+2 UNIT 2 PAGE- 3
DETERMINATION OF IONIC RADII
Ionic radii may be defined as the distance between the nucleus of an ion and its electron charge cloud. The electron charge cloud , may extend theoretically up to very large distance. However, it is important to characterise the radius of each ion by some specific value. For this purpose, the internuclear distance in an ionic compound is determined from X-ray measurements. This distance is then taken as the sum of the radii of the two ions involved.
Knowing the radius of one, that of the other can be calculated. For example, distance between the nuclei of Na+ and F- ions is found from X-ray studies of NaF crystals to be 231 pm. Thus,
The problem remains unsolved unless the radius of one of the ions is known. Several solutions have been suggested. One proposed by Pauling has been mostly followed. He selected four ionic solids viz., NaF, KCl, RbBr and CsI . In each salt, the cation and anion are iso-electronic i.e., they have the same number of electrons which correspond to a noble gas configuration. In NaF, both Na+ and F- ions have 10 electrons each and have 1s22s2p6 configuration which is same as neon. Pauling proposed that in ionic compounds formed by iso-electronic ions, ratio of the two ionic radii should be inversely proportional to the ratio of the effective nuclear charges of the two ions. The effective nuclear charge (Zeff) is obtained by subtracting a screening constant from the actual nuclear charge. The effective nuclear charges of Na+ and F- ions at their peripheries come out to be 6.5 and 4.5 respectively. Hence according to Pauling’s rule,
Also, we know that
Solving (i) and (ii) , we have ,
Following this procedure, the radii of most of cations and anions have been determined. The radii of some of the ions are given in Table.
Cation | Radius(pm) | Anion | Radius(pm) |
Li+ | 60 | F- | 136 |
Na+ | 95 | Cl- | 181 |
K+ | 133 | Br- | 195 |
Rb+ | 148 | I- | 216 |
Cs+ | 160 | O2- | 140 |
Cu+ | 96 | S2- | 184 |
Ag+ | 126 | N3- | 171 |
Mg2+ | 65 | C4- | 260 |
Ca2+ | 99 | ||
Al3+ | 50 |
The radius of a cation is invariably smaller than that of the corresponding atom.
For example, the radius of Na atom is 154 pm, while that of Na+ ion is 95 pm. The reason is that with the elimination of one (or more) electrons from the valency shell of the atom, the effect of nuclear attraction increases because it is now acting on smaller number of electrons. The electron charge cloud is therefore , pulled more inward towards the nucleus than before.
The radius of an anion is invariably larger than that of the corresponding atom.
For example, the radius of Cl atom is 99 pm, while that of Cl- ion is 181 pm. The reason is that with the addition of one (or more) electrons to the valency shell, the effective force of attraction on the electron charge cloud decreases. As a result, electron charge cloud expands and hence ionic radius increases.
Trends in Ionic radii
i) Along a Group
The ionic radius increases in a group with increase in atomic number as :
Li+< Na+ < K+ < Rb+< Cs+
The reason for the increase is the increase in the principal quantum number though the number of electrons in the valency shell remains the same. Similarly in the halogens, the ionic radius increases as :
F - < Cl - < Br- < I -
ii) Along a Period: Along a period , the ionic radii of isoelectronic ions (having the same number of electrons) decrease as the nuclear charge increases.
Na+ > Mg2+ > Al3+
RADIUS RATIO RULE AND THE SHAPE OF IONIC CRYSTALS
Ionic solids are made up of cations and anions which have definite arrangements. Each ion is surrounded by a number of ions of opposite charge. For example, in an ionic solid of the type A+X-, A+ ions are surrounded by a definite number of X- ions. This number is called co-ordination number of A+. similarly , the number of A+ ions which are surrounding a X- ion is called co-ordination number of X- ion. The arrangement of ions in a crystal (shape) and co-ordination number of ions depends upon the ratio of its radius with respect to the radius of the ions or atoms surrounding it. In the case of ionic solids , the ratio of the radius of the cation to the radius of the anion is called radius ratio. Thus ,
In order to study the role played by this ratio in determining the co-ordination number and the shape of an ionic solid, let us suppose that the co-ordination number of ion A+ in an ionic compound AX, is 3. Thus X- ions will be in contact with one A+ ion. This may be represented in Fig 30(a).
