Question

$N_2{H}_4$ reduces $I{O}_3^{-}$ to $I^{+}$ in acidic medium as

$N_2{H}_4+I{O}_3^{-}+H^{+}⟶N_2+I^{+}+H_{2}O$

$N_2{H}_4$ is oxidised to $N_2$. Both $N-$ atom in $N_2$ are from $N_2{H}_4$. It can be verified by :

• A. taking ${}^{15}N$ enriched $N_2{H}_4$,
• B. taking ${}^{2}D$ enriched $N_2{H}_4$,
• C. taking both (A) and (B)
• D. taking none of the above

Answer 1

The correct option is A taking ${}^{15}N$ enriched $N_2{H}_4$,

$N_2{H}_4+I{O}_3^{-}+H^{+}⟶N_2+I^{+}+H_{2}O$

If both $N$ are coming from hydrazine then if we use $N^{15}$ instead of$N^{14}$, then molar mass of $N_2$ found to be $30.$