Problem 95
Question
A compound contains two types of atoms: \(\mathrm{X}\) and \(\mathrm{Y}\). It crystallizes in a cubic lattice with atoms \(\mathrm{X}\) at the corners of the unit cell and atoms \(\mathrm{Y}\) at the body centres. The simplest possible formula of this compound is: (a) XY (b) \(\mathrm{X}_{2} \mathrm{Y}_{2}\) (c) \(\mathrm{XY}_{6}\) (d) \(X_{8} Y^{2}\)
Step-by-Step Solution
Verified Answer
The simplest formula for the compound is XY.
1Step 1: Understand the Lattice Structure
In a cubic lattice, atoms positioned at the corners of the unit cell are shared by eight adjacent unit cells. This setup is used to calculate the number of each type of atom per unit cell.
2Step 2: Calculate Atoms X in the Unit Cell
Atoms X are located at the 8 corners of the unit cell. Since each corner atom is shared by 8 unit cells, the contribution from corner atoms per unit cell is \(\frac{8}{8} = 1\). Thus, there is 1 atom of X per unit cell.
3Step 3: Calculate Atoms Y in the Unit Cell
Atoms Y are exclusively located at the body center of the unit cell, which means this atom is entirely within the single unit cell. Therefore, there is 1 atom of Y per unit cell.
4Step 4: Determine the Simplest Formula
From the previous steps, we have 1 atom of X and 1 atom of Y inside the unit cell. Therefore, the simplest formula representing the compound is XY.
Key Concepts
Cubic Lattice StructureUnit Cell CalculationSimplest Formula Determination
Cubic Lattice Structure
A crystal lattice is essentially the three-dimensional arrangement of atoms within a crystal. When we talk about a cubic lattice structure, this means the unit cell, or the smallest repeating unit of the lattice, forms a cube. The cubic lattice is one of the most common and simplest to understand. Its simplicity makes it a fundamental concept in understanding solid-state chemistry.
In our example of a cubic lattice, atoms located at each corner of the cube are involved. Each corner atom is shared with adjacent cells, eight in this specific case. It is important to note that atoms positioned precisely at the vertices are essentially part of multiple unit cells, emphasizing the concept of shared atoms within crystalline structures. This sharing fundamentally alters the way we count atoms per unit cell, a crucial element for further calculations and formula determination.
By understanding this structure, students can better grasp how atoms are distributed across unit cells and why their placement affects how we interpret the chemical formula of a compound.
In our example of a cubic lattice, atoms located at each corner of the cube are involved. Each corner atom is shared with adjacent cells, eight in this specific case. It is important to note that atoms positioned precisely at the vertices are essentially part of multiple unit cells, emphasizing the concept of shared atoms within crystalline structures. This sharing fundamentally alters the way we count atoms per unit cell, a crucial element for further calculations and formula determination.
By understanding this structure, students can better grasp how atoms are distributed across unit cells and why their placement affects how we interpret the chemical formula of a compound.
Unit Cell Calculation
Calculating how many atoms reside within a unit cell is pivotal for developing a basic understanding of a compound's structure. Let us explore how we accomplish this with our example of the cubic lattice.
In this structure, atoms
On the other hand, for atoms Y, located at the center of the cube, there is only 1 per unit cell. This is because the body-centered atom is not shared with other unit cells, meaning it contributes fully as one atom within that unit structure. The detailed understanding of this arrangement allows for precise chemical formula allocation based on the actual presence of each atom type within a structural unit.
In this structure, atoms
- located at the corners are shared by 8 unit cells simultaneously
- atoms at the center are wholly contained within one unit cell
On the other hand, for atoms Y, located at the center of the cube, there is only 1 per unit cell. This is because the body-centered atom is not shared with other unit cells, meaning it contributes fully as one atom within that unit structure. The detailed understanding of this arrangement allows for precise chemical formula allocation based on the actual presence of each atom type within a structural unit.
Simplest Formula Determination
Understanding and calculating the simplest formula for a compound critically depend on the number of each type of atom within the unit cell. Having calculated the number of atoms X and Y in the cubic unit cell in the prior sections, we determine the simplest, empirical formula by considering their ratio.
From the example given:
The simplicity of this process highlights its significance, ensuring accurate representation of the compound through its empirical formula. Understanding the methodical enumeration of atoms serves as a foundation for accurately speaking to the compound's stoichiometry and overall chemical representation.
From the example given:
- We found 1 atom of X per unit cell, accounted via the contribution from corner positions.
- We also found 1 atom of Y per unit cell, attributed to the body-centered position.
The simplicity of this process highlights its significance, ensuring accurate representation of the compound through its empirical formula. Understanding the methodical enumeration of atoms serves as a foundation for accurately speaking to the compound's stoichiometry and overall chemical representation.
Other exercises in this chapter
Problem 91
The number of atoms per unit cell in a \(\mathrm{BCC}\), a FCC and a simple cubic cell are respectively (a) \(1,4,2\) (b) \(2,4,1\) (c) \(4,1,2\) (d) \(4,2,2\)
View solution Problem 94
Amorphous substances show (i) short and long range order (ii) short range order (iii) long range order (iv) have no sharp melting point (a) (i) and (ii) are cor
View solution Problem 96
Which of the following statements is correct? (1) The coordination number of each type of ion in \(\mathrm{CsCl}\) crystal is 8 (2) A metal that crystallizes in
View solution Problem 97
The edge length of the unit cell of \(\mathrm{NaCl}\) crystal lattice is \(552 \mathrm{pm}\). If the ionic radius of sodium ion is \(95 \mathrm{pm}\). What is t
View solution