Problem 138
Question
An ionic compound has a unit cell consisting of A ions at the corners of a cube and \(\mathrm{B}\) ions on the centres of the faces of the cube. The empirical formula for this compound would be (a) \(\mathrm{AB}\) (b) \(\mathrm{A}_{2} \mathrm{~B}\) (c) \(\mathrm{AB}_{3}\) (d) \(\mathrm{A}_{3} \mathrm{~B}\)
Step-by-Step Solution
Verified Answer
The empirical formula is \(\text{AB}_3\).
1Step 1: Identify the Contributions to the Unit Cell
In a cube, there are 8 corners, with each ion contributing 1/8 to the unit cell since each corner is shared among 8 adjacent unit cells. If A ions are at the corners, then the total contribution of A ions is \(8 \times \frac{1}{8} = 1\).
2Step 2: Calculate the Contribution of B Ions
There are 6 faces of a cube, with each face-centered ion contributing 1/2 to the unit cell since each face is shared with one adjacent unit cell. If B ions are on the centres of the faces, then the total contribution of B ions is \(6 \times \frac{1}{2} = 3\).
3Step 3: Determine the Empirical Formula
Based on the contributions from steps 1 and 2, there is 1 A ion and 3 B ions per unit cell. Thus, the empirical formula of the compound is \(\text{AB}_3\).
Key Concepts
Unit CellIonic CompoundsCrystallography
Unit Cell
A unit cell is the smallest repeating unit that makes up a crystal lattice. It's like the building block of a crystal, showing the arrangement of atoms, ions, or molecules in space. Understanding how these cells stack together helps us explain many of the material's properties.
Let's say you have a cube which serves as a simple example of a unit cell. This cube can contain atoms or ions at its corners, on its edges, or in its center. In the original exercise, A ions sit at the cube's corners, while B ions fill the faces. This specific arrangement gives us a unique perspective of how these particles interact and define the compound's identity.
When you look at a cubic unit cell, you see that the corners each contribute only a fraction to the unit cell (1/8 per corner because 8 cubes share each corner). Similarly, faces share ions with only one neighboring cell, contributing half each (1/2). These different sharing rules are crucial to determine the actual number of each atom or ion in the unit cell and build the correct empirical formula.
Let's say you have a cube which serves as a simple example of a unit cell. This cube can contain atoms or ions at its corners, on its edges, or in its center. In the original exercise, A ions sit at the cube's corners, while B ions fill the faces. This specific arrangement gives us a unique perspective of how these particles interact and define the compound's identity.
When you look at a cubic unit cell, you see that the corners each contribute only a fraction to the unit cell (1/8 per corner because 8 cubes share each corner). Similarly, faces share ions with only one neighboring cell, contributing half each (1/2). These different sharing rules are crucial to determine the actual number of each atom or ion in the unit cell and build the correct empirical formula.
Ionic Compounds
Ionic compounds are formed when positive and negative ions bind together. The forces holding these ions together are called ionic bonds, which are strong due to the attraction between opposite charges. These compounds usually form from a metal and a non-metal, like sodium chloride (NaCl).
In the context of the exercise, A and B ions are part of an ionic compound. Here, A ions (likely metal) sit at cube corners, and B ions (likely non-metal) position themselves on cube faces. Together, they create an electrostatically balanced structure, ensuring the overall electrical neutrality of the compound.
Properties of ionic compounds include:
In the context of the exercise, A and B ions are part of an ionic compound. Here, A ions (likely metal) sit at cube corners, and B ions (likely non-metal) position themselves on cube faces. Together, they create an electrostatically balanced structure, ensuring the overall electrical neutrality of the compound.
Properties of ionic compounds include:
- High melting and boiling points
- Electrical conductivity when melted or dissolved in water
- Often soluble in water
Crystallography
Crystallography is the study of crystals, their arrangement, and structure. It helps us understand both the chemical and physical properties of materials by exploring how their internal structure affects them.
By examining unit cells like in our original problem, we tap into a broader world shown by crystallography. We're not just looking at a single piece, but how it repeats to form large, ordered structures. This uniform repetition is what defines crystals and gives rise to distinct physical properties such as shape, cleavage patterns, and even how they refract light.
A crystal's repeating pattern is largely responsible for its stability and properties. With crystallography, scientists can determine the atomic layout using techniques such as X-ray diffraction. These findings are invaluable for fields ranging from material science to pharmacology, as knowing the precise layout of atoms in a crystal can significantly impact the design and manufacturing of products."
By examining unit cells like in our original problem, we tap into a broader world shown by crystallography. We're not just looking at a single piece, but how it repeats to form large, ordered structures. This uniform repetition is what defines crystals and gives rise to distinct physical properties such as shape, cleavage patterns, and even how they refract light.
A crystal's repeating pattern is largely responsible for its stability and properties. With crystallography, scientists can determine the atomic layout using techniques such as X-ray diffraction. These findings are invaluable for fields ranging from material science to pharmacology, as knowing the precise layout of atoms in a crystal can significantly impact the design and manufacturing of products."
Other exercises in this chapter
Problem 136
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In a compound, atoms of element \(\mathrm{Y}\) from ccp lattice and those of element \(\mathrm{X}\) occupy \(2 / 3^{\text {nd }}\) oftetrahedral voids. The form
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