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Que. from IIT-JEE 8. 10 Yrs. Que. from AIEEE Student’s Name :______________________ Class :______________________ Roll No. :______________________ ADDRESS: R-1, Opp. Raiway Track, New Corner Glass Building, Zone-2, M.P. NAGAR, Bhopal (cid:1) : (0755) 32 00 000, 98930 58881, www.tekoclasses.com THE KEY E Crystalline solids: T A T Crystalline solids are those whose atom, molecules or ions have an ordered arrangement extending overS S a Long Range. example-(Rock salt, NaCl). U O S Amorphous solids: A G Amorphous solids are those whose constitutent particles are randomly arrange and have no ordered2 3 long range structure. example: Rubber, Glass ect. of 2 e TYPES OF CRYSTALLINE SOLIDS: g a P Type of Solid Intermolecular forces Properties Examples Ionic Ion-Ion forces Brittle, hard high Melting NaCl, KCl, MgCl 2 Dispersion Soft, low melting non- L Molecular forces/Dipole-Dipole conducting H2O, Br2, CO2, CH4 PA /H-bond O H m Covalent B s.co network Covalent bonds Hard: High melting C-Diamond SiO2 81 , sse Metallic Metallic bonds Variable hardness and melting Na, Zn, Cu, Fe 88 cla point conducting 0 5 o 3 k TYPES OF UNIT CELL: 9 w.te Collection of lattice points, whose repetition produce whole lattice is called a unit cell. The whole lattice0 98 ww can be considered to be made by repetion of unit cell. 0, 0 0 : 0 e 0 sit 2 b 3 we 1. Unit Cell: 5)- m 75 0 o ( fr H: e P g Unit Cell Parameters ) ka Crystal Systems Bravais Lattice Sir ac Intercepts Crystal Angles K. dy P 1 Cubic PrimiBtiovde,y F Caecnet Cereendtered, a = b = c α = β = γ = 90° S. R. u ( t A S Primitive, Face Centered, Y d 2 Orthorhombic a ≠ b ≠ c α = β = γ = 90° RI a Body Centered, End Centered A o K l wn 3 Rhombohedral Primitive a = b = c α = β = γ ≠ 90° R. o 4 Monoclinic Primitive, End Centered a ≠ b ≠ c α = γ = 90°, β ≠ 90° G D A 5 Triclinic Primitive a ≠ b ≠ c α ≠ β ≠ γ ≠ 90° H E U E 6 Tetragonal Primitive, Body Centered a = b ≠ c α = β = γ = 90° S FR 7 Hexagonal Primitive a = b ≠ c α = β = 90°, γ = 120° or : t c e r Di S, E S S A CL O K E T 1.1 Primitive or simple cubic (PS/SC) unit cell: Spheres in one layer sitting directly on top of those in E previous layer, so that all layers are identical. Each sphere is touched by six other, hence coordinationT A T number is 6. 52% of available space occupied by spheres. S S Example: Polonium crystallises in simple cubic arrangement. U O S A G 2 3 of Z = 1 ; C.N. = 6 3 e g a P 1.2 Body Centered cubic (BCC) unit cell: Spheres in one layer sit in the depression made by first layer in a-b-a-b manner. Coordination number is 8, and 68% of available space is occupied by atoms. Example: Iron, sodium and 14 other metal crystallises in this manner. L A P O H m Z = 2 ; C.N. = 8 B s.co 81 , sse1.3 Face centered cubic (FCC) unit cell: 88 a 5 clExamples : Al, Ni, Fe, Pd all solid noble gases etc. 0 o 3 k 9 w.te 0 98 ww Z = 4 ; C.N. = 12 0, 0 0 : 0 e2. Density of cubic crystals: 0 sit 2 b 3 weTYPE OF PACKING: 5)- m 3. Closest packing of atoms: This is the most efficient way of packing 74% of available space is occupied75 0 fro by spheres and coordination number is 12. H: ( e (i) Hexagonal close pack (A-B-A-B) type packing : Each layer has hexagonal arrangement of touchingP kag sphere and 3rd layer is similar (exactly on top) of first layer. Sir) ac(ii) Cubic close pack (A-B-C-A-B-C): AB layers are similar to HCP arrangement but third layer is offsetK. P from both A and B layers. The fourth layer is exactly on top of first layer. R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A H E U E S FR or : t c e r Di S, E S S A L C O K