You can read my previous posts (Oct 20<\/a> and Oct 25<\/a>) about writing ionic formulas.\u00a0 Assuming you’re up to speed, we’ll discuss how to do it with transition metals.<\/p>\n The transition metals are in the low area in the middle of the periodic table.\u00a0 You’ll recognize iron (Fe), copper (Cu) and others among them.\u00a0\u00a0Most transition metals\u00a0have more than one possible combining capacity – in that way, they are different from the other elements.\u00a0 Of course, we know that to write an ionic formula, we need to know the combining capacity of the metal and of the nonmetal.\u00a0 How do we proceed?<\/p>\n The tell is the Roman Numeral in the written formula.\u00a0 For instance, you might\u00a0encounter iron(III)sulphate.\u00a0 Note the III in the middle of the compound.\u00a0 Only formulas with transition metals have a Roman Numeral.\u00a0 The Roman Numeral tells you the combining capacity of the metal.\u00a0 In the case of iron(III)sulphate, the combining capacity of iron is 3.\u00a0 From the point of view of positive and negative, metals are always positive.\u00a0 Therefore, you can imagine the charge on the iron to be +3.<\/p>\n Knowing (from the Roman Numeral) the combining capacity (i.e., the\u00a0charge)\u00a0of the metal, we proceed as I’ve described in previous posts:\u00a0 positives have to equal negatives.<\/p>\n iron:\u00a0 Fe3+<\/span><\/sup> Iron brings 3+; sulphate brings 2-. The way to make the positives equal the negatives is to go to six each:\u00a0 2×3+=6+, and 3×2-=6-.\u00a0 Therefore, we need two irons and three sulphates:<\/p>\n Fe2<\/span><\/sub>(SO4<\/span><\/sub>)3<\/span><\/sub>.<\/p>\n So, the formula of iron(III)sulphate is Fe2<\/span><\/sub>(SO4<\/span><\/sub>)3<\/span><\/sub>.<\/p>\n 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<\/span><\/sub>. Consider the nitrate (you can look up its charge on a table of ions): Remember, you only use Roman Numerals with transition metals<\/em> – not the others – and only in written<\/em> formulas.<\/p>\n Hope this helps.<\/p>\n
\nsulphate: SO4<\/span><\/sub>2-<\/span><\/sup><\/p>\n
\n\u00a0
\nnitrate: NO3<\/span><\/sub>–
\n<\/span><\/sup>
\nSo 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.<\/p>\n