Complete and balance the followig equations in basic solution :
`Hg_2(CN)_2+Ce^(4+)toCO_3^(2-)+NO_3^(ɵ)+Hg(OH)_2+Ce^(3+)`
(a) By considering C in `+4` oxidation state and N in `+5` oxidation state.
(b) By considering C in `+4` oxidation state and N in `+5` oxidation state
(c) By considering Hg in `+2` and C in `+4` oxidation state.
(d) Explain why the same result is obtained regardless of the choice of oxidation state.

Explanation

`overset(+1)(Hg_(2))overset(-4+3)((CN)_(2)) rarr overset(+4)(C)O_(3)^(2-)+ overset(+5)(N) O_(3)^ɵ+overset(+2)(Hg)(OH)_(2)`
(a) If the oxidation state of of C is `-4` and that of N is `+3` then the oxidation state of Hg is `+1` In the products, oxidation numbers are `+4,+5` and `+2`, respectively. The total increase in oxidation number is 22.
(b) `Hg_(2)(CN)_(2) rarr CO_(3)^(2-)+NO_(3)^ɵ+Hg(OH)_(2)`
If the oxidation state of C is `+4` and that of N is `+5`, then that of Hg must be `-9`
Then the only element oxidised oxidised is Hg, and the total increase in oxidation number is again 22.
The oxidation number of Hg, `-9`, is impossible, but it still gives the same change in oxidation number.
(c) If oxidation state of Hg is `+2` and C is `-4`, then that of N must be `+2` The oxidation still involves `+22` change in oxidation number.
(d) No matter which set of oxidation number is used, as long as they total zero, the charge on the `Hg_(2)(CN)_(2)`, the same number of electrons must be added to the half reaction.
`underset(2x=-8)overset(-4 xx 2)(C_(2)) underset(8-(-8)=16e^(-))(rarr) underset(underset(2x=8)(2x -12=-4))(2CO_(3)^(2-)+16e^(-))`
`underset(2x=6)overset(+3 xx 2)(N_(2)) underset(10-6 =4e^(-))(rarr) underset(underset(2x=10)(2x-12=-2))(2NO_(3)^ɵ+ 4e^(-))`
`underset(2x=2)(overset(+1xx2)(Hg_(2))) underset(4-2=2e^(-))(rarr) underset(underset(2x=4)(2x-4=0))(2Hg(OH)_(2)+2e^(-))`
`ulbar(Hg_(2)(CN)_(2) rarr 2CO_(3)^(2-)+2NO_(3)^ɵ +2Hg(OH)_(2)+22e^(-))`.