Crystal Structure of Some Cubic Ionic Solids

Structure of Sodium Chloride (NaCl)

The formation of NaCl crystal can be explained in the following manner. The two ions form one ion-pair of opposite charges by the electrostatic force of attraction. Each of the ions has a strong residual field around it and will naturally attract another ion-pair much in the same way as two magnets attract each other, i.e., a cluster is formed. Two clusters combine together to give a unit cell. Finally large number of unit cells unite together to form threedimensional cubic crystal.

Piling up of ion pairs in stages: (a) shows an ion-pair, (b) two ion-pairs, (c) four ion-pairs of sodium chloride

An examination of NaCl crystal (figure) makes the following points clear:

(i) Each ${Na}^{+}$ ion is surrounded by six ${Cl}^{-}$ ions at the corners of a regular octahedron, and similarly each ${Cl}^{-}$ ion is surrounded by six ${Na}^{+}$ ions. It is, therefore, termed as $6:6$ arrangement. The radius ratio $\left\lbrack r_{{Na}^{+}}/r_{{Cl}^{-}} = 0.95/1.81 = 0.524 \right\rbrack$ suggests that coordination number of each ion is six.

(ii) In the octahedral structure, ${Cl}^{-}$ ions may be regarded as having a cubic closed packed (ccp) arrangement in which all octahedral holes are filled by ${Na}^{+}$ ions.

(iii) The distance between two adjacent ions of different kinds is equal to $2.815\text{Å}$ . Thus, two ions are not touching each other as the sum of their ionic radii is $2.76\text{Å}(0.95 + 1.81 = 2.76)$ . This type of structure is possessed by most of the alkali metal halides ( $KCl,Nal,RbF,Rbl$ ), alkaline earth metal oxides and AgF , $AgCl,AgBr,{NH}_{4}Cl,{NH}_{4}Br$ , etc.

The unit cell of sodium chloride had four sodium ions and four chloride ions.

(a) Number of sodium ions $= 12$ (at the edge centres) $\times \frac{1}{4} + 1$ (at body centre) $= 4$

(b) Number of chloride ions $= 8$ (at the corners) $\times \frac{1}{8} + 6$ (at face centres) $\times \frac{1}{2} = 4$

Structure of Zinc Sulphide (ZnS)

ZnS crystal has two types of structures:

(1) Zinc blende structure, (2) Wurtzite structure.

(1) Zinc blende structure: The unit cell of zinc blende structure is shown in Fig. In this structure, the zinc ions are represented by hollow circles and the sulphide ions are represented by solid circles. The salient features of this structure are as follows:

(i) The sulphide ions are arranged in ccp arrangement, i.e., at the corners of the 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 close packed lattice, this means that there are two tetrahedral site per $S^{2 -}$ ions. In this arrangement, only half of the tetrahedral site are occupied by ${Zn}^{2 +}$ ions so that each $S^{2 -}$ ions has one ${Zn}^{2 +}$ ion. Therefore, the stoichiometry of the compound is $1:1$ .

Tetrahedral arrangement around $S^{2 -}$ ion

Unit cell representation of zinc blend structure. Hollow circles represent ${Zn}^{2 +}$ ions and solid circles represent $S^{2}$ ions.

(iv) Each zinc ion is surrounded by four sulphide ions which are disposed towards the corners of a regular tetrahedron. Similarly, each $S^{2 -}$ ion is surrounded by four ${Zn}^{2 +}$ ions. The arrangement around one ${Zn}^{2 +}$ ion and one $S^{2 -}$ ion is also shown in fig. Thus, the coordination numbers of ${Zn}^{2 +}$ and $S^{2 -}$ ions in zinc blende are described as $4:4$ .

(v) In zinc sulphide, the radius ratio $r^{+}/r^{-}$ is 0.40 . This suggests that the coordination number is 4 . This supports the structure represented in Fig.

Zinc blende structure is also adopted by some other compounds such as $ZnO,CuCl,CuBr,Cul$ , etc.

(2) Wurtzite structure: It has hexagonal close-packed (hcp) structure. $S^{2 -}$ ions adopt (hcp) arrangement and ${Zn}^{2 +}$ ions occupy half of the tetrahedral sites.

