Solids and Ionic Solids
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written by Rituraj Kalita

[Relevant Syllabus of Gauhati University (G.U., Guwahati, Assam, India), B.Sc. II year Chemistry Major, 2007: 
Types of solids, macroscopic properties of solids, properties of ionic compounds; types of units cells, crystal lattices and Miller indices, crystal systems and Bravais lattices.
Closed-packed structures, ionic radii, radius ratio and structures, spinel and perovskite structures, lattice energy of ionic solids: Born-Haber cycle calculations.]

Suggested Books: G W Castellan, K L Kapur, Negi & Anand

Solids: The term solids means crystalline solids; on the other hand the 'amorphous solids' are structurally very similar to liquids. The crystalline solids are characterized by the regular arrangement of its constituent particles (i.e., atoms, ions or molecules) in a orderly, repetitive pattern in three dimensions, which ordered arrangement extends to long distances within the solid piece. So it is said that crystalline solids may be characterized by the existence of a 'long-range order' of arrangement of its constituent particles. On the other hand, liquids and amorphous solids (e.g., glass) have only a 'short-range order' of similar arrangement of their constituent particles (Fig. 1, a & b respectively).

Types of Solids: The solids may be classified in different ways; here they are classified according to the type of bonds that binds the constituent particles (units) of the solid crystal together. As you already know, here are four types of bonds (metallic bond, ionic bond, van der Walls interactions and covalent bonds) that binds the individual constituent particles forming the solid, and accordingly the solids are classified into: (1) metallic solids e.g., metals and metallic alloys (2) ionic solids e.g., salts such as common salt (3) molecular solids or van der Walls crystals e.g., ice, dry ice, iodine (4) covalent solids e.g., diamond, silicon.

Macroscopic Properties of Solids: All solids are characterized by certain macroscopic, experimentally measurable properties. The most characteristic one is rigidity, which distinguish them from the 'fluids' (i.e., the liquids and the gases). Rigidity is the property that offers a definite shape to a piece of solid: technically it is understood as a very high value of the shear modulus (i.e., the ratio of the shear stress to the shear strain -- shear modulus is somewhat similar to the coefficient of viscosity). Solids also have very high values of the tensile (Young's) modulus, and also shares very high values of the bulk modulus (compressibility) and very low values of the thermal expansion coefficient along with the liquid phase (distinguishing these two 'condensed phases' from the gases). Mathematical explanations: Denoting stresses by S and strains by g, Shear modulus is S_sh/g_sh, tensile modulus is S_te/g_te, bulk modulus is V.(dP/dV)_T, thermal volume-expansion coefficient is (1/V).(dV/dT)_P.

Properties of Ionic Solids: Ionic solids, as you already know, are further characterized by the very high melting and boiling points (both properties shared, however, with covalent solids and some transition metals also), and by a characteristic brittleness that distinguish it from the three other classes of solids. Thus, common salt is brittle, while copper, iodine and diamond are not. Experimentally, brittleness is measured by subjecting a piece of solid to a sudden stress (termed 'shock'), and thus it is observed under what minimum shock (sudden, compressive stress) it cracks/ breaks.

Crystal Lattices and Unit Cells: A piece of (crystalline) solid always consists of a crystal-lattice structure as if generated out of repetitions, in three dimensions, of a fundamental microscopic three-dimensional unit of structure called the unit cell. 
  Fig. 2
The unit cell is a (3-D) parallelepiped containing three definite vectors a, b and c as their sides (Fig. 2) and at the eight corners of which sits the centers of a constituent particle (atom/ ion/ molecule, as is the case) of the solid. These three tiny vectors are also called the primitive translations of the crystal-structure pattern, because starting at any point therein, and then moving (i) integral multiples of the distance a along the direction of the vector a, (ii) integral multiples of the distance b along the direction of the vector b and (iii) that of the distance c along the direction of the vector c, we reach an exactly equivalent point in the crystal-structure pattern. (Here integral multiples of a may mean a, 2a, 5a, -3a or even 0.a i.e., 0 --- and so on for those of b & c). Arbitrarily choosing one corner of one unit cell as the origin, we also designate three crystallographic axes x, y & z (though they are generally denoted as x, y & z crystallographic axes, these three axes are not necessarily perpendicular to one another) respectively in the direction of these three vectors a, b & c. The angles between these three vectors (i.e., between these three axes) are denoted as a, b and g: thus a is the angle between the vectors b & c, b is the angle between the vectors c & a, and g is the angle between the vectors a & b. 

