KEAM 2016 Physics and Chemistry Paper

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