what's the oxidation number of Cr in Al2(Cr2O7)3?

O = 2- * 21 O = 42-
Al = 3+ *2 Al = 6+
So you have a total of 36- which means you have to divide 36+ between 6 molecules of Cr to make it a neutral compound. So each Cr must be 6+

U could also just look at Cr2O7(2-) since it would be its own anion

2*Cr+7*(-2)=-2 => Cr = +6