1. Here, we’ll go through another example of finding the oxidation number of each element in a polyatomic ion.

2. We’re asked to find the oxidation number of each element in the ion with the formula C 5 H 5 minus Find the oxidation number of each element in the ion:.

3. We’ll start by writing the formula up here Find the oxidation number of each element in the ion:. Element Oxidation Number K+1 S+7 O–2

4. And a table here for the oxidation number of each element Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

5. Carbon has variable oxidation numbers, so the oxidation number (click) of carbon in this ion is unknown. Therefore, we’ll call it (click) x Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

6. Hydrogen’s symbol is written (click) to the right of the symbol for C, but carbon is not a metal, so this is not a metallic hydride, therefore the oxidation number of hydrogen is (click) the normal positive 1 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

7. We’ve called the oxidation number of carbon x, Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

8. So the total charge on 5 carbon atoms is (click) 5x Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

9. The oxidation number of a hydrogen atom is plus 1 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

10. So the total charge on 5 hydrogen atoms is (click) 5 times positive 1 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

11. The net charge of this ion, shown on the top right (click) of the formula is negative 1, so the charges on all the atoms add up to (click) negative 1 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

12. We can solve for x in this equation to find the oxidation number of carbon Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

13. So we write 5x Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

14. Plus 5 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

15. Equals negative 1 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

16. Subtracting 5 from both sides, gives us 5x = negative 1 minus 5 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

17. which equals negative 6 Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

18. Dividing both sides by 5 gives us… Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

19. X equals negative 6 5 th ’s Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

20. Or as a mixed number, negative 1 and 1 fifth Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

21. So we can say the the oxidation number of carbon in this ion is Element Oxidation Number Cx H+1 Find the oxidation number of each element in the ion:.

22. Negative 6 fifths or negative 1 and 1 fifth. Even though non-integer oxidation numbers are not as common as integer ones, we see that they are possible, so don’t be alarmed if you occasionally get a fraction for an answer. Element Oxidation Number C H+1 Find the oxidation number of each element in the ion:.

23. So we can summarize (click) by saying that the oxidation number of carbon in this ion is negative 6 fifths or negative 1 and 1 fifth. Element Oxidation Number C H+1 Find the oxidation number of each element in the ion:.

24. And the oxidation number of hydrogen is positive 1. Element Oxidation Number C H+1 Find the oxidation number of each element in the ion:.