Structure of Cesium Chloride (CsCI)

The unit cell of CsCl is shown in figure below. In this structure, cesium ions are represented by hollow circles whereas chloride ions are represented by solid circles. The structure of CsCl may be described as follows:

1. The ${Cl}^{-}$ ions form the simple cubic arrangement and the ${Cs}^{+}$ ions occupy the cubic interstitial sites. In other words, the ${Cl}^{-}$ ions may be regarded as lying at the corners of a cube and cesium ions are present at the body centre of each cube.

2. Each ${Cs}^{+}$ ion is surrounded by eight ${Cl}^{-}$ ions which are disposed towards the corners of a cube. This is clear from Fig. (a). When the unit cell is extended as shown in Fig. (b), it becomes clear that each ${Cl}^{-}$ ion is

The above structure may be described as body centred cubic. But this would be a misnomer because the body centred cubic structure is one which has the same atoms at the body centre as well as the corners. But in CsC l, the atoms at the corners are different from the atoms at the centre. This type of structure is adopted by CsBr, CsF, TlCl, TlBr, tec.

Since coordination numbers of ions in CsCl (8:8) are higher than those in NaCl (6:6), we might expect that cesium chloride lattice would be more stable since each ion would have more ions of opposite charge as its neighbours. This is actually the case.

- The cesium chloride lattice is 1% more stable than sodium chloride lattice.

Hence question arise that if CsCl lattice is more stable then why do the halides of Li, Na, K and Rb not have this lattice? The answer lies in the radius ration effect. The radius ration in the halde3s of Li, Na, K and Rb lies between 0.73 and 0.41 and therefore, the coordination number has to be 6. If an attempt is made to pack 8 halide ions around a smaller positive ion, say Na+ ion then some of the negative ions will touch each other sooner than they can approach a positive ion, as shown in Fig. The lattice would thus be unstable. This difficulty can be overcome by packing 6 negative ions around the central positive ion. Hence, the coordination number of 6 is preferred.

Structure of Calcium Fluoride (CaF2)

The unit cell of CaF2 is shown in figure. The calcium ions are represented by hallow circles and fluride ions are represented by circle. The CaF2 structure may be described as follows.

1. The ${Ca}^{2 +}$ ions are arranged in ccp arrangement so that there are ${Ca}^{2 +}$ ions at all the corners and at the centre of each face of the cube. The fluoride ions occupy all the tetrahedral sites.

2. Since there are two tetrahedral sites for each ${Ca}^{2 +}$ ion and $F^{-}$ ions occupy all the tetrahedral sites, there will be two $F^{-}$ ions for each ${Ca}^{2 +}$ ion. Thus, the stoichiometry of the compound is $1:2$ .

3. In this arrangement, each $F^{-}$ ion is surrounded by four ${Ca}^{2 +}$ ions whereas each ${Ca}^{2 +}$ ion is surrounded by eight $F^{-}$ ions. Thus, the coordination numbers of ${Ca}^{2 +}$ and $F^{-}$ ion are 84. This structure is adopted by a large number of fluorides such as ${BaF}_{2},{SrF}_{2},{PbF}_{2},{HgF}_{2},{CuF}_{2}$ etc.

Antifluorite Structure or Structure of ${Na}_{2}O$

In antifluorite structure the anions are arranged in cubic close packing (ccp) while the cations occupy all the tetrahedral voids, e.g., ${Na}_{2}O$ .

(i) The $O^{2 -}$ ions constitute a ccp type of lattice and the ${Na}^{+}$ ions occupy all the tetrahedral voids.

(ii) The oxide ion, $O^{2 -}$ , is in contact with four oxide ions. Thus, ${Na}_{2}O$ has $4:8$ co-ordination.

