Solid State and its solutions
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G


 IIT–JEE Syllabus
Classification of solids; crystalline state, seven crystal systems (only cell parameters: a, b, c, a, b and g); packing fcc, bcc, hcp; nearest neighbours, simple ionic compounds, point defects.
 Types of Solids
 a) Crystalline solids: In a single crystal the regularity of arrangement of the pattern extends throughout the solid and all points are completely equivalent.
 b) Amorphous solids: An amorphous solid differs from a crystalline substance in being without any shape of its own and has a completely random particle arrangements, i.e. no regular arrangement. Example: Glass, Plastic
Crystalline solid  Amorphous solid 
1. The constituent particles are arranged in a regular fashion containing short range as well as long range order.  1. The constituent particles are not arranged in any regular fashion. There may be at the most some short range order only. 
2. They have sharp melting point  2. They melt over a range of temperature. 
3. They are anisotropic i.e., properties like electrical conductivity, thermal expansion, etc have different values is different direction.  3. They are isotropic i.e., properties like electrical conductivity, thermal expansion, etc have same value in different directions 
4. They undergo a cleavage.  4. They undergo an irregular cut 
(figure 1a)  (figure 1b) 
2.1 What is a crystal
A crystalline solid consist of a large number of small units, called crystals, each of which possesses a definite geometric shape bounded by plane faces. The crystals of a given substance produced under a definite set of conditions are always of the same shape.
 Unit Cells
In this topic we would be studying certain properties of a solid which depend only on the constituents of the solid and the pattern of arrangement of these constituents. To study these properties, it would be convenient to take up as small amount of the solid as possible because this would ensure that we deal with only the minimum number of atoms or ions. This smallest amount of the solid whose properties resemble the properties of the entire solid irrespective of the amount taken is called a unit cell. It is the smallest repeating unit of the solid. Any amount of the solid can be constructed by simply putting as many unit cells as required. 
 External Features of Crystals
Crystals are bounded by surfaces which are usually planar and referred as faces. Edge is formed by the intersection of our adjacent faces. The angle between any two faces is called an interfacial angle. Although the size of the faces or even shapes of the crystals of one and the same substance may vary widely with conditions of formation etc., but the interfacial angles between any two corresponding faces of the crystal remain invariably the same throughout.
 Crystal Systems
From external appearance we come across a variety of crystal forms in many shapes. But on the basis of length of axes and the axial angles, it has been possible to classify the various crystal forms into only seven fundamental systems. These are:
 Cubic
 Orthorhombic
 Tetragonal
 Monoclinic
 Triclinic
 Hexagonal
 Rhombohedral
The crystals belonging to any one system may differ in shape, in size, etc., but their axial ratios as also the axial angles will always be the same. The intercepts on the three axes are denoted by a, b and c and the axial angles by a, b and g.
Seven crystal systems : The seven crystal systems are given below.
Crystal System  Bravais Lattices  Parameters of Unit Cell  
Intercepts  Crystal angle  Example  
1. Cubic  Primitive, Face Centered, Body Centered = 3  a = b = c  a = b = g = 90^{o}  Pb,Hg,Ag,Au
Diamond, NaCl, ZnS 
2. Orthorhombic  Primitive, Face Centered, Body Centered, End Centered = 4  a ¹ b ¹ c  a = b = g = 90^{o}  KNO_{2}, K_{2}SO_{4} 
3. Tetragonal  Primitive, Body Centered =2  a = b ¹ c  a = b = g = 90^{o}  TiO_{2},SnO_{2} 
4. Monoclinic  Primitive, End Centered = 2  a ¹ b ¹ c  a = g = 90^{o},
b ¹ 90^{o} 
CaSO_{4},2H_{2}O 
5. Triclinic  Primitive = 1  a ¹ b ¹ c  a ¹ b ¹ g ¹ 90^{0}  K_{2}Cr_{2}O_{7},
CaSO_{4}5H_{2}O 
6. Hexagonal  Primitive = 1  a = b ¹ c  a = b = 90^{0}, g = 120^{o}  Mg, SiO_{2}, Zn, Cd 
7. Rhombohedral  Primitive = 1  a = b = c  a = g = 90^{o}, b ¹ 90^{o}  As, Sb, Bi, CaCO_{3} 
Total = 14 
 Bravais Lattices
Bravais (1848) showed from geometrical considerations that there are only seven shapes in which unit cells can exist. Moreover he also showed that there are basically four types of unit cells depending on the manner in which they are arranged in a given shape. These are : Primitive, Body Centered, Face Centered and End Centered. He also went on to postulate that out of the possible twenty eight unit cells (i.e. seven shapes ´ four types in each shape = 28 possible unit cells), only fourteen actually would exist. These he postulated based only on symmetry considerations. These fourteen unit cells that actually exist are called Bravais Lattices.
6.1 Primitive Cubic Unit Cell
In a primitive cubic unit cell the same type of atoms are present at all the corners of the unit cell and are not present anywhere else. It can be seen that each atom at the corner of the unit cell is shared by eight unit cells (four on one layer, as shown, and four on top of these). Therefore, the volume occupied by a sphere in a unit cell is just oneeighth of its total volume. Since there are eight such spheres, the total volume occupied by the spheres is one full volume of a sphere. Therefore, a primitive cubic unit cell has effectively one atom. A primitive cubic unit cell is created in the manner as shown in the figure. Four atoms are present in such a way that the adjacent atoms touch each other. Therefore, if the length of the unit cell is ‘a’, then it is equal to 2r, where r is the radius of the sphere. Four more spheres are placed on top of these such that they eclipse these spheres. The packing efficiency of a unit cell can be understood by calculating the packing fraction. It is defined as ratio of the volume occupied by the spheres in a unit cell to the volume of the unit cell and void fraction is given as = (1 – Packing fraction).Therefore, PF = . (This implies that 52 % of the volume of a unit cell is occupied by spheres). \VF » 0.48 
Figure 3(b) 
6.2 Body Centered Cubic Unit Cell
A Body Centred unit cell is a unit cell in which the same atoms are present at all the corners and also at the center of the unit cell and are not present anywhere else.
This unit cell is created by placing four atoms which are not touching each other. Then we place an atom on top of these four. Again, four spheres eclipsing the first layer are placed on top of this. The effective number of atoms in a Body Centered Cubic Unit Cell is 2 (One from all the corners and one at the center of the unit cell). Moreover, since in BCC the body centered atom touches the top four and the bottom four atoms, the length of the body diagonal () is equal to 4r. The packing fraction in this case is = \VF » 0.32 
Figure 4(b) 
6.3 Face Centered Cubic Unit Cell
In a fcc unit cell, the same atoms are present at all the corners of the cube and are also present at the centre of each square face and are not present anywhere else. The effective number of atoms in fcc is 4 (one from all the corners, 3 from all the face centers since each face centered atom is shared by two cubes). Since, here each face centered atom touches the four corner atoms, the face diagonal of the cube () is equal to 4r.  Figure 5(a)  
\
\VF » 0.26 
Figure 5(b)  
6.4 Hexagonal Primitive Unit Cell
Each corner atom would be common to 6 other unit cells, therefore their contribution to one unit cell would be 1/6. Therefore, the total number of atoms present per unit cell effectively is 6.
Figure 6(b) ABCD is the base of hexagonal unit cell 
Figure 6(a) 
\
Hence FE = =
\ The height of unit cell (h) =
The area of the base is equal to the area of six equilateral triangles, = . The volume of the unit cell = .
Therefore
PF = ; \VF » 0.26
Density of crystal lattice
The density of crystal lattice is same as the density of the unit cell which is calculated as
r = =
r =
Illustration 1: Iron (aFe) crystallizes in a b.c.c. system with a = 2.86Å. Atomic weight of iron is 55.85 g mol^{–1}. Calculate the density of iron.
Solution: Since, density r =
Hence n = 2, Atomic weight = 55.85
Volume = V = a^{3} = (2.861 × 10^{–8} cm)^{3}, Av. No. = 6.023 × 10^{23}
Putting the values r = 7.92 g/cm
Exercise 1: Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8Å,
b = 4.4Å and C = 7.2Å. If the molar mass is 21.76g. Calculate density of crystal.
 Close Packing of Spheres
Atoms are space filling entities and structures can be described as resulting from the packing of spheres. The most efficient, called Closest Packing, can be achieved in two ways, one of which is called Hexagonal Close Packing (Hexagonal Primitive) and the other, Cubic Close Packing (ccp or fcc). Hexagonal close packing can be built up as follows:
Place a sphere on a flat surface. Surround it with six equal spheres as close as possible in the same plane. Looking down on the plane, the projection is as shown in Fig. 7 (a). Let us call this layer as the A layer. Now form over the first layer a second layer of equally bunched spheres, Fig 7 (b), so as to nestle into the voids. It will be clearly seen that once a sphere is placed over a void, it blocks the void which are adjacent to that void. Let us call this layer as the B layer. Now, it can be clearly seen that there are two types of voids created by the B layer of spheres. If a sphere is placed on the x – type of voids, it would resemble the A layer of spheres in the sense that it eclipses the spheres of the A layer. This arrangement (i.e., ABAB……) is called the hexagonal close packing (hcp) or hexagonal primitive. On the other hand, if the spheres were to be placed on the y – type of voids, it would neither eclipse the A layer nor the B layer of spheres. This would clearly be a unique layer. Let us call this layer as the C layer (as seen in figure 7(c). This arrangement (i.e, ABCABC…) is called the Cubic Close Packing (fcc). 
Exercise 2: It can be seen now that both fcc and Hexagonal Primitive Structure have the same packing fraction. Moreover this is also the highest packing fraction of all the possible unit cells with one type of atom with empty voids. Can you explain this?
