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Chapter 7: Problem 16

Detailed calculations show that the value of \(Z_{\text { eff }}\) for the outermost electrons in Si and Cl atoms is \(4.29+\) and \(6.12+\) , respectively.(a) What value do you estimate for \(Z\) eff experienced by the outermost electron in both Si and Cl 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 \(Z_{\text { eff? }} ?(\mathbf{d})\) Which method of approximation more accurately accounts for the steady increase in \(Z_{\text { eff }}\) that occurs upon moving left to right across a period? (e) Predict \(Z_{\text { eff }}\) for a valence electron in \(\mathrm{P}\) , phosphorus, based on the calculations for Si and Cl.

Short Answer

Expert verified
The estimated Z_eff values using core and valence electron contributions are 4 for Si and 7 for Cl. Using Slater's rules, the estimated Z_eff values are 3.40 for Si and 5.20 for Cl. Comparing both methods, the core and valence electron contribution method provide a more accurate estimate of Z_eff, as the errors were smaller than Slater's rules. Both methods correctly account for the increase of Z_eff from Si to Cl, moving left to right across a period. Based on these calculations, we predict the Z_eff for a valence electron in phosphorus (P) to be 5.

Step by step solution

01

Estimation using core and valence electron contribution

First, let's estimate the Z_eff values by assuming core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant. The equation for Z_eff can be written as: \[Z_{eff} = Z - S\] where Z is the atomic number and S is the screening constant. For silicon (Si): Z = 14 (10 core electrons and 4 valence electrons) For chlorine (Cl): Z = 17 (10 core electrons and 7 valence electrons) Let's calculate the screening constant (S) using core and valence electron contributions for both Si and Cl.
02

Calculate the screening constant (S) values using core/valence electron contributions

(a) For Si: S(Si) = 10(core electrons) * 1.00 + 4(valence electrons) * 0.00 = 10 For Cl: S(Cl) = 10(core electrons) * 1.00 + 7(valence electrons) * 0.00 = 10 Now, we will calculate the estimated Z_eff values for Si and Cl.
03

Calculate the Z_eff values using core and valance electron contribution method

For Si: Z_eff(Si) = Z(Si) - S(Si) = 14 - 10 = 4 For Cl: Z_eff(Cl) = Z(Cl) - S(Cl) = 17 - 10 = 7 So, the estimated Z_eff using core and valance electron contributions are 4 for Si and 7 for Cl.
04

Estimation using Slater's rules

(b) Now, we will use Slater's rules to estimate the Z_eff values for Si and Cl. For silicon (Si), the electron configuration is: 1s虏 2s虏 2p鈦� 3s虏 3p虏 For chlorine (Cl), the electron configuration is: 1s虏 2s虏 2p鈦� 3s虏 3p鈦� Now, let's calculate the S values using Slater's rules for both Si and Cl.
05

Calculate screening constant (S) values using Slater's rules

For Si: S(Si) = (2 * 0.30) + (8 * 0.85) + (4 * 1) = 10.60 For Cl: S(Cl) = (2 * 0.30) + (8 * 0.85) + (7 * 1) = 11.80 Now, we will calculate the estimated Z_eff values for Si and Cl using Slater's rules.
06

Calculate the Z_eff values using Slater's rules

For Si: Z_eff(Si) = Z(Si) - S(Si) = 14 - 10.60 = 3.40 For Cl: Z_eff(Cl) = Z(Cl) - S(Cl) = 17 - 11.80 = 5.20 So, the estimated Z_eff using Slater's rules are 3.40 for Si and 5.20 for Cl.
07

Compare the two methods' accuracy

(c) The calculated Z_eff values are 4.29 for Si and 6.12 for Cl. Comparing these values, we can say that: Core and valence electron contribution method: Error for Si: |4.29 - 4| = 0.29 Error for Cl: |6.12 - 7| = 0.88 Slater's rules method: Error for Si: |4.29 - 3.40| = 0.89 Error for Cl: |6.12 - 5.20| = 0.92 We can see that the errors for both Si and Cl when using core and valence electron contributions are smaller than when using Slater's rules.
08

Determine which method accounts for Z_eff's increase across a period

(d) When we move from left to right across a period, Z_eff should increase. Let's see if this is the case for the two methods: Core and valence electron contribution method: Z_eff increases from 4 (Si) to 7 (Cl). Slater's rules method: Z_eff increases from 3.40 (Si) to 5.20 (Cl). Both methods show an increase in Z_eff when moving from Si to Cl. However, the core and valence electron contribution method had a more accurate estimation of Z_eff for Si and Cl.
09