The above mentioned crystal structures could be summarized as follows:

Crystal Structure Units	Brief description and Examples	Co-ordination Number	No. of formula per Unit cell
1. Rock Salt (NaCl - type)	${Cl}^{-}$ ions in C.C.P, ${Na}^{+}$ ions occupy all the octahedral voids. Example: Halides of $Li,Na,K$ and Rb , $AgCl,AgBr,{NH}_{4}Cl$ etc.	$N a^{+} - 6$	4
2. CsCl - type	ions at the corner of a cube and (B.C.C.) structure $Cs +$ ions in the cubic void	$C s^{+}$ - 8	1
3. Zinc. Blende $S^{- 2}(ZnS)$	ions in C.C.P., ${Zn}^{+ 2}$ ions occupy alter tetrahedral voids i.e. only half of the total number of tetrahedral voids are occupied Examples: $ZnS,CuCl,CuBr$ , Cul, Agl etc.	$Z n^{+ 2}$ - 4 $S^{- 2} - 4$	4
4. Fluorite Structure ( ${CaF}_{2}$ - type)	ions (+ve ions) in C.C.P. and $F^{-}$ ions (-ve ions) in all the tetrahedral voids. Example: ${CaF}_{2},{SrF}_{2},{BaF}_{2},{BaCl}_{2}$ etc.	$C a^{+ 2}$ - 8 $F^{-} - 4$	4
5. Antifluorte Structure ( ${Li}_{2}O$ - type)	Negative ions means $O^{- 2}$ in CCP and Positive ${Li}^{+}$ ions in all the tetrahedral voids. Example: ${Na}_{2}O$	$N a^{+}$ - 4 $O^{- 2} - 8$	4

Structure of ${Fe}_{3}O_{4}$ (Magnetite)

In ${Fe}_{3}O_{4},{Fe}^{3 +}$ and ${Fe}^{2 +}$ ions are present in the ratio $2:1$ and may be considered as ${FeO}^{\circ}.{Fe}_{2}O_{3}$ . In ${Fe}_{3}O_{4}$ oxide ions are arranged in ccp. ${Fe}^{2 +}$ ions occupy octahedral voids while ${Fe}^{3 +}$ ions are equally distributed between octahedral and tetrahedral voids.

${MgFe}_{2}O_{4}$ also has similar structure to magnetite, in which ${Mg}^{2 +}$ ions are present in place of ${Fe}^{2 +}$ ion in ${Fe}_{3}O_{4}$ .

Normal Spinel Structure

In spinel, a mineral ${MgAl}_{2}O_{4}$ , 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 ${ZnFe}_{2}O_{4}$ ) also possess spinel structure. These are very important magnetic materials and are used in telephones and memory loops in computers.

Magnetite also possess spinel structure. Here, divalent ions ( ${Fe}^{2 +}$ ) are in octahedral voids while the trivalent ions are in a set of tetrahedral voids.

Structure of Calcium Carbide

The structure of ${CaC}_{2}$ is similar to that of sodium chloride in which sodium ions are replaced by calcium ions while chloride ions by carbide ions. The carbon atoms in carbide are associated in pairs which are aligned in parallels. Unlike sodium chloride, the cubic symmetry is distorted in calcium carbide.

The structure of ${FeS}_{2}$ has been found to be similar to that of ${CaC}_{2}$ . However, unlike ${CaC}_{2}$ , the $S_{2}$ units in $FeS_{2}$ , do not align in parallels.

Examples of Some Covalent Solids

Silicates

1. Simple ortho silicates

(i) $\ {SiO}_{4}^{- 4}$ tetrahedron do not share oxygen atom with one another and exist as discrete ${SiO}_{4}^{- 4}$ orthosilicate anion.

(ii) $\ {SiO}_{4}^{- 4}$ ion is linked to $M^{+ 2}$ ions through co-ordinate bond. e.g., ${Mg}_{2}{SiO}_{4},{Fe}_{2}{SiO}_{4}$ ,

2. Pyrosilicates or Islands

(i) Two ${SiO}_{4}^{- 4}$ tetrahedra share a common oxygen.

(ii) Island structure have formula $\left( {Si}_{2}O_{7} \right)^{- 6}$ .

3. Ring or Cyclic Silicate Anion:

(i) Two oxygen atoms of each tetrahedron are shared with other. e.g., $\left( {Si}_{3}O_{9} \right)^{- 6}$ and $\left( {Si}_{6}O_{18} \right)^{- 12}$ . General formula $\left( {Si}_{n}O_{3n} \right)^{- 2n}$

4. Infinite Chain Silicates

(i) Pyroxenes: Which contain single strand chain of composition $\left( {SiO}_{3}^{- 2} \right)_{n}$ .

