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12 -- Part 5: Intermolecular Forces: Liquids and Solids
Intermolecular forces determine the trends in melting points.
The melting points for ionic compounds are dependent on interionic forces.
The interionic forces in CaO would be larger than in KI.
CaO should have a higher melting point.
There are two factors that contribute to the interionic forces.
The first and second are the charges.
You would expect one ionic compound to have a lower melting point than the other.
There are several examples in Table 12.8.
The noble gases are held together by London forces, and thus they are considered to bemolecules.
The electrons in the metal atoms represented by a cube are highly delocalized.
The bonding forces at the center of each face are stronger than those arising.
There are plane surfaces, sharp edges, and regular geometric shapes in crystals.
They have aroused interest from the beginning.
We have only recently come to a fundamental understanding of the state.
Following the discovery of X-rays, this understanding was greatly expanded.
The key idea is that the regularity observed in crystals is due to an underlying pattern in the arrangement of atoms.
There are a number of situations in which you have had to deal with repeating patterns in one or two dimensions.
Projects that include stringing beads to make a necklace, wallpapering a room, or creating a design with floor tiles are included.
The appropriate lattice for other crystals may involve planes that are not equidistant or that intersect at angles other than 90 degrees.
There are seven possibilities for crystal lattices, but only one will be emphasized.
There are three sets of parallel planes.
We arrange the three cube in the threedimensional space lattice so that the centers of the structural particles of the directions are located at lattice points.
In the line-and-ball drawings in the top row, the centers of spheres are shown in the unit cells.
Spheres come into contact with each other in the simple cell.
The spheres are in contact with the cube diagonal.
The diagonal of each face is where contact is located in the face-centeredcc cell.
The colors used for the spheres are only for emphasis.
When spheres are stacked together, there must always be some empty space.
The volume of the holes, or voids, is at a minimum in some arrangements.
Closest packed structures are the basis of a number of crystal structures.
There are holes in the spheres.
A packed pyramid.
The first sphere is placed in the next layer.
The entire pattern of oranges is fixed.
The fruit stand are often packed with two different types of holes.
The structure begins to repeat itself when layer C is the same as layer A.
The spheres in layer C are not in line with those in layer A.
When the fourth layer is added, the structure begins to repeat itself.
Spheres in layer A are red.
Those in layer B are yellow and those in layer C are blue.
There are three spheres in one of the layers.
The hole is formed when a sphere in the upper layer sits in the lower layer.
The hole is made between two groups of spheres.
The closest packed structure has a face-centered unit cell.
The holes account for only 25% of the total volume in the hcp and fcc structures.
The holes account for a higher percentage of the total volume.
The bcc unit cell has holes in it.
The 14 spheres on the left are from a larger array of spheres.
The top and bottom layers have the same number of atoms.
The fcc unit cell is revealed by the rotation of the group of 14 spheres.
Table 12.9 has some examples.
Each atom in a crystal is in contact with other atoms.
The coordination number is 8 for the bcc structure and 12 for the fcc and hcp structures.
Solid lines are used to join the atoms in that cell.
The unit cell is not a cube.
There are examples.
The face-centered cubic is the same as the closest packed one.
The blue-shaded unit cell is the focus.
Only the center of two atoms are pictured.
The blue cell has an atom in the center.
All eight unit cells share the corner atom.
The colors used for the spheres are only for emphasis.
It is wrong to conclude that the bcc unit cell consists of nine atoms because it takes nine atoms to draw it.
Other unit cells share the other atoms.
Eight adjoining unit cells share the corner atoms.
One-eighth of the corner atom should be thought of as belonging to a unit cell.
The corner atoms account for 1 and the central atom is all in the unit cell.
The unit cell contains 8.
The bcc structure can be used to illustrate this.
In other words, the unit cell is occupied and empty.
The percentage of empty space is the same regardless of the sphere's diameter.
We can see things with our eyes and light.
Light that is much shorter wavelength is needed to see how atoms are arranged in a crystal.
The original beam of X-rays is scattered in all directions when it encounters atoms.
The distribution of electronic charge in the atoms and/or molecule is related to the pattern of this scattered radiation.
It is possible to infer the structure of the substance from the visible pattern of the scattered X-rays.
The use of high-speed computers to process large amounts of X-ray data has increased the power of the X-ray method.
The method of scattering X-rays from a crystal is suggested by Max von Laue.
A geometric analysis proposed by W. H. Bragg and the Braggs can explain what happened.
The scattering of X-rays by a crystal is usually from no more than 20 planes deep in a crystal.
The size of the single crystal needs to have enough of the surface available to make a single crystal, yet it is dominated by a few of the surface planes.
The two triangles are the same.
The length of the side opposite the angle is equal to the hypotenuse of the triangle.
Wave b travels farther than wave a by the distance.
With different orientations of the crystal, we can determine atomic spacings and electron densities for different directions through the crystal.
Certain properties can be deterred by calculation once a crystal structure is known.
The edge of the cell corresponding to Figure 12-45 is found to be 287 pm.
A bcc unit cell has nine atoms associated with it.
One atom is located at each of the eight corners of the cube.
The Pythagorean has a formula a2 + b2 + c2.
It is important to know the atomic arrangement of each cell and that the atoms at the corners are shared between the cells.
In the bcc structure, there is Potassium.
We need to sketch or visualize a unit cell of the crystal for both of these calculations.
We need to know which atoms are in contact.
The density of iron is calculated using the data from the example 12-9 and the Avogadro constant.
To calculate the density, we need the mass of the unit cell and its volume.
The mass of the unit cell can be calculated using Fe atoms per bcc unit cell.
1023 Fe atoms are 55.85 g Fe.
The unit cell has a volume of 12.87 liters.
The ratio of mass to volume is called density.
Crystal structure data can be used to determine the density of materials.
This method is useful when we only have a small amount of new compounds to study.
Experimental density and cell edge length can be used to determine the type of unit cell adopted by a metal.
The result of Practice example 12-9A can be used to calculate the density of potassium.
To evaluate the Avogadro constant, NA, use the result of Practice example 12-9B, the molar mass of Al, and its density of 12.6984 g cm-32.
The mass of a unit cell can be determined from the volume and density of the cell.
If we try to apply the packing-of-spheres model to an ionic crystal, we run into two problems: the cations and the anions.
The oppositely charged ion will be discussed in Chapter 9.
We think of them as being in contact.
Because of mutual repulsions, like- charged ion are not in direct contact.
Some ionic crystals are similar to a tightly packed arrangement of one type of ion with holes filled by another type of ion.
In establishing a packing arrangement, the relative sizes of cations and anions are important.
The face-centered cubic arrangement is a common arrangement.
The cation occupies one of the holes between the closest-packed spheres and can be seen as adopting the face-centered cubic structure.
A triangular hole is needed to form the structure.
The hole is 0.225 R and r is 0.155 R. The calculations show that the hole in the closest packed structure is bigger than the hole in the other hole.
The simple cubic arrangement is one of the arrangements adopted by the ionics.
The simple cubic arrangement has larger holes than the closest packed arrangement.
The unit cell has a hole in it.
The largest hole is 0.732 R, which is the size of the cubic hole.
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