Structure of Ionic Compounds

STRUCTURE OF IONIC COMPOUNDS

Simple ionic compounds are of two types i.e. AB and AB₂ type. From the knowledge of close packed structures and the voids developed there in, we can have an idea about the structures of simple ionic compounds.

Among the two ions, constituting the binary compounds, the anions usually constitute the space lattice with hcp or ccp type of arrangements whereas the cations, occupy the interstitial voids

(a) If the anions (B^-) constitute the crystal lattice and all octahedral voids are occupied by cations (A⁺), then the formula of the ionic solid is AB.

(b) Similarly, if half of the tetrahedral voids are occupied by cations, then the formula of the solid crystal becomes A⁺B^-.

(c) When the anions (B^-2) are constituting space lattice and all the tetrahedral voids are occupied by the cations (A⁺), then the formula of the solid crystal will be A₂B.

Ionic compounds of the type AB

Ionic compounds of the type AB have three types of crystalline structures. (A) ZnS type (B) NaCl types (C) CsCl types

1. Sodium chloride (Rock salt) type structure

The sodium chloride structure is composed of Na⁺ and Cl^- ions. The number of sodium ions is equal to that of Cl^- ions. The radii of Na⁺ and Cl^- ions 95 pm and 181 pm giving the radius ratio of 0.524

$\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=\frac{95}{181}=0.524$

The radius ratio of 0.524 for NaCl suggest an octahedral void. Thus the salient features of this structure are as follows:

(i) Chloride ions (In a typical unit cell) are arranged in cubic close packing (ccp). In this arrangement, Cl^- ions are present at the corners and at the centre of each face of the cube. This arrangement is also regarded as face centred cubic arrangement (fcc).

(ii) The sodium ions are present in all the octahedral holes.

(iii) Since, the number of octahedral holes in ccp structure is equal to the number of anions, every octahedral hole is occupied by Na⁺ ions. So that the formula of sodium chloride is NaCl i.e. stoichiometry of NaCl is 1:1.

(iv) Since there are six octahedral holes around each chloride ions, each Cl^- ion is surrounded by 6 Na⁺ ions. Similarly each Na⁺ ion is surrounded by 6 Cl^- ions. Therefore, the coordination number of Cl^- as well as of Na⁺ ions is six. This is called 6:6 coordination.

(v) It should be noted that Na⁺ ions to exactly fit the octahedral holes, the radius ratio $\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}$ should be equal to 0.414. However, the actual radius ratio $\left(\frac{r_{N{a}^{+}}}{r_{C{l}^{-}}}=0.524\right)$ exceeds this value. Therefore to accommodate large Na⁺ ions, the Cl^- ions move apart slightly i.e. they do not touch each other and form an expanded face centred lattice.

(vi) The unit cell of sodium chloride has 4 sodium and 4 chloride ions as calculated below

No of sodium ions = 12 (at edge centres) x1/4 + 1 (at body centre)x1= 4

No of chloride ions = 8(at corner)x1/8+6(at face centres)x1/2 = 4

Thus, the number of NaCl units per unit cell is 4.

(vii) The edge length of the unit cell of NaCl type of crystal is 2(r+R) (r = radii of Na⁺ ion)

\begin{align}  \text{i}\text{.e}\text{.} \\  \text{a = 2}\left( {{\text{r}}_{\text{N}{{\text{a}}^{\text{+}}}}}+{{r}_{C{{l}^{-}}}} \right) \\ \end{align} (R = radii of Cl^- ion)

Thus, the distance between Na⁺ and Cl^- ions = a/2

Most of the halides of alkali metals, oxides and sulphides of alkaline earth metals have this type of structures. Some of the common examples are NaI, KCl, RbI, RbF, NH₄Cl, NH₄Br, AgCl, AgBr and AgI.

Ferrous oxide also has sodium chloride, types structure in which O^-2 ions are arranged in ccp and Fe⁺² ions occupy octahedral ions. However, this oxide is always non – stoichiometric and has the composition$F{e}_{0.95}O$. It can be explained on the assumption that some of the Fe⁺² ion are replaced by 2/3^rd as many Fe⁺³ ions in the octahedral voids.

2. Zinc blende (ZnS) type structures (sphelerite)

The zinc sulphide crystals are composed of equal number of Zn⁺² and S^2- ions. The radii of the two ions (Zn⁺² = 74 pm and S^-2 = 184 pm) led to the radius $\left(r^{+}/r^{-}\right)$ as 0.40 which suggests a tetrahedral arrangement.

