Oxidation number of `N` in `(NH_(4))_(2)SO_(4)` is

Options

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

Explanation

To find the oxidation number of nitrogen (N) 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: Write the oxidation state equation for \(NH_4^+\)
In the ammonium ion \((NH_4^+)\), we need to find the oxidation number of nitrogen (N). Let’s denote the oxidation number of nitrogen as \(X\).

### Step 3: Assign oxidation numbers to other elements
- The oxidation number of hydrogen (H) is +1.
- The total charge of the ammonium ion is +1.

### Step 4: Set up the equation
For the ammonium ion \((NH_4^+)\):
\[ X + 4(+1) = +1 \]
Where \(4(+1)\) accounts for the four hydrogen atoms.

### Step 5: Solve for \(X\)
Now, we can solve for \(X\):
\[ X + 4 = +1 \]
\[ X = +1 - 4 \]
\[ X = -3 \]

### Step 6: Conclusion
The oxidation number of nitrogen in \((NH_4)_2SO_4\) is \(-3\).

### Final Answer
The oxidation number of nitrogen in \((NH_4)_2SO_4\) is \(-3\).

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