 Octahedral and Tetrahedral Voids
(Figure 8(a) Voids in FCC unit cell)
(Effective number of octahedral holes = 4, Effective number of tetrahedral holes = 8)
The close packing system involves two kinds of voids – tetrahedral voids and octahedral voids. The former has four spheres adjacent to it while the latter has six spheres adjacent to it. These voids are only found in either fcc or Hexagonal Primitive unit cells. Let us first consider a fcc unit cell (as shown in the figure 8(a).). Not all the atoms of the unit cell are shown (for convenience). Let us assume that there is an atom (different from the one that forms the fcc) at the center of an edge.
Let it be big enough to touch one of the corner atoms of the fcc. In that case, it can be easily understood that it would also touch six other atoms (as shown) at the same distance. Such voids in an fcc unit cell in which if we place an atom it would be in contact with six spheres at equal distance (in the form of an octahedron) are called octahedral voids (as seen in fig. 8(a).
On calculation, it can be found out that a fcc unit cell has four octahedral voids effectively. The number of effective octahedral voids in a unit cell is equal to the effective number of atoms in that unit cell.
Let us again consider a fcc unit cell. If we assume that one of its corners is an origin, we can locate a point having coordinates (¼,¼,¼). If we place an atom (different from the ones that form the fcc) at this point and if it is big enough to touch the corner atom, then it would also touch three other atoms (as shown in the fig. 8(b) which are at the face centers of all those faces which meet at that corner. Moreover, it would touch all these atoms at the corners of a regular tetrahedron. Such voids are called tetrahedral voids (as seen in fig. 8(b).  
Since there are eight corners, there are eight tetrahedral voids in a fcc unit cell. We can see the tetrahedral voids in another way. Let us assume that eight cubes of the same size make a bigger cube as shown in the fig. 8(c). Then the centers of these eight small cubes would behave as tetrahedral voids for the bigger cube (if it were face centered). The number of tetrahedral voids is double then the number of octahedral voids. Therefore, the number of tetrahedral voids in hcp is 12. 
 Radius Ratio Rules
The structure of many ionic solids can be accounted for by considering the relative sizes of the positive and negative ions, and their relative numbers. Simple geometric calculations allow us to workout, as to how many ions of a given size can be in contact with a smaller ion. Thus, we can predict the coordination number from the relative sizes of the ions.
9.1 Coordination Number 3
The adjacent fig. 9(a) shows the smaller positive ion of radius r^{+} is in contact with three larger negative ions of radii r^{–}. It can be seen that AB = BC = AC = 2r^{–}, BD = r^{–} + r^{+}. Further, the angle ABC is 60^{o}, and the angle DBE is 30^{o}. By trigonometry Cos 30^{o} = (BE / BD).BD = (BE / cos 30^{o}), r^{+} + r^{–} = r^{–} / cos 30^{o} = (r^{–} / 0.866) = r^{–} ´ 1.155, r^{+} = (1.155 r^{–}) – r^{–} = 0.155 r^{–}, Hence (r^{+} / r^{–}) = 0.155. 
9.2 Coordination Number 4 (Tetrahedron)
Angle ABC is the tetrahedral angle of 109^{o} 28′ and hence the angle ABD is half of this, that is 54^{o}64′. In the triangle ABD, sin ABD= 0.8164 = AD / AB = . Taking reciprocals, , rearranging, we get, . 
9.3 Coordination Number 6 (Octahedron) or 4 ( Square Planar)
A cross section through an octahedral site is shown in the adjacent figure and the smaller positive ion (of radius r^{+}) touches six larger negative ions (of radius r^{–}). (Note that only four negative ions are shown in this section and one is above and one below the plane of the paper).
It is obvious that AB = r^{+} + r^{–} and BD = r^{–}. The angle ABC is 45^{o} in the triangle ABD. cos ABD = 0.7071 = (BD / AB) = . Rearranging, we get, = 0.414 

Limiting radius ratio

Co – ordination number  Shape  Example  
x < 0.155  2  Linear  BeF_{2}  
0.155 £ x < 0.225  3  Planar Triangle  AlCl_{3}  
0.225 £ x < 0.414  4  Tetrahedron  ZnS  
0.414 £ x < 0.732  4  Square planar  PtCl_{4}^{2}  
0.414 £ x < 0.732  6  Octahedron  NaCl  
0.732 £ x < 0.999  8  Body centered cubic  CsCl  
Exercise 3: The radius of a calcium ion is 94 pm and of an oxide ion is 146 pm. Predict the crystal structure of calcium oxide.
 Classification of Ionic Structures
In the following structures, a black circle would denote a cation and a white circle would denote an anion.In any solid of the type A_{x}B_{y}, the ratio of the coordination number of A to that of B would be y : x.
10.1 Rock Salt Structure
Cl^{–} is forming a fcc unit cell in which Na^{+} is in the octahedral voids. The coordination number of Na^{+} is 6 and therefore that of Cl^{–} would also be 6. Moreover, there are 4 Na^{+} ions and 4 Cl^{–} ions per unit cell. The formula is Na_{4}Cl_{4} i.e,NaCl. The other substances having this kind of a structure are halides of all alkali metals except cesium halides and oxides of all alkaline earth metals except beryllium oxide.  Figure 10 Unit cell structure of NaCl 
10.2 Zinc Blende Structure
Sulphide ions are face centered and Zinc is present in alternate tetrahedral voids. Formula is Zn_{4}S_{4}, i.e, ZnS. Coordination number of Zn is 4 and that of sulphide is also 4. Other substance that exists in this kind of a structure is BeO. 
10.3 Fluorite Structures
Calcium ions are face centered and fluoride ions are present in all the tetrahedral voids. There are four calcium ions and eight fluoride ions per unit cell. Therefore the formula is Ca_{4}F_{8}, (i.e, CaF_{2}). The coordination number of fluoride ions is four (tetrahedral voids) and thus the coordination number of calcium ions is eight. Other substances which exist in this kind of structure are UO_{2}, and ThO_{2}.  Figure 12 
 AntiFluorite Structure
Oxide ions are face centered and lithium ions are present in all the tetrahedral voids. There are four oxide ions and eight lithium ions per unit cell. As it can be seen, this unit cell is just the reverse of Fluorite structure, in the sense that, the positions of cations and anions is interchanged. Other substances which exist in this kind of a structure are Na_{2}O, K_{2}O and Rb_{2}O. 
10.5 Cesium Halide Structure
Chloride ions are primitive cubic while the cesium ion occupies the center of the unit cell. There is one chloride ion and one cesium ion per unit cell. Therefore the formula is CsCl. The coordination number of cesium is eight and that of chloride ions is also eight. Other substances which exist in this kind of a structure are all halides of cesium. 
10.6 Corundum Structure
The general formula of compounds crystallizing in corundum structure is Al_{2}O_{3}^{.} The closest packing is that of anions (oxide) in hexagonal primitive lattice and twothird of the octahedral voids are filled with trivalent cations. Examples are : Fe_{2}O_{3}, Al_{2}O_{3} and Cr_{2}O_{3}.
10.7 Pervoskite Structure
The general formula is ABO_{3}. One of the cation is bivalent and the other is tetravalent. Example: CaTiO_{3}, BaTiO_{3}. The bivalent ions are present in primitive cubic lattice with oxide ions on the centers of all the six square faces. The tetravalent cation is in the center of the unit cell occupying octahedral void. 
10.8 Spinel and Inverse Spinel Structure
Spinel is a mineral (MgAl_{2}O_{4}). Generally they can be represented as M^{2+}M_{2}^{3+}O_{4}, where M^{2+} is present in oneeighth of tetrahedral voids in a FCC lattice of oxide ions and M^{3+} ions are present in half of the octahedral voids. M^{2+} is usually Mg, Fe, Co, Ni, Zn and Mn; M^{3+} is generally Al, Fe, Mn, Cr and Rh. Examples are ZnAl_{2}O_{4}, Fe_{3}O_{4}, FeCr_{2}O_{4} etc. Many substances of the type also have this structure. In an inverse spinel the ccp is of oxide ions, M^{2+} is in oneeight of the tetrahedral voids while M^{3+} would be in oneeight of the tetrahedral voids and onefourth of the octahedral voids.
Illustration 2: A solid compound AB has NaCl structure. If the radius of the cation is 95 pm. What is the radius of anion (B^{–})?
Solution: Radius into for octahedral coordination = 0.414
Hence radius of the anion (B^{–}) = pm = 229.46 pm
Exercise 4: The unit cell of silver iodide (AgI) has 4 iodine atoms in it. How many silver atoms must be there in the unit cell.
Exercise 5: The coordination number of the barium ions, Ba^{2+}, in barium chloride (BaF_{2}) is 8. What must be the coordination number of the fluoride ions, .
 Imperfections in a Crystal
The discovery of imperfections in an other wise ideally perfect crystal is one of the most fascinating aspects of solid state science. An ideally perfect crystal is one which has the same unit cell and contains the same lattice points throughout the crystal. The term imperfection or defect is generally used to describe any deviation of the ideally perfect crystal from the periodic arrangement of its constituents.
11.1 Stoichiometric Defects
Stoichiometric compounds are those where the numbers of the different types of atoms or ions present are exactly in the ratios indicated by their chemical formulae. They obey the law of constant composition that “the same chemical compound always contains the same elements in the same composition by weight”. Previously these were called Daltonide compounds.