$\text{~e.g., Diopside~}\left( {MgCaSi}_{2}O_{6} \right)$

(ii) Amphiboles: Which contains double strand cross linked chains of $\left( {Si}_{4}O_{11}^{- 6} \right)_{n}$ . e.g., asbestos minerals.

5. Infinite sheet silicates

(i) Three oxygen of each ${SiO}_{4}$ tetrahedron are shared with each other. General formula $\left( {Si}_{2}O_{5}^{- 2} \right)_{n}$ . e.g., Clay.

6. Framework silicate

(i) Four oxygen atoms of each ${SiO}_{4}$ tetrahedron are shared with one another.

(ii) Three dimensional network results by above arrangement. General formula of framework silicate is ${SiO}_{2}$ . e.g., Quartz and Zeolite

Structure of Diamond

Diamond has the same space lattice as that of zinc sulphide with some differences

In diamond carbon atoms occupy the place of zinc and sulphur atoms in zinc sulphide.

Diamond is considered to be built up of two interesting or interpenetrating face centred lattice related to one another, like zinc and sulphur lattices in zinc sulphide.

Carbon atom in diamond is ${sp}^{3}$ hybridized having tetrahedral structure with four carbon atoms located at four corners of tetrahedral at a distance of $1.54\text{Å}$ and bond angle $109^{\circ}28^{'}$ . Since no free valence electrons are left on carbon atom, the diamond possesses a giant molecule having cage like structure and bad conductor nature. The giant molecular structure of diamond is responsible for hardness, high density and high m.pt. $\left( 3600^{\circ}C \right)$ . It should be noted here that diamond is the hardest known substance The high refractive index of diamond is responsible for its glitter when they are properly cut.

The diamond crystal is therefore, regarded as a three dimensional giant molecule, a polymer of carbon with fcc structure with alternate four out of eight tetrahedral holes occupied by carbon atoms as well as $C - C$ bond distance being 1.54 Â.

Structure of Graphite

The structure of graphite crystal is an entirely different type with atoms held in position by Vander Waal's forces of attraction.

Carbon atom in graphite is $sp^{2}$ hybridized having two dimensional sheet like structure. Each carbon atom make use of three of the four valence electrons, forming three covalent bonds with other carbon atoms leaving a singly occupied p-orbital on each carbon atom at right angles to the plane of covalent $\sigma$ bonds. These p-orbitals overlap to form a giant $\pi$ orbital lying below and above the plane of carbon ring.

Thus, graphite possesses layers of atoms having hexagonal nature ( $C - C$ bond length $1.42\text{Å}$ ). These layers are held together by weak Vander Waal's forces and hence they may slide over each other to show lubricating properties in graphite. The pi-bonding electrons can move through the layers and this accounts for good conductor nature of graphite.

Weak binding between the layers results the graphite crystals slippery and flaky conferring on graphite its valuable lubricating property.

The energy required to separate network solids into their constituent atoms is usually large, i.e., about 836.8 kJ ${mol}^{- 1}$ . Therefore, graphite has very high melting point.

Illustration 9: KCI crystallizes in the same type of lattice as NaCI does. Given that

$\frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}} = 0.5\ \text{~and~}\ \frac{r_{{Na}^{+}}}{r_{K^{+}}} = 0.7$

Calculate (a) the ratio of the side of the unit cell for KCl to that for NaCl and (b) the ratio of density of NaCl to that of KCl.

Solution: $\ NaCl$ crystallizes in the fcc unit cell, so

$r_{{Na}^{+}} + r_{Cl} = \frac{a}{2}$

Where $a$ is the edge length of the cell

Thus, $\ \frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}} = 0.5\$ and $\ \frac{r_{{Na}^{+}}}{r_{K^{+}}} = 0.7$

We will have

$\frac{r_{{Na}^{+}}}{r_{Cl}} + 1 = 0.5 + 1 = 1.5$

$\frac{r_{{Na}^{+}} + r_{Cl}}{r_{{Cl}^{-}}} = 1.5$ …………….(i)

$\frac{r_{{Na}^{+}}}{r_{{Cl}^{-}}}/\frac{r_{{Na}^{+}}}{r_{K^{+}}} = \frac{0.5}{0.7}\frac{r_{K^{+}}}{r_{Cl}} = \frac{0.5}{0.7}$

