1 A cubic crystal possesses in all .........elements of symmetry [1] 9 [2] 13 [3] 1 [4/] 23 Sol1. A cubic crystal possesses total 23 elements of symmetry (i) Plane of symmetry (3 + 6) = 9 (ii) Axes of symmetry (3 + 4 + 6) = 13 (iii) Centre of symmetry [1] = 1 Total symmetry = 23 2 In a simple cubic cell, each atom on a corner is shared by [1] 2 unit cells [2] One unit cell [3/] 8 unit cells [4] 4 unit cells sol2. In a simple cubic unit cell the atoms are present only on the corners. Each corner of a cubic unit cell is a part of eight unit cells. Hence each of the eight atom present at the eight corners is shared amongst eight unit cells. 3 Each unit cell of NaCl consists of 13 chlorine atoms and [1] 13 Na atoms [2/] 14 Na atoms [3] 6 Na atoms [4] 8 Na atoms Sol3. NaCl arrangement is a face centred cubic arrangement (cubic close packing) of the anion in which the anions are present at the corners and the face centres of the cube while the cation occupies all the octahedral voids that is all the edge centres and the body centre of each unit cell. In this case the unit cell would consist of 14 anions (8 at corners and 6 at face centres) and 13 cations (12 at edge centres and 1 at body centre).This arrangement can also be considered to be the face centred cubic arrangement (cubic close packing) of the cation in which the catons are present at the corners and the face centres of the cube while the anion occupies all the octahedral voids that is all the edge centres and the body centre of each unit cell. In this case the unit cell would consist of 14 cations (8 at corners and 6 at face centres) and 13 anions (12 at edge centres and 1 at body centre). 4 In cubic close packing (ccp) arrangement, the pattern of the successive layers will be designated as : [1] AB, AB, AB... etc [2] AB, ABC, AB... etc [3/] ABC, ABC, ABC... etc [4] None of these sol4. Cubic close packing is also known as face centred cubic packing. It is ABC arrangement of hexagonal layers. 5 A compound is formed by elements A and B. This crystallizes in the cubic structure when atoms A are the comers of the cube and atoms B are at the centre of the body. The simplest formula of the compounds is [1/] AB [2] AB 2 [3] A 2 B [4] AB 4 Sol5. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 An atom in the body of the cube is not shared by other cells. Hence the contribution of the atom at the centre of the cube = 1 1 = 1 atom A is present at each corner of the cube. As there are 8 corners in a cube, the number of corner atom per unit cell = 8 x 1 8 = 1. Atom B is present at the body centre. As any unit cell can have only one centre, Number of atoms at the body centre per unit cell = 1 Hence, the formula of the solid is A 1 B 1 or A B. 6 A certain metal fluoride crystallizes in such a way that F atoms occupy simple cubic lattice sites, while metal atoms occupy the body centres of half the cubes. The formula of metal fluoride is [1] M 2 F [2] MF [3] MF 2 [4] MF 8 Sol6. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 An atom in the body of the cube is not shared by other cells. Hence the contribution of the atom at the centre of the cube = 1 1 = 1 F atoms is present at each corner of the cube. As there are 8 corners in a cube, the number of corner atom per unit cell = 8 x 1 8 = 1. Metal M is present at half the body centres. As any unit cell can have only one centre, Number of atoms at the body centre per unit cell = 1 2 Hence, the formula of the solid is M 1 2 F 1 or M F 2 . 7 A compound alloy of gold and copper crystallizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of cube and copper atoms occupy the centres of each of the cube faces. The formula of this compound is - [1] AuCu [2] AuCu 2 [3] AuCu 3 [4] None Sol7. