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

The order of increasing atomic size for each group is: - Group a: \(S < Se < Te\) - Group b: \(Br < Ni < K\) - Group c: \(F < Si < Ba\)

Step-by-step solution

Group a: \(\mathrm{Te}, \mathrm{S}, \mathrm{Se}\)

All three elements belong to group 16, and they are all Chalcogens. As we move down a group in the periodic table, the atomic size increases due to the addition of electron shells. Thus, in this group, the order of increasing size will be: S < Se < Te

Group b: \(\mathrm{K}, \mathrm{Br}, \mathrm{Ni}\)

These elements belong to different groups and periods. K is in group 1 and period 4, Br is in group 17 and period 4, and Ni is in group 10 and period 4. Generally, as we move across a period from left to right, the atomic size decreases because of the increasing effective nuclear charge. In this case, we have: K > Ni > Br

Group c: \(\mathrm{Ba}, \mathrm{Si}, \mathrm{F}\)

Ba is in group 2 and period 6, Si is in group 14 and period 3, and F is in group 17 and period 2. As mentioned earlier, atomic size increases as we move down a group (due to the addition of electron shells) and decreases as we move from left to right across a period (due to increased effective nuclear charge). Considering these periodic trends, the order of increasing size in this group is: F < Si < Ba In conclusion, for each group, the order of increasing atomic size is as follows: - Group a: S < Se < Te - Group b: Br < Ni < K - Group c: F < Si < Ba