Solid State Solved Problems

Page # 16 SOLID STATE SOLVED PROBLEMS Q.1. Distinguish between : (i) Hexagonal and monoclinic unit cell. Sol. S.No. Point of Difference Hexagonal Monoclinic 1. Type of unit cells Pr imitive Primitive and end centred 2. Axial distances ab c ab c 3. Axial angles     90º   120º     90º   90º Q.2 A compound forms hexagonal close packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral volds ? Sol. Number of close-pakced spheres, N = 0.5 ml Now number of octahedral voids generated = N = 0.5 mol Number of tetrahedral voids generated = 2N = 2× 0.5 mol = 1.0 mol Total number of voids generated = 0.5 + 1.0 mol = 1.5 mol Thus total number of voids = 1.5 = 6.023 = 1023 = 9.0345 × 1023 Number of tetrahedral voids = 1 = 6.023 × 1023 = 6.023 × 1023. Q.3 A compound is formed by two elements M and N. The element N forms ccp structure and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ? Sol. The number of tetrahedral voids formed is equal to twice thenumber of atoms of elements N and only 1/3rd of these are occupied by the atoms of element M. Hence the ratio of number of atom M and N is 2 × 1 : 1 or 2 : 3 and the formula of the compound is M2 3 N3. Q.4. An element with molar mass 2.7 × 10.7 × 1022 kg mol–1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 103 kgm–3, what is the nature of the cubic unit cell ? Sol. Here,  = 2.7 × 103 kmg–3 M = 2.7 × 10–3 kg mol–1 a = 405 pm = 405 × 10–12 m We know that Density,  = or = Z= ZM a3NA  . a 2 . NA M 2.7  103  ( 405  10 12 )3  6.022  10 23 =4 2.7  10 2 Thus, value of Z(= 4) suggests that the lattice is cop. Page # 17 SOLID STATE Q.5 Calculate the efficiency of packing in case of metal crystal for (a) simple cubic (b) body-centred cubic (c) face-centred cubic (with the assumption that atoms are touching each other) Sol.(a) In simple cubic : A r is the B Suppose the length of the unit cell edge is a and radius of the sphere. Therefore, cell edge become equal to 2r. r C  a = 2r r a D We know that a simple cube has effectively one sphere because the spheres at the corners of the cube are shared by eight unit cells (8 × 1/8 = 1).  Volume of one sphere = 4 r3 3 Volume of the unit cell = (a3) = (2r)3 = 8r3 Packing efficiency = = = ( 4 / 3)r 3 8r 3 = 100 = Volume occupied by one particle in the unit cell  100 Total volume of unit cell  × 100 6 3.1416 × 100 = 52.4 % 6 (b) In body-centred cubic : In EFD, G FD2 = EF2 + ED2 B H b2 = a2 + aa = 2a2 A b= 2 a c F Now, in AFD, AF2 = AD2 + FD2 E c2 = a2 + b2 = a2 + 2a2 = 3a2 c= a a C b a D 3 a The length of the body diagonal c is equal to 4r, where r is the radius of the sphere, therefore, c = 4r, as all the three sphere along the diagonal touch each other. Therefore, 3 a = 4r ; a = 4r 3 ....(i) 3  4  r =  ....(ii)  3  We know, per unit cell in bcc structure, has effectively 2 spheres and total volume of two spheres is 2 × a3 (4/3) r3 and volume of the cube is a3 is (4/ 3 r)3 Page # 18 SOLID STATE Therefore, Pacing efficiency = 2  4 / 3r 3  100 % [( 4 / 3 )r ]3 (c) = = Volume occupied to two sphers in the unit cell  100 Total volume of the unit cell 8 / 3r 3  100 [64 /(3 / 3 )r 3 ] = 68 % In face-centred cubic : ABC, AC2 = BC2 + AB2 In G b2 = a2 + a2 = 2a2 or b= 2a b = 4r = a= r r F =2 2r 2 E ....(i) b r a 2a 4r a A H if r is the radius of the sphere, we find or B C D 3  4r   = or a3 = (2 r )3 =   2 We know, per unit cell in fcc (or ccp) structure, has effectively 4 spheres and total volumes of four a3 spheres is equal to 4 × (4/3) pr3 and volume of the cube is a3 or (2 2r )3 using eq. Therefore, Packing efficienty Q.6 Sol. M= = Sol. 4  ( 4 / 3)r 3  100 % ( 2 2r )3 = (16 / 3)r 3  100% 16 2r 3 = 74 % Silver crystallises in fcc lattice. If edge legnth of thecell is 4.07 × 10 –8 cm and density in 10.5 g cm –3 , calculate the atomic mass of silver. Edge length, a = 4.07 × 10–8 cm  Volume of unit cell, a3 = (4.07 × 10–8)3 = 67.41 × 10–24 cm3 Atoms per unit cell of fcc, Z = 4 = Q.7 = ZM 3 a  NA   NA  a 3 Z 10.5  6.023  1023  67.41 10 24 = 106.57 g mol–1 4 A cubic solid is made of two elements PO and Q are at the corners of the cube and P at the body-centre. What is the formula of the compound ? What are the coordiantion number of P and Q ? We know that in simple cubic number of atoms at corner 1/8 = 8 = 1, and in body centre it is also 1. So the formula of compound is PQ. The structure of the compound formed by P and Q resembles CsCl structure. Thus coordination number of PO and Q is 8 each. Page # 19 SOLID STATE Q.8 Sol. Nobium crystallises in body-centred cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 92.90 u. According to question, Z=2  = 8.55 gm cm–3 M = 92.90 u We know that = ZM 3 a  NA 2  92.90 a3 = 8.55  6.023  1023 a3 = 36.09 × 10–24 cm3 a = 3.3 × 10–8 cm or 3.3 × 10–10 m We know for bcc structure r= 3 3 a= × 3.3 × 10–10 4 4 3 3 a= × 3.3 × 