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7.1: Ionic Compounds

Difficulty Level: At Grade Created by: CK-12
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Lesson Objectives

  • Distinguish between inorganic and organic chemistry.
  • Interpret a molecular formula.
  • Explain why an ionic compound is represented by an empirical formula.
  • Be able to determine the charge of monatomic ions from the position of the element on the periodic table.
  • Use the Stock system to identify the charge of transition metal ions.
  • Name an ionic compound given its formula.
  • Write the correct formula for an ionic compound given its name.

Lesson Vocabulary

  • binary ionic compound
  • empirical formula
  • inorganic chemistry
  • molecular formula
  • monatomic ion
  • organic chemistry
  • polyatomic ion
  • ternary ionic compound

Check Your Understanding

Recalling Prior Knowledge

  • Which subatomic particles are lost or gained in order to form ions?
  • The elements of which group on the periodic table are particularly stable and unreactive?
  • Which groups contain the representative elements and which groups contain the transition elements?

Chemistry can be broadly divided into two main classes based on the identity of the elements present in the chemical compounds. Organic chemistry is the branch of chemistry that deals with compounds containing carbon. Organic chemistry is very complex and will be covered in a later chapter. This chapter is about inorganic chemistry. Inorganic chemistry is the branch of chemistry that deals with compounds that do not contain carbon. There are a few exceptions of small molecules that contain carbon but are not subject to the rules for naming organic compounds. We will begin our study with an examination of two types of formulas. The remainder of this lesson will be about the naming and formula writing of ionic compounds.

Types of Formulas

Recall that a molecule is two or more atoms that have been chemically combined. A molecular formula is a chemical formula of a molecular compound that shows the kinds and numbers of atoms present in a molecule of the compound. Ammonia is a compound of nitrogen and hydrogen as shown below:

An empirical formula is a formula that shows the elements in a compound in their lowest whole-number ratio. Glucose is an important simple sugar that cells use as their primary source of energy. Its molecular formula is C6H12O6. Since each of the subscripts is divisible by 6, the empirical formula for glucose is CH2O. When chemists analyze an unknown compound, often the first step is to determine its empirical formula, as we will see in a later chapter. There are a great many compounds whose molecular and empirical formulas are the same. If the molecular formula cannot be simplified into a smaller whole-number ratio, as in the case of H2O or P2O5, then the empirical formula is also the molecular formula.

Ionic compounds are quite different from molecular compounds such as water. Water and other molecular compounds exist as individual molecules (Figure below).

A water molecule consists of one atom of oxygen bonded to two atoms of hydrogen.

Ionic compounds do not exist as discrete molecular units. Instead an ionic compound such as sodium chloride (NaCl) consists of a large three-dimensional array of alternating Na+ and Cl- ions (Figure below).

A crystal of table salt, sodium chloride, is a large array of alternating positive and negative ions. The purple spheres represent the Na+ ions, while the green spheres represent the Cl- ions.

Watch a video about The Structure of Ionic Solids: http://www.youtube.com/watch?v=TLPY9Z6z4Mg (1:35) .

Because of this structure, the chemical formulas of ionic compounds always represent the lowest whole-number ratio between the two ions. In the case of NaCl, there are equal numbers of sodium ions and chloride ions in the salt crystal. In the case of magnesium chloride, with a formula of MgCl2, there are twice as many chloride ions as magnesium ions in each crystal. The formulas of ionic compounds are always empirical formulas.

Monatomic Ions

An ionic compound consists of a positive metal ion and a negative nonmetal ion combined in a proportion so that the overall compound is electrically neutral. In other words, the total positive charge is exactly canceled out by the total negative charge. So in order to accurately write formulas for all ionic compounds, it is necessary to know the charges of the ions involved. We will begin our study with monatomic ions, which are ions that consist of a single atom with either a positive or negative charge.