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 A face-centred atom in a cube is shared by two unit cells. Hence the contribution of the atom at the centre of the face = 1 2 Gold, Au atom is present at each corner of the cube. As there are 8 corners in a cube, the number of corner atom per unit cell = 8 x 1 8 = 1. Copper, Cu atom is present at each face centre of the cube. As there are 6 faces in a cube, number of face-centred atoms per unit cell = 6 x 1 2 = 3. Hence, the formula of the solid is Au 1 Cu 3 or Au Cu 3 8 In a face centred cubic arrangement of A & B atoms whose A atoms are at the corner of the unit cell & B atoms at the face centres. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is [1] A 7 B 3 [2] AB 3 [3/] A 7 B 24 [4] A 7 B 4 Sol8. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 A face-centred atom in a cube is shared by two unit cells. Hence the contribution of the atom at the centre of the face = 1 2 atom A is present at each corner of the cube. As there are 8 corners in a cube, the number of corner atom per unit cell = 8 x 1 8 = 1. Atom B is present at each face centre of the cube. As there are 6 faces in a cube, number of face-centred atoms per unit cell = 6 x 1 2 = 3. Hence, the formula of the solid is A 1 B 3 or A B 3 But One of the A atom is missing from one corner in unit cell. Hence the contribution of A becomes 1 – 1 8 = 7 8 Hence, the formula of the solid becomes A 7 8 B 3 or A 7 B 24 9 Xenon crystallizes in face centre cubic lattice and the edge of the unit cell is 620 PM, then the radius of Xenon atom is - [1/] 219.20 PM [2] 438.5 PM [3] 265.5 PM [4] 536.94 PM Sol9. In the case of face centred unit cell the atoms at the corner do not touch each other. The atom touching each other are the ones on the face diagonal. So the face diagonal is equal to four times the radius that is d = 4r. Also the diagonal of the square is .2 a hence d = 4r = .2 a a = 2 .2 r 4r = .2 a or r = 1 2 .2 x a = .2 a 4 = 1 2 .2 x a where a = 620 pm hence r = 1 2 .2 x 620 pm = 219.20 pm. 10 The edge length of NaCl type unit cubic cell is 508 pm. If the radius of the cation is 110 pm. the radius of the anion is [1] 285 pm [2] 398 pm [3/] 144 pm [4] 618 pm sol10. This is a sodium chloride (NaCl) arrangement. It is a face centred cubic arrangement (cubic close packing) of the anion in which cation occupies all the octahedral voids that is all the edge centres and the body centre of each unit cell. In this arrangement the anions are present at the corners and the cation in the centre of edge. Hence the edge length becomes equal to a = 2r + +2r − . = 2 ¦r + +r − ) = 2×bond length inter ionic distance, r + +r − = bond length = edge length 2 radius of the anion, r − = bond length – radius of cation = bond length – r + = edge length 2 – r + = a 2 – r + edge length, a =508 pm radius of cation, r + = 110 pm radius of the anion, r − = 508 2 – 110 pm = 144 pm 11 Fraction of total volume occupied by atoms in a simple cube is - [1] n 2 [2] .3 n 2 [3] .2 n 6 [4/] n 6 Sol11. In a simple cubic unit cell the atoms are present only on the corners. Each corner of a cubic unit cell is a part of eight unit cells. Hence each of the eight atom present at the eight corners is shared amongst eight unit cells. The contribution of the corner of a cubic unit cell is 1/8. Hence number of atoms per unit cell for a simple cubic cell = 8 x 1 8 = 1 In case of a simple cubic unit cell the atoms at the corner of unit cell touch each other. This makes the edge of the unit cell twice the radius of the particle making the unit cell. Hence a = 2r. ∴ Packing fraction = Volume occupied by one atom Volume of unit cell = 4 3 nr 3 a 3 = 4 3 nr 3 ¦2r ) 3 = n 6 a = 2r therefore, P.F. = 4 3 nr 3 ¦2r ) 3 = 0.52 % P.F. = 52% 12 If edge of a bcc crystal of an element is a cm, M is the atomic mass and N 0 is Avogadro’s number, then density of the crystal is : [1] 4M a 3 N 0 [2] 2N 0 M a 3 [3/] 2M N 0 a 3 [4] Ma 3 2N 0 Sol12. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 = a 3 ¢ = Z ×M N 0 ×a 3 For body centred cubic lattice z = 2 ¢ = 2×M N 0 ×a 3 13 Lithium borohydride (LiBH 4 ), crystallizes in a orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are : a = 6.81 Å, b = 4.43 Å, c = 7.17Å. If the molar mass of LiBH 4 is 21.76 g mol –1 . The density of the crystal is - [1/] 0.668 g cm −3 [2] 0.585 g cm −3 [3] 1.23 g cm −3 [4] 0.00668 g cm −3 Sol13. