`PbO_(2)rarrPbO,DeltaG_(298)lt0`
`SnO_(2)rarrSnO,DeltaG_(298)gt0`
Most proble oxidation states of `Pb` and `Sn` will be

Options

A. `Pb^(4+) m Sn^(4+)`
B. `Pb^(4+),Sn^(2+)`
C. `Pb^(2+),Sn^(2+)`
D. `Pb^(2+),Sn^(4+)`

Answer

D

Explanation

`overset(+4)(Pb)O_(2) to overset(+2)(Pb)O, DeltaG_(298) lt 0`
For this reaction `DeltaG` is negative , hence `Pb^(2+)` is more stable than `Pb^(4+)`
`overset(+4)(Sn) O_(2) to overset(+2)(Sn) O, " "DeltaG_(298) gt 0`
For this reaction `DeltaG` is positive, hence `Sn^(4+)` is more stable than `Sn^(2+)` because for spontaneous change `DeltaG` must be negative.