Predict the empirical formulas of the ionic compounds formed from the following pairs of elements. Name each compound. a. \(\mathrm{Li}\) and \(\mathrm{N}\) c. \(\mathrm{Rb}\) and \(\mathrm{Cl}\) b. Ga and \(\mathrm{O}\) d. \(\mathrm{Ba}\) and \(\mathrm{S}\)

To predict the empirical formula of each compound, we need to first identify the ionic charges of the elements involved. We can determine their charges based on their position in the Periodic Table of elements. a. Li (lithium) is an alkali metal in Group 1, so it has a +1 charge. N (nitrogen) is a non-metal in Group 15, so it has a -3 charge. c. Rb (rubidium) is an alkali metal in Group 1, so it has a +1 charge. Cl (chlorine) is a halogen in Group 17, so it has a -1 charge. b. Ga (gallium) is a metal in Group 13, so it has a +3 charge. O (oxygen) is a non-metal in Group 16, so it has a -2 charge. d. Ba (barium) is an alkaline earth metal in Group 2, so it has a +2 charge. S (sulfur) is a non-metal in Group 16, so it has a -2 charge.

We use the criss-cross method by taking the absolute value of the charges and using them as subscripts. For example, if an element has a +2 charge and its partner has a -3 charge, the empirical formula will be written as X2Y3 (X and Y being the respective elements). a. Li (+1) and N (-3): The empirical formula is Li3N (3 lithium atoms and 1 nitrogen atom) c. Rb (+1) and Cl (-1): The empirical formula is RbCl (1 rubidium atom and 1 chlorine atom) b. Ga (+3) and O (-2): The empirical formula is Ga2O3 (2 gallium atoms and 3 oxygen atoms) d. Ba (+2) and S (-2): The empirical formula is BaS (1 barium atom and 1 sulfur atom)

Now that we have the empirical formulas, we can name the compounds as follows: a. Li3N: Lithium nitride c. RbCl: Rubidium chloride b. Ga2O3: Gallium oxide d. BaS: Barium sulfide