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Chapter 1: Problem 104

A solid has 3 types of atoms namely \(X, Y\) and \(Z . X\) forms an FCC lattice with \(\mathrm{Y}\) atoms occupying all the tetrahedral voids and \(Z\) atoms occupying half the octahedral voids. The formula of the solid is: (a) \(\mathrm{XYZ}\) (b) \(\mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z}\) (c) \(\mathrm{X}_{4} \mathrm{YZ}_{2}\) (d) \(\mathrm{X}_{4} \mathrm{Y}_{2} \mathrm{Z}\)

Short Answer

Expert verified
\(\mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z}\)

Step by step solution

01

Understanding FCC lattice

In a face-centered cubic (FCC) lattice, there is one atom at each corner of the cube and one atom at the center of each face. The total number of atoms per unit cell can be calculated as: 8 corners × 1/8 atom per corner + 6 faces × 1/2 atom per face = 1 + 3 = 4 atoms. Since X forms the FCC lattice, there are 4 X atoms per unit cell.
02

Counting tetrahedral voids

In an FCC lattice, the number of tetrahedral voids is twice the number of atoms in the lattice. Therefore, in this case, there are 4 X atoms, resulting in 4 × 2 = 8 tetrahedral voids. Since Y atoms occupy all tetrahedral voids, there are 8 Y atoms per unit cell.
03

Counting octahedral voids

The number of octahedral voids in an FCC lattice is equal to the number of atoms in the lattice. Hence, there are 4 octahedral voids in the unit cell. Since Z atoms occupy half the octahedral voids, there will be 4/2 = 2 Z atoms per unit cell.
04

Determining the formula

Combining the count of each type of atom in the unit cell, we have 4 X atoms, 8 Y atoms, and 2 Z atoms. To find the simplest whole number ratio, we divide each count by the smallest number of atoms present, which in this case is 2 (corresponding to Z atoms). This gives a ratio of X:Y:Z = 4:8:2 = 2:4:1. Therefore, the formula of the solid is \(\mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z}\), which simplifies to \(\mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z}\) as the final answer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

FCC lattice structure
The face-centered cubic (FCC) lattice is a common crystal structure found in many metals and alloys. Its distinctive feature is that it consists of atoms located at each corner of a cube and at the center of each of the cube's faces.

This arrangement results in a highly dense packing of atoms, which is why it is so prevalent among metallic elements. Understanding the FCC lattice is critical in solid state chemistry as it dictates many physical properties of the material, such as density, melting point, and how it interacts with other substances.
Tetrahedral voids in crystal
In an FCC lattice, tetrahedral voids are the small, empty spaces surrounded by four atoms, which form a tetrahedron—a pyramid with a triangular base. Each FCC unit cell has eight tetrahedral voids.

The presence of these voids is significant because other atoms can occupy these spaces, effectively becoming part of the crystal structure. This capability allows for the formation of various compounds and alloys with properties distinct from those of the parent FCC metal.
Octahedral voids occupancy
When examining octahedral voids within the FCC lattice, we find that they correspond to the empty spaces at the center of the cube and on the edges of the unit cell. Each FCC has exactly four octahedral voids, which can also be occupied by other atoms.

The proportion of these voids filled can significantly influence the material's chemical and physical behavior. For example, in the context of the exercise, only half of the octahedral voids are occupied, which leads to a specific stoichiometry in the resulting crystalline structure.
Chemical formula determination
To determine the chemical formula of a solid comprising different atoms in various lattice positions, we must consider the number of each atom type per unit cell. By finding the ratio of these atoms and simplifying to the lowest whole numbers, we can derive the empirical formula of the compound.

This process, which involves comparing the occupancy of different types of voids, reveals the stoichiometric relationship between the elements in the solid. It's a crucial step in understanding the composition of crystalline materials and is deeply rooted in the principles of solid state chemistry.

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Most popular questions from this chapter

In calcium fluoride structure, the coordination numbers of calcium and fluoride ions respectively are: (a) 6 and 8 (b) 8 and 8 (c) 4 and 8 (d) 8 and 4

The empty space left in a hexagonal close packing of spheres in three dimensions is: (a) \(64 \%\) (b) \(26 \%\) (c) \(14 \%\) (d) \(52.4 \%\)

The number of atoms (n) contained within a body centred cubic cell is: (a) 4 (b) 3 (c) 2 (d) 1

Which type of solids here are generally good conductors of electricity? (a) Metallic (b) Covalent (c) Ionic (d) Molecular

Number of atoms in the unit cell of Na (BCC type crystal) and \(\mathrm{Mg}\) (FCC type crystal) are respectively: (a) 4,4 (b) 4,2 (c) 2,4 (d) 1,1
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