Question

What is oxidation number of each atom in $ZnN{H}_{4}P{O}_4$?

Answer 1

$ZnN{H}_{4}P{O}_4$ is a neutral compound and hence, the sum of oxidation numbers of the elements will be equal to 0.
This complex is formed of Zn element, $N{H}_4$ ions and $P{O}_4$ ions.
Zn forms ZnO in which oxygen has a -2 oxidation state, so Zn will have a +2 oxidation charge.
$N{H}_4$ forms $N{H}_{4}OH$ where OH has -1 oxidation state, so $N{H}_4$ will have a +1 oxidation state.
Now, $N{H}_4$ has +1 oxidation state and hydrogen always has +1 oxidation state (with non-metals).
So, oxidation state of $N+(+1\times 4)=+1$
Oxidation state of N = -3
Now, $P{O}_4$ forms $H_{3}P{O}_4$ which is a neutral compound. Again H has +1 charge,
So, oxidation state of $P{O}_4+(+1\times 3)=0$
Or, oxidation state of $P{O}_4=-3$
Since the oxidation state of oxygen is -2 (with less electronegative elements),
Oxidation state of $P+(-2\times 4)=-3$
Or, oxidation state of P = +5