1.1 Formulae of Ionic Compounds 1.1.1 Ratio of Ions in Ionic Compounds 1.1.1.1 Example(1) Cross Method(A) Calcium Chloride 1.1.1.1 Example(1) Cross Method(B) Aluminium Sulphate 1.1.1.2 Example(2) Jig-saw Method(A) Calcium Chloride 1.1.1.2 Example(2) Jig-saw Method(B) Aluminium Sulphate 1.1.1.3 Questions on Cross Method Level 1 Questions 1.1.1.3 Questions on Cross Method Level 2 Questions 1.1.1.4 Questions on Jig-saw Method Level 1 Questions 1.1.1.4 Questions on Jig-saw Method Level 2 Questions 1.1.2 Number of Ions from an Ionic Compounds 1.1.2.1 Questions on Total Number of Ions from an Ionic Compounds
1.1.2 Number of ions from an ionic compound Ionic compounds consisting of simple ions: Total number of ions contained in a given ionic compound is equal to the sum of the positive and the negative ions. e.g. one formula unit of CaCl2 consists of 1 Ca2+ ion and 2 Cl– ions, i.e. a total of 3 ions. Ionic compounds consisting of polyatomic ions: Total number of ions contained in a given ionic compound is again equal to the sum of the positive and the negative ions, with polyatomic ions being considered as single units. e.g. one formula unit of Al2(SO4)3 consists of 2 Al3+ ions and 3 (SO4)2– ions i.e. a total of 5 ions. The S and O atoms of SO42– do not form S2– ion or O2–ion.