The Seven Crystal Systems: Thus the three distances a, b, c and the three angles (a, b, g) are the six parameters which distinguish (characterize) the shape of the unit parallelepiped i.e., the primitive unit cell. (As you already know, the primitive unit cell, in contrast to the face-centered or the body-centered unit cell, is a unit cell which has constituent particles placed only at the eight corners of the unit cell and nowhere else). When a unit cell is repeated in three dimensions along the direction of the three vectors a, b, c the crystal-structure pattern is generated. Now, considering the symmetry of the crystal-structure pattern thus generated, we find that there are only seven distinct types of unit parallelepipeds, i.e., only seven distinct types of primitive unit cells possible. Thus, the crystals are classified into seven crystal systems depending on the shape of their constituent unit cells. These seven crystal systems are as given in the table:

The Fourteen Bravais Lattices: If from a crystal-lattice structure each of the constituent particles (atoms/ molecules) is removed (say, hypothetically)  and is instead replaced by a geometric point each, we get a similarly regular lattice of points in space: this lattice of points is called the point lattice or the space lattice. However, all the possible point lattices (i.e., all the possible regularly repeated three-dimensional lattice of points) cannot be obtained if points are placed only at the corners of the seven types of unit cells corresponding to the seven crystal systems. It was shown by Bravais that there are a total of fourteen types of unit cells (not only the above seven primitive types, but also seven more types that places additional points at face-centers, body-centers or end-centers of the unit cells in the former seven types) to generate every possible three-dimensional arrangement of equivalent points in space (i.e., every possible point lattice). These fourteen types of space lattices are the fourteen Bravais lattices. Among them, the additional (i.e., non-primitive) seven types of lattices are described in terms of the seven primitive lattices: thus two of the additional cubic lattices are described as face-centered cubic (fcc) and body-centered cubic (bcc), in terms of the primitive cubic (i.e., simple cubic) lattice. The fourteen Bravais lattices are as tabulated below  (please see relevant illustrations from any good book):

Lattice Planes and their Miller-Indices Designations: Whenever we consider a space (point) lattice, we may easily visualize any set of parallel and (consecutively) equidistant planes that pass through some points in that point lattice. Such a set of parallel and equidistant planes is called  lattice planes (alternatively, crystallographic planes), and such a set of planes is together designated with a particular integer-triplet h, k, l (denoted just as hkl) called the Miller indices. As an example, try to visualize the plane in a primitive cubic crystal that passes through the points (a,0,0), (0,a,0) & (0,0,a). Isn't the plane passing through the points (2a,0,0), (0,2a,0) & (0,0,2a) just parallel to it? What about the plane passing through (3a,0,0), (0,3a,0) & (0,0,3a)? And isn't the distance between the first and the second plane obviously equal to the distance between the second and the third? [If you're interested, may note that this distance is (1/√3)a]. Now, these three aforesaid planes are just three members of a particular set of a great many number of parallel and equidistant lattice planes, this set of planes being denoted by the Miller indices 111 (i.e., we refer to this set of planes as the 111 planes). Now let us try to understand how such a set of parallel and equidistant planes gets designated with just a single particular triplet (i.e., hkl) of Miller indices.
For this purpose, let us consider the intercepts of a lattice plane along the three crystallographic axes x, y & z. For the second plane, the intercepts are obviously 2a, 2b & 2c. We now divide these intercepts, respectively, by the three unit-cell distance parameters a, b, c. This gives the numbers (2,2,2). Now we take the reciprocals of these three numbers, getting (1/2,1/2,1/2). These three numbers are then multiplied with, say, the lowest common multiplier (LCM) of the three denominators, giving the smallest possible set of the three integers i.e., (1,1,1). This set (i.e., 111) is the set of Miller indices for this plane (and also for the whole set of parallel and equidistant planes). To check that any other plane of this set give the same set of Miller indices, let us consider the third plane now. The intercepts here are obviously (3a,3b,3c). Dividing by (a,b,c) respectively, we get (3,3,3). Taking reciprocals we get (1/3,1/3,1/3), and then multiplying with 3 to get the smallest possible set of integers, we arrive just at (111) again. Now, please repeat this procedure yourself for the first plane also to check that we arrive at the same set of Miller indices i.e., at 111.
So, let us now generalize the procedure for getting the Miller indices: For any member-plane of the given set of lattice planes, we find the intercepts (ma,nb,pc) along the three crystallographic axes. Now, we divide these three intercepts with the unit-cell distance parameters (a,b,c) respectively to get a set of three numbers (m,n,p). Now we take their reciprocals, getting the set (1/m,1/n,1/p). Now we multiply them with the smallest necessary integer (say, with the LCM of the three denominators m, n, p) to get the smallest possible set of integers (h,k,l), thus arriving at the Miller indices (hkl).
Note: The Miller indices set with one or more negative indices are written with bars above the numbers: thus (-1,1,1) is written as (111), while (2,-3,-3) is written as (233). The commas, obviously, aren't written.

