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 $N_1{O}_2$)?

• A. $+3$ and $-3$
• B. $-3$ and $+3$
• C. $-3$ and $+1$
• D. $-1$ and $+3$

Answer: (B)