Recall that cations are positive ions that are formed when a metal atom loses one or more electrons. For the representative elements, all of the valence electrons are removed from the atom. Since the valence electrons are constant within a particular group, all we need to know is the group number of the given element and we will be able to know its charge when it becomes a cation. Group 1 elements form ions with a 1+ charge, Group 2 with a 2+ charge, and Group 13 with a 3+ charge. The heavier p-block metals such as tin and lead are a special case and will be discussed with the transition metal ions. Naming monatomic ions is done simply by using the element’s name. For example, the Na+ ion is the sodium ion, while the Al3+ ion is the aluminum ion.


Anions are negative ions that are formed when a nonmetal atom gains one or more electrons. Atoms typically gain electrons so that they will have the electron configuration of a noble gas. All the elements in Group 17 have seven valence electrons due to the outer ns2np5 configuration. Therefore, each of these elements would gain one electron and become an anion with a 1− charge. Likewise, Group 16 elements form ions with a 2− charge, and the Group 15 nonmetals form ions with a 3− charge. Naming anions is slightly different than naming cations. The ending of the element’s name is dropped and replaced with the –ide suffix. For example, F- is the fluoride ion, while O2- is the oxide ion. Table below shows the common monatomic ions for the representative elements.

Common Monatomic Ions
1+ 2+ 3+ 3- 2- 1-
lithium, Li+ beryllium, Be2+ aluminum, Al3+ nitride, N3- oxide, O2- fluoride, F-
sodium, Na+ magnesium, Mg2+ gallium, Ga3+ phosphide, P3- sulfide, S2- chloride, Cl-
potassium, K+ calcium, Ca2+ arsenide, As3- selenide, Se2- bromide, Br-
rubidium, Rb+ strontium, Sr2+ telluride, Te2- iodide, I-
cesium, Cs+ barium, Ba2+

Transition Metal Ions

Most transition metals differ from the metals of Groups 1, 2, and 13 in that they are capable of forming more than one cation with different ionic charges. As an example, iron commonly forms two different ions. It can sometimes lose two electrons to form the Fe2+ ion, while at other times it loses three electrons to form the Fe3+ ion. Tin and lead, though members of the p block rather than the d block, also are capable of forming multiple ions. The names of these types of ions needs to indicate the particular charge that ion carries. The Stock system of naming these ions uses a Roman numeral in parentheses after the name of the ion. The iron ions previously mentioned are named the iron(II) ion and iron(III) ion. In speaking, the name of iron(II) is said, “iron two ion.” Table below lists the names and formulas of some of the common transition metal ions.

Common Transition Metal Ions
1+ 2+ 3+ 4+
copper(I), Cu+ cadmium, Cd2+ chromium(III), Cr3+ lead(IV), Pb4+
gold(I), Au+ chromium(II), Cr2+ cobalt(III), Co3+ tin(IV), Sn4+
mercury(I), Hg22+ cobalt(II), Co2+ gold(III), Au3+
silver, Ag+ copper(II), Cu2+ iron(III), Fe3+
iron(II), Fe2+
lead(II), Pb2+
manganese(II), Mn2+
mercury(II), Hg2+
nickel(II), Ni2+
platinum(II), Pt2+
tin(II), Sn2+
zinc, Zn2+

Notice in Table above that there are three cation names that lack the Roman numeral. Silver, cadmium, and zinc only form ions with one charge, so by convention the Stock system is not used with these ions. The mercury(I) ion is a special case in that consists of a pair of mercury atoms bonded together and a total charge of 2+. So each mercury ion can be thought of as carrying a 1+ charge. In some instances, charges greater than 4+ are seen, but these are not the most common charges and so are not included in the table.