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 volume of orthorhombic unit cell, v = length×breadth×height = a×b×c density, ¢ = mass volume z given for the cell = 4 The unit cell dimensions are : a = 6.81 Å, b = 4.43 Å, c = 717Å molar mass of LiBH 4 is 21.76 g mol –1 The density of the crystal is, ¢ = Z ×M N ×V = 4צ21.76 gmol −1 ) ¦6.022×10 23 mol −1 )¦6.81×4.43×7.17×10 −24 cm 3 ) = 0.668 g cm −3 14 The unit cell of a metallic element of atomic mass 108 and density 10.5 g cm −3 is a cube with edge length of 409 PM. The structure of the crystal lattice is - [1/] FCC [2] bcc [3] hcp [4] None Sol14. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 =a 3 ¢= Z ×M N ×a 3 = z×M 6.022×10 23 mol −1 ×a 3 z = ¢×6.022×10 23 mol −1 ×a 3 M Here, M = 108, N A = 6.023 x 10 23 a = 409 PM = 4.09 x 10 –8 cm, ¢ = 10.5 g/cm 2 Put on these values and solving we get - z = 4 = number of atoms per unit cell So, The structure of the crystal lattice is FCC 15 The density of KBr is 2.75 g cm −3 . Length of the unit cell is 654 pm. K = 39, Br = 80. Then what is true about the predicted nature of the solid [1/] Solid has face centred cubic system with Z = 4. [2] Solid has simple cubic system with Z = 4. [3] Solid has face centred cubic system with Z = 1 [4] Solid has body centred cubic system with Z = 2 Sol15. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 =a 3 ¢ = Z ×M N×a 3 = z×M 6.022×10 23 mol −1 ×a 3 z = ¢×6.022×10 23 mol −1 ×a 3 M For the ionic substance the molar mass is the mass of one mole of formula units. For an ionic compound of the formula A x B y the molar mass is x×molar mass of A+y ×molar mass of B . For KBr x = y = 1, hence the molar mass = molar mass of K + molar mass of Br = 39 + 80 = 74.5 g/mol Hence, M = 74.5, N A = 6.023 x 10 23 a = 654 pm = 6.54 x 10 –8 cm, ¢ = 2.75 g cm −3 Put on these values and solving we get - z = 4 = number of atoms per unit cell So, The structure of the crystal lattice is FCC 16 Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 A. Assuming density of the oxide as 4.0g cm –3 , then the number of Fe 2+ and O 2– ions present in each unit cell will be [1/] Four Fe 2+ and four O 2– [2] Two Fe 2+ and four O 2– [3] Four Fe 2+ and two O 2– [4] Three Fe 2+ and three O 2– Sol16. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 = a 3 ¢ = Z ×M N×a 3 = z×M 6.022×10 23 mol −1 ×a 3 z = ¢×6.022×10 23 mol −1 ×a 3 M For the ionic substance the molar mass is the mass of one mole of formula units. For an ionic compound of the formula A x B y the molar mass is x×molar mass of A+y ×molar mass of B . For FeO x = y = 1, hence the molar mass = molar mass of Fe + molar mass of O = 56 + 8 = 64 g/mol Hence, M = 64, N A = 6.023 x 10 23 a = 5 A = 5 x 10 –8 cm, ¢ = 4 g cm −3 Put on these values and solving we get - z = 4 = number of atoms per unit cell So, The structure of the crystal lattice is FCC Hence the number of Fe 2+ and O 2– ions present in each unit cell will be Four Fe 2+ and four O 2– 17 The number of atoms in 100 gm of an FCC crystal with density d= 10 g cm −3 and cell edge as 200 pm is equal to: [1] 3 × 10 25 [2/] 5 × 10 24 [3] 1 × 10 25 [4] 2 × 10 25 Sol17. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 = a 3 ¢ = Z ×M N×a 3 = z×M 6.022×10 23 mol −1 ×a 3 Mass of a particle = molar mass avogadro' s number = ¢×a 3 z number of particles in a given mass = given mass mass of a particle = given mass×z ¢×a 3 given mass = 100 g density = 10 g cm −3 cell edge, a = 200 pm = 2×10 −8 cm z = 4 (for an FCC lattice) substituting the values and calculating, number of particles in 100 g = 5 × 10 24 18 A compound CuCl has face centred cubic structure. Its density is 3.4 g cm −3 . The length of unit cell is. [Cu = 63.5 , Cl = 35.5] [1/] 5.783 Aº [2] 6.783 Aº [3] 7.783 Aº [4] 8.783 Aº Sol18. