(a)Ti in TiO2
To determine the oxidation number of an atom in a compound, we assign to the most electronegative element in the compound a negative oxidation number equal to its charge in simple ionic compounds. For neutral compounds, the oxidation numbers of all elements must add to give zero; for ionic species, the oxidation numbers must add to give the ionic charge of the compound. Here, O is the more electronegative element, and it gets the -2 oxidation number. There are two oxide ions per Ti ion in the empirical formula; since the compound is neutral, Ti must have an oxidation number of 4. Algebraically, we do the following: Ti + 2(-2) = 0 and solve for Ti.
Sn in SnCl3-
Here, all oxidation numbers must add to -1 (the charge on the ion.) We recognize that there are three Cl- ions per Sn in the formula; since each Cl gives a single negative charge, we must have Sn in the +2 state, or Sn + 3(-1) =-1.
(c)C in C2O42-
Here, we assign oxygen the -2 charge (it is the most electronegative element.) Now, we can add up the oxidation numbers: 2C +(-2)4 = -2. Note that the oxidation numbers add to give the ionic charge. Solving for C, we have C = 3.
(d)N in N2H4
This again a binary compound - we assign H an oxidation number of +1 (it's bonded to a nonmetal), so we have 2N + 4(1) =0, and N=-2.
(e)N in HNO2
We assign O the -2 charge and H the +1 charge, so, adding up the oxidation numbers, 1 + N + 2(-2) = 0 and N = 3.
(f)Cr in Cr2O72-
Give oxygen the -2 charge and add 'em up: 2Cr + 7(-2) = -2. We have Cr in the +6 state.