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Chapter 8: Problem 57

For each of the following groups, place the atoms and/or ions in order of decreasing size. a. \(\mathrm{Cu}, \mathrm{Cu}^{+}, \mathrm{Cu}^{2+}\) b. \(\mathrm{Ni}^{2+}, \mathrm{Pd}^{2+}, \mathrm{Pt}^{2+}\) c. \(\mathrm{O}, \mathrm{O}^{-}, \mathrm{O}^{2-}\) d. \(\mathrm{La}^{3+}, \mathrm{Eu}^{3+}, \mathrm{Gd}^{3+}, \mathrm{Yb}^{3+}\) e. $\mathrm{Te}^{2-}, \mathrm{I}^{-}, \mathrm{Cs}^{+}, \mathrm{Ba}^{2+}, \mathrm{La}^{3+}$

Short Answer

Expert verified
a. Cu > Cu⁺ > Cu²⁺ b. Pt²⁺ > Pd²⁺ > Ni²⁺ c. O²⁻ > O⁻ > O d. La³⁺ > Eu³⁺ > Gd³⁺ > Yb³⁺ e. Te²⁻ > I⁻ > Cs⁺ > Ba²⁺ > La³⁺

Step by step solution

01

a. Cu, Cu⁺, Cu²⁺

The atomic sizes are affected by ionic charges. The more positive the charge, the smaller the size of the ion due to increased effective nuclear charge and less repulsion between electrons. Here, all species belong to the same element (Copper). Therefore, ionic size decreases with an increase in positive charge. So the order is: Cu > Cu⁺ > Cu²⁺.
02

b. Ni²⁺, Pd²⁺, Pt²⁺

In this group, we have three ions with the same charge (+2). We can compare their sizes based on atomic radii in the periodic table. The size of the atom increases as we move down a group in the periodic table. Ni, Pd, and Pt belong to the same group and Pt being the lowest, will have the largest radius. Therefore, the order is: Pt²⁺ > Pd²⁺ > Ni²⁺.
03

c. O, O⁻, O²⁻

In this group, the atomic size increases with an increasing number of negative charges. This is due to decreased effective nuclear charge and more repulsion among the electrons. Thus, the order is: O²⁻ > O⁻ > O.
04

d. La³⁺, Eu³⁺, Gd³⁺, Yb³⁺

These four ions have the same charge (+3). The sizes can be compared based on their positions in the periodic table as they all belong to the same group. As the atomic size increases as we move down a group, and La is the highest and Yb being the lowest, the order is: La³⁺ > Eu³⁺ > Gd³⁺ > Yb³⁺.
05

e. Te²⁻, I⁻, Cs⁺, Ba²⁺, La³⁺

In this group, we have different ions with various charges. First, arrange them by charge: Te²⁻ > I⁻ > Cs⁺ > Ba²⁺ > La³⁺. Then, consider their positions in the periodic table. Te and I are in the same row, so their size will be affected by the charge and shielding effects: Te²⁻ > I⁻. Cs⁺, Ba²⁺, and La³⁺ are located in the same group, so the size increases going up the group: Cs⁺ > Ba²⁺ > La³⁺. Combining the orders, we get: Te²⁻ > I⁻ > Cs⁺ > Ba²⁺ > La³⁺.

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Most popular questions from this chapter

The standard enthalpies of formation for $\mathrm{S}(g), \mathrm{F}(g), \mathrm{SF}_{4}(g),\( and \)\mathrm{SF}_{6}(g)\( are \)+278.8,+79.0,-775,$ and \(-1209 \mathrm{kJ} / \mathrm{mol}\) respectively. a. Use these data to estimate the energy of an \(\mathrm{S}-\) F bond. b. Compare your calculated value to the value given in Table \(8.5 .\) What conclusions can you draw? c. Why are the \(\Delta H_{f}^{\circ}\) values for \(\mathrm{S}(g)\) and \(\mathrm{F}(g)\) not equal to zero, since sulfur and fluorine are elements?

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