Ch.7 - Periodic Properties of the Elements

Problem 35c

Chapter 7, Problem 35c

True or false? c. S2− is larger than K+.

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Identify the ions involved: \( \text{S}^{2-} \) and \( \text{K}^+ \).

Understand that \( \text{S}^{2-} \) is a sulfide ion, which has gained 2 electrons, and \( \text{K}^+ \) is a potassium ion, which has lost 1 electron.

Consider the electron configuration: \( \text{S}^{2-} \) has the electron configuration of \( \text{Ar} \) (18 electrons), and \( \text{K}^+ \) also has the electron configuration of \( \text{Ar} \) (18 electrons).

Recognize that both ions are isoelectronic, meaning they have the same number of electrons.

Compare the nuclear charge: \( \text{S}^{2-} \) has a lower nuclear charge (16 protons) compared to \( \text{K}^+ \) (19 protons), which means \( \text{S}^{2-} \) is larger due to less effective nuclear attraction on the same number of electrons.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Ionic Radius

Ionic radius refers to the size of an ion in a crystal lattice. Cations, like K+, are smaller than their neutral atoms due to the loss of electrons, which reduces electron-electron repulsion and allows the remaining electrons to be pulled closer to the nucleus. Conversely, anions, such as S2−, are larger than their neutral atoms because the addition of electrons increases repulsion among them, causing the electron cloud to expand.

Charge and Size Relationship

The charge of an ion significantly influences its size. Anions (negatively charged ions) are generally larger than their neutral atoms due to increased electron repulsion, while cations (positively charged ions) are smaller due to decreased electron-electron repulsion. This relationship is crucial for comparing the sizes of different ions, as the charge directly affects the ionic radius.

Comparison of Ions

When comparing the sizes of different ions, it is essential to consider both their charges and the number of electrons. In this case, S2− has gained two electrons, resulting in a larger size, while K+ has lost one electron, making it smaller. Understanding these differences allows for accurate comparisons between ions, such as determining whether S2− is larger than K+.

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Related Practice

Textbook Question

Consider S, Cl, and K and their most common ions. (a) List the atoms in order of increasing size.

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Textbook Question

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

Textbook Question

Arrange each of the following sets of atoms and ions in order of increasing size. Se2−,Te2−, Se

Textbook Question

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.

Textbook Question

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?

Textbook Question

Write the electron configurations for the following ions, and determine which have noble-gas configurations.

a. Ru3+

b. As3−

c. Y3+

d. Pd2+

e. Pb2+

f. Au3+