There is an older system for naming some of these cations that is still occasionally used. The Latin root of the metal name is written with one of two suffixes: (1) –ic for the ion with a higher charge, and (2) –ous for the ion with a lower charge. Again, iron will serve as an example. The Latin name for iron is ferrum, so the Fe3+ ion is called the ferric ion, while the Fe2+ ion is called the ferrous ion. The primary disadvantage of this system is that the suffixes do not tell you exactly what the charge is for a given ion. For copper, the Cu2+ is the highest charge, so it is called the cupric ion, while the Cu+ ion is the cuprous ion. The Stock system is a much more specific system and will be used as the primary method of naming transition metal compounds throughout this book.

Binary Ionic Compounds

A binary ionic compound is a compound composed of a monatomic metal cation and a monatomic nonmetal anion.

Naming Binary Ionic Compounds

When examining the formula of a compound in order to name it, you must first decide what kind of compound it is. For a binary ionic compound, a metal will always be the first element in the formula, while a nonmetal will always be the second. The metal cation is named first, followed by the nonmetal cation. Subscripts in the formula do not affect the name. Table below shows three examples.

Examples of Binary Ionic Compounds
Formula Name
KF potassium fluoride
Na3N sodium nitride
Ca3P2 calcium phosphide

Notice that in each of the formulas above, the overall charge of the compound is zero. Potassium ion is K+, while fluoride ion is F. Since the magnitude of the charges is equal, the formula contains one of each ion. This would also be the case for a compound such as MgS, in which the ions are Mg2+ and S2−. For sodium nitride, the sodium ion is Na+, while the nitride ion is N3−. In order to make a neutral compound, three of the 1+ sodium ions are required in order to balance out the single 3− nitride ion. So the Na is given a subscript of 3. For calcium phosphide, the calcium ion is Ca2+, while the phosphide ion is P3−. The least common multiple of 2 and 3 is 6. To make the compound neutral, three calcium ions have a total charge of 6+, while two phosphide ions have a total charge of 6−. The Ca is given a subscript of 3, while the P is given a subscript of 2.

Practice naming binary ionic compounds with this short crossword puzzle: https://sites.google.com/site/pattihowellsciencespot/uploads/binary%20ionic%20compounds.html?attredirects=0&d=1.

Naming Compounds Using the Stock System

Naming compounds that involve transition metal cations necessitates the use of the Stock system. Consider the binary ionic compound FeCl3. To simply name this compound iron chloride would be incomplete because iron is capable of forming two ions with different charges. The name of any iron-containing compound must reflect which iron ion is in the compound. In this case, the subscript in the formula indicates that there are three chloride ions, each with a 1− charge. Therefore, the charge of the single iron ion must be 3+. The correct name of FeCl3 is iron(III) chloride, with the cation charge written as the Roman numeral. Table below some several other examples.

Examples of Naming Using the Stock System
Formula Name
Cu2O copper(I) oxide
CuO copper(II) oxide
SnO2 tin(IV) oxide

The first two are both oxides of copper (shown in Figure below). The ratio of copper ions to oxide ions determines the name. Since the oxide ion is O2−, the charges of the copper ion must be 1+ in the first formula and 2+ in the second formula. In the third formula, there is one tin ion for every two oxide ions. This means that the tin must carry a 4+ charge, making the name tin(IV) oxide.

Copper(I) oxide, a red solid, and copper(II) oxide, a black solid, are different compounds because of the charge of the copper ion.

You can practice naming multivalent metal compounds by completing this online crossword puzzle: https://sites.google.com/site/pattihowellsciencespot/uploads/multivalent%20metals%20compounds.html?attredirects=0&d=1.

Writing Formulas for Binary Ionic Compounds

If you know the name of a binary ionic compound, you can write its formula. Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charge to cancel each other out. Consider the compound aluminum nitride. The ions are:

\begin{align*}\text{Al}^{3+} \ \ \ \ \text{N}^{3-}\end{align*}

Since the ions have charges that are equal in magnitude, one of each will be the lowest ratio of ions in the formula. The formula of aluminum nitride is AlN. The ions for the compound lithium oxide are:

\begin{align*}\text{Li}^+ \ \ \ \ \text{O}^{2-}\end{align*}

In this case, two lithium ions are required to balance out the charge of one oxide ion. The formula of lithium oxide is Li2O.

