Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

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 \(\mathrm{Z}\) atoms occupying half the octahedral voids. The formula of the solid is (a) XYZ (b) \(\mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z}\) (c) \(\mathrm{X}_{4} \mathrm{YZ}_{2}\) (d) \(\mathrm{X}_{4} \mathrm{Y}_{2} \mathrm{Z}\)

Step-by-Step Solution

Answer

The formula of the solid is \( \mathrm{X}_{2} \mathrm{Y}_{4} \mathrm{Z} \).

Step 1: Understanding the FCC Lattice

The FCC (face-centered cubic) lattice is one common type of crystal structure. In this lattice, there is one atom at each corner of the cube and one atom at the center of each face. Thus, an FCC lattice effectively contains 4 atoms per unit cell.

Step 2: Calculating the Number of Y Atoms

In an FCC lattice, there are typically 8 tetrahedral voids per unit cell. Each void can be occupied by a Y atom. Since all tetrahedral voids are occupied by Y atoms, there will be 8 Y atoms per unit cell.

Step 3: Determining the Number of Octahedral Voids

In the FCC structure, there are 4 octahedral voids per unit cell. Since Z atoms occupy half of these voids, there will be 2 Z atoms in the unit cell.

Step 4: Writing the Empirical Formula

With the numbers of X, Y, and Z atoms as 4, 8, and 2 respectively, we need to find the simplest whole number ratio. Simplifying the ratio (4:8:2) by dividing by 2, we get X:4/2=2, Y:8/2=4, Z:2/2=1. Therefore, the empirical formula is X₂Y₄Z.

Key Concepts

FCC lattice

An FCC lattice, or face-centered cubic lattice, is a type of crystal structure where calcium atoms are arranged densely. This structure is unique because each unit cell, which is the smallest repeating unit of the lattice, contains atoms positioned at all corners and the centers of each face of the cube.
This arrangement results in a high packing efficiency, meaning that the atoms fill much of the available space in the lattice.
In such a dense arrangement, an FCC lattice has 12 edges each shared by 4 unit cells and 6 face-centered atoms each shared by 2 unit cells. This effectively means there are

• 8 corner atoms shared by eight unit cells, contributing 1/8th to each cell.
• 6 face-centered atoms shared by two unit cells, contributing 1/2 to each cell.

Therefore, the total number of atoms per unit cell in an FCC lattice is 4.

Tetrahedral voids

In an FCC lattice, tiny empty spaces, known as voids, exist between the atoms. Tetrahedral voids are a type of these spaces. They are formed by four atoms of an FCC structure, thus giving them a tetrahedral shape.
This kind of void is smaller compared to other voids, which makes it suitable for smaller atoms to reside. Each tetrahedral void in an FCC unit cell can be occupied by smaller atoms in the crystal structure, optimizing the packing efficiency of the structure.
Importantly, an FCC lattice contains 8 tetrahedral voids per unit cell. These voids are significant because they directly influence how atoms and ions are packed within the solid, contributing to the stability and properties of the material.

Octahedral voids

The octahedral voids are another type of void found in the FCC lattice. These voids are larger compared to the tetrahedral ones and are formed by six atoms. The shape surrounding each octahedral void resembles an octahedron, hence the name.
In an FCC unit cell, there are a total of 4 octahedral voids.

• One void is located at the center of the unit cell, being shared by several surrounding atoms.
• Other voids are positioned at the edges, shared among neighboring cells.

Unlike smaller tetrahedral voids, octahedral voids can host slightly larger atoms or ions, which makes them crucial in determining the lattice structure's ability to incorporate different types of atoms. These voids help in understanding how the lattice can accommodate varying atom sizes without distorting the structure.

Empirical formula

The empirical formula of a compound offers insight into the simplest whole-number ratio of the constituent atoms. In the context of an FCC lattice with various voids occupied, determining the empirical formula helps identify the stoichiometric combinations of atoms within the solid.
To find the empirical formula, we consider the number and type of atoms occupying the lattice and voids. Based on the given information in the exercise:

• The number of X atoms per unit cell is 4, considering the FCC structure.
• The number of Y atoms, filling all tetrahedral voids, is 8.
• The number of Z atoms, present in half of the octahedral voids, is 2.

These values are put into a ratio and simplified to get the simplest form, which was calculated to be X₂Y₄Z. This simplified ratio helps to understand the proportions of different atoms contributing to the formation of the compound, leading to a clearer understanding of the chemical formula.