{"id":33040,"date":"2022-01-30T20:58:57","date_gmt":"2022-01-30T15:28:57","guid":{"rendered":"https:\/\/infinitylearn.com\/surge\/?p=33040"},"modified":"2025-06-03T13:23:54","modified_gmt":"2025-06-03T07:53:54","slug":"important-questions-for-class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions","status":"publish","type":"post","link":"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/","title":{"rendered":"Important Questions for Class 12 Chemistry Chapter 1 The Solid State Class 12 Important Questions"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_37 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" style=\"display: none;\"><label for=\"item\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input type=\"checkbox\" id=\"item\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1' style='display:block'><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/#Class_12_Solid_State_Important_Questions_with_Answers\" title=\"Class 12 Solid State Important Questions with Answers\">Class 12 Solid State Important Questions with Answers<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/#The_Solid_State_Class_12_Important_Questions_Short_Answer_Type_-_I_SA_-_1\" title=\"The Solid State Class 12 Important Questions Short Answer Type \u2013 I (SA \u2013 1)\">The Solid State Class 12 Important Questions Short Answer Type \u2013 I (SA \u2013 1)<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/#More_Resources_for_Class_12\" title=\"More Resources for Class 12\">More Resources for Class 12<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/#The_Solid_State_Class_12_Important_Questions_Short_Answer_Type_-_II_SA_II\" title=\"The Solid State Class 12 Important Questions Short Answer Type \u2013 II [SA II]\">The Solid State Class 12 Important Questions Short Answer Type \u2013 II [SA II]<\/a><ul class='ez-toc-list-level-4'><li class='ez-toc-heading-level-4'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/infinitylearn.com\/surge\/cbse\/study-materials\/important-questions\/class-12-chemistry-chapter-1-the-solid-state-class-12-important-questions\/#The_Solid_State_Class_12_Important_Questions_Long_Answer_Type_LA\" title=\"The Solid State Class 12 Important Questions Long Answer Type (LA)\">The Solid State Class 12 Important Questions Long Answer Type (LA)<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<p>Find important questions with answers for Class 12 Chemistry, Chapter 1: Solid State. Aligned with <a href=\"https:\/\/infinitylearn.com\/surge\/cbse\/cbse-syllabus\/\"><strong>CBSE syllabus<\/strong><\/a>, these questions help you revise concepts for annual exams and entrance tests like <a href=\"https:\/\/infinitylearn.com\/surge\/neet-exam\"><strong>NEET<\/strong><\/a> and <a href=\"https:\/\/infinitylearn.com\/surge\/iit-jee\"><strong>JEE<\/strong><\/a>. Download the PDF below for easy access.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"Class_12_Solid_State_Important_Questions_with_Answers\"><\/span>Class 12 Solid State Important Questions with Answers<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1. Which point defect in crystals does not alter the density of the relevant solid? (Delhi) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> Frenkel defect.<\/p>\n<p><strong>Question 2. Which point defect in its crystal units alters the density of a solid? (Delhi) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> Schottky defect.<\/p>\n<p><strong>Question 3. Which point defect in its crystal units increases the density of a solid? (Delhi) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> Metal excess defect increases the density of a solid. It is due to presence of extra cations in the interstitial sites.<\/p>\n<p><strong>Question 4. How do metallic and ionic substances differ in conducting electricity? (All India) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> The electrical conductivity in metallic substances is due to free electrons while in ionic substances the conductivity is due to presence of ions.<\/p>\n<p><strong>Question 5. Which point defect of its crystals decreases the density of a solid? (Delhi &amp; All India) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> Schottky defect.<\/p>\n<p><strong>Question 6. What is the number of atoms in a unit cell of a face-centred cubic crystal? (All India) 2009<\/strong><\/p>\n<p><strong>Answer:<\/strong> The number of atoms in a unit cell of fcc-crysta! is 4 atoms.<\/p>\n<p><strong>Question 7. Write a feature which will distinguish a metallic solid from an ionic solid. (Delhi) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> The electrical conductivity in metallic solid is due to free electrons while in ionic solid the conductivity is due to presence of ions.