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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? .
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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 3 2 2 3 Fe2O3
Naming Compounds with MultiOx elements from their formulas. • Naming a compound with a MultiOxelement from its FORMULA requires a bit of working backwards. • 1. Look at the formula and refer to your MultiOxlist 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 • PbCl • 3. Use the NEGATIVE ION to determine the charge of the MultiOx ion: • +4-4 • ( ? ) + 4 ( -1) = 0 • PbCl
+4-4 • ( +4 ) + 4 ( -1) = 0 • PbCl • 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