Two types of defects may be observed in stoichiometric compounds, called Schottky and Frenkel defects respectively. At absolute zero, crystals tend to have a perfectly ordered arrangement. As the temperature increases the amount of thermal vibration of ions in their lattice sites increases and if the vibration of a particular ion becomes large enough, it may jump out of its lattice site. The higher the temperature, the greater the chance that lattice sites may be unoccupied. Since the number of defects depends on the temperature, these are sometimes called thermodynamic effect.
Schottky Defects: The defects rise if some of the lattice points in a crystal are unoccupied. The points which are unoccupied are called lattice vacancies. The existence of two vacancies, one due to a missing Na^{+} ion and the other due to a missing Cl^{–} ion in a crystal of NaCl, is shown in figure 16. The crystal, as a whole remains neutral because the number of missing positive and negative ions is the same. Thus a Schottky defects consists of a pair of holes in the crystal lattice.
Figure 16
Schottky defects appear generally in highly ionic crystals in which the positive and the negative ions do not differ much in size. Sodium chloride and cesium chloride furnish good examples of ionic crystals in which Schottky defects occurs.
Frenkel Defects
These defects arise when an ion occupies an interstitial position between the lattice points. This is shown in figure for the crystal of AgBr.
Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+} 
Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–} 
Ag^{+}  Br^{–}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  
Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–}  Ag^{+}  Br^{–} 
Figure 17: Frenkel Defects in a Crystal 
As can be seen, one of the Ag^{+} ions occupies a position in the interstitial space rather than its own appropriate site in the lattice. A vacancy is thus created in the lattice as shown. It may be noted again that the crystal remains neutral since the number of positive ions is the same as the number of negative ions. The presence of Ag^{+} ions in the interstitial space of AgBr crystal is responsible for the formation of a photographic image on exposure of AgBr crystals (i.e., photographic plate) to light.
ZnS is another crystal in which Frenkel defects appear. Zn^{2+} ions are entrapped in the interstitial space leaving vacancies in the lattice.
Frenkel defects appear in crystals in which the negative ions are much larger than the positive ions. Like Schottky defects, the Frenkel defects are also responsible for the conduction of electricity in crystals and also for the phenomenon of diffusion in solids.
11.2 NonStoichiometric Defects
In nonstoichiometric or Berthollide compounds the ratio of the number of atoms of one kind to the number of atoms of the other kind does not correspond exactly to the ideal whole number ratio implied by the formula. Such compounds do not obey the law of constant composition. There are many examples of these compounds particularly in the oxides and sulphides of the transition elements. Thus in FeO, FeS or CuS the ratio of Fe : O, Fe : S or Cu : S differs from that indicated by the ideal chemical formula. If the ratio of atoms is not exactly 1 : 1 in the above cases, there must be either an excess of metal ions or a deficiency of metal ions. Electrical neutrality is maintained either by having extra electrons in the structure or changing the charge on some of the metal ions. This makes the structure irregular in some way i.e., it contains defects.
Metal Excess: This may occur in two different ways
FCentres: A negative ion may be absent from its lattice site leaving a hole which is occupied by an electron, thereby maintaining the electrical balance. This type of defect is formed by crystals which would be expected to form Schottky defects. When compounds such as NaCl, KCl, are heated with excess of their constituent metal vapours, or treated with high energy radiation, they become deficient in the negative ions and their formulae may be represented by AX_{1–}_{d}, where d is a small fraction. The crystal lattice has vacant anion sites which are occupied by electrons. Anion sites occupied by electrons in this way are called F centres (F is an abbreviation Farbe, the German word for colour).
Interstitial ions and electrons: Metal excess defects also occur when an extra positive ion occupies an interstitial position in the lattice and electrical neutrality is maintained by the inclusion of an interstitial electron. Their composition may be represented by general formula A_{1+}_{d}X. This kind of metal excess defect is much more common than the first and is formed in crystals which would be expected to form Frenkel defects. Examples include ZnO, CdO, Fe_{2}O_{3}.
Crystals with either type of metal excess defect contain free electrons, and if these migrate they conduct an electric current. These free electrons may be excited to higher energy levels, giving absorption spectra and in consequence their compounds are often coloured e.g. nonstoichiometric NaCl is yellow, nonstoichiometric KCl is lilac.
Metal Deficiency Defects: In certains cases, one of the positive ions is missing from its lattice site and the extra negative charge is balanced by some nearby metal ion acquiring two charges instead of one. There is evidently, a deficiency of the metal ions although the crystal as a whole is neutral. This type of defect is generally found amongst the compounds of transition metals which can exhibit variable valency. Crystals of FeO, FeS and NiO show this type of defects. The existence of metal deficiency defects in the crystal of FeO is illustrated.
Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–} 
O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+} 
Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  O^{2–}  Fe^{2+}  O^{2–}  
O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{+3}  O^{2–}  Fe^{+3}  O^{2–}  Fe^{2+} 
Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–}  Fe^{2+}  O^{2–} 
Figure 20 
It is evident from the above discussion that all types of point defects result in the creation of vacancies or ‘holes’ in the lattice of the crystals. The presence of holes lowers the density as well as the lattice energy or the stability of the crystals. The presence of too many holes may cause a partial collapse of the lattice.
 Solution to Exercises
Exercise 1: Since
Density, r = \
Here n = 4, Av. No = 6.023 ´ 10^{23}, &
Volume = V = a ´ b ´ c
= 6.8 ´ 10^{–8} ´ 4.4 ´ 10^{–8} ´ 7.2 ´ 10^{–8} cm^{3}
= 2.154 ´ 10^{–22} cm^{3}
\ density, r = \
= 0.6708 gm/cm^{3}
Exercise2: HCP is of the ABAB… type and FCC is of ABCABC… type. In HCP and FCC both, the layers are formed of same atoms of the same type in the same arrangement (each sphere surrounded by six) but the placement of the layers on top of each other is different.
\both will have same packing fraction.
This is the highest packing fraction possible because in both types, maximum number of atoms of the same size are present around the atom (namely six).
Exercise3: The ratio . The prediction is an octahedral arrangement of the oxide ions around the calcium. Because the ions have equal but opposite charges, there must also been an octahedral arrangement of calcium ions around oxide ions. Thus we would expect a rock
salt (NaCl) structure.
Exercise4: The formula, AgI, tells us that the ratio of silver atoms to iodine atoms is 1:1. Hence, if there are four iodine atoms in the unit cell, there must be four silver atoms.
Exercise5 The coordination number of barium ions tells us that it is surrounded by eight fluoride ions (charge 8 ´ (1) = 8). In order to balance out the eight negative charges, we need four barium ions (charge 4 ´ (+2) = +8). Hence, the coordination number of the fluoride ions must be 4. You will find that the coordination number and the charge on ions always
balance out to neutralify.
 Solved Problems
13.1 Subjective
Problem 1: Copper has the fcc crystal structure. Assuming an atomic radius of 130pm for copper atom (Cu = 63.54):
 What is the length of unit cell of Cu?
 What is the volume of the unit cell?
 How many atoms belong to the unit cell?
 Find the density of Cu.
Solution: As we know
r = ,
 a) for fcc structure
4r = a
a = r
= ´ 130 pm
= 367.64 pm
 b) volume of unit cell = a^{3} = o
= g/m^{3}
= 4.94 ´ 10^{–23} cm^{3}
 c) n = 4
 d) r = = 8.54 gm / cm^{3}
Problem2: Compute the percentage void space per unit volume of unit cell in zincfluoride 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
= =
Q for tetrahedral holes,
= 0.225
= {1 + (0.225)^{3}}
= 0.7493
\ Void volume = 1 – 0.7493
= 0.2507/unit volume of unit cell
% void space = 25.07%
Problem 3: The density of CaO is 3.35 gm/cm^{3}. The oxide crystallises in one of the cubic systems with an edge of 4.80 Å. How many Ca^{++} ions and O^{–2} ions belong to each unit cell, and which type of cubic system is present?
Solution: From equation
r(density) = 3.35 gm/cm^{3}
a = 4.80 Å
M_{m} of CaO = (40 + 16) gm = 56 gm CaO
Q r = where n = no. of molecules per unit cell
\ n = = 3.98
or n » 4
So, 4molecules of CaO are present in 1 unit cell
So, no. of Ca^{+ +} ion = 4
No. of O^{– –} ion = 4
So, cubic system is fcc type.
Problem 4: A metal crystallizes in two cubic phases, face centered cubic (fcc) and body centered cubic bcc whose unit cell length are 3.5 and 3.0A° respectively. Calculate the ratio of density of fcc and bcc. [IIT JEE ‘99]
Solution: r =
For face center cubic (fcc)
n = 4, a = 3.5 Å
\r_{(fcc)} = ————— (I)
For bcc lattice
N = 2, a = 3.0 Å
r_{bcc} = ————— (ii)
From equation (i) equation (ii)
\ = 1.259
Problem 5: A compound formed by elements X & Y, Crystallizes in the cubic structure, where X is at the corners of the cube and Y is at the six face centers. What is the formula of the compound? If side length is 5A°, estimate the density of the solid assuming atomic weight of of X and Y as 60 and 90 respectively.
Solution: From eight corner atoms one atoms (X) contributes to one unit cell.
From six face centres, three atoms (Y) contributes to one unit cell.
So, the formula of the compound is XY_{3}.
As we know that,
r = , here n = 1
Molar mass of XY_{3}
M_{m} = 60 + 3 ´ 90 = 330 gm
r = gm/cm^{3}
a = 5Å = 5 ´ 10^{–8} cm
= gm/cm^{3} = 4.38 gm / cm^{3}
Problem 6: In LiI Crystal, I^{–} ions form a cubical closest packed arrangement and Li^{+} ions occupy octahedral holes. What is the relationship between the edgelength of the unit cells and the radii of the I^{–} ions? Calculate the limiting ionic radii of Li^{+} and I^{–} if a= 600pm.
