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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

[7.113]When magnesium metal is burned in air (Figure 3.6), two products are produced. One is magnesium oxide, MgO. The other is the product of the reaction of Mg with molecular nitrogen, magnesium nitride. When water is added to magnesium nitride, it reacts to form magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment, a piece of magnesium ribbon is burned in air in a crucible. The mass of the mixture of MgO and magnesium nitride after burning is 0.470 g. Water is added to the crucible, further reaction occurs, and the crucible is heated to dryness until the final product is 0.486 g of MgO. What was the mass percentage of magnesium nitride in the mixture obtained after the initial burning? (d) Magnesium nitride can also be formed by reaction of the metal with ammonia at high temperature. Write a balanced equation for this reaction. If a \(6.3-\mathrm{g}\) Mg ribbon reacts with 2.57 \(\mathrm{g} \mathrm{NH}_{3}(g)\) and the reaction goes to completion, which component is the limiting reactant? What mass of \(\mathrm{H}_{2}(g)\) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium nitride is \(-461.08 \mathrm{kJ} / \mathrm{mol} .\) Calculate the standard enthalpy change for the reaction between magnesium metal and ammonia gas.

Which of the following statements about effective nuclear charge for the outermost valence electron of an atom is incorrect? (i) The effective nuclear charge can be thought of as the true nuclear charge minus a screening constant due to the other electrons in the atom. (ii) Effective nuclear charge increases going left to right across a row of the periodic table. (iii) Valence electrons screen the nuclear charge more effectively than do core electrons. (iv) The effective nuclear charge shows a sudden decrease when we go from the end of one row to the beginning of the next row of the periodic table. (v) The change in effective nuclear charge going down a column of the periodic table is generally less than that going across a row of the periodic table.

Consider the following equation: $$\mathrm{Ca}^{+}(g)+\mathrm{e}^{-} \longrightarrow \mathrm{Ca}(g)$$ Which of the following statements are true? (i) The energy change for this process is the electron affinity of the Ca' ion. (ii) The energy change for this process is the negative of the first ionization energy of the Ca atom. (ii) The energy change for this process is the negative of the electron affinity of the Ca atom.

(a) Does metallic character increase, decrease, or remain unchanged as one goes from left to right across a row of the periodic table? (b) Does metallic character increase, decrease, or remain unchanged as one goes down a column of the periodic table? (c) Are the periodic trends in (a) and (b) the same as or different from those for first ionization energy?

(a) One of the alkali metals reacts with oxygen to form a solid white substance. When this substance is dissolved in water, the solution gives a positive test for hydrogen peroxide, \(\mathrm{H}_{2} \mathrm{O}_{2}.\) When the solution is tested in a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal? (b) Write a balanced chemical equation for the reaction of the white substance with water.
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