10–10 4 4 r = 1.43 × 10–10 m r= Q.9 Sol. If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between r and R. An octahedral void is represented by the shaded protion. Suppose the length of the each arm of the square is 2R (R is radius of atom). Let r be the radius of void. For right angled trangle ABC AC = ( g) R = 2 2 R = 2 (r + R) r= D 2R R or AC = R + R + 2r = 2r + 2R =2 (r + R) A 8R r or ( 2B )2  ( 2R )2 = r Also ( AB )2  (BC )2 = B C 2R 2R–R r = ( 2 – 1) R = 0.414 × R Q.10 Non-stoichlometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 : 1. Can you acount for the fact that this substance is ptype semiconductor ? Sol. Cu2O having Cu : O ratio less than 2 : 1 shows metal deficiency due to cation vacanicies. Metal deficient compounds conduct electricity through positive holes conduction mechanism therefore, it is a p-type semiconductor. Q.11 Calssify each of the following as being eiother a p-type of an n-type semiconductor : (i) Ge doped with In. (ii) B doped with Si. Sol. (i) p-type semicoinductor (ii) n-type semicoinductor Page # 20 SOLID STATE Q.12 Sol. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (a) What is the length of the unit cell ? (b) How many unit cells are there in 1.00 cm3 of aluminium ? (a) For ccp structure, radius of atom = edge length 2 2 or r= a 2 2 or a=2 2 r Edge length (a) = 125 × 2 2 = 354 pm (b) Volume of one unit cell (a3) = (354 × 10–12)3 a3 = 4.436 × 10–29 = 4.435 × 10–23 cm3 Number of unit cells of aluminium in 1 cm3 = Q.13 Sol. 1 4.436  10 23 = 3.25 × 1022 units cells If NaCl is doped with 10–3 mol % of SrCl2, What is the concentration of cation vacancies ? We know that, doping of SrCl2 to NaC brings in replacement of two Na+ ions by each Sr2+ ions, but each Sr2+ occupis only one lattice point. This produces one cation valencey. Thus, doping of 10–3 mole of SrCl2 in 100 moles of NaCl will produce cation vacancies = 10–3 mol. or 1.00 mole of NaCl will have cation vacancies after doping = 10 3 = 10–5 mol 100 OR The addition of SrCl2 to NaCl procues cation vacancies equal in number to that of Sr2+ ions. Number of moles of SrCl2 added to 1 mol of NaCl = 10 3 = 10–5 mol 100 Number of Sr2+ added to 1 mol of NaCl = 10–5 × 6.02 × 1023 ions = 6.02 × 1018 ions Concentration of cation vacancies = 6.02 × 1018 Q.14 In a closed packed structure of oxides, one eighth of the tetrahedral voids are occupied by bivalent cations A and half of the octahedral holes are occupied by trivalent cations B. Calculate the molecular formula of oxide. Sol. No. of bivalent cations A = 1 1 ×2= 8 4 No. of trivalent cation B = 1 1 ×1= 2 2 Hence, the formula of the compound is A1/4 B1/2O or AB2O4 Q.15 Define F-centres, Mention some properties of solids which have F-centres in the crystals. Sol. When anions are missing from their lattice sites, leaving holes which are occupied by the electrons to maintain the electrical balance, the vacant sites occupied by electrons are called F–centres. Properties due to F–centre are as follows: (i) The crystals having F–centres are coloured. (ii) When the crystals are exposed to light, they become photoconductors. (iii) The crystals having F-centre are paramagnetic due to the presence of unpaired electrons Page # 21 SOLID STATE Q.16 Sol. CsCl forms a bcc lattice. Cs+ and Cl– ions are in contact with each other long the body diagonal. The radius of Cl– ions is 181 pm and unit cell edge length is 412 pm. Calculatate radius of Cs+. In right-angled ABC AC = AB2  AC2 = a2  a2 = 2a In right-angled ACD CD  = AB2  AC2 = 3a 2 (rCs  rCl ) = = 2 (rCs  2  181) = 1.732 × 412 3a 2rCa = (1.732 × 412 ) – (2 × 181) rCa Q.17 =  2a 2  a2 = 351.584 351.584 = 175.792 pm 2 Examine the illustration of a portion of the defective crystal given below and answer the following questions: (i) What are these type of vacancy defects called ? (ii) How is the density of a crystal affected by these defects? (iii) Name one ionic compound which can show this type of defect in the crystalline state. (iv) How is the stoichiometry of the compound affected? Sol. (i) Schottky defects (ii) decreases Q.17 (a) Why ice has porous structure? (b) How does conductivity of superconductors vary with temperature? (c) What is the effect of Frenkel defect on the electrical conductivity of crystalline solids ? (a) Due to cage like structure on account of intermolecular H-bonding among H2O molecules. (b) It decreases with rise in temperature. (c) Conductivity increases because vacancies are created. Sol. Q.18 Sol. (iii) Sodium chloride (iv) No effect Analysis shows that Nickes oxide has formula Ni0.98 O1.00. What fractions of the nickel exist as Ni+2 and Ni+3 ions. Let Ni+3 be x mol Ni+2 be 0.98 – x (Sum of O. N. of all atoms in a compound is equal to zero) So 3x + 2 (0.98 – x) + (–2) x = 0.04  of Ni3+ = 0.04  100 = 4.08 0.98 % of Ni+2 = 100 – 4.08 = 95.92