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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 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 \(Z_{\text {eff }} ?\) (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 P, phosphorus, based on the calculations for \(\mathrm{Si}\) and \(\mathrm{Cl}\).

Short Answer

Expert verified
The simplified method for estimating the effective nuclear charge (\(Z_{\text{eff}}\)) is more accurate than Slater's rules when comparing values for Silicon (Si) and Chlorine (Cl). Using the simplified method, we predict the \(Z_{\text{eff}}\) value for a valence electron in Phosphorus (P) to be approximately 5.5.

Step by step solution

01

Understand the effective nuclear charge concept

The effective nuclear charge (\(Z_{\text{eff}}\)) is the net positive charge experienced by an electron in an atom. It is the difference between the positive charge of the nucleus and the negative charge of the screening electrons. The shielding effect of these screening electrons reduces the electromagnetic force experienced by the outermost electron, leading to a lower effective nuclear charge.
02

Calculate \(Z_{\text{eff}}\) with the simplified method

According to the simplified method, core electrons contribute 1.00 and valence electrons contribute 0.00 to the screening constant. For Silicon (Si): Atomic number Z = 14 (14 protons in nucleus) Core electrons: 1s, 2s, 2p (total 10) Valence electrons: 3s, 3p (total 4) Screening constant S = 10 x 1.00 + 4 x 0 = 10 \(Z_{\text{eff, Si}} = Z - S = 14 - 10 = 4\) For Chlorine (Cl): Atomic number Z = 17 (17 protons in nucleus) Core electrons: 1s, 2s, 2p, 3s (total 10) Valence electrons: 3p (total 7) Screening constant S = 10 x 1.00 + 7 x 0 = 10 \(Z_{\text{eff, Cl}} = Z - S = 17 - 10 = 7\)
03

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

To calculate \(Z_{\text{eff}}\) using Slater's rules, we need to consider the contributions from all surrounding electrons to the screening constant. Slater's rules state that electrons in the same group shield the electron more effectively than those further away. For Silicon (Si): Following Slater's rules: S = 1s (2 * 0.3) + 2s, 2p (8 * 0.85) + 3s, 3p (4 * 1) = 0.6 + 6.8 + 4 = 11.4 \(Z_{\text{eff, Si}} = Z - S = 14 - 11.4 = 2.6\) For Chlorine (Cl): Following Slater's rules: S = 1s (2 * 0.35) + 2s, 2p (8 * 0.85) + 3s (2 * 1) + 3p (5 * 1) = 0.7 + 6.8 + 2 + 5 = 14.5 \(Z_{\text{eff, Cl}} = Z - S = 17 - 14.5 = 2.5\)
04

Compare the methods and determine the accuracy

Now we have calculated the \(Z_{\text{eff}}\) values using both methods, let's compare them with the given values: Simplified method: \(Z_{\text{eff, Si}}\) = 4 (compared to 4.29+) \(Z_{\text{eff, Cl}}\) = 7 (compared to 6.12+) Slater's rules: \(Z_{\text{eff, Si}}\) = 2.6 (compared to 4.29+) \(Z_{\text{eff, Cl}}\) = 2.5 (compared to 6.12+) The simplified method gives more accurate results, as its values are closer to the given values. Also, it can account for the steady increase in \(Z_{\text{eff}}\) that occurs upon moving left to right across a period.
05

Predict \(Z_{\text{eff}}\) for Phosphorus (P)

To predict the \(Z_{\text{eff}}\) for a valence electron in Phosphorus (P), we use the values calculated for Si and Cl: For Phosphorus (P), which lies between Si and Cl in the periodic table, we can make an approximation based on the simplified method (since it was more accurate): \(Z_{\text{eff, P}} \approx \frac{Z_{\text{eff, Si}} + Z_{\text{eff, Cl}}}{2} = \frac{4 + 7}{2} = 5.5\) Therefore, we predict the \(Z_{\text{eff}}\) value for a valence electron in Phosphorus (P) to be 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 are a set of guidelines used to estimate the effective nuclear charge (\(Z_{\text{eff}}\)) experienced by an electron in a multi-electron atom. These rules help us understand how electrons are affected by both the positive charge of the nucleus and the repulsion from other electrons.

The rules provide a method for calculating the screening constant (\(S\)), which represents the shielding effects of other electrons. This helps in accurately determining \(Z_{\text{eff}}\), giving insight into the electron's net attraction to the nucleus.

  • Identify the electron of interest in the atom and group other electrons based on their principal quantum number (n) and electron configuration.
  • Apply specific shielding contributions based on the group the electrons belong to:
    • Electrons within the same group (n) contribute 0.35 each, except for 1s which contribute 0.30.
    • Electrons in the \((n-1)\) group contribute 0.85 each.
    • Electrons in the \((n-2)\) or lower groups contribute 1.00 each.
Using these contributions, you calculate \(S\) by adding all the shielding effects. Once you have \(S\), you can determine \(Z_{\text{eff}}\) using the formula \(Z_{\text{eff}} = Z - S\), where \(Z\) is the atomic number.