$\frac{r_{K^{+}} + r_{{Cl}^{-}}}{r_{{Cl}^{-}}} = \frac{0.5}{0.7}\ + 1 = \frac{1.2}{0.7}$ …………(ii)

From the Equation (i) and (ii) the relation

$\frac{r_{{Na}^{+}} + r_{{Cl}^{-}}}{r_{K^{+}} + r_{{Cl}^{-}}} = \frac{1.5}{\frac{1.2}{0.7}}$

$\text{or~}\ \frac{a_{KCl}}{a_{NaCl}} = \frac{1.2}{0.7 \times 1.5}\text{~or~}\ \frac{a_{KCl}}{a_{NaCl}} = 1.143$

$\text{Density~}(D) = \frac{ZM}{a^{3} \times N_{0}}\ \left( V = a^{3} \right)$

We will have

$\frac{D_{NaCl}}{D_{KCl}} = \frac{\frac{M_{NaCl}}{\left( a_{NaCl} \right)^{3}}}{\frac{M_{KCl}}{\left( a_{KCl} \right)^{3}}}$

$= \frac{M_{NaCl}}{M_{KCl}} \times \left( \frac{a_{KCl}}{a_{NaCl}} \right)^{3} = \frac{58.5}{74.5} \times (1.143)^{3} = 1.1726$

Illustration 10:

Gold crystallizes in a face centered cubic unit. If the edge length is ' $a$ ', calculate the closest distance between the gold atom and an impurity atom if it occupies (a) a tetrahedral void (b) an octahedral void without any change in volume of the unit cell.

Solution:

(a) If $A,B$ , are the gold atoms and C is impurity atom, r and $r^{'}$ are the radius of gold atom and impurity atom respectively, we have by sin rule

$\frac{sin109.47}{2r} = \frac{sin35.265}{r + r^{'}}$

$\Rightarrow r^{'} = 0.225r$

$\text{Now~}\sqrt{2}a = 4r \Rightarrow r = \frac{\sqrt{2}a}{4}$

$\left( r + r^{'} \right) = \frac{\sqrt{2}a}{4} + 0.225 \times \frac{\sqrt{2}a}{4}$

(b) For octahedral void

$A B = \frac{\sqrt{8}}{2} r$

$A C = r + r^{"} (r^{"} = r a d i u s o f i m p u r i t y a t o m)$

$r^{'} = \frac{\sqrt{8}}{2} r - r = 0.414 r$

$r + r^{"} = 1.414 r = 1.414 \times \frac{\sqrt{2} a}{4} = \frac{a}{2}$

Illustration 11 :

An element A crystalizes in FCC from with atomic radius of 1.5 A. Another element B also crystalizes in FCC form with atomic radius of 1.15 A. An alloy of both the elements is formed which has a BCC structure. Find (i) the density of the alloy and (ii) the percentage change in volume. (Given atomic mass of A = 40 and atomic mas of B = 60)

Solution:

For alloy

In body centred cavity

$\frac{r a d i u s o f a t o m a t b o d y c e n t r e}{r a d i u s o f a t o m a t c o r n e r} \frac{1.15}{1.5} = 0.76$

This means that the atom at body centre is in contact with corner atoms but corner atoms are not in contact with each other.

$∴ \sqrt{3} a = a (1.5 + 1.15) = 5.3$

$a = \frac{5.53}{1.732} = 3.06 A$

Volume of unity cell of alloy $= {(3.06 \times 10^{- 8})}^{3} = 28.65 \times 10^{- 24} c m^{3}$

Density of alloy $= \frac{(1 \times 40) + (1 \times 60)}{6.023 \times 10^{23} \times {(3.06 \times 10^{- 8})}^{3}} = 5.76 g / c m^{3}$

For element A

$\sqrt{2} a = 4 r A$

$∴ a = \frac{4 \times 1.5}{\sqrt{2}} = 4.24 A$

Volume of unit cell of element
$A = {(4.24 \times 10^{- 8})}^{3} = 7.62 \times 10^{- 23} c m^{3}$

For element B,

$\sqrt{2} a = 4 r B$

$∴ a = \frac{4 \times 1.5}{\sqrt{2}} = 3.25 A$

$∴ V o l u m e o f u n i t c e l l o f e l e m e n t B = {(3.25 \times 10^{- 8})}^{3} = 3.43 \times 10^{- 23} c m^{3}$