An alternative way to writing a correct formula for an ionic compound is to use the crisscross method. In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped. Shown below is the crisscross method for aluminum oxide.

The red arrows indicate that the 3 from the 3+ charge will cross over to become the subscript of the O. The 2 from the 2− charge will cross over to become the subscript of the Al. The formula for aluminum oxide is Al2O3.

Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions. In the case of aluminum nitride, the crisscross method would yield a formula of Al3N3, which is not correct. It must be reduced to AlN. Following the crisscross method to write the formula for lead(IV) oxide would involve the following steps:

The crisscross first yields Pb2O4 for the formula, but that must be reduced to the lower ratio and PbO2 is the correct formula.

Click here to watch an animation of ionic bonding: http://www.dlt.ncssm.edu/core/Chapter9-Bonding_and_Geometry/Chapter9-Animations/IonicBonding.html.

Sample Problem 7.1: Binary Ionic Compound Formulas

Write the correct formulas for the following ionic compounds:

  1. barium chloride
  2. chromium(III) oxide

Step 1: Plan the problem.

In each case, write the metal cation, followed by the nonmetal anion. Crisscross the ion charges in order to make the ionic compound neutral. Reduce to the lowest ratio if necessary.

Step 2: Solutions

Step 3: Think about your result.

The formula for barium chloride is BaCl2. Notice that the charge of the chloride ion is written as 1− rather than just – in order to do the crisscross. However, the number 1 is not used as a subscript. The formula for chromium(III) oxide is Cr2O3.

Practice Problems
  1. Write formulas for the binary ionic compounds formed between the following pairs of elements:
    1. cesium and fluorine
    2. calcium and sulfur
    3. aluminum and chlorine
    4. zinc and nitrogen
  2. Write the formula and give the name for the compound formed by the following ions:
    1. Fe3+ and O2-
    2. Ni2+ and S2-
    3. Au+ and Cl-
    4. Sn4+ and I-
  3. Give names for the following compounds:
    1. Ag2S
    2. PdO
    3. PtCl4
    4. V2O5

Ternary Ionic Compounds

Not all ionic compounds are composed of only monatomic ions. A ternary ionic compound is an ionic compound composed of three or more elements. In a typical ternary ionic compound, there is still one type of cation and one type of anion involved. The cation, the anion, or both, is a polyatomic ion.

Polyatomic Ions

A polyatomic ion is an ion composed of more than one atom. The ammonium ion consists of one nitrogen atom and four oxygen atoms. Together, they comprise a single ion with a 1+ charge and a formula of NH3+. The carbonate ion consists of one carbon atom and three oxygen atoms and carries an overall charge of 2−. The formula of the carbonate ion is CO32-. The atoms of a polyatomic ion are tightly bonded together and so the entire ion behaves as a single unit. Figure below shows several models and Table below lists many of the most common polyatomic ions.

(A) The ammonium ion (NH4+) is a nitrogen atom (blue) bonded to four hydrogen atoms (white). (B) The hydroxide ion (OH-) is an oxygen atom (red) bonded to a hydrogen atom. (C) The carbonate ion (CO32-) is a carbon atom (black) bonded to three oxygen atoms.

Common Polyatomic Ions
1- 2- 3- 1+ 2+
acetate, CH3COO- carbonate, CO32- arsenate, AsO33- ammonium, NH4+ dimercury, Hg22+
bromate, BrO3- chromate, CrO42- phosphite, PO33-
chlorate, ClO3- dichromate, Cr2O72- phosphate, PO43-
chlorite, ClO2- hydrogen phosphate, HPO42-
cyanide, CN- oxalate, C2O42-
dihydrogen phosphate, H2PO4- peroxide, O22-
hydrogen carbonate, HCO3- silicate, SiO32-
hydrogen sulfate, HSO4- sulfate, SO42-
hydrogen sulfide, HS- sulfite, SO32-
hydroxide, OH-
hypochlorite, ClO-
nitrate, NO3-
nitrite, NO2-
perchlorate, ClO4-
permanganate, MnO4-