Such a structure will be quite stable because the positive ion (A+) finds three negative ions (X-) immediately in contact with it. The forces of attraction will be quite strong. At the same time, the negative ions are not in contact with one another. The forces of repulsion , therefore will be quite small. The limiting case of stability, with co-ordination number 3 , arises when the negative ions (X-) are in contact with each other , as shown in Fig 30 (b). The positive ion, no doubt, is still touching the negative ions and the force of attraction is quite strong. But the negative ions are also touching one another and , therefore the force of repulsion is stronger than before. In such case, the radius ratio r+/ r- is 0.155. If the ratio r+/ r- falls below this value ( i.e., A+ ion becomes still smaller), the structure will be as shown in Fig 30 c. The positive ion is not touching any of the negative ions now. The force of attraction , therefore is weak. At the same time, the negative ions are in direct contact with one another. The force of repulsion , therefore will be strong. Hence the structure is unstable and therefore , cannot exist.
Evidently, for co-ordination number 3, the lowest , i.e., the limiting radius r+/ r- is 0.155.
Similarly, the limiting radius ratio for the cations to fit in the tetrahedral site is 0.225. In other words, for the radius ratio r+/ r- ranging between 0.155 to 0.225 , trigonal site will be preferred in which co-ordination number of cation is 3.
Similarly, the limiting radius ratio for the cations to fit into an octahedral site of anions is 0.414. Therefore, tetrahedral sites (C.N of cation 4) will be preferred between the radius ratio range 0.225 to 0.414. In the range 0.414 to 0.732, octahedral sites (C.N of cation 6) will be adopted by cations, while above 0.732, cubic site (C.N of cation 8) will be favoured.
Thus, we find that radius ratio is a very important factor for determining the most stable form of arrangement of ions in the crystal lattice of an ionic solid.
The permitted co-ordination numbers and structural arrangements of anions around cations for different r+/ r- ratios are given in the following Table.
Limiting radius ratio r+/r- | Possible co-ordination number | Structural arrangement | Example |
0.155 - 0.225 | 3 | Plane Triangular | B2O3 |
0.225 - 0.414 | 4 | Tetrahedral | ZnS, CuCl, CuI, SiO44-, BaS, HgS |
0.414- 0.732 | 6 | Octahedral | NaCl, MgO, NaBr, CaS, MnO, KBr, CaO |
0.732 - 1.0 | 8 | Cube | CsCl, NH4CI, NH4Br CsBr, TℓBr, |
NOTE
1. In case of crystals of the type AB , if the radius ratio is 0.732 or above, the crystal would have CsCl type structure, i.e., body-centred cubic structure in which each ion has a co-ordination number 8. If the radius ratio is less than 0.732 but equal to or greater than 0.414, the crystal would have NaCl type structure in which each ion has a co-ordination number 6. If the ratio is less than 0.414 but equal to or higher than 0.225, the crystal would have a ZnS type structure in which each ion has a co-ordination number 4. Lastly, if the ratio is less than 0.225 but equal to or higher than 0.155, the crystal would adopt a planar trigonal structure in which the co-ordination number of each ion is 3.
2. In case of crystals of the type AB2 , if the radius ratio is 0.732 or above, the crystal would have fluorite (CaF2) type structure in which co-ordination number of the cation is 8 and that of the anion is 4. If the ratio lies between 0.414 and 0.732 , the crystal would adopt rutile (TiO2) type structure in which co-ordination number of the cation is 6 and that of anion is 3.
Thus, with the help of radius ratios, it is possible to predict the shape of ionic crystals.
Problems
1. Sodium metal crystallises in body-centred cubic lattice with cell edge(a) 429 pm. What is the radius of sodium atom ?
2. A cubic solid made of two elements P and Q , Q atoms are at the corners of the cube and P at the body centre. What is the formula of the compound ? What are co-ordination numbers of P and Q ?
3. MgO has structure of NaCℓ and TℓCℓ has CsCℓ structure . What are co-ordination numbers of the ions in MgO and TℓCℓ ?
4. A compound , formed by elements A and B crystallises in cubic structure, where A atoms are at the corners of a cube, while B atoms are at the face centres. What is the formula of the compound ?
5. Structure of CsBr is the same as that of CsCl. Sketch the unit cell of CsBr showing the positions of Cs+ ions and Br- ions . Find out the total number of Cs+ and Br- ions per unit cell.
6. The length of the unit cell of body-centred cubic metal crystal is 352 pm. Calculate the radius of an atom of the metal.
7. A solid AB has the NaCl structure. If the radius of the cation A+ is 120 pm. Calculate the maximum possible value of the anion B-.
8. If the radius of Cs+ is 169 pm and that of Cl- is 181 pm, predict the structure of CsCl and what is the C.N of Cs+ ?
9. The radii of Na+ and F- are 95 pm and 136 pm respectively. (a) What is the anion arrangement ? (b) What is the C.N of the cation ?