Significance of Lattice Planes and their Miller Indices:  
The concept of lattice planes is not just a theoretical idea of solid-state science. The faces of crystals (crystal faces) that we observe in actual crystals (say, in gems, or in a blue vitriol crystal) must be a lattice plane (i.e., a plane passing through some lattice points, instead of some arbitrary plane imagined by you that goes somewhat midway between the lattice points]. Also, any such observed crystal face is most likely to be such a lattice plane that passes through a large number of lattice points (e.g., 111 or 110 plane, instead of, say, 789 plane). 
The second physical significance of lattice planes is that in x-ray diffraction crystallographic study of solids, it is some set of lattice planes that acts like a diffraction grating, and the interplanar distance d in the corresponding Bragg equation 2d sinq = nl (already familiar to you) is nothing other than the distance d_hkl between two such consecutive lattice planes of the set of lattice planes (i.e., of the set of parallel and equidistant planes, distinguished by the Miller indices hkl). Now, for properly working with the Bragg equation, let us put forward the following simple relation between d_hkl and the Miller indices (h,k,l) for crystal systems with mutually orthogonal axes (i.e., for the cubic, tetragonal and orthorhombic crystal systems and the nine corresponding Bravais lattices -- derivation omitted): 
d_hkl  = 1/√((h/a)² + (k/b)² + (l/c)²) -- so that, for cubic system (i.e., b = c = a), 
d_hkl  = a/√(h² + k² + l²). Thus, for cubic systems, we get d₁₁₁ = a/√3, d₁₁₀ = d₁₀₁ = d₀₁₁ = a/√2 and d₁₀₀ = d₀₁₀ = d₀₀₁ = a

A numerical problem: It is known that elemental iron (Fe) has bcc (body-centered cubic) lattice structure. In x-ray diffraction study of iron crystal, using x-ray of wavelength 1.36 Å, the diffraction angle was found to be 42⁰ for diffraction involving the110 lattice planes with order n = 2. What is the unit cell distance parameter (a) for iron, and what density you would predict for iron?
Solution: For the 110 lattice planes in the cubic crystal, the interplanar spacing 
d₁₁₀ = a/√(1²+1²) = a /√2. Here q = 42⁰, n = 2, l = pm, 
so using Bragg equation 2d sinq = nl (with d = d₁₁₀) gives 
d₁₁₀ = 2*1.36/(2*sin(42⁰)) Å = 2.0325 Å. So, the unit cell parameter 
a = √2d₁₁₀ = 2.874 Å. For bcc lattice, we have density r = 2(M/N_A)/a³ ; 
using M = 0.05585 kg mol^-1, we predict r = 7814 kg m^-3 = 7.814 g cm^-3 

Among the above distance expressions for cubic systems, the first one (for d₁₁₁) was stated somewhere above: please derive, using two-dimensional school-level geometry, one of the second set (d₁₁₀, d₁₀₁, d₀₁₁) of values (see Fig. 3, it's a rather self-explanatory figure).

Fig. 3

Also note that the third set of values is rather obvious to anyone: you may note that the 100 lattice planes are parallel to the YOZ plane, and thus they have infinite intercepts along the y and z axes, giving k = 0 and l = 0). Some lattice planes for cubic system are as illustrated below (Fig. 4); for better learning, it is advisable for you to practice the art of drawing these planes on paper!

