Question

In a face centered cubic arrangement of \(A\) and \(B\) atoms whose \(A\) atoms are at the corner of the unit cell and \(B\) atoms at the face centers. One of the \(B\) atoms missing from one of the face in unit cell. The simplest formula of compound is: (a) \(A B_{3}\) (b) \(A_{8} B_{5}\) (c) \(A_{2} B_{5}\) (d) \(A B_{2 / 5}\)

Short answer

The simplest formula of the compound is (c) A_{2}B_{5}.

Step-by-step solution

Counting A atoms

In a face-centered cubic (FCC) unit cell, there are 8 corner atoms. However, each corner atom is shared by 8 adjacent unit cells, so each unit cell effectively has 1/8 of each corner atom. Multiplying the 8 corner atoms by 1/8 gives us 1 full A atom per unit cell.

Counting B atoms

There are 6 face-centered atoms in the FCC unit cell, but each is shared by 2 unit cells. Normally, this would give us 6/2 = 3 B atoms per unit cell. However, one of the B atoms is missing, so we subtract 1/2 (the contribution of one B atom to the unit cell) from the 3 to get a total of 2.5 B atoms per unit cell.

Determining the Simplest Formula

With 1 A atom and 2.5 B atoms per unit cell, the ratio of A to B atoms is 1 A atom for every 2.5 B atoms. To find the simplest whole number ratio, we can multiply both the number of A atoms and B atoms by 2 to get a whole number ratio, resulting in 2 A atoms for every 5 B atoms. Therefore, the simplest formula is given by A2B5.

Key concepts

Crystal Structure

Understanding the arrangement of atoms in a crystalline solid is essential to grasp the foundation of solid state chemistry. One of the most common and well-studied arrangements is the face-centered cubic (FCC) unit cell. In an FCC unit cell, atoms are tightly packed, occupying every corner and the center of each face of the cube.

The uniqueness of FCC structures lies in how atoms are shared among adjacent unit cells. Corner atoms are shared between eight unit cells, resulting in each unit cell effectively having only one-eighth of each of its eight corner atoms. Meanwhile, face-centered atoms are divided between two unit cells. When studying FCC unit cells, it's integral to accurately count the contribution of each atom to determine the unit cell's composition.

The concept becomes more engaging when a variation is introduced, such as the removal of an atom from the typical structure—this tests our understanding of the three-dimensional arrangement and the importance of considering the entire environment of the unit cell in calculations.

Stoichiometry

Stoichiometry is the quantitative relationship between the elements in a compound, essential for dealing with the chemical composition of substances. When we analyze a crystal structure like an FCC unit cell, the stoichiometry tells us the simplest whole number ratio of the types of atoms present within that structure.

In the provided exercise, we determined the stoichiometry by counting the atoms of element A and element B according to their positions in the unit cell. With one A atom and 2.5 B atoms, the stoichiometry initially appears as a fractional composition. However, to express this in the simplest formula, i.e. to find the ratio of atoms in the smallest whole numbers that maintain the relative proportions, we must find a common factor that lets us transform these fractions into integers. By multiplying both quantities by 2, we get a 2:5 ratio, hence a stoichiometric formula of A2B5.

The stoichiometry serves as a fundamental aspect of crystal chemistry and materials science because it dictates the proportions of elements that make up a compound, influencing the material's properties and behavior.

Solid State Chemistry

Solid state chemistry delves into the study of the synthesis, structure, and properties of solid materials. It lays the groundwork for understanding different materials and their functionalities, which are defined by their crystal structures and stoichiometry.

For example, the FCC unit cell, characterized by its particular atomic arrangement, can influence the hardness, melting point, and conductivity of a material. The removal of an atom in the exercise shifts the stoichiometry, altering the compound's composition and potentially its properties. This demonstrates the interconnectedness between a material's microstructure and its macroscopic traits.

In the educational context, bringing these concepts to life through practical problems, like examining the implications of a missing face-centered atom in an FCC unit cell, can deeply enhance a student's comprehension of solid state chemistry and material science.