10. A solid A+B- has NaCl type close-packed structure. If the anion has a radius of 250 pm, what should be the ideal radius for the cation ? Can C+ having a radius of 180 pm, be slipped into the tetrahedral site of crystal A+B- ? Give reason for your answer.
11. Chromium has body-centred cubic structure. Its edge is 300 pm. What is its density ?
12. An element occurs in BCC structure. It has a cell-edge of 250 pm. Calculate the molar mass , if its density is 8.0 g cm-3 .
13. Calculate the value of Avogadro number from the data. Density of NaCl = 2.165 g cm-3 , distance between Na+ and Cl- ion in NaCl crystal is 281 pm.
14. A body-centred cubic element of density 10.3 g cm-3 has a cell edge of 314 pm. Calculate the atomic mass of the element.
15. The density of a face-centred element ( atomic mass 60.2 amu) is 6.2 g cm-3 . Calculate the length of the unit cell.
16. What is the distance between Na+ and Cl- in NaCl crystal , if density is 2.1650 g cm-3. NaCl crystallises in FCC lattice.
17. An element (density 6.80 g cm-3 )occurs in BCC structure with unit cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element.
18. An element crystallises in a structure having FCC unit cell and an edge of 200 pm. Calculate its density , if 200 g of this element contains 24 x 1023 atoms .
19. Unit cell of an element of atomic mass 96 and density 10.30 g cm-3 is a cube with edge length of 314 pm. Find the structure of the crystal lattice( simple cubic, body centred or face centred).
20. The density of KBr is 2.730 g cm-3. The length of the unit cell is 654 pm. Show that KBr has a FCC structure.
21. A metal (atomic mass 50 has a body-centred cubic crystal structure. The density of the metal is 5.960 g cm-3. Find the volume of the unit cell.
22. The nearest neighbour Ag atoms in the silver crystal are 2.87 x 10 -10 m apart. What is density of silver ? Silver crystallises in FCC form.
23. KCl and CsF have FCC and BCC structure respectively. Molecular mass (CsF) is 2 times the molar mass (KCl) and edge length of CsF unit cell is 1.5 times that of KCl. Calculate the ratio of densities of CsF and KCl crystals.
24. An element of atomic mass 98.5 g/mol occurs in FCC structure. If its unit cell edge length is 500 pm and its density is 5.22 g cm-3, what is the value of Avogadro constant ?
25. The compound CuCl has ZnS structure and edge length of the unit cell is 500 pm. Calculate its density.
26. Sodium chloride has face-centred cubic structure. Its density is 2.163x 10-3 kg m-3 . What is the length of the unit cell?
27. The radius of an octahedral void is ‘r’ and radius of the atom is ‘R’ when these are in close-packing. Derive the relationship between two.
Structures of substances related to close packed lattices.
There are many substances which solidify to form lattices having close packed structures. Substances which crystallise in close packed structures are given below :
1. Metals : The metals such as Be, Cd, Co, Ti and Zn have hcp structure while Al, Ag, Au, Cu, Ni and Pt have ccp structure.
2. Noble gases : Helium shows hcp structure while, Ne, Ar, Kr and Xe show ccp structure.
3. Molecular solids : Solid hydrogen has hcp structure, while molecules having nearly spherical shapes such as solid CH4, HCl and H2S have ccp structure.
4. Ionic solids : Packing of non-uniform spheres
Packing of ionic compounds depends upon the nature and number of constituent ions, cations and anions. The sizes of the cations and anions are different, cations being smaller. In these the lattice is made up of the larger ions (anions) with oppositely charged smaller cations inserted into the voids.
Thus in a close packed structure of ionic solids, cations are surrounded by anions and vice-versa. The number of anions surrounding a cation gives the co-ordination number of the cation and anions vice-versa. The common co-ordination number in ionic crystals are 4, 6 and 8. The ratio of the sizes of cations r+ and r- is called radius ratio r+/ r-. The ratio can predict the type of void in the crystal. This is as follows :
Radius ratio (r+/r-) | Position of cation |
0.414 | Tetrahedral void |
0.732 | Octaheral void |
1.0 | Centre of the body |
STRUCTURE OF SIMPLE IONIC COMPOUNDS
The simple ionic compounds have the formulae AB or AB2 . Each solid tends to have its constituents closely packed to each other. Among the two ions, the anions usually constitute the space lattice with hcp or ccp type of arrangement, whereas cations occupy the interstitial voids. From the knowledge of packed structure and the voids occupied, we can have an idea about the structure of some simple ionic compounds. For example,
i) For a binary compound, if the negative ions constitute the space lattice and all octahedral holes are occupied, then the formula of the ionic compound is AB.
ii) Similarly, if all the tetrahedral holes are occupied, the formula of the ionic compound is A2B.
iii) Now if , half of the tetrahedral holes are occupied, then the formula is AB.