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 =a 3 ¢ = Z ×M N×a 3 = z×M 6.022×10 23 mol −1 ×a 3 a = 3 . ¢×N A z ×M a = 3 . ¢×6.022×10 23 mol −1 z×M density, ¢ = 3.4 g cm −3 face centred structure hence z = 4 molar mass of Cu = 63.5 g mol −1 molar mass of Cl = 35.5 g mol −1 molar mass of CuCl, M = 63.5 + 35.5 = 99 g mol −1 substituting the values and calculating, a = 3 . 3.4×6.022×10 23 mol −1 4×99 g mol −1 = 5.783 Aº 19 The density of KCI is 1.9893 g cm −3 and the length of a side unit cell is 6.29082 A as determined by X-Rays diffraction. The value of Avogadro's-number calculated from these data is [1/] 6.017x 10 23 [2] 6.023 x 10 23 [3] 6.03 x 10 23 [4] 6.017 x 10 19 Sol19. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 =a 3 ¢ = Z ×M N A ×a 3 N A = Z ×M ¢×a 3 KCl has the same structure as NaCl, that is the Face Centred Cubic with z = 4 Molar mass of K = 39 g mol −1 Molar mass of Cl = 35.5 g mol −1 Molar mass of KCl = 74.5 g mol −1 ¢ = 1.9893 g cm −3 Edge length,a = 6.29082 A = 6.29082×10 −8 cm Substituting and calculating: N A = 4×74.5 g mol −1 1.9893 g cm −3 ×6.29082×10 −8 cm 3 N A = 6.017x 10 23 20 The pyknometric density of sodium chloride crystal is 2.165 x 10 3 kg m –3 while its X-rays density is 2.178 x 10 3 kg m –3 . The fraction of unoccupied sites in sodium chloride crystal is [CBSE PMT 2003] [1/] 5.96 x 10 –3 [2] 5.96 [3] 5.96 x 10 –2 [4] 5.96 x 10 –1 sol20. Th e pycnometer (fr o m Gr e e k : π (pukn o s) m e a ni n g "d e n s e "), al s o c all e d υκνός pyknometer o r s pecific gravity bottle, i s a fl a s k, u s u ally m a d e o f gl a s s , wit h a cl o s e - fitti n g gr o u n d gl a s s st o pp e r wit h a c a pill ary tub e thr o u g h it, s o th at air b u b bl e s m a y e s c a p e fr o m th e a pp ar at us . Thi s e n a bl e s th e d e n sity of a flui d t o b e m e a s ur e d a c c ur at ely, b y r ef er e n c e t o a n a ppr o pri at e w o r ki n g flui d s u c h a s w at e r o r m e r c ury, u si n g a n a n al yti c al b al a n c e . If th e fl a s k i s w e i g h e d e m pty, full o f w at er, a n d full of a li qui d w h o s e s p e cifi c gr a vity i s d e sir e d, th e s p e cifi c gr a vity of th e li qui d c a n e a s ily b e c al c ul at e d. Th e p arti cl e d e n sity o f a p o w d e r, t o w hi c h th e u s u al m e t h o d of w e i g hi n g c a n n ot b e a ppli e d, c a n al s o b e d e t e r mi n e d wit h a py c n o m e t er. Th e p o w d e r i s a d d e d t o th e py c n o m e t er, w hi c h i s th e n w e i g h e d, gi vi n g th e w e i g ht o f th e p o w d e r s a m pl e . Th e py c n o m e t e r i s th e n fill e d wit h a li qui d o f k n o w n d e n sity, in w hi c h th e p o w d e r i s c o m pl et el y in s ol u bl e . Th e w e i g ht of th e di s pl a c e d li qui d c a n th e n b e d e t e r mi n e d, a n d h e n c e th e s p e cifi c gr a vity o f th e p o w d er. X- ray density is the theoretical density obtained by X – ray diffraction pattern. It does not account for any defects. The presence of defects makes the pyknometric density different from the X- ray density pyknometric density = 2.165 x 10 3 kg m –3 X-rays density = 2.178 x 10 3 kg m –3 The fraction of unoccupied sites in the lattice = X – ray density – pyknometric density X – ray density = 2.178×10 3 kg m −3 – 2.165×10 3 kg m −3 2.178×10 3 kg m −3 = 5.96×10 −3 21 In a crystal both ions are missing from normal sites in equal number. This is an example of - [1] F-centres [2] Interstitial defect [3] Frenkel defect [4/] Schottky defect Sol21. Schottky defects arise when one positive ion and one negative ion are missing from their respective position leaving behind a pair of holes. These are more common in ionic compounds with high coordination number and having almost similar size of cations and anions. 22 A certain metal crystallizes in a simple cubic structure. At a certain temperature, it arranges to give a body centred structure. In this transition, the density of the metal [1] Decreases [2/] Increases [3] Remain unchanged [4] Changes without a definite pattern sol22. The efficiency of a body centred arrangement is higher than the simple cubic arrangement. Hence the conversion of the simple cubic arrangement to the body centred increases the density. 