A compound formed by elements \(A\) and \(B\) crystallizes in cubic structure where \(A\) atoms are at the corners of a cube and \(B\) atoms are at the face centre. What is the formula of the compound?

Short Answer

Expert verified
The formula of the compound is AB3.

Step by step solution

01

Count the atoms at cube corners

In a cubic structure, there are 8 corners, and each corner atom is shared among 8 different cubes. Therefore, only 1/8 of each corner atom actually belongs to one individual cube. Multiply the number of corner atoms (8) by the fraction that belongs to one cube (1/8) to find the number of A atoms per cube. This equals to 8 x (1/8) = 1 A atom per cube.
02

Count the atoms at face centres

There are 6 faces in a cube and one face-centered atom per face. However, each face-centered atom is shared between 2 cubes, so only 1/2 of each face-centered atom belongs to one cube. Multiply the number of face centres (6) by the fraction that belongs to one cube (1/2). This equals to 6 x (1/2) = 3 B atoms per cube.
03

Determine the formula

Including atoms from the corners and face centres, each cube consists of 1 A atom and 3 B atoms. Therefore, the formula of the compound is AB3.

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Key Concepts

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

Crystallography
Crystallography is the scientific study of crystals and their formation. At its core, crystallography investigates the arrangement of atoms in solid materials and characterizes their structure using mathematical models and physical principles. The arrangement of atoms in a crystal structure is highly ordered and follows repetitive patterns, known as a crystal lattice. Each point in the lattice represents the position of an atom or a group of atoms.

In the context of our exercise, the cubic crystal structure is one of the seven primary crystal systems in crystallography. It is distinguished by a lattice where the unit cells are shaped like a cube, with equal sides and 90-degree angles between them. Inside the cubic lattice, atoms can be positioned at various sites: corners, face centers, edge centers, and body centers. Understanding where atoms are located within a unit cell is crucial to identifying the stoichiometry and properties of the material.
Stoichiometry
Stoichiometry in chemistry is the calculation of relative quantities of reactants and products in chemical reactions. For substances forming crystalline solids, stoichiometry can also refer to the ratio of the different atoms within a crystal lattice. This ratio is an intrinsic property of the compound and determines its chemical formula.

The calculated exercise taps into stoichiometry by seeking to identify the formula of a compound, which is based on the number of atoms of each type (A and B) within the unit cell of its crystal structure. The stoichiometric relationship in a compound with a cubic crystal structure is crucial for understanding the compound's composition. In our example, by identifying the locations of atoms A and B and utilizing the principle of stoichiometry, we deduced the compound's formula as AB3. This reflects the 1:3 ratio of A atoms to B atoms within the unit cell.
Unit Cell Calculations
Unit cell calculations are essential in determining the crystal structure's composition. A unit cell is the smallest repeating unit of a crystal lattice that displays the full symmetry of the crystal's structure. By calculating how many atoms of each element are present in a unit cell, we can derive the empirical formula of the compound.

In the provided exercise, two types of positions are considered: cube corners and face centers. The strategy is to count the atoms occupying each position while considering their sharing with adjacent unit cells. For example, the corner atoms are shared among eight different unit cells, hence, each unit cell effectively contributes 1/8 of every corner atom. Face-centered atoms are shared between two unit cells, counting for a 1/2 contribution per cell. By summing up these contributions, we arrive at the total number of atoms of each kind within a single unit cell and thus the empirical formula. It's a logical process that hinges not only on counting but also on understanding the geometric and shared nature of the crystal lattice.

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Most popular questions from this chapter

The first order diffraction of \(X\) -rays from a certain set of crystal planes oceurs at an angle of \(11.8^{\circ}\) from the planes. If the planes are \(0.281 \mathrm{~nm}\) apart, what is the wavelength of \(X\) -rays?

Silver crystallizes in f.c.c. lattice. If edge length of the cell is \(4.077 \times 10^{-8} \mathrm{~cm}\) and density is \(10.5 \mathrm{~g} \mathrm{~cm}^{-1}\), calculate the atomic mass of silver.

Nitric oxide reacts with bromine and gives nitrosyl-bromide as per reaction given below: $$ 2 \mathrm{NO}_{(\mathrm{g})}+\mathrm{Br}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{NOBr}_{(\mathrm{g})} $$ When \(0.087\) mole of \(\mathrm{NO}\) and \(0.0437\) mole of \(\mathrm{Br}_{2}\) are mixed is a closed container at constant temperature, \(0.0518\) mole of NOBr is obtained at equilibrium. Calculate equilibrium amount of nitric oxide and bromine.

\(100 \mathrm{~g}\) of \(\mathrm{NaCl}\) is stirred in \(100 \mathrm{~mL}\) of water at \(20^{\circ} \mathrm{C}\) till the equilibrium is attained : (a) How much \(\mathrm{NaCl}\) goes into the solution and how much of it is left undissolved at equilibrium? The solubility of \(\mathrm{NaCl}\) at \(20^{\circ} \mathrm{C}\) is \(6.15\) mol/litre. (b) What will be the amount of \(\mathrm{NaCl}\) left undissolved if the solution is diluted to \(200 \mathrm{~mL} ?\)

\(\Delta G^{\circ}\) for \(\frac{1}{2} \mathrm{~N}_{2}+\frac{3}{2} \mathrm{H}_{2} \rightleftharpoons \mathrm{NH}_{3}\) is \(-16.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\) at \(25^{\circ} \mathrm{C}\). Find out \(K_{\mathrm{p}}\) for the reaction. Also report \(K_{\mathrm{p}}\) and \(\Delta G^{\circ}\) for: $$ \mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} \quad \text { at } 25^{\circ} \mathrm{C} $$

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