Slater's Rules offer a more detailed approach to understanding atomic interactions by considering more precise shielding effects.
Screening Constant
The screening constant, denoted as \(S\), plays a crucial role in understanding how electrons within an atom influence one another. It represents the total effect of electron shielding and is an essential factor in calculating the effective nuclear charge (\(Z_{\text{eff}}\)).

Electrons are both attracted to the positively charged nucleus and repelled by other negatively charged electrons. The screening constant quantifies the net effect of these electron-electron interactions, helping to determine the extent to which electrons shield others from the nuclear charge.

To calculate \(S\), each electron's contribution is evaluated based on its position relative to the electron of interest. If you consider Slater's Rules, electrons in the same shell or closer to the nucleus contribute differently to \(S\).

A higher \(S\) value indicates greater shielding, which lowers the effective nuclear charge that an outer electron feels. This means that electrons in larger atoms with more competing electrons experience lower \(Z_{\text{eff}}\), altering their chemical properties as compared to smaller atoms with fewer electrons.

Understanding the screening constant is key for predicting chemical behavior, as it helps describe how variations in \(Z_{\text{eff}}\) affect atomic interactions. This is particularly helpful when analyzing trends across periods and groups in the periodic table.
Shielding Effect
The shielding effect describes how inner-layer electrons repel outer-layer electrons within an atom. This phenomenon impacts how strongly the outer electrons are attracted to the nucleus.

The concept of the shielding effect is essential for understanding why electrons in deeper shells reduce the effective nuclear charge experienced by outer electrons. It is a result of electron-electron repulsion, where the inner electrons act as a shield, diminishing the attractive force of the nucleus on electrons further away.

As the shielding effect increases, the effective nuclear charge decreases. This reduction in attraction allows outer electrons to exist at higher energy levels, as they are less tightly bound to the nucleus.
  • More inner shell electrons lead to a stronger shielding effect.
  • The shielding effect can vary significantly across the periodic table, affecting atomic size and ionization energy.
In practice, this effect explains the trends observed in atomic and ionic radii, ionization energies, and electron affinities across different elements. For example, as you move from left to right across a period, the number of shielding electrons remains relatively constant while the nuclear charge increases, resulting in a stronger effective nuclear charge.

Conversely, going down a group adds more inner electron shells, increasing the shielding effect and reducing the attraction of the outer electrons to the nucleus. Understanding the shielding effect and its impact on \(Z_{\text{eff}}\) is crucial for predicting and explaining the periodic trends that dominate chemistry.

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Most popular questions from this chapter

Which of the following is the expected product of the reaction of \(\mathrm{Mg}(s)\) and \(\mathrm{N}_{2}(g)\) under heat? (i) \(\mathrm{Mg}_{3} \mathrm{~N}(s)\) (ii) \(\mathrm{MgN}_{2}(s)\) (iii) \(\mathrm{Mg}_{3} \mathrm{~N}_{2}(s),\) (iv) \(\mathrm{Mg}(s)\) and \(\mathrm{N}_{2}(g)\) will not react with one another.

Consider the \(\mathrm{A}_{2} \mathrm{X}_{4}\) molecule depicted here, where \(\mathrm{A}\) and \(\mathrm{X}\) are elements. The \(A-A\) bond length in this molecule is \(d_{1}\), and the four \(\mathrm{A}-\mathrm{X}\) bond lengths are each \(d_{2}\). (a) In terms of \(d_{1}\) and \(d_{2},\) how could you define the bonding atomic radii of atoms A and X? (b) In terms of \(d_{1}\) and \(d_{2}\), what would you predict for the \(X-X\) bond length of an \(X_{2}\) molecule? [Section \(\left.7.3\right]\)

(a) What is the trend in first ionization energies as one proceeds down the group 17 elements? Explain how this trend relates to the variation in atomic radii. (b) What is the trend in first ionization energies as one moves across the fourth period from \(\mathrm{K}\) to \(\mathrm{Kr}\) ? How does this trend compare with the trend in atomic radii?

(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?

Detailed calculations show that the value of \(Z_{\text {eff }}\) for the outermost electrons in \(\mathrm{Na}\) and \(\mathrm{K}\) 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 \(\mathrm{Na}\) and \(\mathrm{K}\) 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? (e) Predict \(Z_{\text {eff }}\) for the outermost electrons in the \(\mathrm{Rb}\) atom based on the calculations for \(\mathrm{Na}\) and \(\mathrm{K}\).

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