Question

1. The oxidation of nitrogen in ( mathrm{NH} 4+ ) is ( -3 . ) The hydrogen atoms have +1
and there are four of them. And net charge on the molecule is ( +1 . ) Let
nitrogen's oxidation state be x. Therefore, ( (4 times(+1))+x=+1 . ) Hence, ( x ) is ( -3 . )
2. The oxidation number nitrogen of ( mathrm{NO} 3- ) is ( +5 . ) The oxygen atoms have ( -mathrm{x} )
and there are 3 of them. And net charge on the molecule is - ( 1 . ) Let the
oxidation state of nitrogen be x. Therefore, ( (3 times(-2))+x=-1 . ) Hence, ( x= )
+5
As you can see, both the nitrogen have different oxidation states. For numerical
purposes however, we can assume that each nitrogen is in +1 and overall
oxidation state of nitrogen +2

# IN HY NO3 Find its Onidation no.

Solution