A solid PQ has rock salt type structure in which \(Q\) atoms are the corners of the unit cell. If the body-centred atoms in all the unit cells are missing, the resulting stoichiometry will be (a) \(\mathrm{PQ}\) (b) \(\mathrm{PQ}_{2}\) (c) \(\mathrm{P}_{3} \mathrm{Q}_{4}\) (d) \(\mathrm{P}_{4} \mathrm{Q}_{2}\)

Short Answer

Expert verified
The resulting stoichiometry is \(\mathrm{P}_{3} \mathrm{Q}_{4}\).

Step by step solution

01

Understand the Rock Salt Structure

Rock salt structure (based on NaCl structure) consists of an FCC (face-centered cubic) lattice of Q atoms. P atoms occupy the octahedral sites in the lattice. Each unit cell contains 4 Q atoms from the corners and 12 from the edges (each shared by four unit cells), and 6 P atoms from the faces half shared by other unit cells, along with 1 P atom in the body center.
02

Calculate the Number of Q atoms per Unit Cell

Each corner atom is shared by 8 unit cells, so there is a total contribution of 1 Q atom from the corners (1/8 * 8). The 12 edge atoms each contribute 1/4, so that adds another 3 Q atoms (12 * 1/4). In total, there are 4 Q atoms per unit cell before considering the Q atoms that would have been at body centers.
03

Adjust for Missing Body-Centred Atoms

In the given scenario, the body-centred atoms are missing. Since the body-centred atom would normally contribute 1 whole atom to the unit cell, we do not adjust the count for Q atoms, but we must remove one P atom from each unit cell because P atoms would normally occupy those body-centred positions.
04

Calculate the Number of P atoms per Unit Cell

In a complete rock salt structure, each face contributes 1/2 an atom, with 6 faces, that's 3 P atoms. The body-centred P atom would have contributed 1 whole atom, but since it is missing, we do not count this. Thus, there are only 3 P atoms per unit cell.
05

Determine the Resulting Stoichiometry

Without the body-centred P atom, there are 3 P atoms and 4 Q atoms, which gives us a ratio of P to Q as 3:4. This results in the empirical formula of \(\mathrm{P}_{3}\mathrm{Q}_{4}\).

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Key Concepts

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

Stoichiometry in Solid-State Chemistry
Stoichiometry, a fundamental concept in chemistry, involves the calculation of the relative quantities of reactants and products in chemical reactions. In solid-state chemistry, stoichiometry extends to the analysis of the proportions of elements in crystalline solids. Understanding the stoichiometry of crystalline compounds allows us to predict properties such as reactivity, stability, and electrical behavior.

Crystalline solids, like the rock salt structure discussed in our exercise, have a highly ordered arrangement of atoms. Stoichiometry within these structures is determined by the ratio of the different types of atoms or ions and their specific arrangement within the crystal lattice. When disruptions occur, like the removal of atoms from specific lattice sites, the stoichiometry shifts, and this can lead to changes in the overall chemical formula of the unit cell.

To analyze stoichiometry in solid-state chemistry, we typically start by understanding the perfect, undisturbed structure of the crystal and then account for any defects or missing atoms, as in the case of the rock salt structure with missing body-centred atoms.
Face-Centered Cubic Lattice
The face-centered cubic (FCC) lattice is a common and important type of crystal structure in materials science and solid-state chemistry. It is characterized by atoms located at each of the corners and at the centers of all the cube faces of the unit cell. Each of these face-centered atoms is shared between two unit cells, and each corner atom is shared among eight unit cells.

In an FCC lattice, the coordination number—which represents the number of nearest neighbors an atom has—is 12, indicating a high degree of atomic packing. As a result, FCC structures are typically very dense with a high packing factor. This has implications on the material's properties, such as melting point, density, and ductility.

The FCC lattice structure is also known for its close-packed planes, which allow for easy slippage and deformation, making these materials particularly useful in applications requiring malleability and strength.
Unit Cell Composition
The concept of a unit cell is central to understanding crystalline materials. A unit cell is the smallest repeating pattern that shows the entire structure of the crystal. It acts as a building block for the macroscopic crystal lattice. The composition of a unit cell, therefore, determines the composition of the entire crystal.

In a rock salt structure, which is made up of an FCC lattice, we generally find that the anions (Q atoms in our exercise) occupy the corner and face-centered positions, and cations (P atoms in our exercise) fill the octahedral sites. Knowing this, we can calculate the number of atoms present inside a single unit cell and thus deduce the stoichiometric formula for the compound.

However, if there's a change, such as the missing body-centered atoms in our example, we must adjust our count of P and Q atoms. Such changes in composition can result from defects, impurities, or intentional doping in the material, all of which can alter the physical properties of the crystal. An accurate understanding of unit cell composition is essential for predicting the behavior and characteristics of solid-state materials.

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Most popular questions from this chapter

Which of the following oxide shows electrical properties like metals? (a) \(\mathrm{SiO}_{2}\) (b) \(\mathrm{MgO}\) (c) \(\mathrm{SO}_{2}(\mathrm{~s})\) (d) \(\mathrm{CrO}_{2}\)

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An element (atomic mass = 100 ) having BCC structure has unit cell edge length \(400 \mathrm{pm}\). The density of this element will be \(\left(N_{\Lambda}=6 \times 10^{23}\right)\) (a) \(5.2 \mathrm{~g} / \mathrm{m}\) (b) \(10.4 \mathrm{~g} / \mathrm{ml}\) (c) \(0.42 \mathrm{~g} / \mathrm{ml}\) (d) \(2.6 \mathrm{~g} / \mathrm{ml}\)

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