Problem 13
Question
Detailed calculations show that the value of \(Z_{\text {eff }}\) for \(\mathrm{Na}\) and \(K\) atoms is \(2.51+\) and \(3.49+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both Na and \(K\) by assuming core electrons contribute \(1.00\) and valence electrons contribute \(0.00\) to the screening constant? (b) What values do you estimate for \(Z_{\text {eff }}\) using Slater's rules? (c) Which approach gives a more accurate estimate of \(Z_{\text {eff }} ?\) (d) Does either method of approximation account for the gradual increase in \(Z_{\text {eff }}\) that occurs upon moving down a group?
Step-by-Step Solution
Verified Answer
The methods for estimating \(Z_{\text {eff }}\) in this exercise are not highly accurate. The first method gives a \(Z_{\text {eff }}\) of 1 for both Na and K, while the second method (using Slater's rules) provides a more accurate result of 7.5 for Na and 12.7 for K. However, these estimates are still far from the detailed calculation values (2.51 for Na and 3.49 for K), and neither method accounts for the gradual increase in \(Z_{\text {eff }}\) when moving down a group. Using the trend from the detailed calculations, the predicted effective nuclear charge for the outermost electron in Rb is approximately 4.23.
1Step 1: Understanding the concept of Zeff
The effective nuclear charge (\(Z_{\text {eff }}\)) is the net positive charge experienced by an electron in a multi-electron atom. The outer electrons experience both an attraction to the nucleus and a repulsion from the inner/core electrons. The shielding effect is the phenomenon where the core electrons shield the outer electrons from the full nuclear charge.
2Step 2: Estimate Zeff using core and valence electron contributions
Sodium (Na) has 11 total electrons: 10 core electrons and 1 valence electron. Its actual \(Z_{\text {eff }}\) is 2.51. Potassium (K) has 19 total electrons: 18 core electrons (filled inner shells) and 1 valence electron. Its real \(Z_{\text {eff }}\) is 3.49.
For Na:
\(Z_{\text {eff }} = Z - S = 11 - (10 \times 1 + 1 \times 0) = 1\)
For K:
\(Z_{\text {eff }} = Z - S = 19 - (18 \times 1 + 1 \times 0) = 1\)
3Step 3: Estimate Zeff using Slater's rules
Using Slater's rules for Na (Z = 11):
- For core electrons (1s, 2s, 2p): each contributes 0.35 to the screening constant
- For valence electron (3s): contributes 0.00
Total \(S = 10 \times 0.35 = 3.5\)
\(Z_{\text {eff }} = Z - S = 11 - 3.5 = 7.5\)
Using Slater's rules for K (Z = 19):
- For core electrons (1s, 2s, 2p, 3s, 3p): each contributes 0.35 to the screening constant
- For valence electron (4s): contributes 0.00
Total \(S = 18 \times 0.35 = 6.3\)
\(Z_{\text {eff }} = Z - S = 19 - 6.3 = 12.7\)
4Step 4: Compare the methods and answer the related questions
The first method gives a \(Z_{\text {eff }}\) of 1 for both Na and K. The second method provides a more accurate result (7.5 for Na and 12.7 for K) which is still far from the detailed calculation values (2.51 for Na and 3.49 for K). Therefore, neither method can be considered accurate. Moreover, neither method accounts for the gradual increase in \(Z_{\text {eff }}\) that occurs when moving down a group.
5Step 5: Predict Zeff for the outermost electrons in Rb
Since neither method is accurate, we'll use the trend from the detailed calculation, which suggests that Rubidium (Rb, Z = 37) should have a \(Z_{\text {eff }}\) between those of Na and K. You can estimate it by interpolation:
\(Z_{\text {eff }}_{Rb} = Z_{\text {eff }}_{Na} + ((Z_{\text {eff }}_{K} - Z_{\text {eff }}_{Na}) \times \frac{Z_{Rb} - Z_{Na}}{Z_{K} - Z_{Na}})\)
\(Z_{\text {eff }}_{Rb} = 2.51 + ((3.49 - 2.51) \times \frac{37 - 11}{19 - 11})\)
\(Z_{\text {eff }}_{Rb} \approx 4.23\)
So, the predicted effective nuclear charge for the outermost electrons in Rb is approximately 4.23.
