for each element
4 Answers

N 3
H +1
Cr +6
O 2
The best way to find answers to such problems is to remember
i] the sum of the oxidation numbers in a compound equals zero
ii] the sum of the oxidation numbers in an ions equals the charge on the ion
iii] in normal chemistry oxygen has fixed oxidation number of 2 and hydrogen +1 [except in hydrides]
The normal way is to set up a smple algebra expression for each ion
CrO4 2 Cr + (4 x 2) = 2 hence Cr = +6
NH4+ N + 4 = +1 hence N = 3

H is almost always 1+, and O is almost always 2. It’s easiest if you break the formula into its component ions, which are NH4^+ and and CrO4^2. For the N in ammonia, there are four H+’s, and the overall charge is +1, so the N must be 3. Similarly, in chromate ion, there are 4 O’s, so a total of 8. The overall charge is 2, so the Cr must be +6.

Nh4 2cro4