E T Hexagonal primitive unit cell E T A T S S U O S A G 2 3 of 4 e g a P 4. Types of voids 4.1 Tetrahedral void L A P O H m B s.co 81 , e 8 ss 8 a 5 cl 0 o 3 k 9 w.te 4.2 Octahedral void 0 98 ww 0, 0 0 : 0 e 0 sit 2 b 3 we 5)- m 75 0 o ( fr H: e P g ) ka Sir ac 5. Radius ratio K. P R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A E 5.1 Radius ratio for co-ordination number 3 UH E S FR (Triangular Arrangement): r+ + r– = 32 3r– ; rr+− = 2−33 = 0.155 ctor : e r Di S, E S S A L C O K E T 5.2 Radius ratio for coordination number 4 E T A 3a 3 T (Tetrahedral arrangement): r+ + r– = ; 4r– = 2a = r– S 4 2 US O S A G 2 3 of 5 e g a r+ 3− 2 P = = 0.225 r− 2 L A P O m 5.3 Radius ratio for coordination number 6: r+ + r– = 2 r– BH sses.co (Octahedral Arrangement) or rr−+ = 2–1 = 0.414 8881 , a 5 cl Radius ratio for coordination number 4 0 o 3 k (Square plannar arrangement) 9 w.te 0 98 ww 0, 0 0 : 0 e 0 sit 2 b 3 we 5)- m 75 0 o ( fr H: e P g ) ka 5.4 Radius ratio for coordination number 8 : r+ + r– = 3 a Sir ac 2 K. dy P (Body centered cubic crystal) r+ + r– = 3 r– S. R. u ( t A S Y d RI a A o K l wn R. o G D A E r+ UH FRE r− = 3–1 = 0.732 or : S t c e r Di S, E 6. Types of ionic structures S S A 6.1 Rock salt structure:(NaCl) Larger atom formic ccp L C arrangement and smaller atom filling all octahedral voids. O K E T E 6.2 Zinc blende (sphalerite) structure:(ZnS) Larger T A T atom formic ccp arrangement and smaller atom filling S S half of alternate tetrahedral voids U O S A G 2 6.3 Fluorite structure:(CaF2) Ca2+ forming ccp of 3 arrangement and F– filling all tetrahedral voids. e 6 g a P 6.4 Antifluorite structure :(Li O) O2– ion forming ccp 2 and Li+ taking all tetrahedral voids. L A P O H m B s.co 6.5 Cesium halide structure: (CsCl) Cl– at the corners 81 , sse of cube and Cs+ in the center. 88 a 5 cl 0 o 3 k 9 w.te 6.6 Corundum Structure: (Al2O3) O2– forming hcp and Al3+ filling 2/3 octahedral voids. 0 98 ww 6.7 Rutile structure: (TiO2) O2– forming hcp while Ti4+ ions occupy half of the octahedral voids. 0, 0 : 6.8 Pervoskite structure:(CaTiO ) Ca2+ in the corner of 0 0 e 3 0 sit cube O2– at the face center and Ti4+ at the centre of cube. 2 b 3 we 5)- m 75 6.9 Spinel and inverse spinel structure: (MgAl O )O2– forming fcc, Mg2+ filling 1/8 of tetrahedral voids0 fro and Al3+ taking half of octahedral voids.In an in2ve4rse spinel structure, O2– ion form FCC lattice, A2+ ionsH: ( e P g occupy 1/8 of the tetrahedral voids and trivalent cation occupies 1/8 of the tetrahedral voids and 1/4 of) ka the octahedral voids. Sir ac K. P 7. Crystal defects: R. dy Point defects: When ions or atoms do not hold the theoretical position, this is called point defect. Point S. u ( t defects are of two types: A S Y d (I) Stoichiometric defects: RI a A o (a) Schottky defect: Due to missing of ions from lattice point in pairs. K l wn (b) Frenkel defect: It is caused due to the creation of lattice vacancy as a result of misplaced ion inR. o interstitial site. G D A E (cid:2) Schottky defect common in ionic solid with high coordination number. NaCl, KCl, KBr UH E (cid:2) S FR Frenkel defect :- Solid with low coordination number ZnS, AgBr. or : (II) Non-Stoichiometric defects: Ratio of positive and negative ion differ from that indicated by chemicalt c e formula. r Di (cid:2) Metal-excess defect : S, (a) A negative ion replaced by electron. (F-centre) E S S (b) Extra metal ion present in lattice and electron also present in interstitial site. A L (cid:2) C Metal-deficiency defect caused by : Cation missing from lattice point, electroneutrality maintained byO K metal ions with higher oxidation state