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Wednesday, April 17, 2019


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.
(1) Ionic compounds of the type AB:
Ionic compounds of the type AB have three types of crystalline structures. (1.1) NaCl types (1.2)  ZnS  (1.3) type CsCl types
1.1 .Sodium Chloride (Rock Salt) Type Structure:
1.2  Zince Blende (ZnS) Type Structure (Sphelerite):
1.3 the Wurtize Structure (ZnS):
1.4 Csesium chloride (CsCl) Structure:
2.1  Calcium fluoride (fluorite) structure:
(3) Ionic compound of A2B type:
3.1 Antifluorite structure (Reverse of fluorite) Li2O :

1.1 Sodium Chloride (Rock) type Salt:
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
The radius ratio of 0.524 for NaCl suggest an octahedral void. Thus the salient features of this structure are as follows:

(1) 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).
(2)  The sodium ions are present in all the octahedral (Voids) holes.
(3) 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.
(4) 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.
(A) Nearest neighbor of Na+ and Cl- ions is 6 (Six) at distance a/2.
(B) Next nearest Na+ and Cl- ions is 12 at distance a/root 2
(5)  It should be noted that Na+ ions to exactly fit the octahedral holes, the radius ratio of sodium and chloride ions should be equal to 0.414. However, the actual radius ratio 0.524 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.
(6)  The unit cell of sodium chloride has 4 sodium and 4 chloride ions as calculated below
      No of sodium ions =  12 (at edge centres) ´1/4 + 1 (at body centre)´1= 4
      No of chloride ions = 8(at corner)´1/8+6 (at face centres) ´1/2 = 4
      Thus, the number of NaCl units per unit cell is 4.
(7)  The edge length of the unit cell of NaCl type of crystal is 2(r+R) where r = radii of Na+ ion and R is radii of Cl-
Thus, the distance between Na+ and Cl- ions = a/2
(8) Density and packing efficiency of NaCl are as:
Examples of NaCl type ionic salts:

Most of the halides of alkali metals, oxides and sulphides of alkaline earth metals have this type of structures.
(1) Group 1st Halides ie NaI, KCl, RbI, RbF ( except Cs halides) .
(2) Group 2nd  oxide   MgO, CaO, BaO, SrO ( except BeO)
(3) Ammonium Halides ie NH4Cl ,NH4Br ,NH4I etc
(4) Silver Halides ie AgF, AgCl, AgBr ,( except AgI)
(5) Other examples , TiO ,FeO, NiO etc

Note: 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 Fe0.95 O It can be explained on the assumption that some of the Fe+2 ion are replaced by 2/3rd as many Fe+3 ions in the octahedral voids.


The zinc sulphide crystals are composed of equal number of Zn+2 and S2- ions. The radii of the two ions (Zn+2 = 74 pm and S-2 = 184 pm) led to the radius r+/r- as 0.40 which suggests a tetrahedral arrangement
The salient features of this structure are as follows:
(1)  The sulphide ions are arranged in ccp arrangement, i.e. sulphide ions are present at the corners and the centres of each face of the cube  
(2)  Zinc ions occupy tetrahedral hole. Only half (50 %) 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)
(3)  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] Coordination.
(4)  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)
(5)  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)´1/8 + 6(at face centres)´1/2 = 4
      No. of Zn+2 ions = 4(within the body)´1 = 4
(6) Density, Packing efficiency (PE) and Void % ;
Examples: Thus, the number of ZnS units per unit cell is equal to 4. Some more examples of ionic solids having Zinc blende structures are CuCl, CuBr, CuI, AgI, BeS (beryllium sulphide).

ILLUSTRATIVE EXAMPLE:       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, therefore fraction of tetrahedral voids occupied by Ag+ ions = n/2n = ½ = 50%.

1.3 The wurtize (ZnS) Structure:

It is an alternate form in which ZnS occurs in nature. The main features of this structure are

                                       A unit cell representation of wurtzite structure
(1) Sulphide ions have HCP arrangement and zinc ions occupy tetrahedral voids. 
(2)  Only half the alternate tetrahedral voids are occupied by Zn+2 ions.
(3)  Coordinate no. of Zn+2 ions as well as S-2 ions is 4. Thus, this structure has 4 : 4 coordination
(4)  No. of Zn+2 ions per unit cel 
                                   = 4(within the unit cell) ´1 + 6(at edge centres) ´  = 6
(5) No. of S-2 ions per unit cell 
            = 12(at corners) ´1/6 +2 (at face centres) ½ +3 (within the unit cell)1=6
                   Thus, there are 6 formula units per unit cell.

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 Cs+ to Cl- as 0.93
 which suggest a body centred cubic structure having a cubic hole

The salient features of this structure are as follows:
(1)  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
(2)  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. 
(3)  For exact fitting of Cs+ ions in the cubic voids the ratio r Cs+/rCl-  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.
(4) The unit cell of caesium chloride has one Cs+ ion and one Cl- ion as calculated below
No. of Cl- ion = 8(at corners) ´1/8 = 1
No. of Cs+ ion = 1(at body centre)´1=1
Thus, number of CsCl units per unit cell is 1
(5)  Relation between radius of cation and anion and edge length of the cube,
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.
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
Other common examples  of this type of structure are CsBr, CsI, TlCl, TlBr, TlI and TlCN
      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.   
The salient features of fluorite structure are

(1)  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
(2)  The F ions occupy all the tetrahedral holes.
(3)  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.
(4) 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.
(5) Each unit cell has 4 calcium ions and 8 fluoride ions so formula of unit cell is Ca4F8 which is
explained as  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.
     Radius Ratio, packing efficiency ,void % and density:

Other examples: of structure are SrF2, BaCl2, BaF2, PbF2, CdF2, HgF2, CuF2, SrCl2, etc.
(3) Ionic compound of A2B type:  
3.1 Antiflourite structure (Reverse of fluorite) Na2O):

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.
(1)  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
(2)  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
 Other Examples: – Na2O, K2O, K2S, Na2S, Rb2O, Rb2S

Note: Metals like Al, Ag, Au, Cu, Ni, and Pt have ccp structure and Be, Mg, Co and Zn have a hcp structure. And Noble gases (except He has a hcp structure) have ccp structure.


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