For each pair, indicate which atom loses an electron more easily: (a) \(\mathrm{Na}\) or \(\mathrm{K}\) (b) Na or Al (c) \(\mathrm{Rb}\) or \(\mathrm{Ca}\) (d) \(\mathrm{S}\) or \(\mathrm{Cl}\) (e) Br or Kr

Short Answer

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In summary, the atoms that lose an electron more easily in each pair are: (a) K (b) Na (c) Rb (d) S (e) Br

Step by step solution

01

(a) Na or K

In this pair, we have sodium (Na) versus potassium (K). Both elements are in the same group (Group 1) in the periodic table. When comparing elements in the same group, the one lower in the group has a larger atomic size and thus loses an electron more easily. Therefore, potassium (K) loses an electron more easily than sodium (Na).
02

(b) Na or Al

In this pair, we have sodium (Na) versus aluminum (Al). Sodium is in Group 1, while aluminum is in Group 13. Sodium is further to the left in the periodic table, which means it has a lower ionization energy than aluminum. Therefore, sodium (Na) loses an electron more easily than aluminum (Al).
03

(c) Rb or Ca

In this pair, we have rubidium (Rb) versus calcium (Ca). Rubidium is in Group 1, while calcium is in Group 2. Comparing elements in different groups but along the same period, the one to the left has a larger atomic size and thus loses an electron more easily. Therefore, rubidium (Rb) loses an electron more easily than calcium (Ca).
04

(d) S or Cl

In this pair, we have sulfur (S) versus chlorine (Cl). Both elements are in the same period (Period 3) in the periodic table. When comparing elements in the same period, the one further to the left has a lower ionization energy and loses electrons more easily. Therefore, sulfur (S) loses an electron more easily than chlorine (Cl).
05

(e) Br or Kr

In this pair, we have bromine (Br) versus krypton (Kr). Bromine is in Group 17, while krypton is in Group 18. When comparing elements in different groups in the same period, the one further to the left has a lower ionization energy and loses an electron more easily. Therefore, bromine (Br) loses an electron more easily than krypton (Kr).

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