The vast majority of polyatomic ions are anions, many of which end in –ate or –ite. Notice that in some cases such as nitrate (NO3-) and nitrite (NO2-), there are multiple anions that consist of the same two elements. In these cases, the difference between the ions is in the number of oxygen atoms present, while the overall charge is the same. As a class, these are called oxoanions. When there are two oxoanions for a particular element, the one with the greater number of oxygen atoms gets the –ate suffix, while the one with the fewer number of oxygen atoms gets the –ite suffix. The four oxoanions of chlorine are shown below.

  • ClO-, hypochlorite
  • ClO2-, chlorite
  • ClO3-, chlorate
  • ClO4-, perchlorate

In cases such as this, the ion with one more oxygen atom than the –ate anion is given a per- prefix. The ion with one fewer oxygen atom than the –ite anion is given a hypo- prefix.

Naming Ternary Ionic Compounds

The process of naming ternary ionic compounds is the same as naming binary ionic compounds. The cation is named first, followed by the anion. Some examples are shown in Table below:

Examples of Ternary Ionic Compounds
Formula Name
NaNO3 sodium nitrate
NH4Cl ammmonium chloride
Fe(OH)3 iron(III) hydroxide

When more than one polyatomic ion is present in a compound, the formula of the ion is placed in parentheses with a subscript outside of the parentheses that indicates how many of those ions are in the compound. In the last example above, there is one Fe3+ cation and three OH- anions.

Writing Formulas for Ternary Ionic Compounds

Writing a formula for a ternary ionic compound also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Use the crisscross method to ensure that the final formula is neutral. Calcium nitrate is composed of a calcium cation and a nitrate anion.

The charge is balanced by the presence of two nitrate ions and one calcium ion. Parentheses are used around the nitrate ion because more than one of the polyatomic ion is needed. If only one polyatomic ion is in a formula, parentheses are not used. As an example, the formula for calcium carbonate is CaCO3. The carbonate ion carries a 2− charge and so exactly balances the 2+ charge of the calcium ion.

Sample Problem 7.2: Ternary Ionic Compound Formulas

Write the correct formulas for the following ionic compounds:

  1. potassium sulfate
  2. zinc phosphate

Step 1: Plan the problem.

In each case, write the metal cation, followed by the nonmetal anion. Crisscross the ion charges in order to make the ionic compound neutral. Use parentheses around the polyatomic ion if more than one is present in the final formula. Reduce to the lowest ratio if necessary.

Step 2: Solutions

Step 3: Think about your result.

The formula for potassium sulfate is K2SO4. Two potassium cations with 1+ charge balance out the 2− charge of the sulfate ion. The formula for zinc phosphate is Zn3(PO4)2. Three zinc cations with a 2+ charge balance out two phosphate anions with a 3− charge.

Practice Problems
  1. Name the following compounds:
    1. NH4NO3
    2. Na2Cr2O7
    3. PbCO3
    4. Mg(CH3COO)2
  2. Write formulas for the following compounds:
    1. potassium hydrogen sulfate
    2. iron(III) oxalate
    3. sodium peroxide
    4. tin(IV) chromate

There are two polyatomic ions that produce unusual formulas. The Hg22+ ion is called either the dimercury ion or, preferably, the mercury(I) ion. When bonded with an anion with a 1− charge, such as chloride, the formula is Hg2Cl2. Because the cation consists of two Hg atoms bonded together, this formula is not reduced to HgCl. Likewise, the peroxide ion, O22-, is also a unit that must stay together in its formulas. For example, the formula for potassium peroxide is K2O2.

