# Chemistry

Ok I think I figured this out but im not sure if its right please help. Thanks(=

Determine the oxidation number for nitrogen in each of the following

a)NH3 = -3
b)NO2-1 = 2
c)N2= 2
d)NO2Cl= 2
e)N2H4= 2

The first one is right.
All must add to zero for a compound OR to the ionic charge.
N= -3; H =+1 each for +3. +3 and -3 = 0 so NH3 is ok.
NO2^-1.
You have N = 2. O is -2 each for -4. +2 and -4 = -2 and that ISN'T the ionic charge of -1 so N=2 is not right.

All elements in the free state have an oxidation state of zero. So c is 0.

You need to redo b, c, d, and e.

I don't really understand. The only one I could kinda change was b and I got -1 because O loses an electron. Im really confused

is d -2 and e -4?

It isn't a matter of what loses and what gains.
For NO2^-, O is -2 oxidation state (it is in group VIA or 16 depending upon the system you are using). We want to leave a -1 charge on the ion; therefore, we make N + 3 so that +3 for N + (2*-2 for O) = +3-4=-1. Voila! So N is +3 in nitrite ion.
NOCl. This is a compound, so we want everything to add to zero.
O is -2. Cl is -1. total - charge is -3 so N must be +3 to add to zero. (I don't remember if you had NOCl or another one. If not NOCl, I'll repost it.).

NO2Cl. A compound so it must add to zero. Oxygen is -2. We have 2*-2=-4. Cl is -1. Total - charge is -4 and -1 = -5 so N must be +5 to add to zero.

N2H4. H is +1. There are 4 of them so 4*1=+4. Therefore, N must be -4 for the 2 atoms or -4/2 = -2 for each N.

b)NO2-1
d)NO2Cl
e)N2H4

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