$T o t a l v o l u m e o f A + B = (7.62 \times 10^{- 23}) + (3.43 \times 10^{- 23}) = 11.05 \times 10^{- 23} c m^{3}$

$∴$ Reduction in volume $= (11.05 \times 10^{- 23} - 2.865 \times 10^{- 23}) c m^{3}$

$∴$ Percentage reduction $= \frac{8.185 \times 10^{- 23}}{11.05 \times 10^{- 23}} \times 100 = 74.07 %$

Illustration 12:

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 a cation C+ having a radius of 180 pm be slipped into the tetrahedral site of crystal A+ B- ?

Solution:

The limiting radius for octahedral and tetrahedral holes are:

$\frac{r a d i u s o f c a t i o n (r^{+})}{r a d i u s o f a n i o n (r^{-})} = 0.414$ …………………..(Octahedral hole)

And $\frac{(r^{+})}{(r^{-})} = 0.225$ …………………………………………(Tetrahedral hole)

(a) In NaCl crystal, ${Cl}^{-}$ ions are cubic closed packed and ${Na}^{+}$ ion occupy all the octahedral holes. Since the radius is 250 pm , therefore, the ideal radius of cation to fit into the octahedral hole,

$r^{+} = 0.414 \times r^{-} = 0.414 \times 250pm = 103.5pm$

(b) For a cation to fit exactly (ideally) in tetrahedral hole, the ideal radius of cation,

$r^{+} = 0.225 \times r^{-} = 0.225 \times 250pm = 57.25pm$

But the radius of cation, ' $C^{+}$ ', which is to be slipped into the tetrahedral hole is 180 pm . Thus, radius of cation ' $C^{+}$ ' is much larger than the ideal radius of cation ( 56.25 pm ). Hence, the cation ' $C^{+}$ ' cannot be slipped into the tetrahedral site of crystal. If the cation ' $C^{+}$ ' is slipped into the tetrahedral hole, the ${Cl}^{-}$ ion will no longer remain in cubic closed packed structure. The new structure will thus, be unstable.

Illustration 13:

$A$ solid $AB$ has the NaCl structure, if the radius of the cation is 100 pm (picometer). What is the radius of the anion $B$ ?

Solution:

The co-ordination number ( CN .) of ${Na}^{+}$ ions in NaCl is 6, so the limiting radius ratio is 0.414 to 0.732, i.e.,

$\frac{\text{~radius of~}A^{+}}{\text{radius of~}B^{-}} = 0.414\text{~to~}0.732$

$\text{~or radius of~}B^{-} = \frac{\text{~radius of~}A^{+}}{0.414\text{~to~}0.732} = \frac{100pm}{0.414\text{~to~}0.732} = 241.5\text{~to~}136.6pm$

Thus, the radius of anion $B^{-}$ may vary between 136.6 pm and 241.5 pm .

Imperfections In Solid

Any deviation from the perfectly ordered arrangement constitutes a defect or imperfection. These defects sometimes called thermodynamic defects because the number of these defects depends on the temperature. Crystals may also possess additional defect due to the presence of impurities. Imperfection not only modify the properties of solids but also give rise to new properties.

1. Electronic imperfection:

Generally, electrons are present in fully occupied lowest energy states. But at high temperatures, some of the electrons may occupy higher energy state depending upon the temperature.

For example, in the crystals of pure Si or Ge some electrons are released thermally from the covalent bonds at temperature above 0 K . These electrons are free to move in the crystal and are responsible for electrical conductivity. This type of conduction is known as intrinsic conduction.

The electron deficient bond formed by the release of an electron is called a hole. In the presence of electric field the positive holes move in a direction opposite to that of the electrons and conduct electricity.

2. Atomic imperfection:

The compounds in which the number of irregularity present in the arrangement of atoms or ions are called atomic imperfections. It is of two types:

(i) Stoichiometric Defects:

Those compounds in which the number of positive and negative ions are exactly in the ratio indicated by their chemical formula are called stoichiometric compounds. eg. $NaCl(1:1)$ . These solids show following types of defects:

(a) Schottky defect:

This type of defect is created when same number of positive ion and negative ion are missing from their respective positions leaving behind a pair of holes. Schottky defect is more common in ionic compounds with high co-ordination number and where the size of positive and negative ions are almost equal.

