Multiple Oxidation Elements

Elements that form more than one ion.

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

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 (+).

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

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.

Practice Problems

• Write the formula for these compounds:

• Copper (II) Phosphide Iron (III) Oxide

• (+2) + (-3) = 0 (+3) + (-2) = 0• Cu P Fe O• Cu3P2

3 2 32

Fe2O3

+2 +3

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.

Practice Problems• What is the name of the compound PbCl4 ?• 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

• +4 -4• ( +4 ) + 4 ( -1) = 0• Pb Cl

• Therefore this is LEAD (IV) so the name of the compound is:

• Lead (IV) Chloride

Another Practice Problem

• What would the name of SnI4 be?• +4 -4• ( ? ) + 4 (-1) = 0• Sn I• Therefore this is Tin (IV) so the compound

would be:

• Tin (IV) Iodide