Question

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{V}, \mathrm{V}^{2+}, \mathrm{V}^{3+}, \mathrm{V}^{5+}\) b. \(\mathrm{Na}^{+}, \mathrm{K}^{+}, \mathrm{Rb}^{+}, \mathrm{Cs}^{+}\) c. \(\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}\) d. \(\mathrm{P}, \mathrm{P}^{-}, \mathrm{P}^{2-}, \mathrm{P}^{3-}\) e. \(\mathrm{O}^{2-}, \mathrm{S}^{2-}, \mathrm{Se}^{2-}, \mathrm{Te}^{2-}\)

Short answer

a. V > V²⁺ > V³⁺ > V⁵⁺ b. Cs⁺ > Rb⁺ > K⁺ > Na⁺ c. Te²⁻ > I⁻ > Cs⁺ > Ba²⁺ d. P³⁻ > P²⁻ > P⁻ > P e. Te²⁻ > Se²⁻ > S²⁻ > O²⁻

Step-by-step solution

Compare charges of ions

In this set of species, the protons are the same in number for all species since they are different ions of vanadium. The difference lies in the number of electrons and thus the radii will be affected by the charges of these ions. As the charge increases, the radii decreases because of increased effective nuclear charge experienced by the outermost electrons. Now, arrange them in decreasing size order: V > V²⁺ > V³⁺ > V⁵⁺. b. Na⁺, K⁺, Rb⁺, Cs⁺

Compare atomic radii

In this series, all the species are in their +1 charged states. Since they all have the same charge, we must now consider the atomic radii. For a group of elements on the periodic table, the atomic radius increases as you move down the group. In the case of these species, Cesium has the largest atomic radius and sodium has the smallest. So the order of decreasing size will be: Cs⁺ > Rb⁺ > K⁺ > Na⁺. c. Te²⁻, I⁻, Cs⁺, Ba²⁺ (Given in the order of decreasing size)

Identify same charges and atomic radii trends

In this series, we have 4 species with different charges and elements from different groups. First, we will compare same charged ions. Te²⁻, I⁻ are negatively charged ions, having more electrons and larger atomic radii. With Te²⁻ having more electrons than I⁻, thus it has a larger size. Secondly, let's compare the Cs⁺ and Ba²⁺, which are both positively charged ions. Since they belong to the same group and going down to Ba²⁺, the atomic radii increase. So, the order for these species is Cs⁺> Ba²⁺. Thirdly, let's compare I⁻ and Cs⁺, I⁻ has a greater atomic number and thus a larger size compared to Cs⁺.

Arrange the species in the order of decreasing size

Based on the comparisons in step 1, the order of decreasing size for this group of species is: Te²⁻ > I⁻ > Cs⁺ > Ba²⁺. d. P, P⁻, P²⁻, P³⁻

Compare charges of ions

Like in the first part of the exercise, we will compare the charges of these ions. All are different ions of phosphorus. As the charge becomes more negative, the number of electrons increases and so does the size, because of decreased effective nuclear charge experienced by the outermost electrons. Now, arrange them in decreasing size: P³⁻ > P²⁻ > P⁻ > P. e. O²⁻, S²⁻, Se²⁻, Te²⁻

Compare atomic radii

In this series, all the species have the same -2 charge and belong to the same group in the periodic table. So, we just have to compare their atomic radii. As before, the atomic radius increases as you move down the group. So, the order of decreasing size for these species will be: Te²⁻ > Se²⁻ > S²⁻ > O²⁻.