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Chapter 7: Problem 32

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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Most popular questions from this chapter

An element \(\mathrm{X}\) reacts with oxygen to form \(\mathrm{XO}_{2}\) and with chlorine to form \(\mathrm{XCl}_{4} . \mathrm{XO}_{2}\) is a white solid that melts at high temperatures (above \(1000^{\circ} \mathrm{C} ) .\) Under usual conditions, \(\mathrm{XCl}_{4}\) is a colorless liquid with a boiling point of \(58^{\circ} \mathrm{C}\) . (a) \(\mathrm{XCl}_{4}\) reacts with water to form \(\mathrm{XO}_{2}\) and another product. What is the likely identity of the other product? (b) Do you think that element \(\mathrm{X}\) is a metal, nonmetal, or metalloid? (c) By using a sourcebook such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element \(\mathrm{X}\).

Until the early 1960s, the group 8A elements were called the inert gases. (a) Why was the term inert gases dropped? (b) What discovery triggered this change in name? (c) What name is applied to the group now?

In Table 7.8 , the bonding atomic radius of neon is listed as 0.58 A, whereas that for xenon is listed as 1.40 A. A classmate of yours states that the value for Xe is more realistic than the one for Ne. Is she correct? If so, what is the basis for her statement?

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.

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Cesium is added to water. (b) Strontium is added to water. (c) Sodium reacts with oxygen. (d) Calcium reacts with iodine.
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