In the compound $N_2{H}_4{O}_2$ (often written as $N_2{H}_4{O}_2$), nitrogen occurs in two different ions: $N_1^{+}$ in $N_1{H}_4$ and $N_1^{-}$ in $N_1{O}_2$. What are the oxidation numbers of nitrogen in these two ions (in the order $N_1{H}_4$ then $N_1{O}_2$)?

For NO\_2\^-, assign oxygen an oxidation number of -2 (a common oxidation state for oxygen). Let the oxidation number of nitrogen be y. Set up the equation: y + 2(-2) = -1, then solve for y to find the oxidation number of nitrogen in NO\_2\^-.