Arrange the following groups of atoms in order of increasing size. a. \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) b. \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) c. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

To arrange the elements by increasing atomic size, we need to know their positions in the periodic table. You can find the position of each element using a periodic table. a. \(\mathrm{Te}\) is a group 16, period 5 element; \(\mathrm{S}\) is a group 16, period 3 element; \(\mathrm{Se}\) is a group 16, period 4 element. b. \(\mathrm{K}\) is a group 1, period 4 element; \(\mathrm{Br}\) is a group 17, period 4 element; \(\mathrm{Ni}\) is a group 10, period 4 element. c. \(\mathrm{Ba}\) is a group 2, period 6 element; \(\mathrm{Si}\) is a group 14, period 3 element; \(\mathrm{F}\) is a group 17, period 2 element.

Recall that atomic size decreases across a period from left to right and increases down a group from top to bottom. Keep these trends in mind while arranging the given elements: a. \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\): All three elements belong to group 16, so we need to compare their sizes based on periods. \(\mathrm{S}\) (period 3) < \(\mathrm{Se}\) (period 4) < \(\mathrm{Te}\) (period 5), therefore the order is \(\mathrm{S} < \mathrm{Se} < \mathrm{Te}\). b. \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\): All three elements belong to period 4, so we need to compare their sizes based on groups. \(\mathrm{K}\) (group 1) > \(\mathrm{Ni}\) (group 10) > \(\mathrm{Br}\) (group 17), so the order is \(\mathrm{Br} < \mathrm{Ni} < \mathrm{K}\). c. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\): These elements belong to different groups and periods, so we need to consider both factors. Based on periods, we've \(\mathrm{F}\) (period 2) < \(\mathrm{Si}\) (period 3) < \(\mathrm{Ba}\) (period 6). Since the group trends do not contradict this order, our final arrangement is \(\mathrm{F} < \mathrm{Si} < \mathrm{Ba}\).

a. \(\mathrm{S} < \mathrm{Se} < \mathrm{Te}\) b. \(\mathrm{Br} < \mathrm{Ni} < \mathrm{K}\) c. \(\mathrm{F} < \mathrm{Si} < \mathrm{Ba}\)