Writing Ionic Compound Formulas: Binary & Polyatomic Compounds

  • 0:01 Simple Binary Ionic Compounds
  • 3:20 Ionic Compounds
  • 5:40 Binary Examples
  • 8:15 Polyatomic Compounds
  • 8:45 Polyatomic Examples
  • 11:52 Lesson Summary
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Lesson Transcript
Instructor: Elizabeth (Nikki) Wyman

Nikki has a master's degree in teaching chemistry and has taught high school chemistry, biology and astronomy.

In this lesson, you will learn how to write the chemical formulas for both binary ionic compounds and polyatomic ionic compounds when you are given only the name of the compound. You will see that it is actually quite simple when you learn the steps described in this lesson.

Sample Binary Ionic Compounds

Before you start this video, it is helpful to know how to name an ionic compound based on its chemical formula. In this lesson, you will learn how to start with the name of the ionic compound and turn it in to a chemical formula.

Binary ionic compounds are fairly simple. An ionic compound is a neutrally charged compound that is made up of bonded ions, a cation and an anion. The cation has a positive charge, and the anion has a negative charge. When the two combine into a compound, that compound does not have an overall charge. NaCl is an example. Na+ (a cation with a +1 charge) combines with Cl- (an anion with a -1 charge) to make NaCl (sodium chloride).

Before you learn the steps for writing a chemical formula, I'd like to remind you how to determine the charge on an ion. For the representative elements, the charge of the ion is related to the column or group that the element is in.

  • Group IA elements have only one valence electron, so when they lose that electron they will have a +1 charge.
  • Group IIA elements have two valence electrons. When they lose their two valence electrons they will have a +2 charge.
  • Group IIIA elements have three valence electrons. They lose their three electrons to form +3 ions.
  • Group IVA elements are somewhat of an exception to the trend. Tin (SN) and lead (PB) can lose multiple electrons to form differently charged ions. Carbon, silicon and germanium rarely form ions.
  • Group VA elements have five valence electrons. Instead of losing these five electrons, Group VA elements will gain three electrons to have a -3 charge.
  • Group VIA elements have six valence electrons and gain two electrons to have a -2 charge.
  • Group VIIA elements have seven electrons in their outer shell. These elements gain one electron to have a -1 charge.

Transition metals are elements that live in groups IB to XB. These metals are capable of losing different numbers of electrons and can take multiple ionic forms. The names of transition metal ions contain Roman numerals to indicate the ions' charge.

For example, lead (II) nitrate contains a +2 lead ion: Pb2+. Vanadium (IV) oxide contains a +4 vanadium ion: V4+.

Ionic Compounds

If you have to write the chemical formula of a simple, binary ionic compound given the name of the compound, you follow a set of three steps. Let's go through them using magnesium chloride as an example.

  1. Write the symbols for the cation and the anion: Mg and Cl.
  2. Determine the charge on the cation and anion. If the cation has a Roman numeral after it, that is the charge on that cation. Cations receive Roman numerals when they can take more than one ionic form. If there is no Roman numeral, you can determine the charge from the cation's position on the periodic table. Such is the case with magnesium. Magnesium is in column IIA so it has a charge of +2. Magnesium 2+: Mg2+. Chlorine is in column VIIA so it has a charge of -1: Cl-.
  3. Determine formula. Write the two symbols together and determine how to make the compound neutral by finding the lowest common multiple of the charges on each ion. Then figure out how atoms of each element are needed to make that charge. Mg2+ Cl-. The lowest common multiple is two. To get a charge of two on magnesium, you multiply it by one - (2 * 1 = 2). So, just one magnesium. To get a charge of two on chloride, which has a -1 charge, you need to multiply it by two - (1 * 2 = 2). So you will have two chloride ions, or Cl2.
    1. Alternative step 3: the drop and swap method. The magnitude of the charge on the ion becomes the number of the opposite ion needed. Ignoring + and - signs, drop the number of the charge to subscript position: Mg2 Cl. If the charge on the ion is simply one, you can ignore it. Now swap the subscript's places: Mg Cl2. Now you have your formula.

Binary Examples

Let's try an example: sodium oxide.

  1. Write Na and O.
  2. Write Na+ and O2-.
  3. Na+ O2-. The lowest common multiple is two. To get a charge of two on sodium, which has a +1 charge, you need to multiply it by 2, (1 * 2 = 2). So you will have two sodium atoms or Na2. To get a charge of 2 on the oxygen, you multiply it by one - (2 * 1 = 2). So just one O. The final formula is Na2O.

If we want to do the drop and swap method instead, we go back to step 2, where we had our ions and their charges: Na+ O2-. First we ignore + and - signs, then drop the number of the charge down to subscript position: Na O2. Since the charge on sodium is just one, we don't need to put anything down. Now we swap the subscript's places. Na2 O. Now we have our formula.

Let's try a more difficult one: iron (III) oxide.

  1. Write Fe and O.
  2. Write Fe3+ and O2-. You know the Iron is 3+ because of the Roman numeral III after it.
  3. Fe3+ O2-. The lowest common multiple between three and two is six. So to get a charge of six on the iron, you need to have two iron atoms because 3 * 2 = 6: Fe2. To get a charge of six on the oxygen, you need three of them because 2 * 3 = 6: O3. So the final formula is Fe2O3, and this produces a neutral compound.

We can also do step three using the drop and swap method. Starting with our charged ions Fe3+ and O2-, we ignore the + and - signs, then drop the numbers of our charges down: Fe3 O2. Now we swap the position of these subscripts: Fe2 O3. We arrive with our formula Fe2 O3.

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