N2H4−10e−→X10+ (with two N atoms ) Total oxidation number of two N atoms in N2H4, ⇒ 2x+4=0 2x=−4 N2H4 loses 10 electrons, so total oxidation number of two N-atoms increases by 10, i.e., the total oxidation number of two N-atoms in Y=−4+10=+6 ∴ Oxidation number of each N atom in X10+=+3.