voids or interstitial sites
Radius Ratio in Ionic crystals
In ionic crystal’s it is not possible for both cations and anions to have close packed structures due to their different sizes. However, if one of the ions is bigger than the other, it is common for the bigger ion(usually anions) alone to approach a close-packed structure and smaller ions to fit into voids in this structure. The limiting sizes of cations that can fit into these voids without disturbing the closest packing of anions are as follows:-
Octahedral void radius of cation/radius of anion =0.414
Tetrahedral void radius of cation/radius of anion=0.225
A few common ionic lattices are described below.
- Rock salt structure
In the rock salt structure, anions form a face-centred cubic unit cell and cations occupy octahedral voids. There are 4 anions and four cations per unit cell of this structure and hence the formula of ionic compound is a4b4 or simply ab. Eg NaCl, KCl, MgO, CaO, SrO, etc.
- Anti-fluorite structure
In this case anions form a cubical-closest packing arrangement and cations occupy tetrahedral voids there are 4 anions and 8 cations per unit cell of this structure and hence the formula of ionic compound is a8b4 or simply a2b, eg Li20, Na2O, K2O, Rb2O etc.
- Zinc-blende structure
In zinc-blende structure, anions form a cubical-closest packed structure and cations are present in half of the tetrahedral voids. There are 4 anions and 4 cations per unit cell of this structure and hence the formula of an ionic compound is a4b4 or simply ab. Eg BeO, ZnS etc.
- Body centred cubic structure
In bcc structure, the cation is ocated at the body centre of a cube of anions and the anion is located at the corners of a cube. There is one cation and one anion present per unit cell, and hence the formula of an ionic compound is ab. In this structure each anion has eight positively charged ions as its next nearest neighbours. Hence, coordination number of each cation and anion is eight, e.g. CsCl, CsBr, CsI.
- Fluorite Structure
In fluorite structure, cations form cubical-closest packing and anions occupy tetrahedral voids. Here we have four cations and eight anions. So the formula will be a4b8 or ab2, e.g. VO2, ThO2. CaF2.
- Corundum structure
In corundum structure, anions form hexagonal closest packing and cations are present in two-thirds of the octahedral voids. The general formula of the compound is a2/3b or a2x3, e.g. Fe2O3, Al2O3, Cr2O3 etc.
- Spinel structure
In spinel structure oxides are arranged in cubical-closest packing; one-eighth of the tetrahedral void are occupied by divalent metal ion[a2+](2+charge on a) and one-half of the octahedral voids are occupied by trivalent metal ion[b3+](3+charge on b). In a unit cell. We have
Number of divalent metal ions a2+=(1/8)*8=1
Number of trivalent metal ions b3+=(1/2)*4=2
Number of oxide ions O2-=4
Hence the general formula of the compound is ab20, e.g. ZnAl2O4, MgAl2O4, ZnFe2O4.
Radius cation/radius anion=X
Radius ratio Structure coordination number
X < 0.155 Linear 2
0.155 <= X < 0.255 Planar triangle 3
0.255 <= X < 0.414 Tetrahedronn 4
0.414 <= X <0 .732="" 6="" nbsp="" o:p="" octahedron="">
0.732 <= X bcc 8
Very imporatnt line=For eg chlorides of Li, K and Pb crystallise in the fcc whereas the chlorides of Cs crystallise in the bcc.
Others thongs which have to be added:-
VOIDS OR INTERSTITIAL SITES
It is seen that even in the close packing of spheres, certain hollows are left vacant. These holes or voids or vacant spaces in the crystals are called interstitial sites. 2 imp are:
- Tetrahedral
- Octahedral
Tetrahedral
Those sites where an atom is in contact with four host atoms in the form of a tetrahedron are called tetrahedral sites/voids.
In the fcc unit cell, we have 8 tetrahedral voids, two each on a body diagnol. So effective number of these voids ina unit cell will be eight. Again for hcp we have 12 tetrahedral sites. So we can conclude that the number of tetrahedral voids are equal to double the rank of the unit cell.
Octahedral
Those sites, where an atom is in cotact with 6 host atoms in the form of an octahedron, are called octahedral sites.
In fcc lattice we have 13 such sites. Each one of the 12 at the middle of the edge of the unit cell and one in the centre of the unit cell. So effective number of voids which are at the middle of the edge will be (¼*12)=3(since 1 void wud be shared by four unit cells). So the total octahedral voids will be 3+1=4.
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