Key Concepts
Slater's RulesScreening ConstantShielding Effect
Slater's Rules
Slater's Rules give us a simplified way to calculate the effective nuclear charge (\( Z_{\text{eff}} \)) experienced by an electron in an atom's electron cloud. These rules are useful for helping us understand the effect of electron repulsion on the shielding of nuclear charge.
To use Slater's Rules, we group the electrons into different classes based on their types of orbitals (like 1s, 2p, etc.). Each group contributes differently to the screening constant, used for estimating how much of the nuclear charge is "felt" by an outer electron.
The principled approach is quite structured:
Even though it involves more detailed work, Slater's Rules remain an efficient way to provide a better approximation of \( Z_{\text{eff}} \) by acknowledging these detailed contributions rather than a flat value for all core electrons.
To use Slater's Rules, we group the electrons into different classes based on their types of orbitals (like 1s, 2p, etc.). Each group contributes differently to the screening constant, used for estimating how much of the nuclear charge is "felt" by an outer electron.
The principled approach is quite structured:
- Electrons in the same shell as the electron of interest contribute 0.35 to the screening constant.
- Electrons one shell inside contribute 0.85.
- Electrons further inside contribute a full 1.00 to the screening constant.
Even though it involves more detailed work, Slater's Rules remain an efficient way to provide a better approximation of \( Z_{\text{eff}} \) by acknowledging these detailed contributions rather than a flat value for all core electrons.
Screening Constant
The screening constant helps us calculate the effective nuclear charge by accounting for the shielding effect of inner electrons on outer electrons. It's a vital component in understanding how much of the nuclear charge an electron actually experiences.
In Slater's method, the screening constant (notated as \( S \)) is determined by summing up the contributions of all electrons except the electron in question.
A thoughtfully calculated screening constant allows chemists and physics enthusiasts to gain insights into the behavior of atoms, predicting their interactions and properties based on how well the core electrons shield the outer electrons.
In Slater's method, the screening constant (notated as \( S \)) is determined by summing up the contributions of all electrons except the electron in question.
- The closer an electron is to the nucleus, the more it contributes to the shielding effect.
- Electrons in the outermost shell contribute the least, often 0.00.
A thoughtfully calculated screening constant allows chemists and physics enthusiasts to gain insights into the behavior of atoms, predicting their interactions and properties based on how well the core electrons shield the outer electrons.
Shielding Effect
The shielding effect—or the screening effect—occurs because inner electrons repel outer electrons. In atoms, outer electrons don't experience the full charge of the nucleus; instead, they "feel" a reduced charge due to this effect.
Electrons in lower energy levels effectively block some of the nuclear charge from reaching the electrons further away in the atomic structure.
The more core electrons there are, and their particular arrangement based on energy levels, significantly alter how the shielding effect impacts \( Z_{\text{eff}} \), explaining even complex chemical interactions easily.
Electrons in lower energy levels effectively block some of the nuclear charge from reaching the electrons further away in the atomic structure.
- A `strong` shielding effect means the outer electrons feel less of the nucleus' pull.
- Timely consideration of the shielding effect aids in understanding periodic trends, like ionization energy and atomic radius.
The more core electrons there are, and their particular arrangement based on energy levels, significantly alter how the shielding effect impacts \( Z_{\text{eff}} \), explaining even complex chemical interactions easily.
Other exercises in this chapter
Problem 11
(a) What is meant by the term effective nuclear charge? (b) How does the effective nuclear charge experienced by the valence electrons of an atom vary going fro
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(a) How is the concept of effective nuclear charge used to simplify the numerous electron-electron repulsions in ? many-electron atom? (b) Which experiences a g
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Detailed calculations show that the value of \(Z_{\text {eff }}\) for \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) Wh
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Which will experience the greater effective nuclear charge, the electrons in the \(n=3\) shell in Ar or the \(n=3\) shell in Kr? Which will be closer to the nuc
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