as Fe °O. 0.94 E T THE ATLAS E T A T S S U O S A G 2 3 of 7 e g a P L A P O H m B s.co 81 , e 8 ss 8 a 5 cl 0 o 3 k 9 w.te 0 98 ww 0, 0 0 : 0 e 0 sit 2 b 3 we 5)- m 75 0 o ( fr H: e P g ) ka Sir ac K. P R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A H E U E S FR or : t c e r Di S, E S S A L C O K E T EXERCISE I E T A T Formula of ionic solid from unit cell description S S U O S Q.1 A cubic solid is made up of two elements A and B. Atoms B are at the corners of the cube and A at theA G body centre. What is the formula of compound. 2 3 of 8 Q.2 A compound alloy of gold and copper crystallizes in a cubic lattice in which gold occupy that lattice pointe g a at corners of the cube and copper atom occupy the centres of each of the cube faces. What is theP formula of this compound. Q.3 A cubic solid is made by atoms A forming close pack arrangement, B occupying one. Fourth of tetrahedral void and C occupying half of the octahedral voids. What is the formula of compound. L A P O Q.4 What is the percent by mass of titanium in rutile, a mineral that contain Titanium and oxygen, if structure H m B s.co cWanh abte i sd tehsec roibxeidda atiso an c nloumsebt epra ockf etidta anriruamy ?of oxide ions, with titanium in one half of the octahedral holes. 81 , e 8 ss 8 a 5 clQ.5 Spinel is a important class of oxides consisting of two types of metal ions with the oxide ions arranged in 0 o 3 k CCP pattern. The normal spinel has one-eight of the tetrahedral holes occupied by one type of metal ion 9 w.te and one half of the octahedral hole occupied by another type of metal ion. Such a spinel is formed by0 98 ww Zn2+, Al3+ and O2–, with Zn2+ in the tetrahedral holes. Give the formulae of spinel. 0, 0 0 : Edge length, density and number of atoms per unit cell 0 e 0 sit 2 webQ.6 KF crystallizes in the NaCl type structure. If the radius of K+ ions 132 pm and that of F– ion is 135 pm, 5)- 3 m what is the shortest K–F distance? What is the edge length of the unit cell? What is the closet K–K 75 o distance? (0 fr H: e P gQ.7 A closed packed structure of uniform spheres has the edge length of 534 pm. Calculate the radius of) ka sphere, if it exist in Sir ac (a) simple cubic lattice (b) BCC lattice (c) FCC lattice K. P R. dy S. uQ.8 Calculate the density of diamond from the fact that it has face centered cubic structure with two atoms ( t A S per lattice point and unit cell edge length of 3.569 Å. Y d RI a A o K wnlQ.9 An element crystallizes into a structure which may be described by a cubic type of unit cell having one R. o atom on each corner of the cube and two atoms on one of its body diagonals. If the volume of this unit G D cell is 24×10–24 cm3 and density of element is 7.2 g cm–3, calculate the number of atoms present in A H E U E 200 g of element. S FR or : Q.10 Silver has an atomic radius of 144 pm and the density of silver is 10.6 g cm–3. To which type of cubic t c e crystal, silver belongs? r Di S, Q.11 AgCl has the same structure as that of NaCl. The edge length of unit cell of AgCl is found to be 555 pmE S and the density of AgCl is 5.561 g cm–3. Find the percentage of sites that are unoccupied. S A L C O K E T Q.12 Xenon crystallises in the face-centred cubic lattice and the edge of the unit cell is E 620 pm. What is the nearest neighbour distance and what is the radius of xenon atom? T A T S S Q.13 The two ions A+ and B– have radii 88 and 200 pm respectively. In the closed packed crystal of compoundU O AB, predict the co-ordination number of A+. S A G 2 Q.14 CsCl has the bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl.of 3 9 e g Q.15 Gold crystallizes in a face