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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 1s² 2s² 2p⁴. When it gains 2 electrons to form \(\mathrm{O}^{2-}\) ion, its electron configuration becomes 1s² 2s² 2p⁶. 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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Most popular questions from this chapter

Arrange the following atoms in order of increasing effective nuclear charge experienced by the electrons in the \(n=3\) electron shell: \(\mathrm{K}, \mathrm{Mg}, \mathrm{P}, \mathrm{Rh} , \mathrm{Ti}.\)

Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, including electron configurations, for each process. (b) These two quantities have opposite signs. Which will be positive, and which will be negative? (c) Would you expect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger?

Using only the periodic table, arrange each set of atoms in order from largest to smallest: \((\mathbf{a}) \mathrm{K},\) Li, \(\mathrm{Cs} ;(\mathbf{b}) \mathrm{Pb}, \mathrm{Sn}, \mathrm{Si} ;(\mathbf{c}) \mathrm{F},\) \(\mathrm{O}, \mathrm{N} .\)

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