Find the sum of oxidation number of nitrogen in `(NH_(4))_(2)SO_(4), N_(2)H_(4)` and `N_(2)O_(4)`.

Explanation

To find the sum of the oxidation numbers of nitrogen in the compounds \((NH_4)_2SO_4\), \(N_2H_4\), and \(N_2O_4\), we will calculate the oxidation number of nitrogen in each compound step by step.

### Step 1: Calculate the oxidation number of nitrogen in \((NH_4)_2SO_4\)

1. **Identify the components**: In \((NH_4)_2SO_4\), we have two ammonium ions \((NH_4^+)\) and one sulfate ion \((SO_4^{2-})\).
2. **Charge of sulfate**: The sulfate ion has a charge of -2.
3. **Charge of ammonium**: Each ammonium ion has a charge of +1. Therefore, two ammonium ions contribute a total charge of +2.
4. **Set up the equation**: Let the oxidation number of nitrogen be \(x\). In \((NH_4)_2SO_4\), we have:
\[ 2x + 8 + (-2) = 0 \]
(where +8 comes from the 4 hydrogens in each ammonium ion).
5. **Solve for \(x\)**:
\[ 2x + 8 - 2 = 0 \implies 2x + 6 = 0 \implies 2x = -6 \implies x = -3 \]
Thus, the oxidation number of nitrogen in \((NH_4)_2SO_4\) is -3.

### Step 2: Calculate the oxidation number of nitrogen in \(N_2H_4\)

1. **Identify the components**: In hydrazine \(N_2H_4\), we have two nitrogen atoms and four hydrogen atoms.
2. **Set up the equation**: Let the oxidation number of nitrogen be \(x\). The equation is:
\[ 2x + 4 = 0 \]
(where +4 comes from the four hydrogens).
3. **Solve for \(x\)**:
\[ 2x + 4 = 0 \implies 2x = -4 \implies x = -2 \]
Thus, the oxidation number of nitrogen in \(N_2H_4\) is -2.

### Step 3: Calculate the oxidation number of nitrogen in \(N_2O_4\)

1. **Identify the components**: In dinitrogen tetroxide \(N_2O_4\), we have two nitrogen atoms and four oxygen atoms.
2. **Set up the equation**: Let the oxidation number of nitrogen be \(x\). The equation is:
\[ 2x + 4(-2) = 0 \]
(where -2 is the oxidation number of each oxygen).
3. **Solve for \(x\)**:
\[ 2x - 8 = 0 \implies 2x = 8 \implies x = +4 \]
Thus, the oxidation number of nitrogen in \(N_2O_4\) is +4.

### Step 4: Sum the oxidation numbers

Now, we sum the oxidation numbers of nitrogen from all three compounds:
\[ -3 \, (from \, (NH_4)_2SO_4) + (-2) \, (from \, N_2H_4) + 4 \, (from \, N_2O_4) = -3 - 2 + 4 = -1 \]

### Final Answer

The sum of the oxidation numbers of nitrogen in \((NH_4)_2SO_4\), \(N_2H_4\), and \(N_2O_4\) is \(-1\).