Solution: From question
2r^{+} + 2r^{–} = 600 pm
Þ r^{+} + r^{–} = 300 pm———— (i)
Since, Li^{+} ions occupy octahedral hole
So, for octahedral hole
= 0.414 ——————– (ii)
From equation (I) and equation (ii)
= 300
Þr^{– }´ (0.414 +1) = 300
Þ r^{– }= = 212.164 (Q = 0.414)
\ r^{+} = 212.164 ´ 0.414 = r^{+} = 87.84 pm
Problem 7: Calculate the atomic volume of an element whose atoms have a radius 1.414A° when atoms are cubic closed structure. What is the length of an edge of the unit cell?
Solution: Since atoms are cubic closed structures i.e., for fcc structure
\ 4r = a
Given r = 1.414Å
r = Å
4 ´ Å = a
\ a = 4Å
= 4 ´ 10^{–8} cm = 4Å
Since volume of a unit cell = a^{3}
= (4´10^{–8})^{3}
= 64 ´10^{–24}
= 6.4 ´10^{–23}
Since one unit cell contains 4 atoms
i.e., \ 4 atoms corresponds to volume 6.4 ´ 10^{–23} cm^{3}
\ 1– atom corresponds to volume ´ 10^{–23} cm^{3}
= 1.6 ´ 10^{–23} cm^{3}
Problem 8: Prove that void space in fluorite structure per unit volume of unit cell is 0.251.
Solution: Cations form cubical closest packing and anions occupying the tetrahedral holes.
There are 4 cations per unit cell.
There are 8 anions per unit cell.
Here, face diagonal.
4r_{+} =
where, r_{+}= radius of cation
\a =
\a^{3} = r_{+})^{3}
Volume of unit cell =
Fraction of volume occupied per unit voume of unit cell
i.e. Packing fraction = =
Here in fluorite structure, for tetrahedral hole
=
PF =
= 0.7577
\ void space = 1 – 0.7577
VF = 0.2422/unit volume of the unit cell
Problem 9: 5.35 g of a salt ACl (of weak base AOH) is dissolved in 250 ml of solution. The pH of the resultant solution was found to be 4.827. Find the ionic radius of A+ and Cl– if ACl forms CsCl type crystal having 2.2 g/cm3. Given K_{d}(AOH) = 1.8 ´ 10–5
= 0.732 for this cell unit
Solution: ACl(s) ¾¾® A^{+}(aq) + Cl^{–}(aq)
A^{+}(aq) + H2O AOH(aq) + H^{+}(aq)
at equilibrium C(1–a)
For salts of weak base and strong acid a =
H^{+} = Ca =
antilog (–4.827) =
M = 53.5
For CsCl type structure
r =
a = = 3.43Å
= = 0.866 a = 0.866 ´ 3.43 Å = 2.97 Å
Q = 0.732
on solving = 1.255 Å
= 1.715 Å
Problem 10: In a face centred lattice with all the positions occupied by A atoms the body centred octahedral hole in it is occupied by an atom B of an appropriate size for such a crystal, calculate the void space per unit volume of unit cell. Also predict the formula of the compound.
Solution: Let a be the edge of the cube so that
4r_{A} = a
a =
No since the atoms B has occupied the body central octahedral hole
= a
=
2r_{B} =
= 1.414 – 1 = 0.414
Volume of the cube a^{3} =
Volume occupied by A and B=
Volume occupied per unit volume of unit cell
f =
=
= 0.75
13.2 Objective
Problem 1: In a fcc arrangement of A & B atoms, where A atoms are at the Corners of the unit cell, B atoms at the face centers, two atoms are missing from two corners in each unit cell, then the simplest formula of the compound is
(A) A_{7}B_{6} (B) A_{6}B_{7}
(C) A_{7}B_{24} (D) AB_{4}
Solution: Since there are six atoms (A) in the corner of the unit cell. So, contribution of atoms in 1 unit cell is .
Since 3 facecentered atoms (B) contributes to one unit cell
So, formula is A_{6/8} B_{3}
or A_{6} B_{24}, or AB_{4}
\(D)
Problem 2: In fluorite structure (CaF_{2})
 Ca^{++} ions are ccp & ions are present in all the tetrahedral voids
 Ca^{++} ions are ccp & ions are present in all the octahedral voids
 Ca^{++} ions are ccp & ions are present in all the octahedral voids and half of ions are
 None
Solution: Ca^{2+} ions are ccp and F^{–} ions are present in the tetrahedral voids. So, the no. of Ca^{2+} ions is 4 and no. of F^{–} ions is 8.
So, the formula of the Calcium fluoride Ca_{4}F_{8} or, the simplest formula of calcium fluoride is CaF_{2}
\ (A)
Problem 3: Select the correct statement(s)
 The C.N. of cation occupying a tetrahedral hole is 4
 The C.N. of cation occupying a octahedral hole is 6
 In Schotky defects, density of the lattice decreases.
 In Frenkel defects, density of the lattice increases.
Solution: Since tetrahedral holes are surrounded by 4 nearest neighbours. So, the C.N. of cation occupying tetrahedral hole is 4. Since octahedral hole is surrounded by six nearest neighbours, so, C.N. of cation occupying octahedral is 6.
In schottky a pair of anion and cation leaves the lattice, so density of lattice decreases.
\(A), (B), (C)
Problem 4: A solid is made of two elements P&Q. Atoms P are in ccp arrangement and atoms Q occupy all the octahedral voids and half of the tetrahedral voids, then the simplest formula of the compound is
(A) PQ_{2} (B) P_{2}Q
(C) PQ (D) P_{2}Q_{2}
Solution: Four atoms (P) contributes to one unit cell from ccp arrangement and 4atoms (Q) from the all octahedral voids and 4atoms (Q) from the half of the tetrahedral void contributes one unit cell.
So, formula of solid is P_{4}Q_{8} so, the simplest formula of the solid is PQ_{2}
\(A)
Problem 5: Fraction of total volume occupied by atoms in a simple cubic cell is
(A) (B)
(C) (D)
Solution: In simple cubic arrangement, no. of atoms = 1
a = 2r
\Packing fraction =
=
\ (D)
Problem 6: The limiting radius ratio of the complex
(A) 0.225 – 0.414 (B) 0.414 – 0.732
(C) 0.155 – 0.225 (D) None
Solution: The complex [Ni(CN)_{4}]^{–2} is square planar. So, it has limiting radius ratio as octahedral structure i.e. 0.414 – 0.732
\ (B)
Problem 7: If the ratio of coordination no. P to that of Q be Y:Z, then the formula of
the solid is
(A) P_{y}Q_{z} (B) P_{z}Q_{y}
(C) (D) None
Solution: Since the ratio of the coordination no. P to that of Q is Y : Z
i.e. P is surrounded by Y atoms of Q and Q is surrounded by Z atoms of P
i.e. no. of atoms of P is Z
& no. of atoms of Q is Y
So, the formula is P_{Z}Q_{Y}
\ (B)
Problem 8: Xenon crystallizes in face centre cubic lattice and the edge of the unit cell is 620 pm, then the radius of xenonatom is.
(A) 438.5 pm (B) 219.20 pm
(C) 536.94 pm (D) 265.5 pm
Solution : For fcc lattice
4r = Ö2a where a = 620 pm
or,
pm
= 219.20 pm
\ (B)
Problem 9: The arrangement of Cl^{–} ions in CsCl structure is
(A) hcp (B) fcc
(C) bcc (D) Simple cubic
Solution: Arrangement of atoms or, ions in the corner of the unit cell is simple cubic. So in body centered cubic arrangement, Cl^{–} ions are arranged in the corner of the cube. So, it is a simple cubic.
\(D)
Problem 10: In closest packing of A type of atoms (radius, r_{A}), the radius of atom B that can be fitted into Octahedral void is
(A) 0.155 r_{A }(B) 0.125 r_{A}
(C) 0.414 r_{A} (D) 0.732 r_{A}
Solution: For octahedral void
or, r_{B} = 0.414 r_{A}
\ (C)
Problem 11: In aWCl_{6}
 all Cl^{–} ions are present in cubic close packing
 W occupies, 1/6^{th} of the octahedral holes.
 W occupies 1/3^{rd} of the tetrahedral holes.
 All Cl^{–} ions are present in all octahedral
Solution: In aWCl_{6}, Cl^{–} ions are arranged in cubic close packing, so, there is only one Cl^{–}ion in 1unit cell. So, the formula can be written W_{1/6}Cl, i.e. of the octahedral hole is filled by W.
\(A), (B)
Problem 12: The edge length of a cube is 400 pm. Its body diagonal would be
(A) 600 pm (B) 566 pm
(C) 693 pm (D) 500 pm
Solution: Since in body center cubic, the body diagonal = Ö3a.
= Ö3 ´ 400 pm = 692.82 pm
= 693 pm
\ (C)
Problem 13: The mass of unit cell of CaF_{2} (fluorite structure) corresponds to
 mass of 8Ca^{++} ions & 4 F^{– }ions (B) mass of 4Ca^{++} ions & 8 F^{– }ions
(C) mass of 4Ca^{++} ions & 4 F^{– }ions (D) mass of 1Ca^{++} ions & 2 F^{– }ions
Solution: In CaF_{2} (Calcium fluorite) structure 1unit cell contains 4Ca^{2+} ions and 8F^{–} ions.
So, mass of unit cell of CaF_{2}
= mass of 4 Ca^{2+} ions + mass of 8F ions
\ (B)
Problem 14: Close packing is maximum in the Crystal which is
(A) Simple cube (B) bcc
(C) fcc (D) none
Solution: The close packing in the crystal is 0.52, 0.68 and 0.74 for simple cubic, body centered cubic, and facecentered cubic respectively.
i.e. the close packing is maximum in fcc.
