1. Assign Known Oxidation Numbers:

* Hydrogen (H): +1 (almost always)

* Oxygen (O): -2 (except in peroxides)

* Manganese (Mn): +7 (in permanganate ion, MnO₄⁻)

2. Set Up an Equation:

Since the compound is neutral, the sum of the oxidation numbers of all the atoms must equal zero. Let 'x' represent the oxidation number of nitrogen:

(1 * 4) + (x) + (+7) + (-2 * 4) = 0

3. Solve for x:

* 4 + x + 7 - 8 = 0

* x + 3 = 0

* x = -3

Therefore, the oxidation number of nitrogen (N) in NH₄MnO₄ is -3.