Question

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. \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Short answer

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

Step-by-step solution

Identify the position of the elements in the periodic table

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.

Arrange elements according to trends in atomic size

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 \(\mathrm{F} < \mathrm{Si} < \mathrm{Ba}\).

Final Answer

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