\(C)
Problem 15: An ionic compound AB has ZnS type of structure, if the radius A^{+} is 22.5pm, then the ideal radius of B^{–} is
(A) 54.35pm (B) 100pm
(C) 145.16pm (D) None
Solution: Since ionic compound AB has ZnS type of structure, therefore it has tetrahedral holes, for which,
= 0.225
= 0.225
= 0.225
Hence, r_{–} = 100 pm
\(B)
 Assignments (Subjective Problems)
Level – I
 A solid has a bcc structure, if the distance of closest approach between the two atoms is 1.73Å, what is the edge length of the cell?
 The unit cell of a metal of atomic mass 108 and density 10.5 gm/cm^{3} is a cube with edgelength of 409pm. Find the structure of the crystal lattice.
 A compound alloy of gold and copper crystalizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of a cube and the copper atoms occupy the centres of each of the cube faces determine the formula of this compound.
 The unit cube length of LiCl (NaCl structure) is 5.14A°. Assuming anionanion contact, Calculate the ionic radius of Cl^{–} ion.
 Calculate the value of Avogadro’s number from the following data:
Density of NaCl = 2.165gm/cm^{3}
Distance between Na^{+} & Cl^{– }ion in NaCl Crystal = 281pm
 Aluminium forms face centered cubic crystals. The density of Al is 2.7 gm/cm^{3}. Calculate the length of the side of the unit cell of Al [Al –27]
 The compound CuCl has ZnS structure. Its density is 3.4 gm/cm^{3}. What is the edgelength of unit cell? (Cu – 63.5, Cl – 35.5)
 An unknown metal is found to have a specific gravity of 10.2 at 25°C. It is found to crystallize in a bodycentered cubic lattice with a unit cell edge length of 3.147Å. Calculate the atomic weight.
 Polonium crystallizes in a simple cubic unit cell. It has atomic mass 209 and density = 91.5 kg m^{3}. What is the edgelength of the unit cell?
 Iron occurs as bcc as well as fcc unit cell. If the effective radius of an atom of iron is 124 pm. Compute the density of iron in both these structures.
 Look at the sodium chloride structure and use it to calculate r^{+}/r^{–}.
 RbI crystallizes in bcc structure in which each Rb^{+} is surrounded by eight iodide ions each of radius 2.17A°. Find the length of one side of RbI unit cell.
 An element A (Atomic wt – 100) having bcc structure has unit cell edge length 400 pm. Calculate the density of A and number of unit cells and number of atoms in 10 gm of A.
 Potassium metal crystallizes in a facecentered arrangement of atoms where the edge of the unit cell is 0.574 nm. What is the shortest separation of any two nuclei?
 The density of CaF_{2} is 3.18 gm/cm^{3} at 20°C. Calculate the dimensions of the unit cube of the substance.
Level – II
 The CsCl has cubic structure of ions in which Cs^{+} ion is present in body centere of the cube. Its density is 4.50gm/cm^{3}.
 i) Calculate the edgelength of unit cell.
 What is the distance between Cs^{+} and Cl^{–} ions?
 What is the radius of Cl^{–} ions, if the radius of Cs^{+} ion is 200pm.
 A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB… Any packing of spheres leaves out voids in the lattice. What percentage by volume of this lattice is empty space?
 Chromium metal crystallizes a body centered cubic lattice. The length of the unit cell is found to be 287pm. Calculate the atomic radius. What would be density of Chromium in gm/cm^{3}?
 Potassium metal crystallizes in a face centred arrangement of atoms, where the edge of unit cell is 0.574nm. What is the shortest separation of two potassium nuclei.
 The density of solid argon is 1.65gm per ml at – 233°C. If the Aratom is assumed to be a sphere of radius 1.54 x 108cm, what percentage of solid argon has apparently empty space [Ar = 40]
 The molar mass of NaCl which crystallises with FCC lattice is 58.46 (in grams per mole). The density of the crystal is 2.164 g/cm^{3}. Calculate the distance between Na^{+} and Cl^{–}.
 Calculate the void space as a fraction of the volume of the unit cell for FCC and BCC lattices. Calculate the void space for metallic silver crystallizing in the FCC system, the edge length of the unit cell being 4.07 Å. Calculate also the radius of the silver atom.
 In fcc arrangement of A & B atoms, where A atoms are at corners of the unit cell, B atoms at the facecenters, one of the atoms are missing from the corner in each unit cell then find the percentage of void space in the unit cell.
 Silver iodide crystallizes in the cubic close packed Zinc blende structure. Assuming that the iodide ions occupy the lattice points, what fraction of the tetrahedral sites is occupied by silver ions?
 a) Ferric oxide crystallizes in a hexagenal close – packed aray of oxide (O^{2–}) ions with two out of every three octahedral holes occupied by iron ions. What is the formula of ferric oxide?
 b) In cadium iodide every alternate octahedral hole in an HCP array of iodide I^{–} ions is occupied by a cadmium ion. What is the formula of cadium iodide.
 The metal nickel crystallizes in a face centered cubic structure. Its density is 8.90 gm/cm^{3}. Calculate (a) the length of the edge of a unit cell. (b) the radius of nickel atom. (atomic weight of Ni = 58.89)
 Compute the percentage void per unit volume of unit cell of Corundum.
 KCl Crystallizes in the same type of lattice as does NaCl.
Given that, = 0.5 and = 0.7
Calculate:
 a) The ratio of the side of unit cell for KCl to that for NaCl and
 b) The ratio of density of NaCl to that for KCl.
 An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10^{24}cm^{3} and density of the element is 7.20gm/cm^{3}, calculate no. of atoms present in 200gm of the element.
 Titanium has hexagonal close packing with cell edge length a = b = 295.3 pm, height = c = 472.9 pm. Calculate its density.
Level – III
 If NaCl is doped with 10^{3} mole % of SrCl_{2}, what is the concentration of cation vacancy?
 A tetrahedron can be drawn within a cube. The centre of cube constitutes a tetrahedral void. In closest packing two anions touch each other along the face diagonal of the cube and one one cation (occupying a tetrahedral void) and two anions (occupying two remote corners) touch each other along the body diagonal of the cube show that r/r^{–} = 0.225.
 The ZnS Structure is cubic. The unit cell be described as a facecentered sulphide ion sublattice with Zinc ions in the centers of alternating minicubes made by part on the main cube into eight equal parts.
 a) How many nearest neighbours does each Zn^{2+} have?
 b) How many nearest neigbours does each S^{2– }have?
 c) What angle is made by the lines connecting any Zn^{++} to any two of its nearest neigbours?
 What minimumratio is needed to avoid anionanion contact? Closest cationanion is assumed to touch.
 Prove that the void space percentage in zinc blende structure is 25%.
 BaTiO_{3} crystallizes in the prevoskite structure. This structure may be described as a cubic lattice with barium ions occupying the corner of the unit cell, oxide ions occupying the facecenters and titanium ion occupying the center of the unit cell.
If titanium is described as occupying holes in the BaO lattice, what types of hole does it occupy?
What fraction of this type of hole does it occupy?
Can you suggest a reason why it has certain holes of this type but not other holes of the same type?
 NH_{4}Cl crystallizes in a body centered cubic lattice with a unit cell distance of 387 pm. Calculate
 a) the distance between oppositely charged ions in the lattice and
 b) the radius of the NH_{4}^{+} ion if the radius of Cl^{–} ion is 181 pm.
 In a cubic crystal of CsCl (density = 3.97 gm/cm^{3}) the eight corners are occupied by Cl^{–} ions with Cs^{+}at the centre and visaversa. Calculate the distance between the neighbouring Cs^{+} and Cl^{–} ions. What is the radius ratio of two ions?
 Metallic gold Crystallises in fcc lattice. The length of the Cubic unit cell
is a = 4.07 Å.  a) What is the closest distance between gold atoms.
 b) How many “nearest neighbours” does each gold atom have at the distance calculated in (a).
 c) What is the density of gold.
 d) Prove that the packing fraction of gold is 0.74.
 CsCl crystallises in a cubic and that has a Cl^{–} at each corner and Cs^{+} at the centre of unit cell. If = 1.69Å and = 1.81Å, what is the value of edgelength ‘a’ of the cube? Compare this value of ‘a’ (calculated) from the observed density of CsCl, 3.97 gm/cm^{3} [M(CsCl) = 168.5]
 The composition of a sample of wurtsite is Fe_{0.93}O_{1}._{0} What percentage of iron is present in the form of Fe(III)?
 Pottassium Crystallizes in body centred cubic lattice with a unit cell length a = 5.2Å.
 a) What is the distance between nearest neighbours?
 b) What is the distance between next nearest neighbours?
 How many nearest neighbours does each K atom have?
 How many nextnearest neigbbours does each K atom have?
 e) What is the calculated density of crystalline Potassium?
 Ice crystallizes in a hexagonal lattice. At the low temperature at which the structure was determined, the lattice constants were a = 4.35 Å, and b = 7.00 Å. How many molecules are contained in a unit cell?
[density (ice) = 0.92 gm/cm^{3})
 Using the data given below, find the type of cubic lattice to which the crystal belongs.
Fe V Pd
a in pm 286 301 388
r in gm cm^{3} 7.86 5.96 12.16
 Iron Crystallizes in several modifications. At about 910°C, the body centered cubic aform undergoes a transition to the face centered cubic gform. Assuming that the distance between nearest neighbours is same in the two forms at the transition temperature. Calculate the ratio of density of giron to that of airon at the transition temperature.
