Question

Predict the chemical formula of the ionic compound formed between the following pairs of elements: (a) \(\mathrm{Al}\) and \(\mathrm{F}\), (b) \(\mathrm{K}\) and \(\mathrm{S}\), (c) \(\mathrm{Y}\) and \(\mathrm{O}\), (d) \(\mathrm{Mg}\) and \(\mathrm{N}\).

Short answer

The chemical formulas of the ionic compounds formed between the given pairs of elements are as follows: (a) \(\mathrm{AlF_3}\), (b) \(\mathrm{K_2S}\), (c) \(\mathrm{Y_2O_3}\), and (d) \(\mathrm{Mg_3N_2}\).

Step-by-step solution

Identify the charges on the elements

We can find the charges of the elements by using the periodic table and looking at their groups. Elements in Group 1 form ions with a +1 charge, Group 2 with a +2 charge, Group 13 with a +3 charge, Group 15 with a -3 charge, Group 16 with a -2 charge, and Group 17 with a -1 charge. (a) \(\mathrm{Al}\) and \(\mathrm{F}\): \(\mathrm{Al}\) is in Group 13, so it forms a +3 ion. \(\mathrm{F}\) is in Group 17, so it forms a -1 ion. (b) \(\mathrm{K}\) and \(\mathrm{S}\): \(\mathrm{K}\) is in Group 1, so it forms a +1 ion. \(\mathrm{S}\) is in Group 16, so it forms a -2 ion. (c) \(\mathrm{Y}\) and \(\mathrm{O}\): \(\mathrm{Y}\) (yttrium) is in Group 3, so it forms a +3 ion. \(\mathrm{O}\) is in Group 16, so it forms a -2 ion. (d) \(\mathrm{Mg}\) and \(\mathrm{N}\): \(\mathrm{Mg}\) is in Group 2, so it forms a +2 ion. \(\mathrm{N}\) is in Group 15, so it forms a -3 ion.

Determine the chemical formula

Now, we will determine the chemical formula by satisfying the electrical neutrality in the compound by ensuring that the total positive charge equals the total negative charge. (a) \(\mathrm{Al}\) and \(\mathrm{F}\): We need 3 \(\mathrm{F}\) ions (-1 each) for every \(\mathrm{Al}\) ion (+3) to satisfy neutrality: \(\mathrm{AlF_3}\) (b) \(\mathrm{K}\) and \(\mathrm{S}\): We need 2 \(\mathrm{K}\) ions (+1 each) for every \(\mathrm{S}\) ion (-2) to satisfy neutrality: \(\mathrm{K_2S}\) (c) \(\mathrm{Y}\) and \(\mathrm{O}\): We need 2 \(\mathrm{Y}\) ions (+3 each) for every 3 \(\mathrm{O}\) ions (-2 each) to satisfy neutrality: \(\mathrm{Y_2O_3}\) (d) \(\mathrm{Mg}\) and \(\mathrm{N}\): We need 3 \(\mathrm{Mg}\) ions (+2 each) for every 2 \(\mathrm{N}\) ions (-3 each) to satisfy neutrality: \(\mathrm{Mg_3N_2}\) We have the chemical formulas of the ionic compounds for each pair of elements.