Naming Ionic Compounds

Ionic compounds ► Compounds are created by a combination of charged ions. ► Generally they contain a metal bonded to a nonmetal ► the positive ion (cations) and the negative ion (anions) can be thought of as the first and last names of the compound

► The first part of the name will come from the left side in periodic table ► Followed by naming the second element which will be further to the right

2. Cations keep their name  Ex. Li  Lithium (stays the same!)

Anions change the last few letters to have the suffix –ide ► Ex. O  oxide

4. Positive first, negative last ► Ex. LiO  ► Lithium Oxide

Writing Ionic Formulas ► ► Write the symbols for the elements in the compound. ► ► Remember the ending "ide" is used for fluoride to show that it is a negative ion of fluorine. Example: Barium Fluorine

Barium Fluoride ► Look up the oxidation numbers of the elements involved and write them as superscripts to the right of the elemental symbols. ► Look up the oxidation numbers of the elements involved and write them as superscripts to the right of the elemental symbols.

► Note that when no number accompanies a charge symbol, as in the case of fluoride above, the charge value is understood to be "1". Ba 2+ F -

► Use the correct combination of ions to produce a compound with a net charge of zero. ► Use the correct combination of ions to produce a compound with a net charge of zero. ► Right now the net charge is 1+ (2+) + (1-) = +1 ► F - ► How many more F - do we need to balance out the positive ion?

► Two fluoride ions will cancel out one barium ion. (2+) + 2(1-) = 0 ► we use a subscript of two after the symbol for fluorine to show the ratio of 1:2. BaF 2

Writing Ionic Formulas Mg 2+ N 3- ► What is this compounds net charge? (2+) + (3-) = 1- ► How can we get this to have a net charge of zero? 3(2+) +2(3-) = 0

Cross Over Method ► You can simply take the ionic charge for the positively charged ion and make it the subscript for the negatively charged ion and vice versa.

Ba 2+ F - BaF 2 Cross Over Method 2 1 We then simplify and drop “1s”

Al 3+ Cl - AlCl 3 Cross Over Method 3 1 We then simplify and drop “1s”

Al 3+ O 2- Al 2 O 3 Cross Over Method 3 2