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Consider the isoelectronic ions \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\) . (a) Which ion is smaller? (b) Using Equation 7.1 and assuming that core electrons contribute 1.00 and valence electrons contribute nothing to the screening constant, \(S\) , calculate \(Z_{\text { eff}}\) for these two 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?

Short Answer

Expert verified
The smaller ion is \(\mathrm{K}^{+}\). Using the assumption that valence electrons do not contribute to the screening constant, both \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\) have an effective nuclear charge of 18. However, using Slater's rules, \(\mathrm{Cl}^{-}\) has an effective nuclear charge of 0.65 and \(\mathrm{K}^{+}\) has an effective nuclear charge of 1.65. For isoelectronic ions, as the effective nuclear charge increases, the ionic radius decreases.

Step by step solution

01

Determine which ion is smaller

Isoelectronic ions have the same number of electrons but different numbers of protons. In the case of \(\mathrm{Cl}^{-}\) and \(\mathrm{K}^{+}\), both have 18 electrons. However, \(\mathrm{Cl}^{-}\) has 17 protons, while \(\mathrm{K}^{+}\) has 19. Since the effective nuclear charge is greater for \(\mathrm{K}^{+}\), its electrons will be more attracted to the nucleus, resulting in a smaller ionic radius. Therefore, the smaller ion is \(\mathrm{K}^{+}\).
02

Calculate effective nuclear charge assuming valence electrons don't contribute to the screening constant

Effective nuclear charge, \(Z_{\text { eff}}\), can be calculated as follows: \(Z_{\text { eff}} = Z - S\) Where \(Z\) is the atomic number and \(S\) is the screening constant. Assuming core electrons contribute 1.00 and valence electrons contribute nothing to \(S\), for \(\mathrm{Cl}^{-}\): \(S_\mathrm{Cl^{-}} = 17-18= -1\) \(Z_{\text { eff}}^{\mathrm{Cl}^{-}} = 17 - (-1) = 18\) For \(\mathrm{K}^{+}\): \(S_\mathrm{K^{+}} = 19-18=1\) \(Z_{\text { eff}}^{\mathrm{K}^{+}}= 19 - 1 = 18\)
03

Calculate effective nuclear charge using Slater's rules

According to Slater's rules, the screening constant \(S\) can be calculated using the following formula: \(S = \sum_{i=1}^{n} n_i \times s_i\) Where \(n_i\) is the number of electrons in the i-th shell, and \(s_i\) is the corresponding Slater's rule constant. For \(\mathrm{Cl}^{-}\): \(S_\mathrm{Cl^{-}} = 1(0.35) + 16(1) \) \(S_\mathrm{Cl^{-}} = 16.35\) \(Z_{\text { eff}}^{\mathrm{Cl}^{-}} = 17 - 16.35 = 0.65\) For \(\mathrm{K}^{+}\): \(S_\mathrm{K^{+}} = 1(0.35) + 17(1) \) \(S_\mathrm{K^{+}} = 17.35\) \(Z_{\text { eff}}^{\mathrm{K}^{+}} = 19 - 17.35 = 1.65\)
04

Discuss the relationship between effective nuclear charge and ionic radius for isoelectronic ions

For isoelectronic ions, as the effective nuclear charge increases, the ionic radius decreases. This is because the greater the effective nuclear charge, the stronger the attraction between the positively charged nucleus and the negatively charged electrons, causing the electrons to be pulled in closer to the nucleus. As a result, the ionic radius decreases.

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Key Concepts

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

Effective Nuclear Charge
When discussing isoelectronic ions like \( \mathrm{Cl}^{-} \) and \( \mathrm{K}^{+} \), a key concept is the effective nuclear charge (\(Z_{\text{eff}}\)). This is the net positive charge experienced by electrons in an atom. It takes into account both the total positive charge of the nucleus and the shielding effect of electron clouds between the nucleus and the electron of interest.
The formula for calculating \(Z_{\text{eff}}\) is:\[ Z_{\text{eff}} = Z - S \]- \(Z\): Atomic number (positive charge of nucleus)- \(S\): Screening constant (effect of inner electrons reducing effective nuclear charge)
For isoelectronic ions, the one with higher protons (higher \(Z\)) will generally have a higher effective nuclear charge. This is why \( \mathrm{K}^{+} \), with 19 protons, has a higher \(Z_{\text{eff}}\) than \( \mathrm{Cl}^{-} \), which only has 17 protons, despite both having 18 electrons.
Screening Constant
The screening constant \(S\) represents the extent to which electrons shield or screen each other from the full charge of the nucleus.
Electrons that are closer to the nucleus can shield the outer electrons from the full attractive force of the nucleus. Calculating the screening constant helps to adjust the view of how much nuclear charge an electron "feels." The higher the number of inner electrons, the higher the value of \(S\), resulting in a lower effective nuclear charge experienced by outer electrons.
In simpler calculations as those assuming core electrons provide full screening whereas valence electrons contribute zero, \(S\) is calculated by:
  • Taking each core electron as blocking one unit of charge.
For the ion \( \mathrm{Cl}^{-} \), we assume that each of the inner 17 electrons contributes one unit. This gives \(S=18\) and effective nuclear charge as 18 for \( \mathrm{K}^{+} \), assuming similar conditions for contributions.
Slater's Rules
Slater's Rules offer a more nuanced and detailed approach to computing the screening constant \(S\). It assigns different shielding values to electrons depending on their shell and subshell positions.
Here's a simplified application of Slater's Rules:- Electrons in the same shell (\(n\)) contribute less to \(S\) than electrons in lower shells.- Electrons in \((n,p)\) shells typically contribute 0.35 if they are in the same shell and 0.85 if they are in earlier shells.- Electrons in \((s,d)\) shells contribute even more, often a full shield equivalent.
Using Slater's Rules, you calculate \(S\) for \( \mathrm{Cl}^{-} \) as:\[ S = 1(0.35) + 16(1) = 16.35 \]Thus, \(Z_{\text{eff}}\) becomes:\[ Z_{\text{eff}}^{\mathrm{Cl}^{-}} = 17 - 16.35 = 0.65 \]For \( \mathrm{K}^{+} \), a similar detailed calculation using these rules gives \(\[ S = 17.35 \]\). This precision gives a more accurate picture of the balance of forces at play in atoms and ions.

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Most popular questions from this chapter

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.

Arrange the following oxides in order of increasing acidity: \(\mathrm{CO}_{2}, \mathrm{CaO}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{SO}_{3}, \mathrm{SiO}_{2}, \mathrm{P}_{2} \mathrm{O}_{5} .\)

The following observations are made about two hypothetical elements \(A\) and \(B :\) The \(A-A\) and \(B-B\) bond lengths in the elemental forms of \(A\) and \(B\) are 2.36 and \(1.94 \hat{A},\) respectively. A and \(\mathrm{B}\) react to form the binary compound \(\mathrm{AB}_{2},\) which has a linear structure (that is \(\angle \mathrm{B}-\mathrm{A}-\mathrm{B}=180^{\circ} ) .\) Based on these statements, predict the separation between the two B nuclei in a molecule of \(\mathrm{AB}_{2}\) .

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What is the relationship between the ionization energy of an anion with a \(1-\) charge such as \(F^{-}\) and the electron affinity of the neutral atom, \(\mathrm{F}\) ?

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