 Titanium crystallizes in a face centered cubic lattice. It reacts with C or H interstitially, by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes, but carbon occupies octahedral holes.
 a) Predict the formula of titanium hydride and titanium carbide formed by saturating the titanium lattice with either “foreign” element.
 b) What is the maximum ratio of “foreign” atom radius to host atom radius that can be tolerated in a tetrahedral hole without causing a strain in the host lattice?
 Assignments (Objective Problems)
Level – I
Instructions: Only one choice is correct:
 The coordination number of metal crystallizing in a hexagonal close packed structure is
(A) 12 (B) 4
(C) 8 (D) 6
 The unit cell with crystallographic dimenions
a = b ¹ c, a = b = g = 90° is
(A) Cubic (B) Tetragonal
(C) Monoclinic (D) Hexagonal
 The anions (A) form hexagonal closest packing and atoms (C) occupy only 2/3 of octahedral voids in it, then the general formula of the compound is
(A) CA (B) C_{2}A_{2}
(C) C_{2}A_{3} (D) C_{3}A_{2}
 Silver iodide has the same structure as zinc sulfide, and its density is 5.67gcm^{–3}. The edge length of the unit cell is
(A) 4.50 Å (B) 5.50 Å
(C) 6.50 Å (D) 7.50 Å
 In the closest packing of atom A of radius r_{a}, the radius of atom B that can be fitted in tetrahedral void is
(A) 0.225 (B) 0.155 r_{a}
(C) 0.414 r_{a} (D) 0.732 r_{a}
 If the unit cell length of sodium chloride crystal is 600pm, then its density will be
(A) 2.165gm/cm^{3} (B) 3.247gm/cm^{3}
(C) 1.79 gm/cm^{3} (D) 1.082 gm/cm^{3}
 In a cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one fifth of tetrahedral voids are occupied by divalent (X^{++}) ions, while onehalf of the octahedral voids are occupied by trivalent ions (Y^{+++}), then the formula of the oxides.
(A) XY_{2}O_{4} (B) X_{2}YO_{4}
(C) X_{4}Y_{5}O_{10} (D) X_{5}Y_{4}O_{10}
 If three elements X, Y & Z crystallized in cubic solid lattice with X atoms at corners, Y atoms at cube centre & Zatoms at the edges, then the formula of the compound is:
(A) XYZ (B) XY_{3}Z
(C) XYZ_{3} (D)X_{3}YZ
 A compound XY crystallizes in bcc lattice with unit cell edgelength of 480pm, if the radius of Y^{–} is 225 pm, then the radius of X^{+} is
(A) 190.70pm (B) 225pm
(C) 127.5pm (D) None
 If the edgelength of the unit cell of Sodium chloride is 600pm, and the ionic radius of Cl^{–} ion is 190pm, then the ionic radius of Na^{+} ion is
(A) 310pm (B) 110pm
(C) 220pm (D) None
 In a solid lattice the cation and anion both have left a lattice site. The lattice defect is known as.
(A) interstitial defect (B) valency defect
(C) frenkel defect (D) schottky defect
 The number of tetrahedral and octahedral voids in hexagonal primitive unit cell are
(A) 8,4 (B) 2,1
(C) 12,6 (D) 6,12
 The rank of atoms in the hexagonal until cell is
(A) 4 (B) 3
(C) 5 (D) 6
14 The structure of CsCl Crystal is
(A) bcc lattice (B) fcc lattice
(D) Octahedral (D) None
 In hexagonal primitive unit cell, the Corner is shared by
(A) 4 Unit cells (B) 6unit cells
(C) 3 unit cells (D) 5 unit cells
Level – II
Instruction: In each of the following objective questions, 4/5 alternatives are given. In each one, one or more than one may be correct. It is important to do these questions in one attempt and under a time limit. Therefore you are required to do this assignment in 15 minute.
 An alloy of copper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula.
(A) Cu_{4}Ag_{2}Au (B) Cu_{4}Ag_{4}Au
(C) Cu_{4}Ag_{3}Au (D) CuAgAu
 A substance has density of 2 kg dm^{3} & it crystallizes to fcc lattice with edgelength equal to 700pm, then the molar mass of the substance is
(A) 74.50gm mol^{1} (B) 103.30gm mol^{1}
(C) 56.02gm mol^{1} (D) 65.36gm mol^{1}
 Select the correct statements:
 For CsCl unit cell (edgelength = a), r_{c} + r_{a} =
 For NaCl unit cell (edgelength =), r_{c} + r_{a} =
 The void space in a b.c.c. unit cell is 0.68
 The void space % in a facecentered unit cell is 26%
 A common example of spinel strcuture is
(A) Fe_{2}O_{3} (B) FeO
(C) Fe_{3}O_{4} (D) FeCl_{3}
 In closest packing of atoms
 The size of tetrahedral void is greater than that of the octahedral void.
 The size of the tetrahedral void is smaller than that of the
octahedral void.  The size of tetrahedral void is equal to that of the octahedral void.
 The size of tetrahedral void may be larger or smaller or equal to that of the octahedral void depending upon the size of atoms.
 If a be the edgelength of the unit cell, r be the radius of an atom, then for face centered cubic lattice, the correct relation is
(A) 4r = (B) 4r =
(C) 4a = (D) None
 When an ion leaves its correct lattice site and occupies interstitial sites in its crystal lattice, it is called
(A) Crystal defect (B) Frenkel defect
(C) Schottky defect (D) None
 The 8 : 8 type of packing is present in ?
(A) CsCl (B) KCl
(C) NaCl (D) MgF_{2}
 Select the correct statements
CaCl_{2} doped with NaCl solid solution results in
 Substitutional cation vacancy
 Frenkel’s defect
 Schottky defect
 Decrease of density
 Select the correct statement(s):
 Schottky defect is shown by CsCl
 Frenkel defect is shown by Zns
 Hexagonal close packing (hcp) & Cubic close packing (ccp) structure has same coordination no.12
 At high pressure, the coordination number increases.
 Select the correct statement(s)
 At high pressure, the coordination no. increases from 6:6 to 8:8
 At high pressure, the coordination no. decreases from 8:8 to 6:6
 At high temperature, coordination number decreases
 AT 760K temperature, CsCl structure changes into NaCl.
 A face centre atom corner atom in the hexagonal unit cell structure is shared with
(A) 4unit cells (B) 3unit cells
(C) 6unit cells (D) None
 KBr shows, which of the following defects?
(A) Frenkel defect (B) Schottky defect
(C) Metal excess defeat (C) Metal deficiency
 If in diamond, there is a unit cell of carbon atoms as fcc and if carbon atom is sp^{3} what fractions of void are occupied by carbon atom.
(A) 25% Tetrahedral (B) 50% Tetrahedral
(C) 25% Octahedral (D) 50% Octahedral
 Which of the following doesn’t crystallise in the rocksalt structure?
(A) NaCl (B) KCl
(C) MgCl_{2} (D) CsCl
 Answers to Subjective Assignments
Level – I
 1.997Å 2. FCC
 AuCu_{3} 4. 1.815 Å
 6.08 ´ 10^{23 }6. 4.045 ´ 10^{–8} cm
 5.78Å 8. 95.7
 1.559 ´ 10^{–8} cm 10. BCC = 7.887 g/cm^{3}
FCC = 8.59 g/cm^{3}
 0.414 12. 5.011 Å
 5.188 gm/cm^{3}, 3.01 ´ 10^{22} unit cell 14. 0.406 nm
6.023 ´ 10^{22} atoms of A
 5.46 Å
Level – II
 (i) 85.35 Å, (ii) 73.8275 Å, (iii) 71.82 Å 2. 26%
 r = 124.27 pm, d = 7.3 g/cm^{3} 4. 0.406
 48.612% 6. 2.820 Å
 0.2596, 0.32, 17.54 ´ 10^{–24} cm^{3}, 1.439 Å 8. 43%
 50% 10. (a) Fe_{2}O_{3}, (b) CdsI_{2}
 (a) 3.53 Å, (b) 1.247 Å 12. 22.44%
 A = 1.143, (b) = 1.172 14. 2.3148 ´ 10^{24} atoms
 4.486 gm/c^{3}
Level – III
 6.023 ´ 10^{18} 2. 0.225
 a) 4, (b) 4, (c) 109°28¢, (d) 0.225 5. a) octahedral holes,
 b) 1/4th
 a) 335.15 pm (b) 154.15 pm 7. 0.728
 a) 2.88 Å, (b) 12, (c) 19.4 g/cc, (d) 0.74
 4.13 Å 10. 15.053
 (a) 4.50 Å, (b) 5.20 Å, (c) 8, (d) 6,
(e) 0.921 g/cm^{3}  4 molecules of H_{2}O 13. Fe – bcc, V – bcc, Pd – fcc
 1.0887 15. (a) TiH_{2},TiC, (b) 0.225, (c) 0.414
 6.3697 Ås
 Answers to Objective Assignments
Level – I
 A 2. B
 C 4. C
5 A 6. C
 C 8. C
 A 10. B
 D 12. C
 D 14. A
 B
Level – II
 C 2. B
 A, B, D 4. C
 B 6. B
 B 8. A
 A, D 10. A, B, C, D
 A, C, D 12. C
 B 14. D
 D
Solid State
Hints for Subjective Problems
Level – I
 In bcc closet approach = .
 The structure of LiCl is fcc type.
 For fcc 4r = , and shortest separation between two nuclei = a.
 See fluorite (CaF_{2}) structure.
Level – II
 For CsCl, n = 1.
 See calculation of packing fraction in HCP structure.