23 In a close packed array of N spheres, the number of tetrahedral holes are [1] N 2 [2] 4N [3/] 2N [4] N sol23. In any close packed arrangement of particles (face centred cubic arrangement that is cubic close packing or hexagonal close packing) the number of tetrahedral voids is double the number of particles making the arrangement where as the number of octahedral voids is equal to the number of particles present in the arrangement. 24 Na 2 SeO 4 and Na 2 SO 4 are [1/] Isomorphous [2] Polymorphs [3] Alloys [4] Ferromagnetic sol24. Se and S have very similar sizes hence the sulphate and selenate ions are very similar in size and properties. Hence sodium selenate and sodium sulphate form exactly the same crystal structures and so are isomorphous. 25 The edge length of cube is 400 PM. Its body diagonal would be - [1] 500 PM [2/] 693 PM [3] 600 PM [4] 566 PM Sol25. The body diagonal of a cube, BD = .3×edge length = .3 a = .3 x 400 pm = 692.82 pm or say 693 pm 26 A pure crystalline substance, on being heated gradually, first forms a turbid looking liquid and then the turbidity completely disappears. This behaviour is the characteristic of substances forming [BHU 2000] [1] Isomeric crystals [2/] Liquid crystals [3] Isomorphous’ crystals [4] Allotropic crystals sol26. Liquid crystals are partly ordered materials, somewhere between their solid and liquid phases. Their molecules are often shaped like rods or plates or some other forms that encourage them to align collectively along a certain direction. The order of liquid crystals can be manipulated with mechanical, magnetic or electric forces. Finally, liquid crystals are temperature sensitive since they turn into solid if it is too cold, and into liquid if it is too hot. This phenomenon can, for instance, be observed on laptop screens when it is very hot or very cold. 27 In the Bragg’s equation for diffraction of X-rays, n represents for [MP PMT 2000] [1] Quantum number [2/] An integer [3] Avogadro’s numbers [4] Moles sol27. The bragg's equation for the X -rays is n \ = 2dsin0 . Here n is the order of reflections or the order of diffraction and is always an integer. \ is the wavelength of the X – rays, d is the slit width and 0 is the angle of diffraction. 28 The melting point of RbBr is 682°C, while that of NaF is 988°C. The principal reason that melting point of NaF is much higher than that of RbBr is that [1] The two crystals are not isomorphous [2] The molar mass of NaF is smaller than that of RbBr [3/]’ The internuclear distance, r c + r a is greater for RbBr than for NaF [4] The bond in RbBr has more covalent character than the bond in NaF sol28. The melting point of a crystal is directly proportional to its lattice energy. The lattice energy in case of ionic crystals is itself dependent on the charge of the particles and the distance between them that is the bond length. Lattice energy is proportional to 1 r c +r a . The charge in case of NaCl as well as RbBr is the same but the bond length in case of RbBr is much more than NaCl. Hence the melting point of NaCl is much higher than RbBr. 29 The ionic radii of Rb + and I – are 1.46 and 2.16 A. The most probable type of structure exhibited by it is [1] CsCl type [2/] NaCl type [3] ZnS type [4] CaF 2 type sol29. The structures are generally determined by the relative size of the cation and anion as well as the stoichiometry. RbI is AB kind of arrangement and CaF 2 is AB 2 type. Hence this structure is not possible. Radius of Rb + is 1.46 and I − is 2.16. r + r − = 0.675 which is in between 0.414 – 0.732 hence Rb would occupy the octahedral voids of I. Hence it would be NaCl structure. 30 For tetrahedral co-ordination the radius ratio (r + / r – ) should be [1]. 