centered cubic lattice. If the length of the edge of the unit cell is 407 pm,a P calculate the density of gold as well as its atomic radius assuming it to be spherical. Atomic mass of gold = 197 amu. Q.16 The density of KBr is 2.75 g cm–3 . The length of the edge of the unit cell is 654 pm. Show that KBr has L face centered cubic structure. A P (N = 6.023 ×1023 mol–1 , At. mass : K = 39, Br = 80) O H m B s.co Q.17 An element crystallizes in a structure having FCC unit cell of an edge 200 pm. Calculate the density, if81 , sse 200 g of this element contains 24×1023 atoms. 88 a 5 cl 0 o Q.18 The effective radius of the iron atom is 1.42 Å. It has FCC structure. Calculate its density3 k 9 w.te (Fe = 56 amu) 0 98 ww Q.19 A crystal of lead(II) sulphide has NaCl structure. In this crystal the shortest distance between Pb+2 ion0, 0 and S2– ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? Also calculate the0 : 0 site unit cell volume. 2 0 b 3 we Q.20 If the length of the body diagonal for CsCl which crystallises into a cubic structure with Cl– ions at the5)- m corners and Cs+ ions at the centre of the unit cells is 7 Å and the radius of the Cs+ ion is 1.69 Å, what is75 0 o ( fr the radii of Cl– ion? H: e P g ) ka Sir ac K. P R. dy S. u ( t A S Y d RI a A o K l wn R. o G D A H E U E S FR or : t c e r Di S, E S S A L C O K E T PROFICIENCY TEST E 1. Crystalline solids are isotropic. T A T S S 2. Rhombohedral, triclinic and hexagonal are the unit cells, which have only primitive arrangement possible.U O S A 3. Packing fraction of FCC and HP units cells are same. G 2 3 of 4. The minimum void fraction for any unit cell in any shape having only one type of atom and all voids0 1 e unfilled is 0.26. g a P 5. Packing fraction of a lattice structure depends on the radius of the atom crystallizing in it. 6. The location of tetrahedral voids in FCC unit cell are the centers of 8 minicubes forming a large cube. L A P O 7. Effective number of octahedral voids in a unit cell is equal to the effective number of atoms in the unit cell. H m B s.co 8. Radius ratio for co-ordination number 4 having tetrahedral and square planar geometry is same. 81 , e 8 ss 8 cla 9. The radius ratio value for co-ordination number 4 having square planar geometry and co-ordination0 5 o number 6 having octahedral geometry is same. 3 k 9 w.te 0 98 ww 10. pAa mckeitnagll oicf eslpehmereenst lcerayvsetsa loliuste sv oinidtos ian l tahteti claet tciocen t2a6in%in pge ar cseenqtu beyn vceo loufm laey oefr sth oisf AlatBti AceB is A eBm p..t.y.. .s p. Aacney. 0, 0 0 : 0 site 11. The relation between edge length (a) and radius of atom (r) for BCC lattice is_______. 2 0 b 3 we 12. The relation between edge length (a) and radius of atom (r) for FCC lattice is________. 5)- m 75 0 o ( fr 13. ABCABC......layering pattern is called ________ packing, found in _______ lattice. H: e P g ) ka 14. ABABAB.....layering pattern is called _______packing , found in _______ lattice. Sir ac K. P 15. Height (c) of the hexagonal primitive unit cell in terms of radius of atom (r) is________. R. dy S. u ( St 16. Anions would be in contact with each other only if the cation to anion radius for a given co-ordinationYA d number is_______. RI a A o K l wn 17. The number of tetrahedral voids in hexagonal primitive unit cell is________. R. o G D A H E 18. The limiting radius for co-ordination number 8 is_________. U E S FR 19. For cesium chloride structure, the interionic distance (in terms of edge length, a) is equal to _______. or : t c e r 20. Density of a crystal ___due to Schottky defect and _____due to Frankel defect. Di S, E S S A L C O K E T