In the ionic compounds LiF, NaCl, KBr, and RbI, the measured cation–anion distances are 2.01 Å (Li–F), 2.82 Å (Na–Cl), 3.30 Å (K–Br), and 3.67 Å (Rb–I), respectively. c. What estimates of the cation–anion distance would you obtain for these four compounds using neutral atom bonding atomic radii? Are these estimates as accurate as the estimates using ionic radii?
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객관식
Which of the following elements forms an ion that is smaller in radius than a Ca^{2+} ion?
A
Na (forming Na^{+})
B
Mg (forming Mg^{2+})
C
K (forming K^{+})
D
Sr (forming Sr^{2+})
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검증된 단계별 안내1
Understand that ionic radius depends on the number of protons (nuclear charge) and the number of electrons in the ion. When an atom loses electrons to form a cation, the radius generally decreases because there are fewer electrons repelling each other and the effective nuclear charge per electron increases.
Identify the electron configuration of the Ca^{2+} ion. Calcium (Ca) has atomic number 20, so Ca^{2+} has lost 2 electrons, resulting in 18 electrons, which is the same electron configuration as Argon (Ar).
Compare the ions given (Na^{+}, Mg^{2+}, K^{+}, Sr^{2+}) by considering their number of protons and electrons. All these ions have 18 electrons (isoelectronic with Ar), but different nuclear charges:
- Na^{+} has 11 protons and 10 electrons (actually Na^{+} has 10 electrons, but since Ca^{2+} has 18 electrons, Na^{+} is not isoelectronic; so check carefully). Actually, Na^{+} has 10 electrons, so it is not isoelectronic with Ca^{2+}.
- Mg^{2+} has 12 protons and 10 electrons (also not isoelectronic with Ca^{2+}).
- K^{+} has 19 protons and 18 electrons (isoelectronic with Ca^{2+}).
- Sr^{2+} has 38 protons and 36 electrons (not isoelectronic with Ca^{2+}).
Since Ca^{2+} has 18 electrons, the ions that are isoelectronic with Ca^{2+} are K^{+} (19 protons, 18 electrons) and Ca^{2+} itself (20 protons, 18 electrons).
Among ions with the same number of electrons, the ion with the greater nuclear charge (more protons) will have a smaller radius because the electrons are pulled closer to the nucleus.
Therefore, Mg^{2+} (12 protons, 10 electrons) is smaller than Ca^{2+} (20 protons, 18 electrons) only if they are isoelectronic, but since they are not, we need to consider the actual electron count and effective nuclear charge carefully.
In this problem, the key is to recognize that Mg^{2+} has fewer electrons than Ca^{2+}, but also fewer protons, and the ionic radius decreases with increasing positive charge and decreasing electron count.
Summarize that Mg^{2+} forms a smaller ion than Ca^{2+} because it has fewer electrons and a relatively high nuclear charge per electron, leading to a smaller ionic radius.
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교과서 질문
In the ionic compounds LiF, NaCl, KBr, and RbI, the measured cation–anion distances are 2.01 Å (Li–F), 2.82 Å (Na–Cl), 3.30 Å (K–Br), and 3.67 Å (Rb–I), respectively. b. Calculate the difference between the experimentally measured ion–ion distances and the ones predicted from Figure 7.8.
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