One mole of $${N_2}{H_4}$$  loses ten moles of electrons to form a new compound $$Y$$.  Assuming that all the nitrogen appears in the new compound, what is the oxidation state of nitrogen in $$Y?$$ ( There is no change in the oxidation state of hydrogen ).

Solution :
TIPS/Formulae :
(i) Find oxidation state of $$N$$  in $${N_2}{H_4}.$$
(ii) Find change in oxidation number with the help of number of electrons given out during formation of compound $$Y$$ .
$${N_2}{H_4} \to Y + 10{e^ - },$$     Calculation of $$O.S$$  of $$N$$ in $${N_2}{H_4}:$$
$$2x + 4 = 0 \Rightarrow x = - 2$$
The two nitrogen atoms will balance the charge of $$10e$$ . Hence oxidation state of $$N$$ will increase by + 5, i.e. from -2 to +3.