what´s the change in oxidation state in the corresponding C, iso-propanol is oxidized to acetone with KMno4?

Isopropanol has the molecular formula of C3H8O and acetone has the molecular formula of (CH3)2CO. The KMnO4 really doesn't matter because there is no C in it, so we can ignore that bit of information.

Acetone carbon oxidation states:

Carbon number Attached atoms Sum of oxn nos of attached atoms C oxidation state

1 3H, 1C 3(+1) + 1(0) = +3 -3

2 2C, 1O 2(0) + 1(-2) = -2 +2

3 3H, 1C 3(+1) + 1(0) = +3 -3

2-Propanol carbon oxidation states:

Carbon number Attached atoms Sum of oxn nos of attached atoms C oxidation state

1 3H, 1C 3(+1) + 1(0) = +3 -3

2 2C, 1H, 1OH 2(0) + 1(+1) + 1(-1) = 0 0

3 3H, 1C 3(+1) + 1(0) = +3 -3

So there overall change in the oxidation numbers of C would be +2