What Is The Correct Formula For Chromium(IV)Oxide?

What Is The Correct Formula For Chromium(IV)Oxide?

First of all, we have to realize that this formula is of an ionic compound. The reason it is ionic is because it contains a metal and a non-metal. We have chromium which is a metal and oxide ion or oxygen which is a non-metal. In order to figure out the correct formula for an ionic compound, we have to first figure out the correct oxidation numbers(charges) for the two components of our compound. I have chromium and I have oxygen now what is the oxidation number or charge of chromium. If you have a roman numeral that is the charge of your metal. Transition metals will have roman numerals. Therefore, if you're given a name that has a roman numeral the roman numeral is the charge of your metal. In this case it's four plus. Now what is the charge of oxygen? You can just remember that group 17 will have a charge of minus 1 and group 16 will have a charge of negative 2. Oxygen is in group 16 and it has a charge of negative 2. Once we have these numbers what we can do is we can cross multiply them so that each partner gets its partner's number. We get Cr204. However 2 and 4 can be divided further by 2 to get CrO2