Fig. 4: Lattice Planes:  Left-Top, 011 planes;    Right-Top, 110 Planes;    Left, 111 planes;    Right, 012 planes

Close-Packed Structures for Atomic and Ionic Crystals: In your earlier course, you have learnt about the close-packed structures for atomic crystals, i.e., for crystals containing (one kind of) atoms as the constituent particles, e.g., copper, zinc, iron, diamond etc. You learnt that these atoms, being spherically symmetrical, can be approximated best as spheres, and these spheres could be closest-packed layer after layers, generating either the cubic (fcc, face-centered cubic) close-packing (ccp, layer pattern ABCABCABC..., example Cu metal) or hexagonal close-packing (hcp, layer pattern ABABAB..., example Zn metal), both structures showing co-ordination number (i.e., number of closet-neighbor or touching spheres) of twelve (six in the same layer, three in the upper layer, three in the lower layer). In both of the above cases, 74.04% of the crystal space is occupied by the constituent-particle spheres -- we say that their packing fraction is 0.7404. There's also, you knew, a lesser-packed arrangement of spherical atoms called the body-centered cubic (bcc) structure (example Fe metal) with less efficient packing (with packing fraction 0.6802) and a lower coordination number of eight.

For ionic crystals, however, there are two (e.g., NaCl, CsCl, ZnS) or more (e.g., three each in MgAl₂O₄, CaTiO₃) kinds of ions, and the concept of close-packing of atomic spheres has to be somewhat revised there. The lattice structure of ionic crystals may be visualized as composed or two or more inter-penetrating lattices for each of these ions: thus the CsCl structure consists of two inter-penetrating simple-cubic (SC) lattices: one lattice of the anions (Cl-), and the other of the cations (Cs+). [It is simply erroneous to describe CsCl as having a bcc structure.] The anions, being bigger, are 'almost' close-packed, but, however, the presence of (smaller-sphere) cations in between the anions generally ensure that the close-packing of anions is rather 'approximate', with the nearest-neighbor anions (e.g., six anions in CsCl lattice) aren't exactly touching any given anion (unlike the atoms in an atomic crystal such as Cu). To explain this situation better, we note that within the anion lattice of the spherical anions, there occurs holes or voids in between those spheres, and the cations sit at some of those holes, in this process somewhat pushing the anions away. We may easily visualize the two types of holes - octahedral holes and tetrahedral ones - formed within the anion lattice. For example, in the case of cubic (i.e., fcc) close-packing of anions such as in NaCl, the body centre of the anion-lattice unit cell is an octahedral hole surrounded by six anions, and the centers of each side (edge) is also an exactly similar octahedral hole (to understand why both positions are just similar, consider together the four neighboring unit cells sharing that edge). On the other hand, within one corner anion and the three nearest face-centered anions on a given unit cell, there arises a tetrahedral hole (surrounded by the same four anions) midway amidst these four anions. In the inter-penetrating anion-fcc cum cation-fcc type of lattice in NaCl, all the octahedral holes of the anion lattice are occupied by the Na⁺ cations, while none of the tetrahedral ones are occupied by the cations. [However in spinel (MgAl₂O₄), the Al³⁺ ions occupy octahedral holes and Mg²⁺ ions occupy tetrahedral holes arising in the oxide (O^2–) anion lattice.] In case of the ionic lattice, the term coordination number however means the number of nearest-neighbor oppositely-charged ions for a given ion: thus, in NaCl (s) the coordination number of the cation is 6, while in CsCl it is 8 (in spinel, the coordination number of Mg²⁺ ion is, however, 4).

In deciding which crystal structure to adopt, the ionic solids roughly follows the following guidelines:
(a) The cations must not be held 'loose' in the holes of the anion lattice, i.e., the cation-containing holes in the anion lattice (say, the octahedral holes in the anion lattice of NaCl) must not be larger than the cations themselves. In other words, a cation must keep touching its nearest-neighbor anions. If this condition is not obeyed, the (essentially electrostatic) lattice energy can't become negative enough as then the anion-anion coulombic repulsion would have overshadowed the anion-cation coulombic attraction! Thus NaCl with a radius ratio (i.e., ratio of the radius of cation to the radius of anion, r₊/r_–) of 0.525 (as here r₊ = 0.95 Å, r_– = 1.81 Å) could have never adopted the CsCl structure for which (i.e., for the inter-penetrating sc structure) the radius ratio must be at least 0.732 so as to satisfy the above condition. However, CsCl with a radius ratio of 0.934 (as here r₊ = 1.69 Å, r_– = 1.81 Å) can surely take up this particular lattice structure.
(b) Obeying the former guideline, there's a clear tendency for maximizing the (ionic) coordination number. A higher coordination number offers a greater value of Madelung constant A in the lattice-energy expression, giving rise to a more negative lattice energy. Thus, though radius ratios of both NaCl and CsCl satisfies the above guideline towards adopting the NaCl lattice structure (for this, i.e., for the inter-penetrating fcc structure, radius ratio must be at least 0.414), CsCl adopts a different (i.e., the inter-penetrating sc) structure because this structure offers higher coordination numbers of 8-8 (i.e., 8 for the cation, 8 for the anion -- in contrast to 6-6 in the inter-penetrating fcc NaCl structure).

