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Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Si}\) and \(\mathrm{Cl}\) atoms is \(4.29+\) and \(6.12+\), respectively. (a) What value do you estimate for \(Z_{\text {eff }}\) experienced by the outermost electron in both \(\mathrm{Si}\) and \(\mathrm{Cl}\) 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 \(Z_{\text {eff? }}\) ? (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 \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Short Answer

Expert verified
The more accurate method for estimating the effective nuclear charge (\(Z_{\text{eff}}\)) for silicon (Si) and chlorine (Cl) is method (a), which takes core and valence electrons into consideration; it provides an estimated \(Z_{\text{eff}}\) of 4 for Si and 7 for Cl. Following this method, we predict \(Z_{\text{eff}}\) for a valence electron in phosphorus (P) to be approximately 5.5, considering the steady increase in \(Z_{\text{eff}}\) that occurs upon moving left to right across a period.

Step by step solution

01

(a) Estimate \(Z_{\text{eff}}\) taking core and valence electrons into consideration

To estimate \(Z_{\text{eff}}\), first calculate the screening constant, S, where S = number of core electrons + 0 * (number of valence electrons). Then, calculate the \(Z_{\text{eff}}\) for each atom. For Silicon (Si) - Atomic Number (Z) = 14: Core electrons: 10 (1s, 2s, 2p) Valence electrons: 4 (3s, 3p) \(S_{Si} = 10 + 0 * 4 = 10\) \(Z_{\text{eff (Si)}} = Z - S = 14 - 10 = 4\) For Chlorine (Cl) - Atomic Number (Z) = 17: Core electrons: 10 (1s, 2s, 2p) Valence electrons: 7 (3s, 3p) \(S_{Cl} = 10 + 0 * 7 = 10\) \(Z_{\text{eff (Cl)}} = Z - S = 17 - 10 = 7\)
02

(b) Estimate \(Z_{\text{eff}}\) using Slater's rules

For Silicon (Si) - Electronic Configuration (1s, 2s, 2p, 3s, 3p): For the 3s and 3p electrons: \(S_{3s,3p(Si)} = 1 * (1s) + 1 * (2s) + 1 * (2p) + 0.35 * (electrons \ in \ 3s,3p)\) \(S_{3s,3p(Si)} = 1 + 1 + 1 + 0.35 * 3 = 4.05\) (since the 4th electron is not involved in shielding) \(Z_{\text{eff (Si)}} = Z - S_{3s,3p(Si)} = 14 - 4.05 = 9.95\) For Chlorine (Cl) - Electronic Configuration (1s, 2s, 2p, 3s, 3p): For the 3s and 3p electrons: \(S_{3s,3p(Cl)} = 1 * (1s) + 1 * (2s) + 1 * (2p) + 0.35 * (electrons\ in\ 3s,3p)\) \(S_{3s,3p(Cl)} = 1 + 1 + 1 + 0.35 * 6 = 4.10\) (since the 7th electron is not involved in shielding) \(Z_{\text{eff (Cl)}} = Z - S_{3s,3p(Cl)} = 17 - 4.10 = 12.9\)
03

(c) Determining which approach is more accurate

We are given that detailed calculations show the values of \(Z_{\text{eff}}\) for Si and Cl as 4.29+ and 6.12+. Comparing this to our calculated values: Method (a): Si: 4 Cl: 7 Method (b): Si: 9.95 Cl: 12.9 The given values are closer to the results obtained from method (a). Therefore, method (a) gives a more accurate estimate of \(Z_{\text{eff}}\) for Si and Cl.
04

(d) Evaluating the more accurate method for the steady increase in \(Z_{\text{eff}}\)

Method (a) is better at predicting the steady increase in \(Z_{\text{eff}}\) from Si to Cl. This is because method (b) gives fairly larger values as compared to the observed ones and do not closely follow the observed trend.
05

(e) Predicting \(Z_{\text{eff}}\) for Phosphorus (P)

To calculate the \(Z_{\text{eff}}\) for phosphorus (P) based on calculations for Si and Cl, we can first find the difference between their calculated \(Z_{\text{eff}}\) values using method (a): Increase in \(Z_{\text{eff}}\) from Si to Cl: \(7 - 4 = 3\) Phosphorus (P) - Atomic Number (Z) = 15: Core electrons: 10 (1s, 2s, 2p) Valence electrons: 5 (3s, 3p) Using method (a): \(S_P = 10 + 0 * 5 = 10\) \(Z_{\text{eff (P)}} = Z - S = 15 - 10 = 5\) The regular increase in \(Z_{\text{eff}}\) from Si to P would result in the value: Estimated value for $Z_{\text{eff (P)}} = Z_{\text{eff (Si)}} + \dfrac{3}{2} = 4 + \dfrac{3}{2} = 5.5\) So, the predicted \(Z_{\text{eff}}\) for a valence electron in phosphorus (P) is around 5.5.

