Problem 34 - Chemistry The Central Science

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?

Step-by-Step Solution

Answer

(a) Na+ ion is smaller than F- ion due to its higher effective nuclear charge. (b) Using Equation 7.1, \(Z_{\text {eff,F-}} = 7\) and \(Z_{\text {eff,Na+}} = 1\). (c) Using Slater's rules, \(Z_{\text {eff,F-}} = 2.00\) and \(Z_{\text {eff,Na+}} = 0.85\). (d) For isoelectronic ions, ionic radii and effective nuclear charges are inversely proportional; as effective nuclear charge increases, ionic radius decreases.

Step 1: (a) Compare ion sizes

To determine which ion is smaller, we need to consider the effective nuclear charge experienced by the outermost electrons. Since isoelectronic ions have the same number of electrons, the one with a higher effective nuclear charge will have its electrons pulled closer to the nucleus and thereby have a smaller ionic radius. F- and Na+ have 10 electrons each. F- has 9 protons in its nucleus, while Na+ has 11 protons. Therefore, Na+ will have a higher effective nuclear charge than F- and will be smaller in size.

Step 2: (b) Calculate Z_eff for both ions using Equation 7.1

Equation 7.1 is given as: \(Z_{\text {eff}} = Z - S\), where Z is the atomic number and S is the screening constant. Here, we are told to assume that core electrons contribute 1.00, and valence electrons contribute 0.00 to the screening constant, S. For F-, Z = 9, and S = 2 (two core electrons in 1s). For the Na+ ion, Z = 11, and S = 10 (2 electrons in 1s and 8 electrons in 2s and 2p). Now we can calculate \(Z_{\text {eff}}\) for each ion: \(Z_{\text {eff,F-}} = 9 - 2 = 7\) \(Z_{\text {eff,Na+}} = 11 - 10 = 1\)

Step 3: (c) Calculate Z_eff using Slater's rules

According to Slater's rules, for ns and np electrons, the total screening constant S can be calculated as follows: 1. 0.35 for each electron to the right in the same group (ns, np) 2. 0.85 for each electron in one shell lower than the electron being considered (n-1) 3. 1.0 for each electron in all shells lower than n-1 For F-, there are 2 electrons in the 2s shell and 7 electrons in the 2p shell including the electron being considered. According to Slater's rules: \(S_{\mathrm{F}^{-}} = (1 \times 2) + (0.85 \times 2) + (0.35 \times 6) = 7.00\) Now, we can calculate the \(Z_{\text {eff,F-}}\): \(Z_{\text {eff,F-}} = 9 - 7.00 = 2.00\) For Na+, there are 2 electrons in the 2s shell and 5 electrons in the 2p shell. According to Slater's rules: \(S_{\mathrm{Na}^{+}} = (1 \times 10) + (0.85 \times 2) + (0.35 \times 5) = 10.15\) Now, we can calculate the \(Z_{\text {eff,Na+}}\): \(Z_{\text {eff,Na+}} = 11 - 10.15 = 0.85\)

Step 4: (d) Effective nuclear charge and ionic radius relationship

For isoelectronic ions, their ionic radii are inversely proportional to their effective nuclear charges. As the effective nuclear charge increases, the electrons are pulled closer to the nucleus, resulting in a smaller ionic radius. In this case, Na+ has a higher effective nuclear charge than F-, so it has a smaller ionic radius.

Key Concepts

Isoelectronic Ions

Isoelectronic ions refer to ions that have the same number of electrons. An example of this would be fluoride ion \(\mathrm{F}^{-}\) and sodium ion \(\mathrm{Na}^{+}\). Although these ions have the same electron count, their properties aren't identical. This is because they have different numbers of protons in their nuclei.
The protons determine the positive charge of the nucleus, thereby influencing the effective nuclear charge experienced by the electrons. In the case of \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\):

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

Due to this difference, \(\mathrm{Na}^{+}\) exerts a greater pull on its electrons compared to \(\mathrm{F}^{-}\), which influences its physical size. Even though both ions have 10 electrons, the size is dictated by how these electrons interact with the nucleus.

Ionic Radius

The ionic radius is the measure of an ion's size. It is influenced by the number of protons in the nucleus and their pull on the electrons. In isoelectronic ions like \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\), a greater number of protons means a higher effective nuclear charge.
This causes the electrons to be drawn closer to the nucleus, resulting in a smaller ionic radius. Hence, despite having the same number of electrons:

• \(\mathrm{Na}^{+}\), with 11 protons, has a smaller radius than \(\mathrm{F}^{-}\).
• \(\mathrm{F}^{-}\) has only 9 protons, so its radius is larger.

The general rule is that for isoelectronic ions, the higher the effective nuclear charge, the smaller the ionic radius will be. This relationship adds an intriguing layer to understanding ion sizes beyond just counting electrons.

Slater's Rules

Slater's Rules provide a method for estimating the shielding effect or screening constant \(S\), which is used to determine the effective nuclear charge \(Z_{\text{eff}}\). These rules are essential for considering how electron interactions within atoms affect size and reactivity.
Slater’s Rules provide different contributions to the screening constant based on electron configuration:

• Each electron in the same \(n\) shell group as the electron under study contributes \(0.35\).
• Each electron in the \((n-1)\) shell contributes \(0.85\).
• Electrons in shells lower than \((n-1)\) contribute \(1.00\).

By applying these rules, you can predict the effective nuclear charge experienced by valence electrons with more accuracy.
In our example of \(\mathrm{F}^{-}\) and \(\mathrm{Na}^{+}\):

• For \(\mathrm{F}^{-}\), the screening constant \(S\) is calculated to be around \(7.00\), leading to a \(Z_{\text{eff}}\) of 2.00.
• For \(\mathrm{Na}^{+}\), \(S\) is approximately \(10.15\), giving a \(Z_{\text{eff}}\) of 0.85.

This calculation shows how the fewer protons in \(\mathrm{F}^{-}\) lead to greater effective shielding compared to \(\mathrm{Na}^{+}\), affecting their ionic properties.