Oxidation number ofnitrogen in `(NH_4)_2SO_4` is

Options

A. -1/3
B. -1
C. +l
D. -3

Explanation

To find the oxidation number of nitrogen in the compound \((NH_4)_2SO_4\), we can follow these steps:

### Step 1: Identify the components of the compound
The compound \((NH_4)_2SO_4\) consists of two ammonium ions \((NH_4^+)\) and one sulfate ion \((SO_4^{2-})\).

### Step 2: Assign oxidation states
- The oxidation state of hydrogen (H) is +1.
- Let the oxidation state of nitrogen (N) be \(X\).
- The oxidation state of sulfur (S) in sulfate \((SO_4^{2-})\) is +6, and the oxidation state of oxygen (O) is -2.

### Step 3: Calculate the total charge from the ammonium ions
Each ammonium ion \((NH_4^+)\) has a charge of +1. Since there are two ammonium ions:
\[ \text{Total charge from } (NH_4^+) = 2 \times (+1) = +2 \]

### Step 4: Calculate the total charge from the sulfate ion
The sulfate ion \((SO_4^{2-})\) has a charge of -2.

### Step 5: Set up the equation for the overall charge
The overall charge of the compound is neutral (0). Therefore, we can set up the equation as follows:
\[ \text{Total charge from ammonium ions} + \text{Charge from sulfate} = 0 \]
Substituting the values we have:
\[ 2 \times (+1) + (-2) + 2 \times X + 8 \times (+1) = 0 \]

### Step 6: Simplify the equation
We know that there are 8 hydrogen atoms in total from the two ammonium ions:
\[ 2 + (-2) + 2X + 8 = 0 \]
This simplifies to:
\[ 2X + 8 = 0 \]

### Step 7: Solve for \(X\)
Rearranging the equation gives:
\[ 2X = -8 \]
\[ X = -4 \]

### Step 8: Conclusion
The oxidation number of nitrogen in \((NH_4)_2SO_4\) is \(-3\).