Thus, from the above discussion , it is clear that the idea about arrangement of ions quite reasonably and satisfactorily explains the structures of many ionic solids.
STRUCTURES OF THE IONIC COMPOUNDS OF THE TYPE AB
In this type cations and anions are present in the ratio 1 : 1 . Ionic compounds of this type have any one of the following three type of structures:
i) Rock salt (NaCl) type structure.
ii) Caesium chloride(CsCl) type structure.
iii) Zinc blende (ZnS) type structure.
STRUCTURE OF SODIUM CHLORIDE (NaCl) - ROCK SALT STRUCTURE
The unit cell of sodium chloride is shown in Fig 31 .
Fig 31 Unit cell representation of NaCl structure. Hollow circles represent Na+ ions and solid circles represent Cl- ions.
In this structure Na+ ions are represented by hollow circles and chloride ions by solid circles. The salient features of this structure are as follows :
i) The chloride ions (Cl- ) are arranged in cubic close-packing (ccp) arrangement. In this arrangement , there are Cl- ions at the corners of the cube as well as at the centre of each face (face-centred cubic arrangement).
ii) The sodium ions (Na+) occupy all the octahedral sites.
iii) There is only one octahedral site per atom in a close packed lattice. Therefore there will be one Na+ ion for every Cl- ion. This corresponds to 1 : 1 stoichiometry of sodium chloride crystal.This stoichiometry can also be understood by counting the number of Na+ and Cl- ions which contribute to the unit cell as discussed below : It is clear from Fig 31 that there are eight Cl- ions at the corners and six at the faces of the unit cell. Each ion at the corner is equally shared between eight unit cells and its contribution per unit cell is 1/8 . Similarly, each ion on the face is shared by two unit cells, so that its contribution per unit cell is ½ . Therefore the number of Cl- ions for the unit cell is equal to 8(1/8) + 6 ( ½ ) = 4. Sodium ions can also be calculated in a similar way. There are 12 Na+ ions on the edges and one inside the cell. It can be easily seen that each ion on the edge is shared by four unit cells and its contribution per unit cell is ¼ , while the ion at the body centre contributes only to one unit cell. Therefore the number of Na+ ions for the unit cell is equal to 12 (1/ 4 ) + 1 = 4. Thus the stoichiometry is 1 : 1.
iv) In this structure , each Na+ ion surrounded by six Cl- ions, which are disposed towards the corners of a regular octahedron. This has also been shown in Fig 31. Similarly, each chloride ion is surrounded by six Na+ ions. Thus the co-ordination number of Na+ as well as that of Cl- ion is 6.
v) For cations to fit into octahedral sites , the radius ratio r+/ r-should be equal to 0.414 . The radius of Na+ is 95 pm and that of Cl- ion is 181 pm, so that the radius ratio rNa+ / rCl- is 0.524. Therefore, to accommodate large Na+ ions, the Cl- ions move apart slightly. In other words , the ccp arrangement of Cl- ions is slightly opened up. This type of structure is adopted by most of the alkali metal halides, alkaline earth metal oxides and AgF, AgCl, AgBr, NH4Cl , NH4Br etc.
CAESIUM CHLORIDE STRUCTURE
The unit cell of CsCl is shown in Fig 32. The structure of CsCl may be described as follows :
i) In this structure, the Cl- ions are at the corners of a cube whereas Cs+ ion is at the centre of the cube or vice versa as shown in Fig 32.
ii) This structure has 8 : 8 co-ordination, i.e., each Cs+ ion is touching eight Cl- ions and each Cl- ion is touching eight Cs+ ions.
Fig 32 Caesium Chloride structure
iii) For exact fitting of Cs+ ions in the cubic voids rCs+ / rCl- should be equal to 0.732 ; however, actually the ratio is slightly larger (0.93) . Therefore packing of Cl- ions slightly opens up to accommodate Cs+ ions.
iv) The cell of ceasium chloride has one Cs+ ion and one Cl- ion as calculated below:
Number of Chloride ions = 8 ( At corners) x (1/8 ) = 1
Number of Cs+ ion = 1(At body centre) x 1 = 1
Thus, number of CsCl units per unit cell is 1.
Examples of the compounds having this type of structure are : CsBr, CsI, TℓCℓ and TℓBr.
