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Chapter 7: Problem 32

Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\). (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S,\) calculate \(Z_{\text {eff }}\) for these two ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, $S .(\mathbf{d})$ For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Short Answer

Expert verified
(a) The smaller ion is \(\mathrm{K}^+\). (b) Using Equation 7.1, the effective nuclear charge (\(Z_{\text {eff }}\)) for both \(\mathrm{Cl}^-\) and \(\mathrm{K}^+\) is 1. (c) Using Slater's rules, the effective nuclear charge (\(Z_{\text {eff }}\)) for \(\mathrm{Cl}^{-}\) is 7.5, and for \(\mathrm{K}^+\) is 9.5. (d) For isoelectronic ions, the effective nuclear charge (\(Z_{\text {eff }}\)) is inversely proportional to the ionic radius. As the effective nuclear charge increases, the ionic radius decreases.

Step by step solution

01

Identifying the smaller ion

The \(\mathrm{Cl}^-\) ion has gained an electron in comparison to the neutral \(\mathrm{Cl}\) atom, while the \(\mathrm{K}^+\) ion has lost an electron compared to the neutral \(\mathrm{K}\) atom. Since they are isoelectronic, they both have the same number of electrons, however, \(\mathrm{K}^+\) has a higher nuclear charge due to an additional proton. This higher nuclear charge pulls the surrounding electrons closer, making the \(\mathrm{K}^+\) ion smaller than the \(\mathrm{Cl}^-\). (a) Therefore, the smaller ion is \(\mathrm{K}^+\).
02

Calculating \(Z_{\text {eff }}\) using Equation 7.1

Equation 7.1 states that: \(Z_{\text {eff }} = Z - S\) where \(Z\) is the atomic number and \(S\) is the screening constant. The assumption is that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant. For \(\mathrm{Cl}^{-}\): \(Z = 17\) and \(S = 16\) since all 16 core electrons contribute 1.00. So, \(Z_{\text {eff }} = 17 - 16 = 1\) For \(\mathrm{K}^{+}\): \(Z = 19\) and \(S = 18\) since all 18 core electrons contribute 1.00. So, \(Z_{\text {eff }} = 19 - 18 = 1\) (b) Hence, \(Z_{\text {eff }}\) for \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^+\) is 1 using Equation 7.1.
03

Calculating \(Z_{\text {eff }}\) using Slater's rules

Now, we will use Slater's rules to estimate the screening constant, \(S\), and repeat the calculations for \(Z_{\text {eff }}\). For \(\mathrm{Cl}^{-}\): According to Slater's rules, \(S = 0.85(10) + 1\) for the \(3s\) electron. So, \(S = 9.5\) and \(Z_{\text {eff }} = 17 - 9.5 = 7.5\) For \(\mathrm{K}^{+}\): According to Slater's rules, \(S = 0.85(10) + 1\) for the \(3s\) electron. So, \(S = 9.5\) and \(Z_{\text {eff }} = 19 - 9.5 = 9.5\) (c) Hence, \(Z_{\text {eff }}\) for \(\mathrm{Cl}^{-}\) is 7.5, and for \(\mathrm{K}^+\) is 9.5 using Slater's rules.
04

Discussing the relationship between \(Z_{\text {eff }}\) and ionic radius

(d) For isoelectronic ions, the effective nuclear charge (\(Z_{\text {eff }}\)) is inversely proportional to the ionic radius. As the effective nuclear charge increases, the positive charge in the nucleus pulls the electrons closer, resulting in a smaller ionic radius. In our example, \(\mathrm{K}^+\) has a higher \(Z_{\text {eff }}\) and a smaller ionic radius compared to \(\mathrm{Cl}^-\).

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Most popular questions from this chapter

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Potassium metal is exposed to an atmosphere of chlorine gas. (b) Strontium oxide is added to water. (c) A fresh surface of lithium metal is exposed to oxygen gas. (d) Sodium metal reacts with molten sulfur.

Consider \(\mathrm{S}, \mathrm{Cl},\) and \(\mathrm{K}\) and their most common ions. (a) List the atoms in order of increasing size. (b) List the ions in order of increasing size. (c) Explain any differences in the orders of the atomic and ionic sizes.

(a) The five most abundant elements in the Earth's crust are $\mathrm{O}, \mathrm{Si}, \mathrm{Al}, \mathrm{Fe},\( and Ca. Referring to Figure \)7.1,$ are any of these elements among those known before \(1700 ?\) If so which ones? (b) Seven of the nine elements known since ancient times are metals. Referring to Table \(4.5,\) are these metals mostly found at the bottom or top of the activity series?

Copper and calcium both form +2 ions, but copper is far less reactive. Suggest an explanation, taking into account the ground-state electron configurations of these elements and their atomic radii.

The first ionization energy of the oxygen molecule is the energy required for the following process: $$ \mathrm{O}_{2}(g) \longrightarrow \mathrm{O}_{2}{ }^{+}(g)+\mathrm{e}^{-} $$ The energy needed for this process is \(1175 \mathrm{~kJ} / \mathrm{mol}\), very similar to the first ionization energy of Xe. Would you expect \(\mathrm{O}_{2}\) to react with \(\mathrm{F}_{2}\) ? If so, suggest a product or products of this reaction.
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