 Calculate volume of all the atoms in 1.65g of Argon.
 After removing one of the corner atom.
No. of A atoms =
No. of B atom = 3
Also a = 2r_{A} + 2r_{B}
 Corundum is Al_{2}O_{3}
Here oxide anions are in HCP arrary where as trivalent cations (Al^{3+}) occupy only 2/3^{rd} of octahedral voids. Try to calculate packing fraction first.
 Try to calculate from the given equations
Again
Similarly,
 Here n = ´ 8 + 2 ´ = 2
Level – III
 Two Na^{+}ions are replaced by one Sr^{2+} ion
 Here 2r^{–} = and 2(r_{+} + r_{–}) = a
 For BCC, 2(r^{+}+ r^{–}) = a
 For FCC closest distance between atoms =
 Use this concept, in a compound
Total positive charge = Total negative charge
 The distance between nearest neighbours = and the distance between next nearest neigbhours = a
Subjective Problems
Level – I
 For BCC
= 1.73Å
\ a = Å
a = 1.997Å
 r = , here M_{m} = 108
N_{A} = 6.023 ´ 10^{23}
Putting these values and solving, we get a = 409 pm
= 4.09 ´ 10^{–8} cm
r = 10.5 gm /cm^{3}
n = 4 = number of atoms per unit cell
So, fcc lattice
 One – eighth of each corner atom (Au) and one – half of each face – centred atom (Cu) are contained within the unit cell of the compound.
Thus, number of Au atoms per unit cell = = 1 and number of Cu atoms per unit cell = 3. The formula of the compound is AuCu_{3}.
 The distance between Li^{+} and Cl^{–} ion can be derived as half of the edge length of cube
= Å
=
= 3.63Å
So, the radius of Cl^{–} is = Å
 Using,
r = …(1)
From question, number of molecules per unit cell n = 4
Molar mass, M_{m} = 58.5
Here, = 2.81 pm = 2.81 ´ 10^{–8} p
\ a = 2 = 2´ 2.81´ 10^{–8} cm
Putting these values in equation (i)
N_{A} = 6.08 ´ 10^{23}
 r =
For FCC, n = 4, Atomic weight of Al = 27
a^{3} =
a = 4.050 ´ 10^{–8} cm
 From question
For Zns structure, n = 4
Molar mass of CuCl i.e. M_{m} of CuCl = 99
r = 3.4 gm/cm^{3}
\ r =
a =
a = 5.78 ´ 10^{–8} cm = 5.78 Å
 \ r =
From question, n = 2, M_{m} = ?
a = 3.147Å = 3.147 ´ 10^{–8} cm
\ a^{3} = (3.14 ´ 10^{–8} cm)^{3} = 3.116 ´ 10^{–23} cm^{3}
r = 10.2 gm/cm^{3}
\ M_{m} = = 95.7 gm
\ Atomic weight = 95.7
 r =
For simple cubic type n = 1
r =
a^{3}= cm^{3}
a = 1.559 ´ 10^{–8} cm
 As we know that
Density = , here n = no. of atoms in one unit cell
M_{m }= molar mass of atom or molecule
N_{A}^{ }= Avogadro’s number.
V = Volume of one unit cell.
Case (i)
When iron occurs as bcc.
Here n = 2 atoms per unit cell
For Fe, M_{m} = 55.8 gm/mole
In body centered unit cell
4r = , where a = edgelength of unit cell
or,
a^{3 }=
a = = ´ 124 pm = 286.4 pm = 286.4 x 10^{10}cm
\density (r) ==7.887 g/cc
Case (ii)
When iron occurs in fcc, 4 r =
or, a =
Volume of 1 unit cell = a^{3} =
For this unit cell, n = 4 atoms per unit cell
So, density (r) = = = 8.59 gm/cm^{3}
 For bcc lattice
2 = where a = edge length of the cube
From question,
= 2.17 Å
= 2.17 Å
\ Å = a
or, a = 5.011Å
 a = 400 x 10^{12}m
= 400 x 10^{12} x 100 cm = 4 x 10^{8}cm
So, volume of unit cell edge length = a^{3} = (4 x 10^{8})^{3}
As we know that
density (r) =, here, n = no. of atoms per unit cell = 2
M_{m} = mass of 1 mole atom = 100 gm
N_{A} = Avogadro’s no. = 6.023 x 10^{23}
= = = 5.188 gm/cm^{3}
100 gm of A contains 6.023 ´ 10^{23} atoms of A
\10 gm of A contains 6.023 ´10^{23} ´
= 6.023 x 10^{22} atoms of A
2 atoms remain present in 1unit cell of bcc type
\ 6.0 23 ´10^{22} remains present in unit cells
= 3.0115 x 10^{22} unit cells
 For fcc lattice
4r = Ö2a
\ r = ´ a = ´ 0.574 nm
The shortest separation between the two nuclei
= 2r = ´ 0.574 nm = 0.406 nm
 The structure of CaF_{2} lattice is such that Ca^{++} ions occupy c.c.p. and F^{–} ions occupy all the tetrahedral voids in the cubic unit cell.
So, per cubic unit cell
Total no. of Ca^{++} ions = 4
Total no. of F^{–} ions = 8
\ total no. of molecules of CaF_{2} = 4
\ n = 4
Since, we know that
r =
Since, n = 4, M_{m} of CaF_{2} = 78 gm
r = 3.18 gm/cm^{3} a = ?
\ r =
\ 3.18 =
a = = [162.897 ´ 10^{–24}]^{1/3} = 5.46 ´ 10^{–8} cm = 5.46 Å
Level – II
 As we know,
 i) r = , here number of molecule of CsCl = n = 1
4.50 =
Where a = edge length of the cube
M_{m}(C_{S}Cl) = 168.5
or, a^{3} = cm^{3}
a^{3} = 6.21689 ´ 10^{–23}
a, a^{3} = 0621.69 ´ 10^{–21} cm^{3}
a = 8.535 ´ 10^{–7} cm = 85.35 Å
 ii) \ = ´35 Å
= 73.8275 Å
iii) = 200 pm = 2 Å given
\ = 71.82Å
 For bcc structure
4r =
Þ 4r = ´ 287 pm
Þ r = pm
r = 124.27 pm
Since, r = , here n = 2
r = = 7.30 gm / cm^{3}
 For FCC lattice
4r = a
or, the shortest separation = 2r = a
2r =´ 0.574 nm = 0.4058 nm = 0.406 nm
 Volume of all the atoms in 1.65 gm Argon
= ´ 6.02 ´ 10^{23} ´´ (2.54 ´ 10^{–8})^{3}
= 0.518 cm^{3}
\ Volume of solid Ar = 1.0 cm^{3}
\percentage of space = (1 – 0.518) ´ 100
= 48.16%
 a^{3} = cm^{3}
a = 5.64 ´ 10^{–8} cm
\ Distance between Na^{+} and Cl^{–} = Å = 2.82Å
 First two parts see theory
For metallic Ag crystallizing in the FCC system
4r^{+} =
\ r^{+} = 1.43Å
Total volume of cube = (4.07Å)^{3} = 67.419 ´ 10^{–24} cm^{3}
\void space = 67.419 ´ 10^{–24} ´0.26 cm^{3} = 17.53 ´ 10^{–24} cm^{3}
 There areA atoms and 3 B atoms per unit cell
Also, 2 r_{A} + 2 r_{B }=
or, a =
Volume of unit cell = a^{3} = =
Fraction of volume occupied per unit volume of the unit cell is given by
PF = =
= =
= = 0.570
So, void space = 1 – packing fraction
= 1 – 0.570 = 0.430/unit volume of unit cell
\percentage of void space = 43%
 a) Number of oxide ions per unit cell = 6
\ Number of octahedral voids = 6
Since two out of every three octahedral voids are occupied by iron
\ number of iron ions per nit cell = ´ 6 = 4
\ formula is Fe_{4}O_{6} or Fe_{2}O_{3}
 b) Number of I^{–} ions per unit cell = 6
Number of voids occupied by Cd^{2+ }= = 3
\ formula of CaCl_{3} or CdI_{2}
 a^{3} = cm^{3}
= 4.3943 ´ 10^{–23} cm^{3}
or, a^{3} = 43.943 ´ 10^{–24} cm^{3}
or, a = 3.53 Å
For fcc lattice
4r =
\r = = 1.247Å
 Anions form hexagonal closest packing and catioins occupy only two thirds of octahedral holes
Volume of hexagon = 24
There are six anions and six cations per unit cell
Fraction of volume occupied per unit volume of unit cell is
f =
or, f =
or, f =
[Q for octahedral holes, = 0.414]
f = 0.7756
\ volume of void = 1 – f
= 1 – 0.7756 = 0.2244 / unit volume of unit cell
\ % void = 22.44%
 a) Since NaCl Crysallizes into face centred cubic unit cell, such that
, where a = edge length of unit cell
Since given, ………… (i)
= 0.7 ………… (ii)
Equation (i) divided by (ii),
or,
or, =
Since, or,
or,
or,
 b) r =
= ´ = (1.143)3 ´
 Number of atoms contributed in one unitcell
= one atom from the eight corners + one atom from the two face diagonals
= 1+1 = 2 atoms
Mass of one unit vell = volume ´ its density
= 24 ´ 10^{–24} cm^{3} ´ 7.2 gm cm^{3} = 172.8 ´ 10^{–24} gm
\ 172.8 10^{–24} gm is the mass of one – unit cell i.e., 2 atoms
\ 200 gm is the mass = atoms = 2.3148 ´ 10^{24} atoms
 Area of parallelepiped = a^{2}sin60° =
Area of base in hcp = 3 ´ = cm^{2}
= 22.58 ´ 10^{–16} cm^{2}
Volume of hcp = 22.58 ´ 10^{–16} ´ 4.72 ´ 10^{–8} cm^{2}
= 106.59 ´ 10^{–24} cm^{3}
\ r = g/cc
= 4.486 gm/c^{3}
Level – III
 Due to addition of SrCl_{2}, each Sr^{++} ion replaces two Na^{+} ions, but occupies only one lattice point. So, one cationic vacancy is created.