Predict Z_eff for phosphorus (P)

(e) Now we will predict the Z_eff for phosphorus (P) based on the calculations for Si and Cl. We can try to find a trend using the core and valence electron contribution method since it was more accurate for Si and Cl. P has an atomic number (Z) of 15 (10 core electrons and 5 valence electrons). Using the core and valence electron contribution method, we can estimate Z_eff(P) to be: Z_eff(P) = Z(P) - S(P) = 15 - 10 = 5 So, based on the calculations for Si and Cl, we predict the effective nuclear charge Z_eff for a valence electron in phosphorus to be 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Slater's Rules
Slater's Rules offer a systematic way to estimate the screening constant (S) for electrons in an atom. They help approximate how much the inner electrons shield the valence electrons from the nucleus's full charge. This shielding is crucial for calculating the effective nuclear charge (\(Z_{eff}\)) a valence electron experiences.

According to Slater's Rules, we categorize the electrons into different groups based on their energy level. Within each group:
  • Electrons in the same group as the electron of interest each contribute 0.35 to S.
  • Electrons one shell closer to the nucleus contribute 0.85 per electron.
  • For s- and p-electrons, those in even closer shells contribute 1 to S.
By applying these rules, we can sum the contributions to estimate S. Though a useful approximation, Slater's Rules may not always provide pinpoint accuracy due to variations in electron interactions. Nevertheless, they offer an informed approach for predicting electron behavior in multi-electron systems.
Screening Constant
The screening constant is a pivotal concept in understanding how electrons interact in atoms. It is essentially a measure of the extent to which inner electrons block or 'screen' the outer, valence electrons from the nucleus's attractive force. The greater the screening, the less the outer electrons feel the nuclear charge.

The concept comes into play when calculating the effective nuclear charge (\(Z_{eff}\)), formulated as:\[Z_{eff} = Z - S\]where \(Z\) is the actual atomic number, and \(S\) is the screening constant. The screening constant (\(S\)) depends on the arrangement and number of electrons and can be calculated using approaches like direct estimation or Slater's Rules.

These calculations play a role in predicting various aspects of atomic behavior, including chemical reactivity and bonding. Understanding the screening constant helps us appreciate why different elements exhibit unique properties even within the same period or group on the periodic table.
Valence Electrons
Valence electrons reside in the outermost shell of an atom and are fundamental to chemical bonding and reactions. Their behavior and interactions largely determine an element's properties and its place in the periodic table.

Due to their position, valence electrons are less tightly held by the nucleus compared to core electrons. This makes them more active participants in interactions with other atoms. The effective nuclear charge (\(Z_{eff}\)), which they experience, helps dictate how closely these electrons can approach other atoms during chemical reactions.

In exercises dealing with effective nuclear charge, knowing the number of valence electrons is crucial. By estimating the effective charge these electrons feel, it becomes easier to predict trends, such as atomic size, ionization energy, and electronegativity, as one moves across a period or down a group in the periodic table.
For instance, as seen in the example given, calculating \(Z_{eff}\) for the valence electrons in elements like Si, Cl, and P helps us understand the gradual increase in nuclear charge across periods, offering insights into periodic trends.

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Most popular questions from this chapter

Zincin its \(2+\) oxidation state is an essential metal ion for life. \(\mathrm{Zn}^{2+}\) is found bound to many proteins that are involved in biological processes, but unfortunately \(\mathrm{Zn}^{2+}\) is hard to detect by common chemical methods. Therefore, scientists who are interested in studying \(\mathrm{Zn}^{2+}\) -containing proteins frequently substitute \(\mathrm{Cd}^{2+}\) for \(\mathrm{Zn}^{2+},\) since \(\mathrm{Cd}^{2+}\) is easier to detect. (a) On the basis of the properties of the elements and ions discussed in this chapter and their positions in the periodic table, describe the pros and cons of using \(\mathrm{Cd}^{2+}\) as a \(\mathrm{Zn}^{2+}\) substitute. (b) Proteins that speed up (catalyze) chemical reactions are called enzymes. Many enzymes are required for proper metabolic reactions in the body. One problem with using \(\mathrm{Cd}^{2+}\) to replace \(\mathrm{Zn}^{2+}\) in enzymes is that \(\mathrm{Cd}^{2+}\) substitution can decrease or even eliminate enzymatic activity. Can you suggest a different metal ion that might replace \(Z n^{2+}\) in enzymes instead of \(C d^{2+} ?\) Justify your answer.

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