Problem 31 - Chemistry: The Central Science in SI Units, Global Edition

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, \(S .(\mathbf{d})\) For isoelectronic ions, how are effective nuclear charge and ionic radius related?

Step-by-Step Solution

Answer

(a) The \(\mathrm{Na}^{+}\) ion is smaller than the \(\mathrm{F}^{-}\) ion. (b) Using the given values for the screening constant, we find that \(Z_{\text {eff }}=7\) for a \(2p\) electron in \(\mathrm{F}^{-}\) and \(Z_{\text {eff }}=9\) for a \(2p\) electron in \(\mathrm{Na}^{+}\). (c) Using Slater's rules, we obtain \(Z_{\text {eff }}=5.2\) for a \(2p\) electron in \(\mathrm{F}^{-}\) and \(Z_{\text {eff }}=6.5\) for a \(2p\) electron in \(\mathrm{Na}^{+}\). (d) For isoelectronic ions, the ionic radius and effective nuclear charge are inversely related.

Step 1: (a) Comparing Ionic Radii of Isoelectronic Ions

For isoelectronic ions, the ion with more protons will have a larger nuclear charge, resulting in a stronger attraction between the nucleus and the electrons. This causes the electron cloud to be pulled in more tightly, making the ion smaller. The \(\mathrm{F}^{-}\) ion has 9 protons while the \(\mathrm{Na}^{+}\) ion has 11 protons. Therefore, the \(\mathrm{Na}^{+}\) ion is smaller.

Step 2: (b) Calculate \(Z_{\text {eff }}\) using given values for \(S\)

We will use Equation 7.1: \(Z_{\text {eff }}=Z-S\), where \(Z\) is the atomic number (number of protons) and \(S\) is the screening constant. For \(\mathrm{F}^{-}\): \(Z = 9\) For the \(2p\) electron, there are 2 core electrons contributing 1.00 to \(S\) and 6 valence electrons contributing 0.00 to \(S\). So, \(S = 2(1.00) + 6(0.00) = 2\). \(Z_{\text {eff }}=9-2=7\) for a \(2p\) electron in \(\mathrm{F}^{-}\). For \(\mathrm{Na}^{+}\): \(Z = 11\) For the \(2p\) electron, there are 2 core electrons contributing 1.00 to \(S\) and 8 valence electrons contributing 0.00 to \(S\). So, \(S=2(1.00) + 8(0.00) = 2\). \(Z_{\text {eff }}=11-2=9\) for a \(2p\) electron in \(\mathrm{Na}^{+}\).

Step 3: (c) Calculate \(Z_{\text {eff }}\) using Slater's Rules

According to Slater's rules for \(2p\) electrons, the screening constant can be determined as follows: \(S=0.85\times\)(number of electrons in the same shell with principal quantum number \(n-1\)) + \(0.35\times\)(number of electrons in the same shell) For \(\mathrm{F}^{-}\): \(S=0.85\times(2) + 0.35\times(6) = 1.7 + 2.1 = 3.8\) \(Z_{\text {eff }}=9-3.8=5.2\) for a \(2p\) electron in \(\mathrm{F}^{-}\) using Slater's rules. For \(\mathrm{Na}^{+}\): \(S=0.85\times(2) + 0.35\times(8) = 1.7 + 2.8 = 4.5\) \(Z_{\text {eff }}=11-4.5=6.5\) for a \(2p\) electron in \(\mathrm{Na}^{+}\) using Slater's rules.

Step 4: (d) Relationship between Effective Nuclear Charge and Ionic Radius

For isoelectronic ions, as the effective nuclear charge \(Z_{\text {eff }}\) increases, the attraction between the nucleus and the electrons becomes stronger. This results in the electron cloud being pulled in more tightly, reducing the ionic radius. Therefore, the ionic radius and effective nuclear charge are inversely related for isoelectronic ions.

Key Concepts

Isoelectronic Ions

Isoelectronic ions are ions that have the same number of electrons but different numbers of protons. In the case of \( \mathrm{F}^{-} \) and \( \mathrm{Na}^{+} \), both are isoelectronic with the electron configuration of \([\mathrm{Ne}]\).
Since they have the same number of electrons (10 in each), their differences in properties arise from their nuclear charges.
Despite having similar electron shells, these ions exhibit distinct behaviors because the number of protons in each nucleus is different.

• \( \mathrm{F}^{-} \) has 9 protons
• \( \mathrm{Na}^{+} \) has 11 protons

The ion with more protons, such as \( \mathrm{Na}^{+} \), exerts a stronger pull on the surrounding electrons. This makes the electron cloud more tightly bound, resulting in a smaller ionic radius compared to \( \mathrm{F}^{-} \).

Effective Nuclear Charge

The effective nuclear charge, often symbolized as \(Z_{\text{eff}}\), is a concept that helps explain the pull experienced by electrons in an atom or ion.
Despite the total number of protons, not all the positive charge impacts the electron experience equally due to electron screening.
\(Z_{\text{eff}}\) is given by the formula:\[ Z_{\text{eff}} = Z - S \]Where:

• \( Z \) is the atomic number, indicating the total positive charge (number of protons).
• \( S \) is the screening constant, representing the average repulsive effects of other electrons.

For isoelectronic ions like \( \mathrm{F}^{-} \) and \( \mathrm{Na}^{+} \), we calculate \(Z_{\text{eff}}\) to understand how nuclear charge influences electron behavior. The higher the \(Z_{\text{eff}}\), the stronger the nucleus pulls on the electrons, making the atom smaller. The calculations highlight that \( \mathrm{Na}^{+} \) has a greater effective nuclear charge than \( \mathrm{F}^{-} \).

Ionic Radius

The ionic radius is a measure of the size of an ion in a crystal lattice. For isoelectronic ions, differences in ionic radius are due to variations in effective nuclear charge.
As mentioned, for ions like \( \mathrm{F}^{-} \) and \( \mathrm{Na}^{+} \), the ionic radius is influenced by how strongly the nuclear charge pulls the electron cloud inward.
This results in ions with a higher \(Z_{\text{eff}}\), like \( \mathrm{Na}^{+} \), having smaller ionic radii.

• A larger \(Z_{\text{eff}}\) means electrons are held more closely, tightening the electron cloud and reducing size.
• A smaller \(Z_{\text{eff}}\) allows the electron cloud to be more diffuse, increasing size.

Understanding the relationship between \(Z_{\text{eff}}\) and ionic radius is crucial for predicting and explaining physical and chemical properties of isoelectronic ions.

Slater's Rules

Slater's rules provide a method to estimate the screening constant \(S\), which is crucial for calculating the effective nuclear charge.
These rules help account for how electron shielding varies with different arrangements of electron shells.
Slater's rules take into consideration:

• The number of electrons in the same shell.
• The electrons in lower shells that contribute differently to screening.

Applying Slater's rules to \( \mathrm{F}^{-} \) and \( \mathrm{Na}^{+} \) ions, we use specific coefficients:

• 0.35 for electrons in the same shell.
• 0.85 for electrons in \(n-1\) shells.

For \( \mathrm{F}^{-} \), the calculated \(S\) using Slater's rules provides a \(Z_{\text{eff}}\) of 5.2, while \( \mathrm{Na}^{+} \) gives a \(Z_{\text{eff}}\) of 6.5. This nuanced approach highlights the complexity of electron interactions and refines our understanding of atomic structure beyond simple models.