Concept explainers
(a)
Interpretation:
The formula of perovskite for a unit cell should be determined.
Concept introduction:
A unit cell is the smallest repeating unit of a crystal lattice. Unit cells are classified as primitive and non-primitive. Primitive unit cells are those which have atoms at only corners and non-primitive unit cells are those which have atoms at positions other than corners.
Non- primitive unit cells are of various types:
Face centered unit cell (fcc), body centered unit cell (bcc), edge centered unit cell, end centered unit cell.
Contribution of an atom at a position is as follows:
The number of particles per unit cell, also denoted by Z is as follows:
The elements or substances exist more in solid than liquid and gases. These solids can be two types amorphous and crystalline. The crystalline solids are solids that contain their constituent particle in arranged manner. While amorphous solids are solid in which constituent particles get arranged randomly such as rubber.
(b)
Interpretation:
The oxidation state of titanium ion of given perovskite unit cell should be determined.
Concept introduction:
The elements or substances exist more in solid than liquid and gases. These solids can be two types amorphous and crystalline. The crystalline solids are solids that contain their constituent particle in arranged manner. While amorphous solids are solid in which constituent particles get arranged randomly such as rubber.
(c)
Interpretation:
The geometry around each titanium, oxygen, and calcium atom in given unit cell should be determined.
Concept introduction:
The elements or substances exist more in solid than liquid and gases. These solids can be two types amorphous and crystalline. The crystalline solids are solids that contain their constituent particle in arranged manner. While amorphous solids are solid in which constituent particles get arranged randomly such as rubber.
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EBK CHEMISTRY
- The CsCl structure is a simple cubic array of chloride ions with a cesium ion at the center of each cubic array (see Exercise 69). Given that the density of cesium chloride is 3.97 g/cm3, and assuming that the chloride and cesium ions touch along the body diagonal of the cubic unit cell, calculate the distance between the centers of adjacent Cs+ and Cl ions in the solid. Compare this value with the expected distance based on the sizes of the ions. The ionic radius of Cs+ is 169 pm, and the ionic radius of Cl is 181 pm.arrow_forwardAn amorphous solid can sometimes be converted to a crystalline solid by a process called annealing. Annealing consists of heating the substance to a temperature just below the melting point of the crystalline form and then cooling it slowly. Explain why this process helps produce a crystalline solid.arrow_forwardWhat is the coordination number in the cesium chloride cubic structure?arrow_forward
- Describe the unit cell of lithium (see Figure).arrow_forwardOne of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?arrow_forward(a) Determining an Atom Radius from Lattice Dimensions: Gold has a face-centered unit cell, and its density is 19.32 g/cm3. Calculate the radius of a gold atom. (b) The Structure of Solid Iron: Iron has a density of 7.8740 g/cm3, and the radius of an iron atom is 126 pm. Verify that solid iron has a body-centered cubic unit cell. (Be sure to note that the atoms in a body-centered cubic unit cell touch along the diagonal across the cell. They do not touch along the edges of the cell.) (Hint: The diagonal distance across the unit cell = edge 3.)arrow_forward
- 8.16 Iridium forms a face-centered cubic lattice, and an iridium atom is 271.4 pm in diameter. Calculate the density of iridium.arrow_forwardCalculate the percent of volume that is actually occupied by spheres in a body-centered cubic lattice of identical spheres You can do this by first relating the radius of a sphere, r, to the length of an edge of a unit cell, l. (Note that the spheres do not touch along an edge but do touch along a diagonal passing through the body-centered sphere.) Then calculate the volume of a unit cell in terms of r. The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere (4r3/3).arrow_forwardWhat is a lattice? What is a unit cell? Describe a simple cubic unit cell. How many net atoms are contained in a simple cubic unit cell? How is the radius of the atom related to the cube edge length for a simple cubic unit cell? Answer the same questions for the body-centered cubic unit cell and for the face-centered unit cell.arrow_forward
- Consider the three types of cubic units cells. (a) Assuming that the spherical atoms or ions in a primitive cubic unit cell just touch along the cubes edges, calculate the percentage of occupied space within the unit cell. (Recall that the volume of a sphere is (4/3)r3, where r is the radius of the sphere.) (b) Compare the percentage of occupied space in the primitive cell (pc) with the bcc and fcc unit cells. Based on this, will a metal in these three forms have the same or different densities? If different, in which is it most dense? In which is it least dense?arrow_forwardThe free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?arrow_forwardCalculate the percent of volume that is actually occupied by spheres in a face-centered cubic lattice of identical spheres. You can do this by first relating the radius of a sphere, r, to the length of an edge of a unit cell, l. (Note that the spheres do not touch along an edge but do touch along the diagonal of a face.) Then calculate the volume of a unit cell in terms of r. The volume occupied by spheres equals the number of spheres per unit cell times the volume of a sphere (4r3/3).arrow_forward
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