# Making Ionic Compounds With Multiple Charges Worksheet Page 2

Back to the Procedure:
3. Now use a +2 ion piece and cause it to join with -1 ion piece(s).
Think to remember how many positive and negative pieces were needed to achieve neutrality.
4. Next use a +2 ion piece and cause it to join with a -2 ion.
Make a mental note how many positive and negative pieces were needed to achieve neutrality.
5. Finally use a +2 ion piece(s) and cause it to join with -3 ion piece(s).
Make a mental note how many positive and negative pieces were needed to achieve neutrality.
Follow-up Questions:
Q5. In each case neutrality has to be accomplished. Write the formula for three generic compounds
using X
2+
, and Y
1-
, Y
2-
, and Y
3-
respectively. Think about how many positive and negative pieces
would be needed to achieve neutrality
Q6. Complete the following table by writing the needed formula.:
1-
2-
3-
Divalent cation
monovalent anion (Y
)
divalent anion (Y
)
trivalent anion (Y
)
2+ )
(X
Cl 1-
S 2-
N 3-
Ba 2+
Ca 2+
There are also polyatomic ions that have -1, -2, and -3 charges. Notice that ammonium is the only
common positive polyatomic ion. Here are three more polyatomic ions you need to memorize.
3 2-
4 2-
Write these down and memorize both the formula and name: carbonate CO
, sulfate SO
, and
phosphate PO
4 3-
. When you use a polyatomic ion to write a formula simply write the entire formula
(of course since the charge has been neutralized do not list the charge.) If more than one polyatomic
ion is needed to write an neutrally charged formula in making the compound then enclose the
polyatomic ion in parentheses and write the appropriate subscript to the right of the parentheses.
2+
1-
3 2-
4 3-
For example Ca
forms compounds with OH
, CO
, and PO
to form compounds
Ca(OH) 2 , CaCO 3 and Ca 3 (PO 4 ) 2 respectively. If a polyatomic ion is enclosed with parentheses it
says that more than one ion is needed and the number needed is expressed as a subscript to the right
of the parenthesis.
Q7. Complete the following table by writing the correct formula:
Divalent cation
1-
2-
3-
monovalent anion (Y
)
divalent anion (Y
)
trivalent anion (Y
)
2+ )
(X
NO 3 1-
SO 4 2-
PO 4 3-
Mg 2+
Zn 2+
Cu 2+
Back to the Procedure:
6. Now use a -2 ion piece and cause it to join with +1 ion piece(s).