$\frac{r_{Z{n}^{+2}}}{r_{S^{-2}}}=\frac{74}{184}=0.40$

The salient features of this structure are as follows

(i) The Zinc ions are arranged in ccp arrangement, i.e. sulphide ions are present at the corners and the centres of each face of the cube

(ii) Zinc ions occupy tetrahedral hole. Only half of the tetrahedral holes are occupied by Zn⁺² so that the formula of the zinc sulphide is ZnS i.e. the stoichiometry of the compound is 1:1 (Only alternate tetrahedral holes are occupied by Zn⁺²)

(iii) Since the void is tetrahedral, each zinc ion is surrounded by four sulphide ions and each sulphide ion is surrounded tetrahedrally by four zinc ions. Thus zinc sulphide has

4:4 Co – ordination.

(iv) For exact fitting of Zn⁺² in the tetrahedral holes, formed by close packing of S^-2 ions, the ratio Zn⁺²/S^-2 should be 0.225. Actually this ratio is slightly large (0.40)

(v) There are four Zn⁺² ions and four S^-2 ions per unit cell as calculated below:

No. of S^-2 ions = 8(at corners)x1/8 + 6(at face centres)x1/2 = 4

No. of Zn⁺² ions = 4(within the body)x1 = 4

Thus, the number of ZnS units per unit cell is equal to 4. Some more examples of ionic solids having Zinc blende structures are CuC, CuBr, CuI, AgI, beryllium sulphide.

Illustration 1. If silver iodide crystallizes in a zinc blende structure with I^- ions forming the lattice then calculate fraction of the tetrahedral voids occupied by Ag⁺ ions.

Solution: In AgI, if there are nI^- ions, there will be nAg⁺ ions. As I^- ions form the lattice, number of tetrahedral voids = 2n. As there are nAg⁺ ions to occupy these voids,

No. of S^-2 ions per unit cell

= 12(at corners) x$\frac{1}{6}+2\left(atfacecentres\right)\times \frac{1}{2}+3\left(withintheunitcell\right)\times 1$= 6

Thus, there are 6 formula units per unit cell.

Illustration 2. The co-ordination number of a metal crystallising in hcp structure is

(A) 12 (B) 10

(C) 8 (D) 6

Solution: (A)

4. Caesium chloride (CsCl) structure

The caesium chloride crystal is composed of equal number of caesium (Cs⁺) and Chloride Cl^- ions. The radii of two ions (Cs⁺ = 169 pm and Cl^- = 181 pm) led to radius ratio of $r_{C{s}^{+}}to{r}_{C{l}^{-}}$ as 0.93 which suggest a body centred cubic structure having a cubic hole

$\frac{r_{C{s}^{+}}}{r_{C{l}^{-}}}=\frac{169}{181}=$0.93

The salient features of this structure are as follows:

(i) The chloride ion form the simple cubic arrangement and the caesium ions occupy the cubic interstitial holes. In other words Cl^- ions are at the corners of a cube whereas Cs⁺ ion is at the centre of the cube or vice versa

(ii) Each Cs⁺ ion is surrounded by 8 Cl^- ions and each Cl^- ion in surrounded by 8 Cs⁺ ions. Thus the Co – ordination number of each ion is eight.

(iii) For exact fitting of Cs⁺ ions in the cubic voids the ratio $\frac{r_{C{s}^{+}}}{r_{C{l}^{-}}}$ should be equal to 0.732. However, actually the ratio is slightly larger (0.93). Therefore packing of Cl^- ions slightly open up to accommodate Cs⁺ ions.

(iv) The unit cell of caesium chloride has one Cs⁺ ion and one Cl^- ion as calculated below

No. of Cl^- ion 8(at corners) x1/8 = 1

No. of Cs⁺ ion = 1(at body centre)x1=1

Thus, number of CsCl units per unit cell is 1

(v) Relation between radius of cation and anion and edge length of the cube, $r_{Cs}+r_{Cl}=\frac{a\sqrt{3}}{2}$

Other common examples of this type of structure are CsBr, CsI, TlCl, TlBr

Higher coordination number in CsCl(8:8) suggest that the caesium chloride lattice is more stable than the sodium chloride lattice in which Co – ordination number is 6:6. Actually the caesium chloride lattice is found to be 1$\%$ more stable than the sodium chloride lattice. Then the question arises why NaCl and other similar compounds do not have CsCl type

lattice – This is due to their smaller radius ratio. Any attempt to pack 8 anions around the relatively small cation (Li⁺, Na⁺, K⁺, Rb⁺) will produce a state in which negative ions will touch each other, sooner they approach a positive ion. This causes unstability to the lattice.

Effect of temperature on crystal structure

Increase of temperature decreases the coordination of number, e.g. upon heating to

760 K, the CsCl type crystal structure having coordination 8:8 changed to NaCl type crystal structures having coordination 6:6.

