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Chapter 7: Problem 35

Provide a brief explanation for each of the following: \((\mathbf{c}) \mathrm{O}^{2-}\) is larger than O.\((\mathbf{b}) S^{2-}\) is larger than \(\mathrm{O}^{2-} .(\mathbf{c}) \mathrm{S}^{2-}\) is larger than \(\mathrm{K}^{+} .(\mathbf{d}) \mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}.\)

Short Answer

Expert verified
In summary, (a) \(\mathrm{O}^{2-}\) is larger than O because gaining electrons causes the electron cloud to expand due to increased electron-electron repulsion. (b) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{O}^{2-}\) because sulfur has one more electron shell than oxygen, resulting in a larger size. (c) \(\mathrm{S}^{2-}\) is larger than \(\mathrm{K}^{+}\) because the increase in effective nuclear charge in forming \(\mathrm{K}^{+}\) leads to a smaller ionic size. (d) \(\mathrm{K}^{+}\) is larger than \(\mathrm{Ca}^{2+}\) because potassium loses 1 electron while calcium loses 2, resulting in a smaller ionic size for \(\mathrm{Ca}^{2+}\).

Step by step solution

01

1. Oxygen Ion (\(\mathrm{O}^{2-}\)) is larger than Oxygen Atom (O)

Oxygen atom (O) has the electron configuration 1s虏 2s虏 2p鈦�. When it gains 2 electrons to form \(\mathrm{O}^{2-}\) ion, its electron configuration becomes 1s虏 2s虏 2p鈦�. Gaining electrons causes the electron cloud to expand due to increased electron-electron repulsion. Therefore, anions are generally larger than their neutral atoms. Thus, \(\mathrm{O}^{2-}\) is larger than O.
02

2. Sulfide Ion (\(\mathrm{S}^{2-}\)) is larger than Oxygen Ion (\(\mathrm{O}^{2-}\))

In the periodic table, as we move down a group, the atomic size increases due to an increase in the number of electron shells. Both sulfur (S) and oxygen (O) are in group 16, with sulfur being right below oxygen. Therefore, sulfur has one more electron shell than oxygen. After gaining 2 electrons, both \(\mathrm{S}^{2-}\) and \(\mathrm{O}^{2-}\) have the same charge, but the added electron shell in \(\mathrm{S}^{2-}\) causes it to be larger than \(\mathrm{O}^{2-}\).
03

3. Sulfide Ion (\(\mathrm{S}^{2-}\)) is larger than Potassium Ion (\(\mathrm{K}^{+}\))

Sulfur (S) and potassium (K) are in different groups of the periodic table, with sulfur in group 16 and potassium in group 1. Both elements form ions with a charge difference of 3 units (\(\mathrm{S}^{2-}\) and \(\mathrm{K}^{+}\)). The increase in effective nuclear charge is more significant when losing an electron in forming \(\mathrm{K}^{+}\), leading to a smaller ionic size. Also, potassium's electron configuration becomes similar to that of the noble gas argon, with stronger attraction for remaining electrons. Therefore, \(\mathrm{S}^{2-}\) is larger than \(\mathrm{K}^{+}\).
04

4. Potassium Ion (\(\mathrm{K}^{+}\)) is larger than Calcium Ion (\(\mathrm{Ca}^{2+}\))

Potassium (K) and calcium (Ca) are both in the same period (Period 4). As we move left to right across the same period in the periodic table, the atomic size decreases, as does the ionic size due to an increase in the effective nuclear charge. While both ions lose electrons, calcium loses 2 while potassium loses 1. The loss of electrons in \(\mathrm{Ca}^{2+}\) leads to a stronger attraction between the remaining electrons and the nucleus, resulting in a smaller ionic size. This makes \(\mathrm{K}^{+}\) larger than \(\mathrm{Ca}^{2+}\).

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Key Concepts

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

Anions and Cations
Anions and cations are two types of ions, which are charged particles formed when atoms gain or lose electrons.
  • Anions are negatively charged ions. They form when an atom gains one or more electrons, resulting in more electrons than protons.
  • Cations are positively charged ions. They occur when an atom loses one or more electrons, resulting in more protons than electrons.

When atoms become ions, their sizes change due to electron-electron repulsion and the strength of the pull from the nucleus.
  • Anions, like \( \text{O}^{2-}\), are larger than their parent atoms because adding electrons increases repulsion among electrons, causing the electron cloud to expand.
  • Cations, such as \(\text{K}^+\), are smaller than their neutral atoms because losing electrons reduces repulsion and the remaining electrons feel a stronger pull from the nucleus.
Electron Configuration
Electron configuration is the arrangement of electrons around the nucleus of an atom. It follows specific rules and is essential for understanding chemical behavior and ionic sizes.
  • Electrons fill orbitals in a specific order, typically following the Aufbau principle, which states they fill the lowest energy levels first.
  • The electron configuration of an element shows how electrons are distributed among different orbitals.

When atoms gain or lose electrons to form ions, their electron configurations change. For instance:
  • The oxygen atom, with the configuration \(1s^2 2s^2 2p^4\), gains two electrons to form \(\text{O}^{2-}\), achieving a noble gas configuration \(1s^2 2s^2 2p^6\).
  • Potassium, losing one electron to become \(\text{K}^+\), also mimics a noble gas, resulting in a configuration \(1s^2 2s^2 2p^6 3s^2 3p^6\), similar to argon.
Periodic Table Trends
Understanding periodic table trends helps predict atomic and ionic sizes. These trends are influenced by electron configurations and the structure of the table.
  • Atomic Size: Generally increases down a group due to the addition of electron shells, and decreases across a period due to increasing nuclear charge pulling electrons closer.
  • Ionic Size: Depends on whether the ion is an anion or cation. Anions are generally larger than their atoms, whereas cations are smaller.

Consider sulfur (S) and oxygen (O), both in group 16:
  • \(\text{S}^{2-}\), having more electron shells, is larger than \(\text{O}^{2-}\).

Similarly, elements like potassium (K) and calcium (Ca) in the same period:
  • \(\text{K}^+\) is larger than \(\text{Ca}^{2+}\) because the cations formed have different proton-electron interactions, with calcium having more effective nuclear charge after losing two electrons.

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Most popular questions from this chapter

Use electron configurations to explain the following observa tions: (a) The first ionization energy of phosphorus is greater than that of sulfur. (b) The electron afnity of nitrogen is lower (less negative) than those of both carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of fluorine. (d) The third ionization energy of manganese is greater than those of both chromium and iron.

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.

Consider \(S, C 1,\) and \(K\) 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.

It is possible to define metallic character as we do in this book and base it on the reactivity of the element and the ease with which it loses electrons. Alternatively, one could measurehow well electricity is conducted by each of the elements to determine how "metallic" the elements are. On the basis of conductivity, there is not much of a trend in the periodic table: silver is the most conductive metal, and manganese the least. Look up the first ionization energies of silver and manganese; which of these two elements would you call more metallic based on the way we define it in this book?

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