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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 provide a method to estimate the effective nuclear charge, denoted as \( Z_{\text{eff}} \), that an electron experiences in an atom. This concept is crucial in understanding how electrons shield each other from the full charge of the nucleus. The rules involve a series of steps to calculate a screening constant, which is used to then determine the \( Z_{\text{eff}} \).

Each electron in the same or lower energy levels contributes to the shielding effect, and their contributions are weighted differently depending on their location relative to the electron of interest. For example, electrons in the same subgroup shield more effectively than those in orbitals with a lower principal quantum number. This allows us to estimate how strongly the nucleus holds onto its electrons, which is especially valuable when discussing transition metals or calculating ionization energies.

Applying Slater's rules involves considering the electron configuration of the element, dividing electrons into groups based on their principal quantum number and type of orbital, and then subtracting the weighted screening constant from the atomic number to obtain \( Z_{\text{eff}} \).
Screening Constant
The screening constant, symbolized by \( S \), is a value representing the extent to which electrons in an atom shield each other from the nuclear charge. Essentially, it's the effect that the inner shell electrons have in reducing the nuclear charge for the outer shell electrons. The idea here is that not all protons in the nucleus contribute equally to the force experienced by an outer electron.

The greater the screening constant, the more an electron is shielded from the nucleus and the lower the effective nuclear charge it experiences. This shielding effect lies at the heart of many trends observed in the periodic table, such as atomic size and ionization energy. In simple calculations, core electrons may be considered to contribute a whole unit to the screening constant, while valence electrons contribute none, providing a rough estimate of \( Z_{\text{eff}} \). However, this is a simplification and does not account for the varying degrees of screening between different shells and subshells of electrons, which is where Slater's rules come into play.
Periodic Trends
Periodic trends refer to patterns observed within the periodic table that show how different properties of elements change systematically across periods and down groups. As you move from left to right across a period, the effective nuclear charge typically increases. This is because the number of protons (and hence the nuclear charge) increases, but the shielding by inner electrons does not increase as substantially. Consequently, the attraction between the nucleus and the valence electrons strengthens, causing the atomic radius to decrease and ionization energies to increase.

Understanding periodic trends is fundamental when predicting the chemical behavior of elements. For instance, the increase in \( Z_{\text{eff}} \) across a period can explain why nonmetals have higher electronegativities compared to metals. Trends are essential in making informed predictions about reactivity, bonding, and other chemical properties.
Atomic Structure
Atomic structure describes the arrangement and behavior of electrons, protons, and neutrons within the atom. At the center of the atom is the nucleus, comprising protons and neutrons. Electrons move around the nucleus in specific regions called orbitals, each with a particular energy level and shape.

The understanding of atomic structure is pivotal in explaining why atoms exhibit certain chemical and physical properties. Properties such as atomic radius, ionization energy, and metallic character all depend on how electrons are distributed in these energy levels and how they interact with the nucleus. The notions of screening and effective nuclear charge are direct consequences of the electron-nucleus interactions, which can be predicted through the use of theoretical models such as Slater's rules. This forms the basis of quantum mechanics and is crucial for explaining the behavior of atoms in different chemical contexts.

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Most popular questions from this chapter

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?

(a) What is meant by the terms acidic oxide and basic oxide? (b) How can we predict whether an oxide will be acidic or basic based on its composition?

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.

Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like nonmetals. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds. If not, correct it. If it is, explain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. (d) The hydride ion is \(\mathrm{H}\). Write out the process corresponding to the first ionization energy of the hydride ion. (e) How does the process in part (d) compare to the process for the electron affinity of a neutral hydrogen atom?

Although the electron affinity of bromine is a negative quantity, it is positive for Kr. Use the electron configurations of the two elements to explain the difference.

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