Question

Why the oxidation number of ( H ) is -1 in
1) N2H4
2) CaH2
while ( N ) is different from ( Ca ) ?

Answer 1

Becasue H is in group one and it loose electron to become octet. so it has -1 one charge

Answer 2

But it is -1 when bonded with a non metal (N) , and it's also -1 when bonded with a metal ( Ca )

Answer 3

Here is a simplified set of oxidation state rules.
http://www.chemteam.info/Redox/Redox-Rules.html

You must understand that oxidation rules are arbitrary rules mostly for bookkeeping. When H is with a metal, such as CaH2, I always give it an oxidation state of -1 because it's obvious this is a hydride. In N2H4, technically we may call this a hydride,too, but I always give H a +1 in anything other than a metal hydride. So I would give H in N2H4 a +1 each which makes N = -2 each. You will find in balancing redox equations that it simplifies the problem if H can be given +1 each and O a -2 each. So in NH3 I would call N -3, etc.

Answer 4

But there are other sources that gave -1 to H in N2H4 abd +2 to N !