Exampes and Problems with no answers
Problem #1: What is the oxidation state of carbon in calcium oxalate?
The oxidation state of carbon in the oxalate ion is +3, based on two things: 1. the oxidation state of oxygen is usually -2 2. the sum of the oxidation states for a compound is zero, and for an ion, the sum is the charge on the ion.
Problem #2: C in CO32¯
The O is −2 and three of them makes −6. Since −2 is left over, the C must be +4
Problem #3: Cr in CrO42¯
The O is −2 and four of them makes −8. Since −2 must be left over, the Cr must be +6
Problem #4: Cr in Cr2O72¯
The O is −2 and seven of them makes −14. Since −2 is left over, the two Cr must be +12.What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr2 being +12. Each Cr atom is considered individually.
Problem #5: Fe in Fe2O3
The O is −2 and three of them makes −6. Each Fe must then be +3
Problem #6: Pb in PbOH+
The O is −2 and the H is +1. In order to have a +2 for the formula, the Pb must be a +2.
Problem #7: V in VO2+
Two O gives a total of −4. To make the formula have a +1 charge, the V must be +5
Problem #8: V in VO2+
The V is +4. This comes from the one O being −2 and the fact that a +2 must be present on the ion.
Problem #9: Mn in MnO4¯
Four O totals to −8. The Mn is +7, leaving −1 left over.
Problem #10: Mn in MnO42¯
Four O totals to −8. The Mn is +6, leaving −2 left over.
Bonus Problem: What is the oxidation state of Re in potassium nonahydridorhenate, which has the formula K2ReH9
We know that potassium always takes on a +1 oxidation state. Eliminate the potassium to get this:ReH92¯That looks very much like a hydride, when hydrogen takes on a −1 oxidation state.
The fact that there are nine of them means Re must have an oxidation state of +7.
Another compound with Re = +7 is NaReO4.