Note
It may be noted that the temperature and pressure also affect the structure of ionic solid. For example at ordinary temperatures and pressures, chlorides, bromides and iodides of lithium,sodium, potassium and rubidium possess the NaCl structure with 6 : 6 co-ordination. It is observed that on application of high pressure they transform to the CsCl structure with 8 : 8 co-ordination. Thus high pressure increases the co-ordination number. On the other hand , CsCl on heating transforms to NaCl structure at 760 K. Thus, at higher temperature, the co-ordination number decreases.
ZINC BLENDE STRUCTURE
The unit of Zinc blende structure is shown in Fig 36. In this structure, the zinc ions are represented by hollow circles and the sulphide ions are represented by solid circles. The salient features of the structure are as follows:
i) The sulphide ions are arranged in ccp arrangement ie S2- ions are present at the corners of a cube and at the centre of each face.
ii) The zinc ions occupy half of the tetrahedral sites.
iii) Since, there are two tetrahedral sites per atom in a closed packed lattice, this means that there are two tetrahedral sites per S2- ion. In this arrangement, only half of the tetrahedral sites are occupied, so that each S2- ion has one Zn2+ ion. Therefore, the stoichiometry of the compound is 1:1.
Fig 36 Unit cell representation of Zinc blende structure. Hollow circles represent Zn2+ ions and solid circles represent S2- ions
iv) Each zinc ion is surrounded by four sulphide ions which are disposed towards the corners a regular tetrahedron. Similarly, each S2- ion is surrounded by four Zn2+ ions. The arrangement around one Zn2+ ion is also shown in Figure. The co-ordination numbers of Zn2+ ion and S2- ion in zinc blende are described as 4 : 4.
v) In zinc sulphide , the radius ratio r+/ r- is 0.40. This suggests that the co-ordination number is 4. This supports the structure presented in Fig 36.
Zinc blende structure is also adopted by compounds such as ZnO, CuCl, CuBr, CuI etc.
STRUCTURES OF IONIC COMPOUNDS OF THE TYPE AB2
These are ionic compounds having cations and anions in the ratio 1 : 2. Most of the compounds have CaF2 structure.
CALCIUM FLOURIDE STRUCTURE
(FLUORITE STRUCTURE)
The unit cell of calcium fluoride is shown in Fig 37 .
Fig 37 Unit cell representation of CaF2 . The Ca2+ ions adopt ccp arrangement and F - ions occupy all tetrahedral sites.
The calcium ions are represented by hollow circles and fluoride ions are represented by solid circles.
The CaF2 structure may be described as follows :
i) the Ca2+ions are arranged in ccp arrangement so that there are Ca2+ions at the corners and at the centre of each face of the cube. The fluoride ions occupy all the tetrahedral sites.
ii) Since there are two tetrahedral sites for each Ca2+ ion and F- ions occupy all the tetrahedral sites , there will be two F - ions for each Ca2+ ion. The stoichiometry of the compound is 1 : 2.
iii) In this arrangement each F- is surrounded by four Ca2+ ions whereas each Ca2+ ions is surrounded by eight F- ions. Thus the co-ordination number of Ca2+ and F- ions are 8 : 4.
This structure is adopted by a large number of fluorides such as BaF2, SrF2, PbF2, BaF2, HgF2, CuF2 etc.
ANTIFLUORITE STRUCTURE
In antifluorite structure the anions are arranged in cubic close packing (ccp) while the cations occupy all the tetrahedral voids. Na2O has an antifluorite structure. In this case:
(i) The O2- ions constitute a ccp type lattice and Na+ ions occupy all the tetrahedral voids.
(ii) The oxide ion O2- is in contact with eight sodium ions whereas in turn, each Na+ ion is in contact with four oxide ions. Thus Na2O has 4 : 8 co-ordination.
Examples : Li2O, Na2O, K2O, Cl2O, Na2S, K2S, Rb2S etc.
STRUCTURE OF THE OXIDES OF IRON
Iron forms three important oxides. These are FeO, Fe2O3 and Fe3O4. These compounds are non-stoichiometric and are easily oxidised and reduced into one another.
i) Structure of FeO
The oxide ions O2 - , form a face centred cubic (fcc) arrangement. Fe2+ ions occupy the octahedral holes. This will result in a perfect rock salt structure. However, it is non-stoichiometric with the composition Fe0.95 O called wustite. We can obtain this composition if a small number of Fe2+ ions are replaced by two-thirds as many Fe3+ ions in the octahedral voids.
ii) Structure of Fe2O3: When all the Fe2+ ion in FeO are replaced by Fe3+ions, the oxide of the composition Fe2O3
( i.e ., Fe : O = 2 : 3 ) is obtained
iii) Structure of Fe3O4. (Magnetite) : In Fe3O4, Fe3+ and Fe2+ ions are present in the ratio 2 : 1. It may be considered as having the composition FeO.Fe2O3. In Fe3O4 , the oxide ions are arranged in ccp . Fe2+ ions occupy octahedral voids while Fe3+ ions are equally distributed between octahedral and tetrahedral voids.
