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.

  1329_Unit cell representation.JPG

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

      No. of Ca+2 ions = 8(at corners)×1/8 + 6 (at face centres)´1/2  

      No. of F ions = 8 (within the body)×1 = 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

  1788_Anti-fluorite structure.JPG

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  

Crystal structure

Brief description

Coordination number

No. of atoms per unit cell

Examples

1.

Rock salt (NaCl – type)

Cl- ions in ccp Na+ ions occupy all octahedral voids

Na+ = 6 Cl- = 6

4

Li, Na, KI, and Rb halides NH4Cl, NH4Br, NH4I, AgF, AgCl, AgBr, MgO, CaO, TiO, FeO, NiO

2.

Zinc blende (ZnS – types)

S-2 ions in ccp Zn+2 ions occupy alternate tetrahedral voids

Zn+2 = 4 S-2 = 4

4

ZnS, BeS, CuCl, CuBr, CuI, AgI, HgS

3.

Wurtzite (ZnS – type)

S-2 ions in hcp Zn+2 ion occupy alternate tetrahedral voids

Zn+2 = 4 S-2 = 4

4

ZnS, ZnO, CdS, BeO

4.

Caesium chloride (CsCl type)

Cl- ions in bcc Cs+ ions in the body of cube

Cs+2 = 8 Cl- = 4

1

CsCl, CsBr, CsI, CsCN,CaS

5.

Fluorite (CaF2type)

Ca+2 ions in ccp, F- ions occupy all tetrahedral voids

Ca+2 = 8 F- = 4

4

CaF2, SrF2, BaF2, BaCl2, SrCl2, CdF2, HgF2

6.

Anti fluorite (Li2O – type)

O-2 ions in ccp, Li+ ions occupy all tetrahedral sites

Li+ = 4 O-2 = 8

4

K2O, Na2O, K2S, Na2S

Illustration 42. 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  

                        = 4 × (4/3 πra3) + 1/2 × 8 × (4/3πrc3) = 16√2πra3= π/3√2 {1 + (rc/ra)3}

                         for tetrahedral holes,  

                       rc / ra = 0.225 = π/3√2 {1 + (0.225)3} = 0.7496  

                        ∴ Void volume = 1 – 0.7496 = 0.2504/unit volume of unit cell  

                        % void space = 25.04%  

Illustration 43. Select the correct statements:-  

(A)   For CsCl unit cell (edge-length = a), rc + ra = 3√2 a  

(B)   For NaCl unit cell (edge-length =), rc + ra = l/2  

(C)   The void space in a b.c.c. unit cell is 0.68  

(D)   The void space % in a face-centered unit cell is 26%  

Solution:          In bcc structure are r+ + r = √3 / 2 a  

                        Hence, for CsCl, rC + ra = √3 / 2 a  

                        ∴ (A)  

                        Since, NaCl crystallises in fcc structure  

                        ∴ 2rC + 2ra = edge length of the unit cell  

                        Hence, rC + ra = l/2  

                        ∴ (B)                                                                

                        Since packing fraction of a bcc unit cell is 0.68    

                        ∴ void space = 1–0.68 = 0.32  

                        ∴ (C)                                                      

                        In fcc unit cell PF = 74%  

                        ∴ VF = 100–74 = 26                              \(D)      

Illustration 44.    Addition of CdCl2 to AgCl yields solid solutions, where the divalent cations Cd+2occupy the Ag+ sites which one of the following statements is true  

                        (A) The no. of cationic vacancies is one half of the no. of that of divalent ions added.  

                        (B) The no. of cationic vacancies is one half of the no. of that of divalent ions added.  

                        (C) The no. of anionic vacancies is equal in no. to that of divalent ions added.  

                        (D) No cationic or anionic vacancies are produced.  

Solution:          To maintain the electrical neutrality, the no. of cationic vacancies is equal to the no. of divalent ions added.  

Illustration 45.  CsCl has bcc structure with  at the centre and  ion at each corner. If rCs+ = 1.69Å and rCl = 1.81Å, what is the edge length “a” of the cube?  

                        (A) 3.50Å                                              (B) 3.80Å  

                        (C) 4.04Å                                              (D) 4.50Å  

Solution: (C) Assuming the closest approach between  and  ions, the internuclear separation is one-half of the cubic diagonal i.e.

                        .69 + 1.81 = 3.50 = a√3 / 2

                         ∴ a = 2 × 3.5 / √3 = 4.04Å

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