Problem 31 - 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⁺ is smaller than F⁻. (b) Using Equation 7.1, \(Z_{\text {eff}}\) for F⁻ is 8 and for Na⁺ is 10. (c) Using Slater's rules, \(Z_{\text {eff}}\) for F⁻ is 5.5 and for Na⁺ is 8. (d) For isoelectronic ions, effective nuclear charge and ionic radius are inversely related: as \(Z_{\text {eff}}\) increases, ionic radius decreases, and vice versa.

Step 1: (a) Compare the size of the isoelectronic ions F⁻ and Na⁺

For isoelectronic ions, the one with the greater positive charge will have a stronger attractive force pulling the electrons closer to the nucleus, making the ion smaller. In this case, Na⁺ has a charge of +1 whereas F⁻ has a charge of -1. Since Na⁺ has a higher positive charge, Na⁺ is smaller than F⁻.

Step 2: (b) Calculate Zeff for the 2p electrons in F⁻ and Na⁺ using Equation 7.1

Equation 7.1 states that \(Z_{\text {eff}} = Z - S\), where Z is the nuclear charge and S is the given screening constant. For both ions, core electrons contribute 1.00, and valence electrons contribute 0.00 to the screening constant. For F⁻: Z = 9 (there are 9 protons in the nucleus) S = 1.00 (from 1 core electron) + 0.00 (6 valence electrons) = 1.00 Thus, \(Z_{\text {eff}} = Z - S = 9 - 1 = 8\) For Na⁺: Z = 11 (there are 11 protons in the nucleus) S = 1.00 (from 1 core electron) + 0.00 (10 valence electrons) = 1.00 Thus, \(Z_{\text {eff}} = Z - S = 11 - 1 = 10\)

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

Slater's rules are a more accurate way to estimate the screening constant. For n=2 (2p electrons), we need to evaluate the core electrons (n-1=1) and the other valence electrons within the same shell. For F⁻: S = 0.30(6) + 0.85(2) = 1.80 + 1.70 = 3.50 Thus, \(Z_{\text {eff}} = Z - S = 9 - 3.5 = 5.5\) For Na⁺: S = 0.30(10) + 0.85(0) = 3.00 Thus, \(Z_{\text {eff}} = Z - S = 11 - 3 = 8\)

Step 4: (d) Relationship between effective nuclear charge and ionic radius for isoelectronic ions

For isoelectronic ions, as the effective nuclear charge (\(Z_{\text {eff}}\)) increases, the attractive force between the nucleus and the electrons also increases. This results in a smaller ionic radius. Therefore, there is an inverse relationship between effective nuclear charge and ionic radius: as \(Z_{\text {eff}}\) increases, ionic radius decreases, and vice versa.

Key Concepts

Ionic Radius

The ionic radius is a measure of an ion's size. When you have isoelectronic ions, which are ions with the same number of electrons, the ionic radius tends to vary based on the nuclear charge. To understand how, consider two key points:

• The more positive the charge of an ion, the smaller the radius. This is because a greater number of protons in the nucleus pull more strongly on the electrons, drawing them closer.
• Conversely, the more negative the charge, the larger the ion. Here, the reduced positive pull on electrons allows them to spread out more.

In the case of ( Na^+ ) and ( F^- ), both are isoelectronic with 10 electrons. However, ( Na^+ ), having 11 protons compared to 9 in ( F^- ), is smaller as the electrons are drawn more tightly towards the nucleus due to the higher positive charge.

Effective Nuclear Charge

The effective nuclear charge ( ( Z_{eff} )) is an important concept in chemistry that helps explain how tightly electrons are held by the nucleus. It's calculated using the formula ( Z_{eff} = Z - S ), where ( Z ) is the nuclear charge (number of protons) and ( S ) is the screening constant (which accounts for the shielding effect of electrons closer to the nucleus).
For any given electron, ( Z_{eff} ) reflects the actual positive charge 'felt' by that electron after considering repulsion from others.

• In ions like ( F^- ) and ( Na^+ ), ( Z_{eff} ) explains the extent to which electrons are attracted to the nucleus.
• A higher ( Z_{eff} ) indicates stronger attraction, leading to a smaller effective size or ionic radius.

For ( F^- ), ( Z = 9 ) and assuming basic screening, ( S = 1 ), gives a ( Z_{eff} ) of 8. For ( Na^+ ), with ( Z = 11 ), ( Z_{eff} ) is 10. Despite both ions having identical electron configurations, the significantly larger effective nuclear charge of ( Na^+ ) makes it smaller.

Slater's Rules

Slater’s rules provide a more refined approach for calculating the screening constant, ( S ), which helps determine the effective nuclear charge. The rules account for other electrons' shielding effect, dividing them into core and valence categories to adjust how the nucleus' charge is perceived by any given electron.
Let's look at Slater's approach:

• For valence electrons ( ( n = 2 ) level electrons, for instance), electrons in the same shell partially shield each other by a small amount (usually 0.35 each), while inner-shell electrons shield much more effectively (up to 1.00 each for core electrons).
• In assessing ( F^- ), Slater's method uses ( S = 1.80 + 1.70 = 3.50 ) from more detailed computations of electron contributions to give ( Z_{eff} = 5.5 ).
• In ( Na^+ ), the screening is smaller, giving ( S = 3.00 ), resulting in ( Z_{eff} = 8 ).

By breaking down shielding into more nuanced contributions, these rules offer a better understanding of electron-nucleus interactions and thus the behavior and size of ions.