MgFe2O4 also has a structure similar to magnetite. In this Mg2+ ions are present in places of Fe2+ ion in Fe3O4.
SPINELS AND FERRITES
A spinel is an important class of oxides consisting of two types of metal ions with oxide ions arranged in cubic-closest packing. The normal spinel has one eighth of the tetrahedral holes occupied by divalent metal ion (A2+) and one half of the octahedral holes occupied by trivalent metal ion (B3+). In a unit cell , we will have :
Number of divalent metal ions A2+ = (1/8) 8 = 1
Number of trivalent metal ions B3+ = (1/2)4 = 2
Number of O2- ions (oxide) = 4
Hence formula of normal spinel is AB2O4
Examples are ZnAl2O4, MgAl2O4 and ZnFe2O4
Ferrites are compounds having the general formula AFe2O4 where A is a divalent cation such as Mg2+ or Zn2+. These are obtained by replacing Fe2+ ions in Fe3O4 by divalent cations. ZnFe2O4 and MgFe2O4 are examples of ferrites. Many of the ferrites which possess spinel-type structures are important magnetic materials and are used in memory-loops in a computer.
Inverse Spinel Structure
The compounds in which the dipositive ions are present in the octahedral holes and the tripositive ions are distributed equally between tetrahedral and octahedral holes. Such compounds are called inverse spinals. The mineral magnetite (lode stone) Fe3O4 is an example of inverse spinals. Its structure can be explained as follows :
The formula Fe3O4 can be written as Fe2+ Fe23+ O42- to indicate the presence of Fe2+ and Fe3+ ions. In Fe3O4 , the O2- ions form the (FCC) close packed structure and Fe2+ ions are present in octahedral holes, whereas Fe3+ ions are equally distributed between tetrahedral and octahedral holes.
Number of O2- ions in one molecule of Fe3O4 = 4
Therefore, the number of octahedral holes = 4
Number of tetrahedral holes = 8
Number of Fe2+ ions in octahedral holes = 1
Number of Fe3+ ions in octahedral holes = 1
Number of Fe3+ ions in tetrahedral holes = 1
Therefore , in four octahedral holes only two ions (one each Fe2+ and Fe3+) are placed and in eight tetrahedral holes only one ion (Fe3+) is placed. Thus, again we find that only one half of the octahedral holes are occupied and one eighth of the tetrahedral holes are occupied. As Fe2+ and Fe3+ both are present in octahedral holes, an electron can jump from Fe2+ to an Fe3+ that is also in an octahedral site. This give rise to electrical conductivity,
intense light absorption responsible for black colour and frromagnetism arising from magnetic interaction between structurally equivalent ions. Examples of inverse spinal structure :
MgFe2O4, MgIn2O4 etc.
TABLE : Some Ionic Solids
Crystal Structure | Brief description and examples | Co-ordination number | Number of formula units per unit cell |
1. Rock salt (NaCl- type) | Cl- ions in ccp , Na+ ions occupy all the octahedral voids ( or vice versa) Examples : Halides of Li, Na, K and Rb, AgCl, AgBr, NH4Cl etc. | Na+ - 6 Cl- - 6 | 4 |
2. CsCl-type | Cl- ions at the corners of a cube and Cs+ ions in the cubic void ( or vice versa) Examples : CsCl, CsBr, CsI etc. | Cs+ - 8 Cl- - 8 | 1 |
3.Zinc-blende (ZnS-type) | S2- ions in ccp, Zn2+ ions occupy alternate tetrahedral voids, i.e., only half of the total number of tetrahedral voids are occupied. Examples : ZnS, CuCl, CuBr, CuI, AgI, etc. | Zn2+ - 4 S2- - 4 | 4 |
4.Fluorite structure (CaF2-type) | Ca2+ ions (+ ve ions) in ccp and F- ions (-ve ions ) in all tetrahedral voids. Examples : CaF2, SrF2, BaF2, BaCl2, PbF2 etc. | Ca2+ - 8 F- - 4 | 4 |
5. Antifluorite structure. (Na2O-type) | Negative ions in ccp and positive ions in all the tetrahedral voids. Examples : Na2O, Li2O, K2S. | Na+ - 4 O2- - 8 | 4 |
IMPERFECTIONS OR DEFECTS IN SOLIDS
Ionic crystals have well defined order of constituent ions. An ionic crystal which has the same lattice points throughout the whole of the crystal is known as ideal crystal. However, such ideal crystals exist only at absolute zero(0 K) temperature. At any temperature above 0 K , the crystals have some departure from complete ordered arrangement. Any deviation from completely ordered arrangement in a crystal constitutes a disorder or a defect. The crystal may have additional defects due to the presence of some impurities. The term disorder or imperfection is generally used to denote departure from perfectly ordered state of the constituents of the crystals. Many properties of crystalline solids such as electrical conductivity and mechanical strength can be explained in terms of imperfections. Imperfections not only modify the properties of solids but also gives rise to new properties.