$\underset{\text{(8:8}\text{coordination)}}{\text{CsCl type crystal}}\stackrel{\text{high}\text{temp}}{\to }\underset{(6:6Co-ordination)}{\mathrm{NaCltypecrystal}}$

Effect of pressure on crystal structure

Increase of pressure increases the Co – ordination number during crystallization e.g. by applying pressure, the NaCl type crystal structure having 6:6 coordination number changes to CsCl type crystal having coordination number 8:8

$\underset{\text{(8:8}\text{coordination)}}{\text{NaCl type crystal}}\stackrel{\text{high}\text{pressure}}{\to }\underset{(6:6Co-ordination)}{\mathrm{CsCltypecrystal}}$

Illustration 3. Out of NaCl and CsCl, which one is more stable and why?

Solution: CsCl is more stable than NaCl. This is because higher the coordination number, greater are the forces of attraction between the cations and the anions in the close-packed arrangement. As CsCl has co-ordination number of 8 : 8 while NaCl has a coordination number of 6 : 6, therefore CsCl is more stable.

IONIC COMPOUND OF THE TYPE AB₂

Calcium fluoride (Fluorite) structure

The salient features of fluorite structure are

(i) The Ca⁺² ions are arranged in ccp arrangement, i.e. these ions occupy all the corners and the centres of each face of the cube

(ii) The F^– ions occupy all the tetrahedral holes.

(iii) Since there are two tetrahedral holes for each Ca⁺² ion and F^- ions occupy all the tetrahedral holes, there will be two F^- ions for each Ca⁺² ions, thus the stoichiometry of the compound is 1:2.

(iv) Each Ca⁺² ion is surrounded by 8F^- ions and each F^- ions is surrounded by 4Ca⁺² ions. The Coordination number of Ca⁺² ion is eight and that of F^- ion is four, this is called 8:4 Coordination.

(v) Each unit cell has 4 calcium ions and 8 fluoride ions as explained below

No. of Ca⁺² ions = 8(at corners)x1/8 + 6 (at face centres)x1/2

No. of F ions = 8 (within the body)x1 = 8

Thus the number of CaF₂ units per unit cell is 4.

Other examples of structure are SrF₂, BaCl₂, BaF₂, PbF₂, CdF₂, HgF₂, CuF₂, SrCl₂, etc.

Ionic compound of A₂B type

The compound having A₂B formula are compounds having anti fluorite structure

Anti fluorite structure is having arrangement of cations and anions opposite to the fluorite structure Li₂O has an anti fluorite structure.

(i) In the crystal structure of Li₂O, the O^-2 ions constitute a cubic close packed lattice (fcc structure) and the Li⁺ ions occupy all the tetrahedral voids

(ii) Each oxide ion, O^-2 ion is in contact with 8 Li⁺ ions and each Li⁺ ions having contact with 4 oxide ion. Therefore, Li₂O has 4:8 coordination

Examples – Na₂O, K₂O, K₂S, Na₂S, Rb₂O, Rb₂S

Normal spinel structure

Spinel is a mineral MgAl₂O₄. In it oxide ions are arranged in ccp with Mg⁺² ions occupying tetrahedral voids and Al⁺³ ions in a set of octahedral voids. Many ferrites (such as ZnFe₂O₄) also possess spinel structure. These are very important magnetic materials and are used in telephone and memory loops in computers.

Structure of Fe₃O₄ (Magnetite)

In Fe₃O₄, Fe⁺² and Fe⁺³ ions are present in the ratio 2:1. it may be considered as having composition FeO.Fe₂O₃. In Fe₃O₄ Oxide arranged in ccp. Fe⁺² ions occupy octahedral voids while Fe⁺³ ions are equally distributed between octahedral and tetrahedral voids

MgFe₂O₄ also has structure similar to magnetite. In this Mg⁺² ions are present in place of Fe⁺² ion in Fe₃O₄. Magnetite has inverse spinet structure.

SUMMARY OF VARIOUS STRUCTURES OF IONIC SOLIDS

Illustration 4. Compute the percentage void space per unit volume of unit cell in zinc-fluoride structure.

Solution: Since anions occupy fcc positions and half of the tetrahedral holes are occupied by cations.

Since there are four anions and 8 tetrahedral holes per unit cell, the fraction of volume occupied by spheres/unit volume of the unit cell is

= $\frac{4\times \left(\frac{4}{3}\pi r_a^3\right)+\frac{1}{2}\times 8\times \left(\frac{4}{3}\pi r_c^3\right)}{16\sqrt{2}{r}_a^3}$ = $\frac{\pi }{3\sqrt{2}}\left\{1+{\left(\frac{r_c}{r_a}\right)}^3\right\}$

$\because$ for tetrahedral holes,

$\frac{r_c}{r_a}$= 0.225 = $\frac{\pi }{3\sqrt{2}}$ {1 + (0.225)³} = 0.7496

$\therefore$ Void volume = 1 – 0.7496 = 0.2504/unit volume of unit cell

$\%$ void space = 25.04$\%$