## Chemistry: Molecular Approach (4th Edition)

a. $CaO$ b. $ZnS$ c. $RbBr$ d. $Al_2O_3$
Use the table 3.2, and the figure 3.7 (Page 98), to see the charges of the ions: a. Calcium is from the second column of the periodic table, and its symbol is "Ca", so its ion will be: $Ca^{2+}$ (See figure 3.7) The oxygen ion is "$O^{2-}$" (See table 3.2) Put these together, where the subscript number will be the charge of the other ion. In this case, the charge of $Ca^{2+}$ is 2, so the subscript number for oxygen is 2. The charge of $O^{2-}$ is -2, so the subscript of calcium will also be 2. $Ca_2O_2$, which simplifies to: $CaO$. b. Zinc has a charge of $2+$ (See figure 3.7), so its ion will be: $Zn^{2+}$ The sulfur ion is "$S^{2-}$" (See table 3.2) Put these together, where the subscript number will be the charge of the other ion: $Zn_2S_2$, which simplifies to: $ZnS$. c. Rubidium is from the first column of the periodic table, and its symbol is "Rb", so its ion will be: $Rb^{+}$ (See figure 3.7) The bromine ion is "$Br^{-}$" (See table 3.2) Put these together, where the subscript number will be the charge of the other ion. In this case, the charge of $Ca^{2+}$ is 2, so the subscript number for oxygen is 2. The charge of $O^{2-}$ is -2, so the subscript of calcium will also be 2. $Rb_1Br_1$ = $RbBr$. d. Aluminium has a charge of $3+$ (See figure 3.7) , so its ion will be: $Al^{3+}$ The oxygen ion is "$O^{2-}$" (See table 3.2) Put these together, where the subscript number will be the charge of the other ion. In this case, the charge of $Al^{3+}$ is 3, so the subscript number for oxygen is 3. The charge of $O^{2-}$ is -2, so the subscript of aluminium will be 2. $Al_2O_3$.