Answer
(a) Sn = +4; Cl = -1;
(b) Cr = +6; O = -2;
(c) V = +5; O = -2; Cl = -1;
(d) V = +3; O = -2;
(e) H = +1, N = +5; O = -2;
(f) Fe = +2; S = +6; O = -2;
Work Step by Step
(a) Cl is an halogen, so its oxidation number is likely to be "-1". Since the compound has 4 of them, the total is "-4". To compensate for this, the oxidation number of $Sn$ one should be "+4" (In order to sum up to 0).
(b) The oxygen usually has a "-2" oxidation number, which sum up to "-6" (because the compound has 3 oxygens). Therefore, the "Cr" oxidation number must be equal to "+6".
(c) Each "Cl" usually gives a "-1" oxidation number, and an oxygen gives "-2", which gives a total of ((-1)*3 + (-2) = -5). To compensate for this, the oxidation number of the "V" should be equal to "+5".
(d) The oxygen usually has a "-2" oxidation number, which sum up to "-6" (because the compound has 3 oxygens). Therefore, the total "V" oxidation number must be equal to "+6", since the compound has 2 of them, the oxidation number for each "V" must be "+3" (6/2).
(e) Normally, the hydrogen gives "+1", and each oxygen gives "-2", which gives us a total of (+1 + 3*(-2) = 1 - 6 = -5). To compensate for this, the "N" oxidation number must be equal to "+5".
(f) In this case, we have a "$SO_4^{2-}$" anion,, and a "$Fe^{2+}$" cation (The charge of the iron ion should be 2, in order to sum up to 0 with the "-2" from the sulfate)
The Iron oxidation number is "2+", because it is the only atom in the $Fe^{2+}$ ion.
Since each oxygen gives a "-2" oxidation number, and this compound contains 4"O", the total value is equal to "-8". The sum must be equal to "-2", which is the charge of the anion $SO_4^{2-}$. Therefore, the "S" oxidation number must be equal to "+6".