0.414 - 0.732 [2] > 0.732 [3] 0.156 - 0.225 [4/] 0.225 - 0.414 Sol30. For tetrahedral coordination the ratio is between 0.225 to 0.414 31 Certain crystals produce electric signals on application of pressure. This phenomenon is called [1] Pyro electricity [2] Ferroelectricity [3/] Piezoelectricity [4] Ferrielectricity Sol31. The crystals which produce electricity on application of pressure are called piezoelectric crystals. Example is quartz. 32 A binary solid (A + B – ) has B – ions constituting the the face centred cube that is the cubic close packing and A + ions occupying 25% tetrahedral holes. The formula of solid is [1] AB [2] A 2 B [3/] AB 2 [4] AB 4 Sol32. the C.C.P. Lattice is the face centred cubic arrangement in which the atoms are present at the corners of the cube and the face centres. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 A face-centred atom in a cube is shared by two unit cells. Hence the contribution of the atom at the centre of the face = 1 2 If a face centred cube is divided in eight cubes then each cube contains one tetrahedral void. Since these voids are within the cube so each tetrahedral void has a contribution of one. The total contribution of the tetrahedral voids in a face centred cube becomes eight. It can be seen that the contribution of the particles making the cube is four and tetrahedral voids is eight. Hence there are two tetrahedral voids for each particle or we can say that: In any close packed arrangement the number of tetrahedral voids is twice the number of particles making that arrangement. B – ion is present at the the corners of the cube and the face centres. Hence the contribution of B – ions = contribution of corner + contribution of the face As there are 8 corners in a cube, the number of corner atom per unit cell = 8 x 1 8 = 1. As there are 6 faces in a cube, number of face-centred atoms per unit cell = 6 x 1 2 = 3. Total number of B – ions = 1 + 3 = 4 A + ions occupying 25% tetrahedral holes, hence the number of A + ions is 25 100 ×8 = 2 The formula of solid is A 2 B 4 or AB 2 33 The unit cell cube length for LiCl (just like NaCl structure) is 5.14 A. Assuming anion-anion contact, the ionic radius for chloride ion is [1/] 1.815 Å [2] 2.8 Å [3] 3.8 Å [4] 4.815 Å sol33. If we assume the anion anion contact then it would become a normal face centred cubic unit cell only and all the formula will be applicable. In the case of face centred unit cell the atoms at the corner do not touch each other. The atom touching each other are the ones on the face diagonal. So the face diagonal is equal to four times the radius that is d = 4r. Also the diagonal of the square is .2 a hence d = 4r = .2 a a = 2 .2 r 4r = .2 a or r = 1 2 .2 x a = .2 a 4 = 1 2 .2 x a where a = 5.14 Å or r = 1 2 .2 x 5.14 Å = 1.815 Å 34 Copper metal has a face-centred cubic structure with the unit cell length equal 0.361 nm. Picturing copper ions in contact along the face diagonal. The apparent radius of a copper ion is - [1/] 0.128 [2] 1.42 [3] 3.22 [4] 4.22 Sol34. In the case of face centred unit cell the atoms at the corner do not touch each other. The atom touching each other are the ones on the face diagonal. So the face diagonal is equal to four times the radius that is d = 4r. Also the diagonal of the square is .2 a hence d = 4r = .2 a a = 2 .2 r 4r = .2 a or r = 1 2 .2 x a = .2 a 4 = 1 2 .2 x a where a = 0.361 nm hence r = 1 2 .2 x 0.361 nm = 0.128 nm. 35 X + Y – ionic compound keeps bcc structure. The distance between two nearest ions is 1.73 Å. What would be edge length of the unit cell ? [1/] 200 pm [2] .3 .2 pm [3] 142.2 pm [4] .2 pm sol35. This is the CsCl arrangement. In this arrangement it is the simple cubic lattice of anion in which the cation occupies all the body centred voids that is all the centres of the unit cell. Since the cation is always bigger than the void it occupies the anions are not allowed to touch each other. The ions which touch each other are on the body diagonal. There are two anions and one cation between them. So the body diagonal is 2 times the bond length. Body diagonal, BD = 2r + +2r − .