Thus, we find that if the radius ratio of a 1:1 binary-compound (i.e., equal numbers of cations & anions, one kind of cation, one kind of anion) type of ionic solid lies in between 0.414 and 0.732, the NaCl structure is preferred! The 1:1 binary ionic solids may in fact adopt one more pair of (nearly-similar) types of lattice structures, which are the cubic ZnS (zinc blende) and hexagonal ZnS (wurzite) structures. [In the zinc blende structure, there is an fcc lattice of S^2– anions, and four out of the eight tetrahedral holes (per unit cell) in this anion lattice is occupied by the Zn²⁺ cations] In both these structures, the coordination numbers are 4-4, and the radius ratio must be at least 0.225. Thus if the radius ratio of an ionic solid is in between 0.225 and 0.414, the pair of ZnS structures is preferred. Thus for ZnS, the radius ratio r₊/r_– is 0.402 (as here r₊ = 0.74 Å, r_– = 1.84 Å). The following figures show the CsCl, NaCl and zinc blende structures respectively:

Now let us derive the aforesaid 'least-value' criteria for the radius ratio in the CsCl structure and the NaCl structure (doing the same for a ZnS structure will be a more complicated job). As we clearly see from the CsCl figure above, the two extreme corner anions (say, the left-bottom and the right-top anions in the above figure) in the CsCl unit cell must be both touching the cation lying midway between them, so that we have √3a = r_– + 2r₊ + r_– = 2r_– + 2r₊. Satisfying this condition, the minimum possible value of the radius ratio r₊ /r_– (i.e., corresponding to the maximum possible radius of anion) arises when the anions along an edge are also touching, i.e., when a = r_– + r_– = 2r_–. This means r_– = a/2 and r₊ = (√3–1)a/2, giving r₊ /r_– = √3–1 = 0.732, which is the above-mentioned 'least-value' for CsCl (interpenetrating sc) structure! Similarly for the NaCl or rock-salt (interpenetrating fcc) structure, we see that the cation lying at the center of an edge of the unit cell must be touching the two anions at the two corners of that edge, so that a = r_– + 2r₊ + r_– = 2r_– + 2r₊. Similarly considering the two anions on a diagonal of a face of the unit cell to be both touching the anion at that face-center, we get √2a = r_– + 2r_– + r_– = 4r_–. Combining both relations, we get r_– = √2a/4, r₊ = (1–√2/2)a/2 = (2–√2)a/4, and r₊ /r_– = (2–√2)/√2 = 0.414, which is the 'least-value' for the radius ratio for this lattice structure! 