The presence of large number of schottky defects in crystal results in significant decrease in its density.

Eg. $NaCl,KCl,CsCl,KBr$ etc.

(b) Interstitial Defect:

This type of defect is caused due to the presence of ions in the normally vacant interstitial sites in the crystal.

(c) Frenkel Defect:

This type of defect is created when an ion leaves its correct lattice site and occupies an interstitial site. Frenkel defects are common in ionic compounds which have low co-ordination number and in which there is large difference in size between positive and negative ions. Eg. ZnS, AgCl, AgBr, Agl etc.

(ii) Non Stoichiometric Defect:

There are many compound in which the ratio of positive and negative ions present in the compound differs from that required by ideal formula of the compound. Such compounds are called Non stoichiometric compounds.

eg. ${VO}_{x}$ (Where x can vary between 0.6 to 1.3.)

In these compounds balance of positive and negative charges is maintained by having extra electrons or extra positive charge.

So, these defects are of following types:

(a) Metal excess defects due to anion vacancies: A compound may have excess metal ion if a negative ion is absent from its lattice site, leaving a hole which is occupied by electron to maintain electrical neutrality.

The holes occupied by electrons are called $F$ - centres and are responsible for the colour of the compound.

Eg. - The excess sodium in NaCl makes the crystal appears yellow.

Excess potassium in KCl makes it violet.

Excess lithium in LiCl makes it Pink.

Greater the number of F-centres greater is the intensity of colour. This type of defects are found in crystal which are likely to possess schottky defects.

(b) Metal excess defects due to interstitial cations: It may occur if an extra positive ion is present in an interstitial site. Electrically neutrality is maintained by the presence of an electron in the interstitial site.

This type of defects are exhibited by the crystals which are likely to exhibit "Frenkel defects". Eg. Yellow colour of ZnS.

(c) Metal deficiency due to cation vacancies: The non-stoichiometric compound may have metal deficiency due to the absence of a metal ion from its lattice site. The charge is balanced by an adjacent ion having higher positive charge.

This type of defects are generally shown by compounds of transition metals.

Calculation of number of point defects

A point defect may result from:

(a) the absence of an atom (vacancy),

(b) displacement of an atom to an interstitial site (self-interstitial),

(c) presence of an impurity atom at an interstice (interstitial impurity atom),

(d) presence of an impurity atom at a lattice site (substitutional) impurity atom).

The number of Schottky defects formed per ${cm}^{3}(\text{ }ns)$ is given by

$n_{s} = Nexp\left( \frac{- W_{s}}{2kT} \right)$

Where, N is the number of sites per ${cm}^{3}$ that could be left vacant, $W_{s}$ is the work necessary to form a Schottky defect, k is the Boltzman's constant and T is the absolute temperature.

The number of Frenkel defects formed per ${cm}^{3}\left( n_{f} \right)$ is given by:

$n_{f} = \sqrt{N^{'}}exp\left( \frac{- W_{f}}{2kT} \right)$

Where, N is the number of sites per ${cm}^{3}$ that could be left vacant and $N^{'}$ is the number of alternate interstitial positions per ${cm}^{3},{\text{ }W}_{f}$ is the work necessary to form a Frenkel defect, k is the Boltzmann's constant and T is the absolute temperature.

Illustration 14:

If $NaCl$ is doped with $10^{- 3}$ mol percent of ${SrCl}_{2}$ , what is the concentration of cation vacancies?

Solution:

Doping of NaCl with ${SrCl}_{2}$ means that two ${Na}^{+}$ ions are replaced by one ${Sr}^{2 +}$ ion. Thus, one cation vacancy is produced as result of doping in one molecule of NaCl .

Thus, 100 moles of NaCl have cation vacancies after doping $= 10^{- 3}\text{ }mol$

1 mole of NaCl will have cation vacancies after doping $= \frac{10^{- 3}}{100} = 10^{- 5}$

Total no. of cation vacancies $= 10^{- 5} \times N_{0}$

$= 10^{- 5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18}$

Solid State & Surface

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