=2¦r + +r − ) = 2×bond length body diagonal of a cube = .3×edge length = .3 a a = body diagonal .3 = 2×bond length .3 The bond length given is, r + +r − = 1.73 Å hence a = 2×1.73 1.73 = 2 nm = 200 pm 36 In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3 rd of tetrahedral voids. The formula of the compound will be [1] X 4 Y 3 [2] X 2 Y 3 [3] X 2 Y [4] X 3 Y 4 Sol36. Since Y forms the ccp lattice, its contribution per unit cell is 4 The total contribution of the tetrahedral voids in any ccp unit cell is 8 X occupies 2/3 rd of the tetrahedral voids hence its contribution is 2 3 ×8 = 16 3 The formula of the compound will be X 16 3 Y 4 or X 4 Y 3 37 An element (atomic mass = 100 g/mol) having bcc structure has unit cell edge 400 pm. Then density of the element is . [CBSE 1996; AIIMS 2002] [1] 10.376 g cm −3 [2/] 5.188 g cm −3 [3] 7.289 g cm −3 [4] 2.144 g cm −3 Sol. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edgelength) 3 =a 3 ¢= Z×M N A ×a 3 The element has body centred cubic lattice hence z = 2 edge length, a = 400 pm = 4×10 −8 cm M = 100 g mol −1 ¢= 2×100 g mol −1 6.022×10 23 mol −1 ×4×10 −8 cm 3 = 5.188 g cm −3 38 Which of the following is incorrect statement: [1] ZnO on heating becomes yellow due to metal excess defect. [2] Density gets reduced due to Schottky defect [3/] Farbe centre is produced due to cation vacancy. [4] Both (1) and (2) Sol. Farbe centres are produced due to the presence of free electrons in the vacancy of anions. These are formed in anion deficiency defects. 39 How many unit cells are present in a cube- shaped ideal crystal of NaCl of mass 1.00 g [Atomic masses: Na = 23. Cl = 35.5] [AIEEE 2003] [1/] 2.57 x 10 21 unit cells [2] 5.14 x 10 21 unit cells [3] 1.28 x 10 21 unit cells [4] 1.71 x 10 21 unit cells Sol39. Number of formula units of NaCl present in 58.5 g NaCl = 6.022×10 23 Number of formula units of NaCl present in 1 g NaCl = 6.022×10 23 58.5 4 formula units of NaCl form 1 unit cell 6.022×10 23 58.5 units of NaCl form 1 4 × 6.022×10 23 58.5 unit cells = 2.57×10 21 unit cells 40 How many number of atoms are there in a cube based unit cell having one atom on each corner and two atoms on each body diagonal of cube [1] 8 [2] 6 [3] 4 [4/] 9 sol40. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 the diagonal of the cube is within it hence the contribution of the atoms on the diagonal is one each. The number of body diagonal in a cube = 4. total contribution of the atoms present on the diagonals = 2×4 = 8 contribution of the atoms at the corners = 1 8 ×8 = 1 Total contribution = 1 + 8 = 9 41 Which of the following statements is not true about NaCl structure [DCE 2001] [1] Cl – ions are in FCC arrangement [2/] Na + ions has coordination number 4 [3] Cl – ions has coordination number 6 [4] Each unit cell contains 4NaCl molecules Sol41. NaCl has sodium in octahedral voids of chloride. Hence its coordination number is 6 and not 4. 42 The number of atoms/molecules contained in one face centred cubic unit cell of a mono atomic substance is - [1] 1 [2] 2 [3/] 4 [4] 6 sol42. the C.C.P. Lattice is the face centred cubic arrangement in which the atoms are present at the corners of the cube and the face centres. Each corner of a cubic unit cell is a part of eight unit cells. Hence each atom present at the corners is shared amongst eight unit cells. Hence the contribution of the atom at the corner of the cube = 1 8 A face-centred atom in a cube is shared by two unit cells. Hence the contribution of the atom at the centre of the face = 1 2 The total contribution of the corners = 8× 1 8 = 1 The total contribution of the face centres = 6× 1 2 = 3 The total contribution in the cube = 1 + 3 = 4 43 Sodium metal crystallizes as a’ body centred cubic lattice with the cell edge 4.29 Å. What is the radius of sodium atom [AllMS 1999] [1/] 1.857x 10 –8 cm [2] 2.371 x 10 –7 cm [3] 3.817 x 10 –8 cm [4] 9.312 x 10 –7 cm Sol43. In a body centred cubic arrangement a particle