For ternary ionic solids containing three kinds of ions (e.g., MgAl₂O₄, CaTiO₃) two particular kinds of lattice structures are quite common. The spinel structure is named after the compound spinel (MgAl₂O₄), some other examples are chromite (FeCr₂O₄) and gahnite (ZnAl₂O₄). All the members of this type of ionic solids has the ionic formula AB₂O₄, where A²⁺ is a metal ion with a +2 valence (e.g., Mg²⁺) and B³⁺ is a metal ion with a +3 valence (e.g., Al³⁺). This structure may be visualized as a combination of the rock salt (NaCl) and zinc blende structures. The oxide (O^2–) anions are in face-centered cubic (fcc) lattice. A²⁺ and B³⁺ cations occupy tetrahedral and octahedral sites (holes) within the anion lattice. Spinel structure may be of two types: normal spinel and inverse spinel. In normal spinels, A²⁺ ions are on tetrahedral sites and the B³⁺ ions are on octahedral sites (we say that there are BO₆ octahedra with B³⁺ ions at their centers, and AO₄ tetrahedra with A²⁺ ions at their centers). In inverse spinels (e.g., magnetite, Fe₃O₄), the A²⁺ (e.g., Fe²⁺) ions and half the B³⁺ (e.g., Fe³⁺) ions are on octahedral sites; the other half of the B³⁺ ions are on tetrahedral sites. The perovskite structure is named after the compound perovskite (CaTiO₃), some other examples are BaTiO₃ , CaZrO₃ , YAlO₃ etc.  The members of this type of ionic solids has the ionic formula ABO₃, where A²⁺ is a metal ion with a +2 valence (e.g., Ca²⁺) and B⁴⁺ is a metal ion with a +4 valence (e.g., Ti⁴⁺). Here the B⁴⁺ cations may be visualized to be in a simple cubic lattice. The A²⁺ cations occupy the body-centers within this cation lattice. [Thus, obviously, the numbers of A²⁺ and B⁴⁺ cation per unit cell are both 1.] The oxide (O^2–) anions remain at the mid-points of the unit-cell edges (see figure below, the greyish bigger spheres), every anion being thus shared with three other unit cells, the twelve anions on the unit cell contribute 12/4 = 3 anions per unit cell (thus, the stoichiometry is ABO₃). Here A²⁺ is 12-coordinated by O^2– (cuboctahedral coordination, with AO₁₂ cuboctahedra), and B⁴⁺ is 6-coordinated by O (octahedral coordination, with BO₆ octahedra).

Lattice Energy (also called cohesive energy) in ionic solids:
The lattice energy U_L of a ionic solid is defined as the energy required to separate one mole of the ionic solid into its constituent ions in gaseous phase: MX (s) ® M⁺ (g) + X^– (g)   DU = U_L   (considering only monovalent ions, for simplicity)
Though U_L can't be directly determined experimentally, but it may be easily estimated using the Hess Law in the well-known form of Born-Haber cycle, in terms of the experimental quantities of sublimation energy DU_sub, bond dissociation energy DU_dis, ionization potential I, electron affinity E and energy of formation  DU_form (see note below). For the reverse process of formation of a mole of the ionic crystalline solid from its constituent ions in gaseous states, the energy of reaction is termed the energy of crystallization, E_crys (thus,  E_crys = –U_L ). Below, in two steps, we discuss a procedure leading to a theoretical expression for the lattice energy, which may then be compared with this experimental estimate.
Note: Thus, for the specific case of NaCl, tracking the (hypothetical) production of one mole of each of the gaseous ions (i.e., Na⁺ & Cl^–), starting from Na(s) and Cl₂(g), via two alternative paths (do draw up yourself the detailed Born-Haber cycle involved herein) we find that DU_sub + ½DU_dis + I – E = DU_form + U_L, finally giving U_L = DU_sub + ½DU_dis + I – E – DU_form. 

