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

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: Determine the oxidation state of the sulfate ion
The sulfate ion \((SO_4^{2-})\) has an overall charge of -2. In this ion, the oxidation state of sulfur (S) is +6, and each oxygen (O) has an oxidation state of -2. The calculation can be represented as:
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