For each of the following atomic numbers, use the periodic table to write the formula (including the charge) for the simple ion that the element is most likely to form in ionic compounds. a. 13 b. 34 c. 56 d. \(\underline{7}\) e. 87 f. 35

Use the periodic table to find the elements corresponding to the atomic numbers. Here are the elements corresponding to the given atomic numbers: a. 13: Aluminum (Al) b. 34: Selenium (Se) c. 56: Barium (Ba) d. \(\underline{7}\): Nitrogen (N) e. 87: Francium (Fr) f. 35: Bromine (Br)

Using the periodic table, we can make predictions about the charge of the ion that each element is most likely to form. Elements in Groups 1, 2, and 13 tend to lose electrons, whereas elements in Groups 15, 16, and 17 tend to gain electrons. The charges of the simple ions usually correspond to the number of electrons gained or lost to achieve a stable noble gas configuration (a full outer electron shell). a. Aluminum (Al) is in Group 13, so it is likely to lose 3 electrons and form a +3 ion: Al^{3+} b. Selenium (Se) is in Group 16, so it is likely to gain 2 electrons and form a -2 ion: Se^{2-} c. Barium (Ba) is in Group 2, so it is likely to lose 2 electrons and form a +2 ion: Ba^{2+} d. Nitrogen (N) is in Group 15, so it is likely to gain 3 electrons and form a -3 ion: N^{3-} e. Francium (Fr) is in Group 1, so it is likely to lose 1 electron and form a +1 ion: Fr^{+} f. Bromine (Br) is in Group 17, so it is likely to gain 1 electron and form a -1 ion: Br^{-}

Now that we have determined the charge for each ion, we can write the formula for the simple ion of each element. a. Aluminum: Al^{3+} b. Selenium: Se^{2-} c. Barium: Ba^{2+} d. Nitrogen: N^{3-} e. Francium: Fr^{+} f. Bromine: Br^{-}