Watch a humorous video lecture about Naming Ionic Compounds from BossChemDude: http://www.youtube.com/watch?v=q2s8hQ5NIpE (8:30)

Lesson Summary

  • Inorganic chemistry is the study of chemical compounds that do not contain carbon.
  • Molecular formulas show the type and number of atoms that occur in a molecule. Empirical formulas show the atoms or ions in their lowest whole-number ratio in a compound. Because ionic compounds have an extended three-dimensional structure, all formulas of ionic compounds are empirical formulas.
  • Atoms of representative elements form monatomic ions by losing or gaining electrons in order to attain the electron configuration of a noble gas.
  • Most transition metals are capable of forming multiple cations with different charges. The Stock system is used to show the charge as a Roman numeral in the name.
  • The nomenclature of ionic compounds is always to name the cation first, followed by the name of the anion.
  • Ionic compounds must have no net electrical charge, so the total amount of positive charge must balance the total amount of negative charge. The crisscross method can be used to write correct formulas for ionic compounds.
  • Some ionic compounds contain polyatomic ions. A set of parentheses is used to designate more than one polyatomic ion in a formula.

Lesson Review Questions

Reviewing Concepts

  1. What element is present in all organic compounds?
  2. Write the molecular formula of a compound whose molecules contain one atom of nitrogen and three atoms of fluorine.
  3. For which three transition elements is Roman numeral not used in the names of its compounds?
  4. Give an example of each of the following: (1) a monatomic cation, (2) a monatomic anion, (3) a polyatomic cation, and (4) a polyatomic anion.
  5. What is the overall charge of an ionic compound?
  6. When are parentheses used in the formula of an ionic compound?


  1. Write the empirical formula for the following compounds:
    1. C2H6
    2. Hg2Cl2
    3. Sb2O5
    4. C8H16O2
  2. Using only the periodic table, write the symbol of the common ion formed by each of the following elements:
    1. Sr
    2. I
    3. Se
    4. Ba
    5. P
    6. Rb
    7. Al
    8. Br
  3. Name the following monatomic ions:
    1. O2-
    2. Li+
    3. W3+
    4. Cu2+
    5. Ga3+
    6. F-
  4. Write formulas for binary ionic compounds formed from each of the following pairs of elements:
    1. potassium and sulfur
    2. silver and chlorine
    3. calcium and oxygen
    4. aluminum and iodine
    5. barium and nitrogen
    6. sodium and selenium
  5. Name the following polyatomic ions:
    1. SO32-
    2. MnO4-
    3. CO32-
    4. ClO-
    5. CH3COO-
    6. H2PO4-
  6. Write the symbol and charge for the following polyatomic ions:
    1. hydrogen sulfide ion
    2. nitrate ion
    3. perchlorate ion
    4. chromate ion
    5. phosphate ion
    6. oxalate ion
  7. Name the following ionic compounds:
    1. KClO3
    2. Cd(NO3)2
    3. CuCl
    4. Ca3(PO4)2
    5. NaCN
    6. (NH4)2SO3
    7. Li2O2
    8. PbS2
    9. Rb2SO4
    10. Mn3P2
    11. NiCO3
    12. Co(OH)3
    13. Hg2Br2
    14. Zn(NO2)2
  8. Write correct formulas for the following ionic compounds:
    1. copper(II) bromide
    2. aluminum hydroxide
    3. silver sulfide
    4. barium acetate
    5. mercury(II) nitrate
    6. lead(II) chromate
    7. potassium permanganate
    8. sodium hydrogen carbonate
    9. calcium silicate
    10. tin(IV) sulfate
    11. ammonium phosphate
    12. gold(III) fluoride
    13. magnesium bromate
    14. chromium(VI) oxide

Further Reading / Supplemental Links

Points to Consider

The large class of inorganic chemical compounds can be sub-classified into ionic compounds and molecular compounds.

  • How is the structures of a molecular compound different from the structure of an ionic compound?
  • Relate molecular compounds to the Law of Multiple Proportions.
  • What is the nomenclature of molecular compounds?

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