is present in the centre of the cube. The particle in the centre of the cube is identical to the particle on the corners. The particles which are touching each other are the particles at the body diagonal. Four particles are in straight line. So the length of the body diagonal becomes body diagonal = 4×radius = 4r the body diagonal of a cube, BD = .3×edge length = .3 a hence body diagonal, BD = 4r = .3 a r = .3 a 4 edge length a = 4.29 Å r = .3 4.29 4 = 1.857Å = 1.857x 10 –8 cm 44 What type of crystal defect is indicated in the diagram shown below? Na + Cl − Na + Cl − Na + Cl − Cl − | ¦ Cl − Na + | ¦ Na + Na + Cl − | ¦ Cl − Na + Cl − Cl − Na + Cl − Na + | ¦ Na + [1] Frenkel defect [2/] Schottky defect [3] Interstitial defect [4] Frenkel and Schottky defects sol44. In this structure equal number of cation and anion are missing from their respective positions. Hence the charges are equal and stoichiometry does not change. This is the schottky defect. 45 Fraction of total volume occupied by atoms in a body centred cube is - [1] n 2 [2/] .3 n 8 [3] .2 n 6 [4] n 6 Sol45. In a body centred cubic unit cell the atoms are present on the corners and the body centres. Each corner of a cubic unit cell is a part of eight unit cells. Hence each of the eight atom present at the eight corners is shared amongst eight unit cells. The atoms at the centre of the cube is a part of only one cube hence its contribution is 1. The contribution of the corner of a cubic unit cell is 1/8. Hence number of atoms per unit cell for a body centred cubic cell = 8 x 1 8 + 1 = 2 In case of a body centred cubic unit cell the atoms at the body diagonal of unit cell touch each other. This makes the body diagonal of the unit cell four times the radius of the particle making the unit cell. Hence body diagonal = 4r = .3 a a = 4 .3 r ∴ Packing fraction = Volume occupied by one atom Volume of unit cell = 4 3 nr 3 a 3 = 4 3 nr 3 ¦ 4 .3 r ) 3 = .3n 8 46 If the r Na + = 95 pm, and r Cl − = 181 pm then the edge length of the NaCl unit cell is [1] 319 pm[2/] 552 pm[3] 390.2 pm [4] 276 pm sol46. NaCl is a face centred cube of Chloride ions in which Sodium ions occupy all the octahedral voids that is all the edge centres and the body centre. Hence the edge length is 2×bond length = 2צr Na ++r Cl −) = 2צ95+181) = 552 pm 47 At room temperature, sodium crystallizes in a body centred cubic lattice with a = 4.24 A. The theoretical density of sodium (At wt. of Na = 23) is [1/] 1.002 g cm −3 [2] 2.002 g cm −3 [3] 3.002 g cm −3 [4] None Sol47. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 = a 3 ¢ = Z ×M N A ×a 3 Na has body centred cubic lattice hence z = 2 edge length, a = 4.24 A = 4.24×10 −8 cm M = 23 g mol −1 ¢ = 2×23 g mol −1 6.022×10 23 mol −1 ×4.24×10 −8 cm 3 = 1.002 g cm −3 48 A solid having no definite shape is called [1/] Amorphous solid [2] Crystalline solid [3] Anisotropic solid [4] None 49 Which of the following magnetic properties are explained by the alignment of magnetic domains as ! ! . . ! . [1] Diamagnetic [2/] Paramagnetic [3] Ferromagnetic [4/] Antiferromagnetic Sol. The specie does contain the unpaired electrons but the number of upspin electrons are exactly same as the number of electrons having down spin. Hence this substance would behave as a diamagnetic substance and would be known as antiferromagnetic substance. 50 Potassium has a bcc structure with nearest neighbor distance 4.52 A. Its atomic weight is 39. Its density (in kg m –3 ) will be [AllMS 1991] [1] 454 [2] 804 [3] 852 [4/] 908 Sol50. Mass of a particle = molar mass avogadro' s number = M 6.022×10 23 mol −1 The total contribution in the unit cell = z mass of the unit cell = total contribution×mass of one particle = z × M 6.022×10 23 mol −1 Volume of a cubic unit cell = ¦edge length) 3 =a 3 ¢ = Z ×M N A ×a 3 K has body centred cubic lattice hence z = 2 edge length, a = 4.52 A = 4.52×10 −10 m M = 39×10 −3 kg mol −1 ¢ = 2×39×10 −3 kg mol −1 6.022×10 23 mol −1 ×4.52×10 −10 m 3 = 908 kg m −3