The energy of crystallization (negative of the lattice energy) in ionic solids is mostly electrostatic energy, arising simply out of the Coulombic attractions and repulsions among all the positive and negatively charged ions in the solid piece, as per the Coulomb's law for electrostatic potential energy u_ij = q_iq_j/(4pe_or_ij) among any two ions i & j considered as point charges. However, there is a small, non-electrostatic part (around 10%) of the lattice energy as discussed later in this section. Further, for simplicity, let us consider only the case of binary ionic solids with magnitudes of all ionic charges equal (including, for example, NaCl or CaO, but not MgAl₂O₄ or even CaCl₂ etc.) i.e., |q_i| = |q_j|. 
Let us note, however, that the distance r_ij between ions i & j can be expressed as r_ij = a_ijr, where r is the smallest interionic distance i.e., the distance between a cation and its nearest anions (e.g., for the NaCl structure, this r is a/2) and a_ij is a numerical factor dependent only on the geometry of the ionic-crystal structure. Furthermore, q_i = z_ie & q_j = z_je where z_i & z_j are the ionic charges in atomic (proton-charge) unit (i.e., +1, –1, +2 etc., e.g., +2 for Ca²⁺). So we now have 
u_ij = z_iz_je²/(4pe_oa_ijr). For a given ion i within the lattice, we now sum u_ij over all other ions j (j ≠ i) to get the electrostatic potential u_i of the ion i. Thus we get u_i = {z_i e²/(4pe_o r)} S_j z_j /a_ij. As for the simple case mentioned above, we have z_j = ±z_i, so each term in the sum S_j z_j /a_ij = S_j ±z_i /a_ij is simply a number determined by crystal geometry, and so this sum can be written as z_iS_i. This gives u_i = z_i² S_i e² /(4pe_or).
Summing the electrostatic energies u_i of all the ions i and noting that the interaction between any two pair of ions should be counted only once, we get the purely electrostatic energy of crystallization (called the Madelung energy named after the scientist E. Madelung) as U_M = ½ S_i u_i = z² e² /(4pe_or) S_i (½ S_i) where z = |z_i| so that z_i² = z². The factor ½ arises here to compensate for having counted every pair of interionic interaction twice within the straightforward summation. This sum S_i (½ S_i) is a sum of numbers: this sum is negative (this means that here the attraction terms predominates over the repulsion terms), and is proportional to N, the total number of ions of any kind (or, in other words, the total number of formula units) in the crystal piece. So we have S_i(½S_i) = –NA where A is a numerical factor dependent on the geometry of the crystal -- this factor A is called the Madelung constant (value of A is 1.763 for CsCl, 1.748 for rock-salt structure etc.). Thus we have 
U_M = –NA z² e² /(4pe_or), and that the per-mole Madelung energy Ü_M = –N_AA z² e² /(4pe_or) where N_A is the Avogadro number. [Let us find Ü_M for NaCl, given that here the smallest interionic distance r (= r₊ + r_–) is 2.76 Å = 2.76 x 10^–10 m: 
putting values N_A = 6.02214 x 10²³ mol ^–1, A = 1.748, z = 1, e = 1.60218 x 10^–19 C & 1/4pe_o = 8.98755 x 10⁹ Nm² C^–2, we get Ü_M = –879.9 kJ mol^–1.] However, the general formula for Madelung energy (note that it's the negative of purely- electrostatic lattice energy) of binary ionic solids with cation and anion charges not necessarily equal (e.g., CaF₂) is 
U_M = –NA |z₊ z_–| e² /(4pe_or) along with Ü_M = –N_AA |z₊ z_–| e² /(4pe_or). 
Problem: Calculate the molar purely-electrostatic lattice energy for MgO (rock-salt structure, but with z = 2), given that 
r = 2.86 Å (Answer: 3396.6 kJ mol^–1 -- almost four times than 879.9 kJ mol^–1, the same quantity for NaCl).
Such high values of lattice energy (of which the negative of Madelung energy is a good working approximation) for divalent-divalent salts explains why they - e.g., alkaline-earth metal sulfides - are generally much less soluble than similar monovalent-monovalent salts - e.g., alkali metal chlorides. Greater the lattice energy, more difficult it becomes for the solvent to break up and dissolve the crystal.

Let us now discuss the non-electrostatic (i.e., non-Coulombic) part of the lattice energy so as to finalize our lattice energy expression. In addition to the negative Madelung energy, the energy of crystallization consists of non-Coulombic (positive) repulsion terms arising among the ions lying at close distances: as per the expression by M. Born the repulsion at distance r_ij between any two ions i & j is b/r_ijⁿ, b being the Born constant and n is a large power, another constant value lying usually between 6 & 12 (depending on the type of ions). So, we can similarly show that for the crystal as a whole, the Born repulsion term is B/rⁿ, B being another constant related to the Born constant b. So the energy of crystallization E_crys actually is 
E_crys = –N_AA |z₊ z_–| e² /(4pe_or) + B/rⁿ          [note that a repulsion contribute a negative term to E_crys]
This gives that the lattice energy U_L = –E_crys  is U_L = N_AA |z₊ z_–| e² /(4pe_or) – B/rⁿ

However the smallest interionic distance r of any given crystal is always of such a value so that (see figure) E_crys is a minimum (and U_L a maximum). This means that dU_L/dr = 0 giving –N_AA |z₊ z_–| e² /(4pe_or²) + nB/rⁿ⁺¹ = 0 implying 
B/rⁿ = N_AA |z₊ z_–| e² /(4pe_or n) which means that U_L = [N_AA |z₊ z_–| e² /(4pe_or)].(1 – 1/n). Or in other words, the introduction of non-Coulombic repulsion simply introduces a small correction-factor (1 – 1/n) to the lattice energy.
Problem: Calculate the (molar) lattice energy for NaCl, given that for NaCl the power n is estimated as 9.1 (Answer: 783.2 kJ mol^–1, only somewhat different from the purely electrostatic one 879.9 kJ mol^–1.)