Consequently, stable structures for ionic compounds result 1 when ions of one charge are surrounded by as many ions as possible of the opposite charge and 2 when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.
In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed. The smaller cations commonly occupy one of two types of holes or interstices remaining between the anions.
The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole.
The larger type of hole is found at the center of six anions three in one layer and three in an adjacent layer located at the corners of an octahedron; this is called an octahedral hole.
Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in [link]. Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array.
The larger cations can then occupy the larger cubic holes made possible by the more open spacing. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of ; all of the tetrahedral holes are filled at this ratio.
Compounds with a ratio of less than may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant. Occupancy of Tetrahedral Holes Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint.
Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions.
What is the formula of zinc sulfide? Thus, the formula is ZnS. Check Your Learning Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes.
What it the formula of lithium selenide? Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of Ratios of less than are observed when some of the octahedral holes remain empty. Stoichiometry of Ionic Compounds Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?
Check Your Learning The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide? In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied.
Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries the two principal features that determine structure are similar.
Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures. When an ionic compound is composed of cations and anions of similar size in a ratio, it typically forms a simple cubic structure. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride.
Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion. We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions.
In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell.
The two unit cells are different, but they describe identical structures. When an ionic compound is composed of a ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in [link].
We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell.
The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the stoichiometry required by the formula, NaCl. The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in [link].
This structure contains sulfide ions on the lattice points of an FCC lattice. The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS. A calcium fluoride unit cell, like that shown in [link] , is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice.
All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of , as required by the chemical formula, CaF 2.
Close examination of [link] will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs. Calculation of Ionic Radii If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.
Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in [link] , calculate the ionic radius for the chloride ion. Solution On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:. From the Pythagorean theorem, we have:.
Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1. It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best.
Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations. The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. C diamond. N diatomic.
As tetrahedral. S S8 ring. Kr monatomic. Which one of the following substances crystallizes as a molecular solid? SiO2 C. CH3OH E. Al2 SO4 3 Butter melts over a range of temperature, rather than with a sharp melting point. Which one of the following substances crystallizes as a covalent crystal? CaO B. CO2 D. KMnO4 Which one of the following crystallizes in a metallic lattice? NaMnO4 C. LiClO4 E. K2Cr2O7 The number of atoms in a body-centered cubic unit cell is A.
The number of atoms in a face-centered cubic unit cell is A. Potassium crystallizes in a body-centered cubic lattice. How many atoms are there per unit cell? A liquid boils when its A. The heat capacity of liquid water is 4. How many kilojoules of heat must be provided to convert 1. Use the graph of vapor pressure to determine the normal boiling point of CHCl3.
Which of the following is not an endothermic process? Use the graph of vapor pressure to determine the normal boiling point of O2. O2 doesn't boil because it is always a gas. Calculate the amount of heat that must be absorbed by Calculate the amount of heat needed to melt 2. The vapor pressure of a liquid in a closed container depends upon A. What mass of water will evaporate? Which one of the following elements would have the lowest melting point?
Br2 C. All metallic elements except Cs, Ga, and Hg are crystalline solids at room temperature. Like ionic solids, metals and alloys have a very strong tendency to crystallize, whether they are made by thermal processing or by other techniques such as solution reduction or electroplating. Metals crystallize readily and it is difficult to form a glassy metal even with very rapid cooling.
Molten metals have low viscosity, and the identical essentially spherical atoms can pack into a crystal very easily. Glassy metals can be made, however, by rapidly cooling alloys, particularly if the constituent atoms have different sizes. The different atoms cannot pack in a simple unit cell, sometimes making crystallization slow enough to form a glass. Most metals and alloys crystallize in one of three very common structures: body-centered cubic bcc , hexagonal close packed hcp , or cubic close packed ccp, also called face centered cubic, fcc.
In all three structures the coordination number of the metal atoms i. We can contrast this with the low coordination numbers i. In the bcc structure, the nearest neighbors are at the corners of a cube surrounding the metal atom in the center. In the hcp and ccp structures, the atoms pack like stacked cannonballs or billiard balls, in layers with a six-coordinate arrangement. Each atom also has six more nearest neighbors from layers above and below.
Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell. What is the coordination number of a chromium atom in the body-centered cubic structure of chromium? What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum? Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom? Nickel metal crystallizes in a cubic closest packed structure.
What is the coordination number of a nickel atom? Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3. Calculate the edge length of the face-centered cubic unit cell and the density of platinum. Barium crystallizes in a body-centered cubic unit cell with an edge length of 5. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum. The density of aluminum is 2. Explain why Si has the lower density even though it has heavier atoms.
The crystal structure of Si shows that it is less tightly packed coordination number 4 in the solid than Al coordination number The 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? Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions.
What is the formula of cadmium sulfide? Explain your answer. In a closest-packed array, two tetrahedral holes exist for each anion.
If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS. A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions.
What is the formula of the compound? What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions? A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions.
Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound? A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array.
The ratio of thallium to iodide must be ; therefore, the formula for thallium is TlI. Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest-packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al? What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes?
What is the oxidation number of titanium? Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions.
Suggest an explanation. Both ions are close in size: Mg, 0. This similarity allows the two to interchange rather easily. Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. One 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. NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4. Thallium I iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4. A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.
What is the spacing between crystal planes that diffract X-rays with a wavelength of 1. A diffractometer using X-rays with a wavelength of 0. Determine the spacing between the diffracting planes in this crystal. A metal with spacing between planes equal to 0. What is the diffraction angle for the first order diffraction peak? Gold crystallizes in a face-centered cubic unit cell. The wavelength of the X-rays is 1. What is the density of metallic gold? When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted.
These X-rays are diffracted at an angle of 7. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first order diffraction? Skip to content Liquids and Solids. Learning Objectives By the end of this section, you will be able to: Describe the arrangement of atoms and ions in crystalline structures Compute ionic radii using unit cell dimensions Explain the use of X-ray diffraction measurements in determining crystalline structures.
The Structures of Metals We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. Unit Cells of Metals The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. A unit cell shows the locations of lattice points repeating in all directions.
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