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

Expert verified
Arrange the groups of atoms in order of increasing size as follows: 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

01

a. Arrange \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\) in order of increasing size

We can find these elements in the same group (Group 16) of the periodic table. Since atomic size increases from top to bottom in a group, we can arrange them in the order of their position in the periodic table: S < Se < Te So the order of increasing size for these atoms is \(\mathrm{S}, \mathrm{Se}, \mathrm{Te}\).
02

b. Arrange \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\) in order of increasing size

These elements belong to different groups but they are in the same period (Period 4). Since atomic size decreases from left to right in a period, we can arrange them in the order of their position in the periodic table: K > Ni > Br So the order of increasing size for these atoms is \(\mathrm{Br}, \mathrm{Ni}, \mathrm{K}\).
03

c. Arrange \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\) in order of increasing size

These elements are neither in the same group nor in the same period. Therefore, we have to use both the group and period trends to determine their order. Ba is in Group 2 and Period 6. Si is in Group 14 and Period 3. F is in Group 17 and Period 2. Comparing Ba and Si: Ba has more energy levels than Si since it is below Si in the periodic table. Thus, Ba has a larger atomic size than Si. Comparing Si and F: Si is left of F in the periodic table within Period 3. Thus, Si has a larger atomic size than F. From the above comparisons, the order of increasing size for these atoms is \(\mathrm{F}, \mathrm{Si}, \mathrm{Ba}\).

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