Problem 32 - Chemistry: The Central Science in SI Units, Global Edition

Question

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?

Step-by-Step Solution

Answer

(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 1: 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 \(\mathrm{K}^+\).

Step 2: 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.

Step 3: 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.

Step 4: 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 \(\mathrm{Cl}^-\).

Key Concepts

Isoelectronic Ions

Isoelectronic ions are ions that have the same number of electrons. Despite having an identical electron count, these ions can have different properties, such as their size. This is because the number of protons, or the nuclear charge, differs between them. For instance, in the exercise,

• \(\mathrm{Cl}^{-}\) has 17 protons and 18 electrons.
• \(\mathrm{K}^{+}\) has 19 protons but also 18 electrons.

The differing number of protons means that while both ions are isoelectronic, their effective nuclear charge is not the same. The nuclear charge impacts how tightly the electrons are held, affecting properties like ionic radius.Understanding isoelectronicity is essential in predicting and comparing properties such as size and reactivity of ions, especially in chemistry contexts involving ionic compounds.

Ionic Radius

The ionic radius is a measure of the size of an ion in a crystal lattice. In isoelectronic ions, despite having the same electronic structure, the ionic radius can differ due to the nuclear charge.When comparing ions like

• \(\mathrm{Cl}^{-}\), which has a lower nuclear charge.
• \(\mathrm{K}^{+}\), which has a higher nuclear charge.

The \(\mathrm{K}^{+}\) ion is smaller because the extra protons pull the electrons closer to the nucleus. Thus, higher positive nuclear charge results in a smaller ionic radius as electrons are more strongly attracted.Understanding ionic radius is crucial in explaining the physical properties of materials, such as melting points, and how ions will pack in lattice structures.

Slater's Rules

Slater's rules provide a way to estimate the screening constant \(S\) that adjusts for the repulsive effect of inner electrons when calculating the effective nuclear charge \(Z_{\text{eff}}\). The rules take into account:

• Electron configuration
• The number of shielding inner electrons
• The relative influence of electrons in different shells

Using Slater's rules involves:

• Grouping electrons by principal quantum number \(n\).
• Applying specific shielding values - such as 0.85 for electrons in the same shell.
• Summing up these contributions to find the screening constant \(S\).

By using Slater’s rules, as in the exercise:

• For \(\mathrm{Cl}^{-}\), \(Z_{\text{eff}}\) is 7.5.
• For \(\mathrm{K}^{+}\), \(Z_{\text{eff}}\) is 9.5.

These calculations show how Slater’s rules refine our understanding of electron pull and ionic radius in contrast to simple assumptions.