Cu+HNO_(3)toCu(NO_(3))_(2)+NO_(2)+H_(2)O The number of Nitrogen atoms, water molecules and the total charge on the product side are respectively when above reaction is balanced ?

Cu+HNO_(3)toCu(NO_(3))_(2)+NO_(2)+H_(2)O The number of Nitrogen atoms,

1.	Cu+HNO_(3)toCu(NO_(3))_(2)+NO_(2)+H_(2)O The number of Nitrogen atoms, water molecules and the total charge on the product side are respectively when above reaction is balanced ?
Answer» `6,3,0` `4,2,2` `4,2,0` `3,2,0` Solution :(x) Oxidation half : `CutoCu^(2+)+2NO_(3)^(-)" "...("Oxidation")` `CutoCu^(2+)+2bare" "...(x)("Oxidation half")` (y) Reduction half : (i) `underset(+5)underset(darr)(NO_(3)^(-))tounderset(+4)underset(darr)(NO_(2))" "...("O.N. of N")` (ii) `NO_(3)^(-)+baretoNO_(2)" "..."(Reduction and " BARE)` (iii) `NO_(3)^(-)+bare+2H^(+)toNO_(2)` ... (Charge balance by `H^(+)`) `NO_(3)^(-)+bare+2H^(+)toNO_(2)+H_(2)O` ... (Balance of O, .H. by `H_(2)O`) (iv) `2NO_(3)^(-)+2bare+4H^(+)to2NO_(2)+2H_(2)O" "...(y)` At product side in the balanced equation, No. of nitrogen atoms = 4 No. of water molecules = 2 Total charge = Zero [SUPPOSE you stop at ionic balance equation (z)] At product side in (z) N = 2, `H_(2)O` = 2 and charge = +2 [But as per data, you can not stop at this stage, you are required to state molecular form from ionic]