1. Multiple Oxidation Elements Elements that form more than one ion.

2. I CAN write formulas for and name compounds containing Multiple Oxidation elements.

3. What is a MultiOx element? Certain elements, particularly from the TRANSITION ELEMENTS and a few HEAVIER AMN ELEMENTS tend to lose both VALENCE electrons as well as one or more from the NEXT ENERGY LEVEL below valence. These are known as MULTIPLE OXIDATION STATE ELEMENTS or MULTIOX for short. Since these come from the METALS of the periodic table, they are always CATIONS (+).

4. There are quite a few elements that exhibit multiple oxidation. REFER TO THE HANDOUT PROVIDED.

5. Compounds with MultiOx Elements Its easy to recognize a MultiOx element in the NAME OF A COMPOUND. – Following the name of the POSITIVE ELEMENT there will be a PARENTHESIS WITH A ROMAN NUMERAL. – The ROMAN NUMERAL indicates the POSITIVE CHARGE of the element IN FRONT OF IT. EXAMPLE Iron (III) Sulfide The roman numeral indicates the IRON has a +3 charge in this compound. – This makes it easy to determine the formula for a compound containing a MultiOx element.

6. Practice Problems Write the formula for these compounds: Copper (II) PhosphideIron (III) Oxide (+2) + (-3) = 0 (+3) + (-2) = 0 Cu P Fe O Cu 3 P 2 3 3 2 2 3 3 2 2 Fe 2 O 3 +2 +3

7. Naming Compounds with MultiOx elements from their formulas. Naming a compound with a MultiOx element from its FORMULA requires a bit of working backwards. 1. Look at the formula and refer to your MultiOx list if needed to see if there is a MultiOx element in the compound. 2. Work BACKWARDS to determine the name of the compound.

8. Practice Problems What is the name of the compound PbCl 4 ? 1. Check for a MultiOx element: LEAD 2. Break the formula apart: ( ? ) + 4 ( -1) = 0 Pb Cl 3. Use the NEGATIVE ION to determine the charge of the MultiOx ion: +4 -4 ( ? ) + 4 ( -1) = 0 Pb Cl

9. +4 -4 ( +4 ) + 4 ( -1) = 0 Pb Cl Therefore this is LEAD (IV) so the name of the compound is: Lead (IV) Chloride

10. Another Practice Problem What would the name of SnI 4 be? +4 -4 ( ? ) + 4 (-1) = 0 Sn I Therefore this is Tin (IV) so the compound would be: Tin (IV) Iodide