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Chapter 7: Problem 36

Consider \(\mathrm{S}, \mathrm{Cl}\), and \(\mathrm{K}\) and their most common ions. (a) List the atoms in order of increasing size. (b) List the ions in order of increasing size. (c) Explain any differences in the orders of the atomic and ionic sizes.

Short Answer

Expert verified
The atomic size order of \(\mathrm{S}\), \(\mathrm{Cl}\), and \(\mathrm{K}\) is: $$\mathrm{K} < \mathrm{S} < \mathrm{Cl}$$ The ionic size order of their common ions (\(\mathrm{S^{2-}}\), \(\mathrm{Cl^{-}}\), and \(\mathrm{K^{+}}\)) is: $$\mathrm{K^{+}} < \mathrm{Cl^{-}} < \mathrm{S^{2-}}$$ The differences in atomic and ionic size orders can be attributed to changes in electron-electron repulsion and effective nuclear charge experienced by the electrons when atoms form ions.

Step by step solution

01

Determine the atomic sizes

To determine the atomic sizes of \(\mathrm{S}\), \(\mathrm{Cl}\), and \(\mathrm{K}\), let's first find their atomic numbers: - \(\mathrm{S}\) (sulfur) has the atomic number 16. - \(\mathrm{Cl}\) (chlorine) has the atomic number 17. - \(\mathrm{K}\) (potassium) has the atomic number 19. These elements are all in the same period (row) of the periodic table, which means they have the same number of electron shells. As we move from left to right across a period, the atomic size generally decreases, primarily due to the increased effective nuclear charge experienced by the electrons, which pulls them closer to the nucleus. Therefore, we can determine that the order of increasing atomic size should be: $$\mathrm{K} < \mathrm{S} < \mathrm{Cl}$$
02

Determine the common ions and their sizes

Now, let's identify the common ions for each element and determine their sizes: - \(\mathrm{S}\) typically forms the \(\mathrm{S^{2-}}\) ion. - \(\mathrm{Cl}\) typically forms the \(\mathrm{Cl^{-}}\) ion. - \(\mathrm{K}\) typically forms the \(\mathrm{K^{+}}\) ion. When an atom gains or loses electrons to form an ion, its size will change. For instance, when an atom loses an electron to create a positively charged ion (like \(\mathrm{K^{+}}\)), the overall electron-electron repulsion decreases, causing the remaining electrons to be pulled closer to the nucleus, resulting in a smaller ionic radius. Conversely, when an atom gains electrons to form a negatively charged ion (like \(\mathrm{S^{2-}}\) and \(\mathrm{Cl^{-}}\)), the electron-electron repulsion increases, causing the electrons to spread out and the ionic radius to increase. By following this logic, we can determine the order of increasing ionic size: $$\mathrm{K^{+}} < \mathrm{Cl^{-}} < \mathrm{S^{2-}}$$
03

Explain the differences in atomic and ionic size orders

The differences in the orders of atomic and ionic sizes can be attributed to the changes in electron-electron repulsion and effective nuclear charge experienced by the electrons when these atoms form their respective ions. Since potassium loses an electron to form the \(\mathrm{K^{+}}\) ion, its electron-electron repulsion decreases, causing the ion to have a smaller radius than its parent atom. Sulfur and chlorine gain electrons to form \(\mathrm{S^{2-}}\) and \(\mathrm{Cl^{-}}\) ions, respectively, leading to increased electron-electron repulsion and a corresponding increase in ionic radius. Thus, the atomic size order is: $$\mathrm{K} < \mathrm{S} < \mathrm{Cl}$$ while the ionic size order is: $$\mathrm{K^{+}} < \mathrm{Cl^{-}} < \mathrm{S^{2-}}$$

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Key Concepts

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

Periodic Trends
Periodic trends describe how certain properties of elements change in a predictable way as you move across or down the periodic table. One key trend is atomic size, which is largely influenced by the number of protons in the nucleus and the number of electrons surrounding it.

As you move from left to right across a period, atomic size tends to decrease. This happens because each successive element has more protons, which means a stronger positive charge in the nucleus. This increased nuclear charge pulls electrons closer to the nucleus, reducing the atomic radius.
  • Across a period: Atomic size decreases
  • Down a group: Atomic size increases (as more electron shells are added)
Potassium (K), sulfur (S), and chlorine (Cl) are in the same period. Thus, the trend will result in potassium having the largest atomic size due to its position farthest to the left, and chlorine having the smallest size, positioned farther to the right.
Electron-Electron Repulsion
Electron-electron repulsion is a key factor in determining the size of ions compared to their neutral atoms. When electrons are added or removed, the balance of repulsive forces among electrons changes. This affects the overall size of the atom or ion.

In the case of potassium ( K ), forming a K^{+} ion involves losing an electron. This reduces electron-electron repulsion because there are fewer electrons available to repel each other, allowing the remaining electrons to be pulled closer to the nucleus. This results in a smaller ionic size compared to its atomic size.

Conversely, sulfur ( S^{2-} ) and chlorine ( Cl^{-} ) gain electrons when they form ions. This increase in electron count raises the electron-electron repulsion within the electron cloud, causing the particles to spread out more. As a result, these ions have a larger ionic radius compared to their atomic size, displaying a contrast to potassium. **In summary**:
  • Positive ions: reduced size due to less repulsion
  • Negative ions: increased size due to more repulsion
Effective Nuclear Charge
Effective nuclear charge ( Z_{eff} ) is the net positive charge experienced by an electron in a multi-electron atom. It considers both the positive charge of the nucleus and the negative charge of the electrons located between the nucleus and the valence electrons.

The increase in effective nuclear charge across a period is a crucial reason why atomic size decreases. While the number of protons (+) increases, adding to the nuclear pull, the additional electrons being added do not sharply increase repulsion because they are added to the same shell and thus don鈥檛 shield one another effectively.

When considering ions, for instance, potassium losing an electron to become K^{+} means electrons experience a higher Z_{eff} due to decreased electron shielding, as there are fewer electrons between the nucleus and the outer electrons. This increases the pull inward, thus decreasing the size of the ion.

By contrast, when sulfur or chlorine gains electrons, their increased repulsion leads to a smaller Z_{eff} in comparison, resulting in a larger ion from the added repulsive forces. **To recap**:
  • Higher Z_{eff} : smaller atom/ion (more nuclear pull)
  • Lower Z_{eff} with increased electrons: larger ion (reduced pull)

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Most popular questions from this chapter

Explain the structure of the periodic table- two columns on the left, a block of ten for the transition metals, a block of six on the right, and a pair of 14 -member rows below, with reference to the orbitals we discussed in Chapter \(\underline{6}\).

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 \(Z_{\mathrm{eff}}\) 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?

Write balanced equations for the following reactions: (a) barium oxide with water, (b) iron(II) oxide with perchloric acid, (c) sulfur trioxide with water, (d) carbon dioxide with aqueous sodium hydroxide.

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. coo (Section 6.2 ) In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength \(58.4 \mathrm{nm}\). (a) What is the energy of a photon of this light in eV? (b) Write an equation that shows the process corresponding to the first ionization energy of \(\mathrm{Hg}\). (c) The kinetic energy of the emitted electrons is measured to be \(10.75 \mathrm{eV}\). What is the first ionization energy of Hg in kJ/mol? (d) Using Figure 7.9 , determine which of the halogen elements has a first ionization energy closest to that of mercury.

How are metallic character and first ionization energy related?
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