There are commonly two types of imperfections:
i) Electronic imperfections
ii) Atomic imperfections
1. Electronic Imperfections
This corresponds to defects in ionic crystals due to electrons. The perfectly ionic or covalent crystals at 0 K have electrons present in fully occupied lowest energy states. But at higher temperatures (above 0 K) , some of the electrons may become free to move and therefore , they are responsible for the electrical conductivity. The bonds from which electrons have been removed become electron deficient and these are referred to as holes. The holes give rise to electronic imperfections. These electronic imperfections are responsible for electrical conductance.
2. Atomic Imperfection or Point defect
These corresponds to defects in ionic crystals due to atoms or ions. If imperfections or defects in a crystal are caused by a departure from the periodic arrangement in the vicinity of an atom or a group of atoms, the imperfections are called point defects.
The point defects in crystal may be discussed under two heads.:
i) Defects in stoichiometric crystals.
ii) Defects in non-stoichiometric crystals.
1. Point defects in Stoichiometric crystals
Stoichiometric Compounds are those in which the number of positive and negative ions are exactly in the ratios indicated by their chemical formulae. For simplicity, we can consider the compounds of the type AB having A+ and B- ions. The defects in these type of compounds are called stoichiometric defects. In these compounds , two types of defects are generally observed. They are :
1) Schottky Defect. 2) Frenkel Defect.
1. Schottky defect
Shottky defect arises if some of the atoms or ions are missing from their normal lattice sites. The lattice sites which are unoccupied are called lattice vacancies or holes. Since the crystal is to remain electrically neutral, equal number of cations and anions are missing. The ideal AB crystal is shown in Fig 38.
Fig 38 An ideal crystal
The existence of two holes one due to a missing cation and the other due to a missing anion is shown in Fig 39.
Fig 39 The Schottky defect in crystals
It may be noted that the crystal , with Schottky defect is electrically neutral because the number of missing cations and anions is the same.
Conditions causing Schottky defects
This type of defect is usually observed in strongly ionic compounds having :
i) high co-ordination number, and
ii) ions ( cations and anions ) of almost similar sizes.
For example NaCl and CsCl ionic solids have Schottky defects.
2. Frenkel defect
This defect arises when an ion is missing from its own position and occupies an interstitial site between the lattice points. The existence of one hole due to missing a cation from its normal position and occupying an interstitial position is shown in Fig 40.
Fig 40 A Frenkel defect
In this case also , the crystal remains electrically neutral because the number of anions and cations remains the same.
Conditions causing Frenkel defects
This defect generally occurs in compounds in which :
i) Co-ordination number is low.
ii) Anions are much larger in size than the cations.
In pure alkali metal halides, these defects are not very common because the ions cannot get into interstitial positions due to their large sizes. These defects can be found in silver halides such as AgCl because of the small size of the Ag+ ion, it can go into the interstitial sites.
The above two defects are called intrinsic defects or thermodynamic defects.
Consequences of Schottky and Frenkel defects
Schottky and Frenkel defects in crystals leads to the following consequences :
i) Because of the presence of these defects, the electrical conductivity of crystals increases.
ii) Due to the presence of holes in the crystal, its density decreases. However, the density decreases only for crystals having Schottky defects.
iii) The presence of 'holes' also decreases the lattice energy or the stability of the crystal.
iv) The Frenkel defects tend to increase the dielectric constant of the crystals.
