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An ionic solid crystallizes in a cubic lattice. The anions are located at the corners and 4 faces of the unit cell. The cations are located on 4 edges of the unit cell. What is the empirical formula for this compound? (A is cation, B is anion) A. A3B B. AB C. AB2 D. AB3

An ionic solid crystallizes in a cubic lattice. The anions are located at the corners and 4 faces of the unit cell. The cations are located on 4 edges of the unit cell. What is the empirical formula for this compound? (A is cation, B is anion) A. A3B B. AB C. AB2 D. AB3

Chemistry
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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**Question:** An ionic solid crystallizes in a cubic lattice. The anions are located at the corners and 4 faces of the unit cell. The cations are located on 4 edges of the unit cell. What is the empirical formula for this compound? (A is cation, B is anion) **Options:** A. A\(_3\)B B. AB C. AB\(_2\) D. AB\(_3\) **Explanation:** To determine the empirical formula of the compound, we need to analyze the contributions of anions (B) and cations (A) to the unit cell: 1. **Anions (B):** - There are 8 corners in a cubic lattice. Each corner is shared by 8 unit cells. Therefore, each corner contributes 1/8 of an anion to a single unit cell. Contribution of corners: \(8 \text{ corners} \times \frac{1}{8} \text{ anion/corner} = 1 \text{ anion}\) - There are 4 faces in this unit cell, with each face being shared by 2 unit cells. Therefore, each face contributes 1/2 an anion to a single unit cell. Contribution of faces: \(4 \text{ faces} \times \frac{1}{2} \text{ anion/face} = 2 \text{ anions}\) - Total anions (B) per unit cell = \(1 + 2 = 3 \text{ anions}\) 2. **Cations (A):** - There are 12 edges in a cubic lattice. Each edge is shared by 4 unit cells. Therefore, each edge contributes 1/4 of a cation to a single unit cell. Contribution of edges: \(4 \text{ edges} \times \frac{1}{4} \text{ cation/edge} = 1 \text{ cation}\) - Total cations (A) per unit cell = \(1 \text{ cation}\) Thus, the empirical formula of the compound is determined by the ratio of the cations to anions in the unit cell, which is \(A_1B_3\) or simply AB\(_3\). Therefore, the correct answer is: **D. AB\
Transcribed Image Text:**Question:** An ionic solid crystallizes in a cubic lattice. The anions are located at the corners and 4 faces of the unit cell. The cations are located on 4 edges of the unit cell. What is the empirical formula for this compound? (A is cation, B is anion) **Options:** A. A\(_3\)B B. AB C. AB\(_2\) D. AB\(_3\) **Explanation:** To determine the empirical formula of the compound, we need to analyze the contributions of anions (B) and cations (A) to the unit cell: 1. **Anions (B):** - There are 8 corners in a cubic lattice. Each corner is shared by 8 unit cells. Therefore, each corner contributes 1/8 of an anion to a single unit cell. Contribution of corners: \(8 \text{ corners} \times \frac{1}{8} \text{ anion/corner} = 1 \text{ anion}\) - There are 4 faces in this unit cell, with each face being shared by 2 unit cells. Therefore, each face contributes 1/2 an anion to a single unit cell. Contribution of faces: \(4 \text{ faces} \times \frac{1}{2} \text{ anion/face} = 2 \text{ anions}\) - Total anions (B) per unit cell = \(1 + 2 = 3 \text{ anions}\) 2. **Cations (A):** - There are 12 edges in a cubic lattice. Each edge is shared by 4 unit cells. Therefore, each edge contributes 1/4 of a cation to a single unit cell. Contribution of edges: \(4 \text{ edges} \times \frac{1}{4} \text{ cation/edge} = 1 \text{ cation}\) - Total cations (A) per unit cell = \(1 \text{ cation}\) Thus, the empirical formula of the compound is determined by the ratio of the cations to anions in the unit cell, which is \(A_1B_3\) or simply AB\(_3\). Therefore, the correct answer is: **D. AB\
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