Question

Order the atoms in each of the following sets from the least exothermic electron affinity to the most. a. $\mathrm{S},\mathrm{Se}$ b. $\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{I}$

Short answer

a. From least exothermic electron affinity to the most: $\mathrm{Se}<\mathrm{S}$ b. From least exothermic electron affinity to the most: $\mathrm{I}<\mathrm{Br}<\mathrm{Cl}<\mathrm{F}$

Step-by-step solution

Analyze the sets of atoms

We are given two sets of atoms: a. S, Se b. F, Cl, Br, I Set (a) contains sulfur (S) and selenium (Se), which are present in the same group (Group 16). Since they are in the same group, we can use the trend of electron affinity in a group to compare their electron affinities. Set (b) contains fluorine (F), chlorine (Cl), bromine (Br), and iodine (I), which are present in the same group (Group 17). We can also use the trend of electron affinity in a group to compare their electron affinities.

Apply the trend of electron affinity in groups

According to the trend, electron affinity becomes less exothermic as we move down a group. Based on this trend, we can now order the given atoms. Set (a): S is above Se in Group 16. So, S has a more exothermic electron affinity than Se. Therefore, the order is Se < S. Set (b): Since F, Cl, Br, and I are in the same group, we can arrange them in the order of least to most exothermic electron affinity from bottom to top: I < Br < Cl < F.

Present the final order

Now that we have ordered the atoms in the given sets, here's the final order: a. From least exothermic electron affinity to the most: Se < S b. From least exothermic electron affinity to the most: I < Br < Cl < F