Difference between Schottky and Frenkel defects
Schottky defect | Frekel defect |
(i) It is due to equal number of cations and anions missing from the lattice sites. (ii) This results in the decrease in the density of the crystal. (iii) This type of defect is found in highly ionic compounds with high co-ordination number and having cations and anions of similar sizes , e.g. NaCl, CsCl etc. | (i) It is due to missing of ions (usually cations) from the lattice sites and these occupy the interstitial sites. (ii) It has no effect on the density of the crystal. (iii) This type of defect is found in crystals where the difference in the size of cations and anions is very large e.g. AgCl, AgBr, ZnS etc. |
POINT DEFECTS IN NON-STOICHIOMETRIC COMPOUNDS
The compounds in which the ratio of positive and negative ions present in the compound differs from that required by ideal chemical formula of the compound called non-stoichiometric compounds. The defects in these compounds are called non-stoichiometric defects. Vanadium oxide , for example has the formula VOx where x lies between 0.6 and 0.13. Similarly, iron(II) oxide samples contain more oxygen atoms than iron atoms while samples of zinc oxide ZnO usually has excess of zinc atoms than oxygen atoms. There are two types of defects known as metal excess defects and metal deficiency defects.
1. Metal excess defects
In these defects , the positive ions are in excess. These arise due to the following two ways
(i) Anion vacancies : In this case, negative ions may be missing from their lattice sites leaving holes in which the electrons remain trapped to maintain the electrical neutrality. This is shown in Fig 41.
Fig 41 Metal excess defects due to anion vacancy
Thus there is an excess of positive (metal) ions although the crystals as a whole is electrically neutral. For example, in alkali metal halides, anion vacancies are produced when alkali metal halide crystals are heated in the atmosphere of the alkali metal vapours. The metal atoms get deposited on the surface of the alkali metal halide crystal, halide ions move into the surface and combine with metal atoms. The electrons thus produced by the ionisation of the metal atoms, diffuse into the crystal and get trapped at the anion vacancies. These trapped electrons in the anionic vacancies are referred to as F-centres and are responsible for interesting properties. Solids containing F-centres ( farben means colour) are paramagnetic because the electrons occupying the vacant sites are unpaired. Moreover such solids when irradiated with light become photo conductors. Non-stoichiometric compounds having metal atoms in excess contain free electrons and if these electrons migrate they conduct an electric current. Thus they are n-type semiconductors. The free electrons in these compounds may be excited to higher energy levels by absorption of radiations in the visible region of the spectrum, thus imparting colour to such compounds. For example, the excess of potassium in KCl makes the crystal appear violet, excess of lithium in LiCl makes the crystal appear pink.
ii) Excess cations occupying Interstitial sites
In this case , there are extra positive ions occupying interstitial sites and the electrons in another interstitial site to maintain electrical neutrality. This is shown in Fig 42
Fig 42 Metal excess defects due to extra cation
The defect may be visualised as the loss of non-metal atoms which leave their electrons behind. The excess metal ions occupy interstitial positions. The common example is zinc oxide (ZnO). Zinc oxide loses oxygen reversibly at high temperatures and turns yellow in colour due the following reaction.
ZnO ® Zn2+ + ½ O2 + 2 e
The excess of Zn2+ions thus formed get trapped in interstitial sites and equal number of electrons are trapped in the neighbouring interstial sites to balance the electrical charge. These electrons give rise to enhanced electrical conductivity.
Consequences of Metal excess defects
1. The crystals with excess defects conduct electricity due to the presence of free electrons. These compounds are also called n-type semi-conductors, since the current is carried by electrons in the normal way.
2. The crystal with metal excess defects are generally coloured. This is due to the presence of free electrons. For example, non-stoichiometric sodium chloride is yellow, non-stoichiometric potassium chloride is lilac.
2. Metal deficiency defects
These contain less number of positive ions than negative ions. These arise due to two ways:
(i) Cation vacancies
In some cases , the positive ions may be missing from their lattice sites. The extra negative charge may be balanced by some nearby metal ion acquiring two positive charges instead of one. This is shown in Fig 43. This type of defect is possible in metal which show variable oxidation states. The common examples of compounds having this defect are ferrous oxide, nickel oxide etc. In case of iron pyrites (FeS ), for example, two out of three ferrous ions in a lattiice may be converted into Fe3+ state and the third Fe2+ ion may be missing from its lattice site.
Fig 43 Metal deficiency defect due
to cation vacancy
(ii) Extra Anions Occupying Interstitial Sites
-Fig 44 Metal Deficient defect due to extra anion
In this case the extra anions may be occupying interstitial positions. The extra negative charge is balanced by extra charges ( oxidation of equal number of cations to higher oxidation states) on the adjacent metal ions. This is shown in Fig 44. Such types of defect is not common because the negative ions are usually very large and they cannot easily fit into the interstitial sites.
Consequences of Metal deficiency defects