Given, 1molofH2N−NH2 (hydrazine), it loses 10 moles of electrons to form a new compound that contains both the N-atoms with same oxidation number means: N2H4⟶10e−+X‌ (product) ‌ (Oxidation number of N -atom in N2H4=−2 ) ∵ Oxidation number of both the N - atoms are same. New total oxidation number of new compound (X) =4−10+x=0‌‌ (due to 4H - atoms ) ∴x=+6
Alternate method
N2H4⟶X Number of electrons lost per N - atom =5 ∴ New oxidation state of N in X=−2+5+x=0 x‌‌=+3