The answer to each of the following questions is a non- negative integer.
Find the oxidation state of N-atom in `[N_2H_6]SO_4`

Answer

To find the oxidation state of the nitrogen atom in the compound \([N_2H_6]SO_4\), we can follow these steps:

### Step 1: Identify the components of the compound
The compound consists of nitrogen (N), hydrogen (H), sulfur (S), and oxygen (O). We need to determine the oxidation states of these elements in the compound.

### Step 2: Assign known oxidation states
- The oxidation state of hydrogen (H) is typically +1.
- The oxidation state of oxygen (O) is typically -2.
- The oxidation state of sulfur (S) in sulfuric acid \((SO_4^{2-})\) is +6.

### Step 3: Set up the equation
Let the oxidation state of nitrogen (N) be \(X\). In the compound \([N_2H_6]SO_4\), we have:
- 2 nitrogen atoms contributing \(2X\)
- 6 hydrogen atoms contributing \(6 \times (+1) = +6\)
- 1 sulfur atom contributing \(+6\)
- 4 oxygen atoms contributing \(4 \times (-2) = -8\)

The overall charge of the compound is neutral (0), so we can set up the equation:
\[ 2X + 6 + 6 - 8 = 0 \]

### Step 4: Simplify the equation
Simplifying the equation gives:
\[ 2X + 4 = 0 \]

### Step 5: Solve for \(X\)
Now, we can solve for \(X\):
\[ 2X = -4 \\ X = -2 \]

### Conclusion
The oxidation state of the nitrogen atom in \([N_2H_6]SO_4\) is \(-2\).

---