\ 10^{–3} mol % SrCl_{2} = 10^{–3} mol % cationic vacancy
= 10^{–3} ´ mol cationic vacancy
= 10^{–5} mol cationic vacancy
= 10^{–5} ´ 6.023 ´ 10^{23} cationic vacancy
= 6.023 ´ 10^{18} cationic vacancy
 a) 4 because each Zn^{2+} ion is surrounded by 4S^{2– }ions
 b) 4, because each S^{2–} ions is also surrounded by 4 Zn^{2+} ions
 c) Since Zn^{++} ions occupies the tetrahedral position, so its angle is 109°28¢
 d) Body diagonal of the little cube
2r^{+} + 2r^{–} = … (i)
Face diagonal of the little cube = 2r^{–} =
\
2r^{+} + 2r^{–} =
\
or, = 0.225
 Anions are in fcc positions and half of the tetrahedral holes are occupied by cations. There are 4 anions and eight tetrahedral holes per unit cell.
Here, facediagonal =
4r =
or, a =
r_{–} = radius of anion
r_{+} = radius of cation
a = edge length of the cell
Volume of unit cell = a^{3}
= =
\Packing Fraction =
=
Since, for tetrahedral holes
\PF =
= 0.7492 / unit volume of unit cell
\void space = 1 – 0.7496
= 0.2508% unit volume of unit cell
\percentage of void space = 25.08%
» 25%
 a) Octahedral holes
 b) The octahedral holes at the centres of the unit cell constitutes one fourth of all the octahedral holes in a FCC lattice
 c) An octahedral hole at the center of the unit cell, occupied by a titanium ion, has 6 nearest neighbhour oxide ions. The other octahedral holes located at the centre of edges of the unit cell, have 6 nearest neighbours as in case with any octahedral hole. But 2 of the 6 neighbours as Ba^{2+} ions (at the unit cell corners terminating the given edge) and 4 are oxide ions. The proximity of two cations, Ba^{++} and Ti^{+4} would electrostatically be unfavourable.
 a) In bcc lattice, oppositely charged ions touch each other along cross diagonal of the cube
\ 2r^{+} + 2r^{–} =
= ´ 387 pm
\ r^{+} + r^{–} = ´ 387 pm = 335.15 pm
 b) Now since, r^{–} = 181 pm
\r^{+} = 335.15 – 181.00 = 154.15 pm
 As we know
Density (r) = here, n = 1
M_{m} = 132.9 + 35.5 = 168.4 gm
or, a^{3} =
or, a = 4.13 x 10^{8}= 4.13 Å
As it is BCC with Cs^{+} at center (radius r^{+}) and Cl^{–} at corner (radius r^{‑})
So, body diagonal =
or, 2 r^{+} + 2r^{–} =
or, r^{+} + r^{–} = = 4.13 Å
r^{+} + r^{–} = 3.57 Å
Now assume that two Cl^{–} touch each other
i.e. r^{–} + r^{–} = a
or, 2r^{–} = a = 4.13 Å
\ r^{–} = 2.065 Å
or,
or,
or,
 a) For fcc lattice nearest distance between two neighbours = 2r
As we know that
4r =
or, 2r = = Å
= 2.88 Å
 If we consider a face centred gold atom, it has four corner and eight other adjacent face centre atoms present at Therefore there are 12 nearest neighbours.
 c) density (r) = = = 4 g/cc
 d) PF = » 74
 We assume that the closest Cs^{+} to Cl^{–} distance is the sum of ionic radii of Cs^{+} and Cl^{–}
or, + = 1.69 + 1.81 = 3.50 Å
Q a = = 4.04Å
Since the density, r =
Where n = no. of molecules in 1unit cell
M_{m} = molar mass of the unit cell
a = edgelength of the cube.
From equation, n = 1
M_{m} of CsCl = 168.5 gm, r = 3.97 gm/cm^{3}
Since, r =
3.97 =
or a^{3} =
or, a = 41.3 ´ 10^{–8} cm = 4.13Å
 Fe_{0.93}O_{1.0} is a nonstoichiometric compound. It is a mixture of Fe^{+2} and Fe^{+3} ions
Let x atoms of Fe^{+++} ions are present in the compound
So, (0.93x) atoms of Fe^{++} ions are present in the compound
In the compound,
Total positive charge = Total negative charge
x ´ (+3) +(0.93 – x) ´ (+2) = +2
or, 3 x + 0.93 ´ 2 – 2 x = +2
or, x + 2 x 0.93 = 2
x = 0.14
\ % of Fe^{+3} ions present = = 15.053
 a) The distance between nearest neighbours
= ´ length of the body diagonal =
= Å = 4.50 Å
 b) Distance between the next nearest neighbours
a = 5.20 Å
 c) Number of K atoms (nearest neighbours) present at a distance of from the body centred atom = 8
 d) Number of K atoms (next nearest neighbours) present at distance of
edge length = 6  e) density=where, n = no. of atoms in 1 unit cell of b.c.c. = 2
M_{m} = molar mass of K
= = 0.921 gm/cm^{3}
 Volume of the unit cell
V = (area of the rhombus base) ´ b
= a^{2} sin 60° ´ b
=
= ´ 132.4 ´ 10^{–24} cm^{3}
^{ }Thus mass of the unit cell = V ´ d
= 132.4 x 10^{24} x 0.92
= 121.808 ´ 10^{–24}gm
Since, 18 ´ 1.66 ´ 10^{24} gm is the weight of one molecule of H_{2}O
\ 121.808 ´10^{–24 }gm is the weight of molecules of H_{2}O
= 4.07 molecules of H_{2}O
» 4 molecules of H_{2}O
 As we know
Density (r) = where, n = no. of atoms per unit cell
M_{m} = molar mass
N_{A} = 6.023 x 10^{23}
or, n =
To predict the type of cubic lattice, we have to know the value of “n”.
Therefore
For Fe, n =
n = 1.985 » 2
Hence cubic lattice of Fe is bcc.
For V, n = = 1.92 » 2
Hence the cubic lattice of V is also bcc.
Similarly for Pd,
n = = 4.02 » 4
Hence the cubic lattice for Pd is face centered.
 In bcc structure (a – form) of iron
4 r =
n = 2 atoms per unti cell from (one atom from 8–corners, one atom from the body centre)
So, r (g–form) = —————————– (i)
In fcc structure (r–form) of iron
4 r = a
\ a = r
n = 4 atoms per unit cell
\ r (g–form)= —————————— (ii)
from equation (ii) equation (i)
= 1.0887
 a) There are 4 – atoms per unit cell and 8tetrahedral sites per unit cell.
So, the ratio of atoms to tetrahedral sites = 1:2
So, the formula would be TiH_{2}
For carbide
Since carbon occupies octahedral holes
So, ratio of octahedral hole to atom = 1:1
The formula of carbide is TiC
 b) Since for tetrahedral hole
The limiting ratios radius
= 0.225 – 0.414
here r^{+} = foreign atom radius
r^{–} = host atom radius
Without causing a strain in the host lattice
i.e., = 0.225 (minimum valune)
 c) 414
 d) The hydrogen atom is small enough to fit into a tetrahedral hole, but carbon atom is not.
Objective Problems
Level – I
 r = = = 1.79 g/cc
\(C)
 In ccp anions occupy primitives of the cube while cations occupied voids. In ccp there are two tetrahedral voids and one octahedral holes.
For one oxygen atom there are two tetrahedral holes and one octahedral hole.
Since one fifth of the tetrahedral voids are occupied by divalent cations (X^{2+})
\ number of divalent cations in tetrahedral voids = .
Since half of the octahedral voids are occupied by trivalent cations (Y^{3+})
\ number of trivalent cations = .
So the formula of the compounds is
or ,
or X_{4}Y_{5}O_{10}
\ (C)
 Atom X is shared by 8 corners
1 atoms of Z is shared by 4 unit cells
Atom Y is present at centre of the unit cell
Hence, effective number of atoms of X per unit cell = = 1
Effective number of atoms of Z per unit cell = = 3
Effective number of atoms of Y per unit cell = 1
Hence, the formula of the compound is XYZ_{3}
\ (C)
 For bcc structure,
2 (r_{+}+r_{–}) = a
or r_{+} + 225 = ´ 480
\ r_{+} = 190.7 pm
\(A)
 = = 300
= 300 – 190 = 110 pm
\(B)
Level – II
 r =
2 = (since, effective number of atoms in unit cell = 4 )
on solving we get M_{m} = 103.03 gm / mol
\(B)
 In bcc structure are r_{+} + r_{–} =
Hence, for CsCl, r_{C} + r_{a} =
\(A)
Since, NaCl crystalise in fcc structure
\ 2r_{C} + 2r_{a} = edge length of the unit cell
Hence, r_{C} + r_{a} =
\ (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)
 For tetrahedral voids
= 0.225 , r_{+} = 0.225 r_{–} —————– (i)
Similarly for octahedral voids
r_{+} = 0.414 r_{–} —————–(ii)
From equation (i) and (ii) it is clear that size of octahedral void larger than that of tetrahedral voids.
\(B)
 Highly ionic compounds show Schottky defect
\(B)
 CsCl has a bcc structure
\(D)
6