Name Binary Compounds and with Binary Compounds with
Name Binary Compounds and with Binary Compounds with Multiple Charge Cations
• I CAN determine the formula/write names for compounds containing multiple charge cations.
Multiple Charge Ions • Some metals form CATIONS with MORE THAN ONE POSSIBLE CHARGE. – Metals from the TRANSITION METALS – Some HEAVY ATOMIC MASS NUMBER elements. • The elements are referred to as MULTIPLE OXIDATION STATE elements. • Only METALS form these ions…there are NO ANIONS (- ions) with multiple charges!
**Some Common Multiple Oxidation Elements**
INDICATING THE PRESENCE OF A MULTIPLE CHARGE CATION. • In the NAMES of compounds, these elements are identified by a ROMAN NUMERAL INSIDE A PARENTHESIS following the name of the CATION ELEMENT. • For example Iron (III) Oxide indicated that Iron is a MULTIPLE CHARGE CATION.
WRITING FORMULAS with Multiple Charge Cations • The ROMAN NUMERAL in the formula’s name actually makes it a little easier to WRITE A FORMULA. – The ROMAN NUMERAL is to indicate the POSITIVE CHARGE of the cation in this particular compound. • For Example Iron (III) Oxide indicates that Iron has a +3 CHARGE in this compound.
Practice Problem • Write the formula for Iron (III) Oxide. +3 • Indicates a POSITIVE cation charge Iron (III) Oxide = Oxygen (ending changed to -ide)…always get the negative charge from the Periodic Table.
• • (+3 ) + (-2 ) = 0 Fe O Fe 2 O 3 Fe 2 O 3
More Practice • What would be the formula for Lead (IV) Nitride? – 1. Look at list to see if you have a cation with more than one charge. – 2. Write symbols for both ions and assign charge. – 3. Balance charges and write the formula.
Naming Compounds with Multiple Oxidation Cations • Follow the same rules as for a BINARY IONIC COMPOUND except: • After the name of the multiox cation, place a Roman Numeral in a parenthesis indicating which charge state the ion is in, then add the name of the anion, changing the ending to IDE.
PRATICE PROBLEMS Name these binary ionic compounds containing multiple oxidation cations. 1. Pb. O 2. Sn. F 4 3. Cu. S 4. Fe 3 P 2 5. Cr 2 O 3 6. Hg 2 I 2
EXAMPLE • Take a look at Pb. O. • Look at the cation…can it have more than one oxidation state (charge)? • YES it could be Pb+2 or Pb+4 • Which state is Pb in in this compound? • Look at the ANION…determine the total NEGATIVE charge. • O is -2, and there’s only 1 atom, so the total NEGATIVE charge is -2, therefore the total POSITIVE charge must be +2. • Since there is only one atom of Pb, it must be in the +2 state.
• • • (+2) + (-2) = 0 Pb O Pb 2 O 2 • Can you reduce the subscripts? • YES Pb 2 O 2 Pb. O • Remember an EMPIRICAL FORMULA is the simplest whole number ratio of atoms in the compound.
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