
The compound \( NH_4NO_2 \) can be separated into two parts:
- \( NH_4^+ \) (ammonium ion)
- \( NO_2^- \) (nitrite ion)

### Step 2: Calculate the oxidation number of nitrogen in \( NH_4^+ \)
Let the oxidation number of nitrogen in \( NH_4^+ \) be \( X \).
- The oxidation number of hydrogen is \( +1 \).
- There are 4 hydrogen atoms, contributing \( 4 \times (+1) = +4 \).

Setting up the equation for the ammonium ion:
\[ X + 4(+1) = +1 \]
\[ X + 4 = +1 \]
\[ X = +1 - 4 \]
\[ X = -3 \]

Thus, the oxidation number of nitrogen in \( NH_4^+ \) is \( -3 \).

### Step 3: Calculate the oxidation number of nitrogen in \( NO_2^- \)
Let the oxidation number of nitrogen in \( NO_2^- \) be \( Y \).
- The oxidation number of oxygen is \( -2 \).
- There are 2 oxygen atoms, contributing \( 2 \times (-2) = -4 \).

Setting up the equation for the nitrite ion:
\[ Y + 2(-2) = -1 \]
\[ Y - 4 = -1 \]
\[ Y = -1 + 4 \]
\[ Y = +3 \]

Thus, the oxidation number of nitrogen in \( NO_2^- \) is \( +3 \).

### Step 4: Summarize the oxidation numbers
In \( NH_4NO_2 \):
- The oxidation number of nitrogen in \( NH_4^+ \) is \( -3 \).
- The oxidation number of nitrogen in \( NO_2^- \) is \( +3 \).

### Final Answer
The oxidation numbers of nitrogen in \( NH_4NO_2 \) are \( -3 \) and \( +3 \).

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