Question

Consider the isoelectronic ions \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+} .\) (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant, \(S,\) calculate \(Z_{\text { eff }}\) for the 2\(p\) electrons in both ions. (c) Repeat this calculation using Slater's rules to estimate the screening constant, \(S\) .(d) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Short answer

(a) Na+ ion is smaller. (b) For F- ion, \(Z_\mathrm{eff} = 8\), and for Na+ ion, \(Z_\mathrm{eff} = 10\). (c) Using Slater's rules, for F- ion, \(Z_\mathrm{eff} = 1.85\), and for Na+ ion, \(Z_\mathrm{eff} = 3\). (d) For isoelectronic ions, the effective nuclear charge (Z_eff) and ionic radius (r) are inversely related: \(Z_\mathrm{eff} \propto \frac{1}{r}\).

Step-by-step solution

Comparing the size of F- and Na+ ions

F- and Na+ are isoelectronic, which means they have the same number of electrons. For F- ion, we have 10 electrons and the same for Na+ ion. To compare their sizes, we need to consider their effective nuclear charge (Z_eff). The ion with the higher Z_eff will have a stronger attraction to the electrons and thus be smaller.

Calculate Z_eff for F- and Na+ ions using Equation 7.1 and given S values

According to Equation 7.1, we can calculate the effective nuclear charge as: \(Z_\mathrm{eff} = Z - S\) For F- ion: Number of protons, Z = 9 Screening constant, S = 1 (core electrons) + 0 (valence electrons) = 1 So, \(Z_\mathrm{eff} = 9 - 1 = 8\) For Na+ ion: Number of protons, Z = 11 Screening constant, S = 1 (core electrons) + 0 (valence electrons) = 1 So, \(Z_\mathrm{eff} = 11 - 1 = 10\) Since Na+ ion has a higher Z_eff, it is smaller than the F- ion.

Calculate Z_eff for F- and Na+ ions using Slater's rules

Slater's rules are a set of empirical rules developed to estimate S values. For F- ion, according to Slater's rules: S = (0.35) × (1 2s electron) + (0.85) × (8 remaining electrons) = 0.35 + 6.8 = 7.15 \(Z_\mathrm{eff} = 9 - 7.15 = 1.85\) For Na+ ion, according to Slater's rules: S = (0.35) × (1 2s electron) + (0.85) × (9 remaining electrons) = 0.35 + 7.65 = 8 \(Z_\mathrm{eff} = 11 - 8 = 3\) Using Slater's rules, we still find Na+ to have a higher Z_eff, and therefore, it is smaller than F- ion.

Determine the relation between effective nuclear charge and ionic radius for isoelectronic ions

For isoelectronic ions, the relation between the effective nuclear charge (Z_eff) and the ionic radius (r) is an inverse relationship. As Z_eff increases, the attraction between the nucleus and the electrons becomes stronger, leading to a decrease in the ionic radius. And, as Z_eff decreases, the ionic radius increases due to weaker attraction between the nucleus and the electrons. This can be summarized as: \(Z_\mathrm{eff} \propto \frac{1}{r}\)

Key concepts

Ionic Radius

The ionic radius is a measure of the size of an ion in a crystal lattice. Ions are atoms or molecules that have lost or gained electrons, thus acquiring a charge. The ionic radius is affected by the charge of the ion, the number of electrons, and the effective nuclear charge (\( Z_{\text{eff}} \)). Generally, when an atom loses an electron to form a cation, its radius decreases because the remaining electrons are drawn closer to the nucleus. On the opposite end, anions, which gain electrons, tend to have a larger radius because the additional electron-electron repulsion slightly expands the electron cloud.

In the context of isoelectronic ions—ions with the same number of electrons but different nuclear charges—the one with the higher effective nuclear charge will usually have a smaller ionic radius. This occurs because the greater charge of the nucleus is able to pull the same number of electrons closer, reducing the size of the ion.

Isoelectronic Ions

Isoelectronic ions are ions that have the same number of electrons, making them have identical electronic structures. However, they may and often do differ in size, which is primarily due to their effective nuclear charge. For example, in the exercise, \( \text{F}^{-} \) and \( \text{Na}^{+} \) ions have the same number of electrons but differ in their ionic radii because of their differing nuclear charges.

The concept of isoelectronic ions allows us to see the correlation between the charge of the nucleus and the size of the ion. The ion with more protons in the nucleus will hold the electrons more tightly, leading to a smaller radius for the ion. This comparison is particularly enlightening when studying periodic trends and the behavior of ions in different chemical environments.

Slater's Rules

Slater's rules provide a method to calculate the effective nuclear charge, \( Z_{\text{eff}} \), that an electron feels from the nucleus, accounting for the screening effect from other electrons. According to these rules, different electrons shield each other to varying degrees based on their positions in orbitals. Electrons in the same orbital shield each other poorly, while those in inner shells shield outer electrons more effectively.

The rules assign values to electrons based on their orbital type and group them accordingly. These values are then summed to find the screening constant, \( S \) used in the equation \( Z_{\text{eff}} = Z - S \), where \( Z \) is the atomic number. Applying Slater's rules reveals more accurate values for \( Z_{\text{eff}} \) compared to simpler calculations that might not differentiate between electron shells. The exercise demonstrates how \( Z_{\text{eff}} \) influences ionic size, which aligns with the experimentally observed ionic radii in isoelectronic ions.