One mole of N_(2)H_4loses 10 mole electrons to from a new compound X Assuming that at the N_2 appears in new compound There is no change in oxidation state of H. What is tha oxidation state on N in X ?

One mole of N_(2)H_4loses 10 mole electrons to from a new compound X A

1.	One mole of N_(2)H_4loses 10 mole electrons to from a new compound X Assuming that at the N_2 appears in new compound There is no change in oxidation state of H. What is tha oxidation state on N in X ?
Answer» Solution :`N_2H_4rarr(X) +10 e^(-)` `:.`X CONTAINS all N-atoms `:. (N^(-2))_(2) rarr (2N)^(a) +10E^(-)` Therefore `2a-(-4) =10 IMPLIES a =+3` (or) `NO_3^(-) +7e^(-) rarr 1//2 N_2 H_4` Therefore , number of ELECTRONS involved in the reduction of `NO_3^(-)`ions is 7 .