Ionic compounds are formed from a positive cation interacting with a negative anion.  Before we jump into how to write ionic compounds, let’s first look at our element symbol and identify the oxidation state number and the subscript.

In the above example, Fe indicates iron.  The 3+ that is superscript to the Fe, identifies the oxidation state.  When the oxidation state is positive, we are dealing with a cation, an element that has lost one or more electrons and is now positively charged.  In this example Fe has lost electrons and therefore has a positive charge of three.  The number 2, written as a subscript to Fe indicates the number of Fe ions present.  In this case Fe2 indicates that there are 2 Fe ions.

When writing ionic compounds, the overall compound has a net charge of zero.  In other words the number of positive charges from the cations equal the number of negative charges on the anions.  Let’s try a simple, straightforward example.  Write the ionic formula for potassium chloride:

In the above example, potassium (K) has 1 positive charge and chlorine (Cl) balances that positive charge with its single negative charge.  Let’s try a more difficult problem, Tin (IV) oxide.  For a review on naming binary compounds, see this post.  Our first step is to write the symbols for tin and oxygen as well as identify the oxidation state for each.  The Roman numeral tells us the oxidation state for Sn is +4.  Oxygen, being a group 6 element, has an oxidation state of -2.

Our next step is to balance the number of positive and negative charges.  We first draw a circle and write our symbols with their oxidation state inside.  All the charges inside the circle must add up to zero.

Since the charges do not balance and the sum does not equal zero, we must add another Oxygen with its two negative charges to the circle.  This will bring the total number of negative charges to four.  These negative charges would then balance the four positive charges supplied by Tin.

Once our charges balance inside the circle, we count the number of each type of ion inside the circle.  In this example we have one tin ion and two oxygen ions.  The number of ions will be our subscript values in our final answer.  There is no need to write the oxidation state superscripts in our final answer.

Here is a practice problem.  Try writing the ionic formula for Aluminum (with a 3+ oxidation state) and Oxygen (with its 2- oxidation state.)  Try the circle method and if you have problems, post here or on the forums.

1 Comment

  • Linh says:

    Cool insights! I have already been searching for something such as this for a little bit now. Thanks for the tips!