AP EAMCET Engineering 2012 Solved Paper

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.