Chemistry Practice Problems
Compounds: Nomenclature of Ionic Compounds
Solutions: 1. Provide the missing name or formula for each ionic compound.
a. TiCl4
titanium(IV) chloride
The charge on titanium is 4+ and it is a transition metal so the Roman numeral must be
included in the name to represent the charge on the metal. The end of chlorine is changed to
–ide because it is a monatomic ion.
b. magnesium nitrate
Mg(NO3)2
Magnesium has 2+ charge while nitrate has a 1− charge so one Mg2+ ion will combine with
two NO3− ions to form a neutral compound with the name magnesium nitrate.
c. potassium carbonate
K2CO3
Potassium has a 1+ charge while carbonate has a 2− charge so two K+ ions combine with one
CO32− ion to form a neutral compound.
d. ammonium nitrite
NH4NO2
Ammonium has a 1+ charge and nitrite has a 1− charge so one of each ion will combine to
form a neutral compound.
e. AgNO3
silver nitrate
There is no Roman numeral because silver, a transition metal, only has one possible charge.
The anion is a polyatomic ion so the end of the name is not changed.
f. ZnCl2
zinc chloride
There is no Roman numeral because zinc only has one possible charge. Chlorine forms a
monatomic ion so the ending of the name is changed to –ide.
g. Fe2O3
iron(III) oxide
Iron is a transition metal so the Roman numeral should be included in the name to represent
the charge on the iron. Its charge is 3+ which is determined from looking at the three oxygen
atoms which each have a 2− charge. Oxygen forms a monatomic ion so the end of the name
is changed to –ide.
Chemistry Practice Problems
h. FeO
iron(II) oxide
In FeO, iron has a 2+ charge because there is only one of each atom and oxygen has a 2−
charge. It is a transition metal so the Roman numeral should be included in the name to
represent the charge on the iron.
i. MnO2
manganese(IV) oxide
Manganese has a charge of 4+ which is determined by looking at the two oxygen atoms
which each have a charge of 2−. Oxygen forms a monatomic ion so the end of the name is
changed to –ide.
j. strontium phosphate
Sr3(PO4)2
Strontium ion has a 2+ charge and phosphate has a 3− charge so three Sr3+ ions and two
PO43− ions are combined to form a neutral compound. Parentheses are included around the
phosphate ion because there are two of the ions in the formula. The subscript belongs to the
ion not to just one atom in the ion.
k. copper(I) oxide
Cu2O
The Roman numeral shows the charge on the copper ion is +1 while oxide has a 2− charge.
Two Cu+ ions combine with one O2− ion to form a neutral compound.
l. copper(II) nitride
Cu3N2
The Roman numeral shows the charge on the copper ion is +2 while nitride has a 3− charge
so two Cu2+ ions combine with two N3− ions to form a neutral compound.
m. aluminum phosphate
AlPO4
Aluminum has a 3+ charge and phosphate is a polyatomic ion with a 3− charge so one of
each ion combine to form a neutral compound.
n. iron(II) carbonate
FeCO3
Iron has a 2+ charge indicated by the Roman numeral. Since carbonate has a 2− charge, one
of each ion combine to form a neutral compound.
o. barium bromide
BaBr2
Barium ion has a 2+ charge and bromide has a 1− charge so one Ba2+ ion and two Br− ions combine to form a neutral compound.
Chemistry Practice Problems
p. LiOH
lithium hydroxide
Lithium has a 1+ charge and hydroxide has a 1− charge so one of each ion combine to form a
neutral compound.
q. MgSO4
magnesium sulfate
The name of the metal atom is unchanged in ionic compounds and sulfate is a polyatomic ion
so it is included in the name without any changes. A Roman numeral is not needed because
magnesium is an alkaline earth metal with only one possible charge.
r. scandium chloride
ScCl3
Scandium only has one possible charge so no Roman numeral is included in the
name. Chlorine forms a monatomic ion so the ending is changed to –ide.