In N2H5Cl, the oxidation number of nitrogen can be determined as follows:

Let the oxidation number of nitrogen be x.

Since the oxidation number of hydrogen is +1 and that of chlorine is -1, we can write:

2x + 5(+1) + 1(-1) = 0

2x + 5 - 1 = 0

2x + 4 = 0

2x = -4

x = -2

Therefore, the oxidation number of nitrogen in N2H5Cl is -2.