To determine the oxidation number of N, we must consider the following rules:

1. The oxidation number of an element in its pure form is 0.

2. The oxidation number of a monatomic ion is equal to its charge.

3. The sum of the oxidation numbers of all atoms in a molecule or polyatomic ion is equal to the overall charge of the molecule or ion.

In N2H4, we have two nitrogen atoms and four hydrogen atoms. Since the overall charge of the molecule is 0, the sum of the oxidation numbers of N and H atoms must be 0. Let's assume the oxidation number of N is x.

2x + 4(-1) = 0

2x - 4 = 0

2x = 4

x = 2

Therefore, the oxidation number of N in N2H4 is -2.