What is the oxidation state of Cr in each of the following compounds?

Compound A = II + (two plus)
Compound B = VI + ( six plus)
Compound C = III +  ( three plus)

In the second part, Sulphur has an oxidation state of 5 plus in the first and 7 plus in the second.

Correction to above post: see below for reasoning

Sulfur oxidation state is +4 in HSO3-.
Sulfur oxidation state is +6 in HSO4-.

Compound A: Oxygen's oxidation state is usually always 2-. Therefore, for the ion to have an overall charge of zero, Cr must be 2+.

Compound B: Using the same concept, oxygen has an oxidation state of 2-. This time there are 3 oxygen's. Therefore the charge contributed by oxygen is (3x-2)=-6. To get an uncharged ion, Cr must have an oxidation state of 6+.

Compound C: 3 oxygen's result in a 6- charge from oxygen. However there are 2 Cr atoms this time. Cr must be 3+ because (2x+3)=+6

for HSO3-, the overall charge on the molecule is -1. The oxidation numbers of each atom in the molecule must add to equal the overall charge. . -6 comes from oxygen and +1 comes from hydrogen (because it is always +1). -6+1+x(S oxidation #) must equal -1. Therefore,

Same math for HSO4-. Overall molecule has a -1 charge, so the oxidation numbers of all atoms must add to -1. 4 oxygen atoms contribute -8 charge. Hydrogen contributes +1 charge. Sulfur must be +6.

-8+1+6=-1