In this ion the oxidation state of sulfur is 6+ and the oxidation state of each oxygen is 2-
+6
K = +1 oxidation state Cl = +3 oxidation state O = -2 oxidation state
O = -2 oxidation state H = +1 oxidation state
S = +4 oxidation state O = -2 oxidation state
The oxidation state is +3.
C = +2 oxidation state O = -2 oxidation state
As SO42- has an overall -2 charge, The Oxygen has a -2 oxidation state, so to balance and give an overall -2 charge, the Sulfur has to have a +6 oxidation state. (-2 x 4) + (s) = -2 s = +6
SO42- (our goal is to get -2 charge) Oxygen by default has a -2 oxidation number, and we have 4 total, -2 X 4 = -8, our goal is -2, so S is +6.
The sulfate ion is SO42 -. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion. The sulfite ion is SO32-. The oxidation state of the sulfur is +4.
The sulfate ion is SO42 -. The oxidation state of the sulfur is +6 (work it out!); therefore, the ion is more properly named the sulfate(VI) ion. The sulfite ion is SO32-. The oxidation state of the sulfur is +4.
YES...CH4 + SO42- → HCO3- + HS- + H2O, is only one formula.
-2 oxidation state
K = +1 oxidation state Cl = +3 oxidation state O = -2 oxidation state
Fe = +3 oxidation state Cl = -1 oxidation state
O = -2 oxidation state H = +1 oxidation state
Mg = +2 oxidation state P = +5 oxidation state O = -2 oxidation state
It depends on what form it is in. It can be in the 4+ oxidation state, 4- oxidation state and every oxidation state in between.
0, 1 and 2 oxidation states