Structure+of+Solids;+Lattice+Energies

Solids can be either: v Amorphous v Crystalline
 * __ Structures of Solids __**

Amorphous Solids (non-crystalline): · An amorphous solid is one in which there is no particular order in the arrangement of particles. · Familiar amorphous solids include glass and rubber. · Their intermolecular forces vary in strength because their particles do not have any long-range order. Therefore, they have no fixed boiling point or melting point.

Crystalline Solids: · Crystalline solids are solids in which their particles are highly ordered in well-defined arrangements. · Usually have flat faces that make definite angles with one another. · They have high boiling and melting points.

In ionic crystals, ions pack themselves so as to maximize the attractions and minimize repulsions between the ions. On the left is an amorphous solid with no orderly structure as opposed to the crystalline one on the right. [] Unit Cells: · Crystalline solids consist of repeating patterns of arrangement called the unit cell. · A crystalline solid can be represented by a three-dimensional array of points called a crystal lattice. Each point in the lattice is called a lattice point. · There are three types of cubic unit cells:

The **primitive cubic** (or **simple cubic**) is the unit cell in which the lattice points are at the corners only. The **body-centered cubic** is the unit cell in which lattice points are at the corners and a single lattice point is located at the center of the unit cell. The **face-centered cubic** is the unit cell in which the cell contains lattice points at the center of each face as well as every corner.

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The table below shows the fraction of an atom that occupies a unit cell when atoms are shared between unit cells: [] Example : How many Niobium and Oxygen atoms are there in a unit cell of Niobium (II) Oxide? (Sorry if this picture is not clear, I tried to make it as clear as possible.) Oxygen Niobium

First of all, let us find out how many Oxygen atoms there are in this unit. There are 12 blue Oxygen atoms on the **edges**. According to the table above, each atom on the edge of a unit cell has a fraction of ¼. Therefore, multiply 12 by ¼: 12 x ¼ = 3. Therefore, there are 3 Oxygen atoms in the unit. Next, we find how many Niobium atoms there are. The unit contains 6 black Niobium atoms on the **faces**. Referring to the table above again, we find that each atom on the face of a unit cell has a fraction of ½. Therefore, multiply 6 by ½: 6 x ½ = 3. Therefore, there are also 3 Niobium atoms in the unit. The table must be memorized as it will not be given during the AP exam. Also, make sure to count every single atom. Here’s an example from the AP Chemistry book : The geometric arrangement of ions in crystals of LiF is the same as that in NaCl. The unit cell of LiF is 4.02 Å on an edge. What is the density of LiF? (Remember: density = mass/unit volume) The unit cell of LiF consists of 4 Li+ atoms and 4 F- atoms. To find the mass contained in one unit cell, multiply the mass of Li with the number of atoms. Then, add it with the mass of F multiplied by the number of atoms. It will be much easier to understand once it is written out: The volume of a cube of length //a// on an edge is //a//3, so the volume of the unit cell is (4.02 Å)3. We can now calculate the density and convert to the common units of g/cm3: Lattice Energy: __** · Lattice Energy is the energy required to completely separate one mole of a solid ionic compound into gaseous ions. · The formation of ionic compounds is an exothermic process since the energy released by the attraction between ions of different charge more than makes up for the endothermic nature of ionization energies. · The strong attractions cause most ionic material to be hard and brittle and have high melting points. · The amount of the lattice energy depends on the charges of the ions, their sizes, and their arrangement in the solid. · Where E is the lattice energy, Q+ is the charge on the cation, Q- is the charge on the anion, k is a constant (8.99 x 109 J-m/C2), and r is the distance between the ions. <span style="font-family: Symbol; mso-list: Ignore; msofareastfontfamily: Symbol; msobidifontfamily: Symbol; msolist: Ignore;">· The higher the charge, the more the energy. As distance decreases, the energy increases.
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Example : Why does MgO have a larger amount of lattice energy than the amount of lattice energy that MgF2 has? Remember: When writing down the charges, the subscripts and coefficients DO NOT count. It is only the element’s charge that is counted. (For example, K3 would have a charge of +1) To do this, write down each of the compounds’ charges. MgF2 has Q = +2,-1 (Mg has a charge of +2 and F has a charge of -1). MgO has Q= +2,-2 (O has a charge of -2). When the two charges are multiplied together, MgO results in a larger amount of lattice energy since it has a higher charge.
 * ** Compound ** || ** Lattice Energy ** ||
 * MgF2 || 2957 ||
 * MgO || 3938 ||

Another example : Why does LiCl have a smaller amount of lattice energy than the amount of lattice energy that LiF has?
 * ** Compound ** || ** Lattice Energy ** ||
 * LiF || 1036 ||
 * LiCl || 853 ||

Fluorine has a smaller radius than Chlorine. As distance decreases, the amount of lattice energy increases. Therefore, LiF would have a greater amount of lattice energy than LiCl.