Chemistry: Ionic Compounds with the Transition Metals

Tutoring chemistry, ionic compounds with the transition metals often need extra attention because of the Roman Numerals.

You can read my previous posts (Oct 20 and Oct 25) about writing ionic formulas.  Assuming you’re up to speed, we’ll discuss how to do it with transition metals.

The transition metals are in the low area in the middle of the periodic table.  You’ll recognize iron (Fe), copper (Cu) and others among them.  Most transition metals have more than one possible combining capacity – in that way, they are different from the other elements.  Of course, we know that to write an ionic formula, we need to know the combining capacity of the metal and of the nonmetal.  How do we proceed?

The tell is the Roman Numeral in the written formula.  For instance, you might encounter iron(III)sulphate.  Note the III in the middle of the compound.  Only formulas with transition metals have a Roman Numeral.  The Roman Numeral tells you the combining capacity of the metal.  In the case of iron(III)sulphate, the combining capacity of iron is 3.  From the point of view of positive and negative, metals are always positive.  Therefore, you can imagine the charge on the iron to be +3.

Knowing (from the Roman Numeral) the combining capacity (i.e., the charge) of the metal, we proceed as I’ve described in previous posts:  positives have to equal negatives.

iron:  Fe3+
sulphate: SO42-

Iron brings 3+; sulphate brings 2-. The way to make the positives equal the negatives is to go to six each:  2×3+=6+, and 3×2-=6-.  Therefore, we need two irons and three sulphates:

Fe2(SO4)3.

So, the formula of iron(III)sulphate is Fe2(SO4)3.

Note that, when writing the word formula from the symbolic formula (the opposite way from what we’ve just done), you need to include the Roman Numeral. For example, suppose you have CuNO3. Consider the nitrate (you can look up its charge on a table of ions):
 
nitrate: NO3

So we know that the charge on nitrate is 1-. Since there is one nitrate and one copper, the charge on the copper must be 1+ so that the positives and negatives are equal. Therefore, we would write this formula as copper(I)nitrate: the Roman Numeral shows the numeric charge on the copper.

Remember, you only use Roman Numerals with transition metals – not the others – and only in written formulas.

Hope this helps.

Jack of Oracle Tutoring By Jack and Diane, Campbell River, BC.

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