<\/p>\n<p><strong>Question 8. Which point defect in crystals of a solid does not change the density of the solid? (Delhi) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> Frenkel defect.<\/p>\n<p><strong>Question 9. Which point defect in crystals of a solid decreases the density of the solid? (Delhi) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> Schottky defect.<\/p>\n<p><strong>Question 10. What type of interactions hold the molecules together in a polar molecular solid? (All India) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> Dipole-dipole forces of attractions hold the molecules together in a polar molecular solid.<\/p>\n<p><strong>Question 11. What type of semiconductor is obtained when silicon is doped with arsenic? (All India) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> n-type semiconductor.<\/p>\n<p><strong>Question 12. Write a distinguishing feature of metallic solids. (All India) 2010<\/strong><\/p>\n<p><strong>Answer:<\/strong> Metallic solids possess high electrical and thermal conductivity due to presence of free electrons.<\/p>\n<p><strong>Question 13. \u2018Crystalline solids are anisotropic in nature.\u2019 What does this statement mean? (Delhi) 2011<\/strong><\/p>\n<p><strong>Answer:<\/strong> It means that crystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.<\/p>\n<p><strong>Question 14. Which stoichiometric defect in crystals increases the density of a solid? (Delhi) 2011<\/strong><\/p>\n<p><strong>Answer:<\/strong> Interstitial defect in crystals increases the density of a solid.<\/p>\n<p><strong>Question 15. What is meant by \u2018doping\u2019 in a semiconductor? (Delhi) 2012<\/strong><\/p>\n<p><strong>Answer:<\/strong> Addition of a suitable impurity to a semiconductor to increase its conductivity is called doping.<\/p>\n<p><strong>Question 16. Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre. (Delhi) 2012<\/strong><\/p>\n<p><strong>Answer:<\/strong> Metallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution or metals conduct electricity through electrons and ionic substances through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.<\/p>\n<p><strong>Question 17. How may the conductivity of an intrinsic semiconductor be increased? (All India) 2012<\/strong><\/p>\n<p><strong>Answer:<\/strong> The conductivity is increased by adding an appropriate amount of suitable impurity. This process is called as intrinsic doping.<\/p>\n<p><strong>Question 18.<\/strong><br \/>\nWhich stoichiometric defect increases the density of a solid? (All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nInterstitial defect increases the density of a solid.<\/p>\n<p><strong>Question 19.<\/strong><br \/>\nWhat are n-type semiconductors? (All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nn-type semiconductor : They are obtained by doping silicon with an element of group15, like P, As etc.<\/p>\n<p style=\"text-align: center;\"><strong>Click Here: <a href=\"https:\/\/infinitylearn.com\/surge\/cbse\/cbse-class-12-syllabus\/\">CBSE Syllabus For Class 12<\/a><\/strong><\/p>\n<p><strong>Question 20.<\/strong><br \/>\nWhat type of stoichmetric defect is shown by AgBr and Agl ? (Comptt. All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nAgBr shows both Frenkel defect and Schottky defect whereas Agl shows Frenkel defect.<\/p>\n<p><strong>Question 21.<\/strong><br \/>\nWhat type of defect can arise when a solid is heated ? (Comptt. All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nVacancy defects can arise when a solid is heated.<\/p>\n<p><strong>Question 22.<\/strong><br \/>\nWhy does LiCl acquire pink colour when heated in Li vapours? (Comptt. All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nThis is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).<\/p>\n<p><strong>Question 23.<\/strong><br \/>\nHow many atoms constitute one unit cell of a face-centered cubic crystal? (Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n4 atoms constitute one unit cell of a fee crystal.<\/p>\n<p><strong>Question 24.<\/strong><br \/>\nWhat type of stoichiometric defect is shown by AgCl? (Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFrenkel defect is shown by AgCl.<\/p>\n<p><strong>Question 25.<\/strong><br \/>\nWhat type of substances would make better Permanent Magnets: Ferromagnetic or Ferrimagnetic? (Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFerromagnetic substances would make better I permanent magnets<br \/>\nExample : Fe, Co, Ni etc.<\/p>\n<p><strong>Question 26.<\/strong><br \/>\nCalculate the number of atoms in a face centred cubic unit cell. (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nIn face centered cubic arrangement, number of lattice points are : 8 + 6.<br \/>\n\u2234 Lattice points per unit cell = \\(8 \\times \\frac{1}{8}+6 \\times \\frac{1}{2}\\) = 4<\/p>\n<p><strong>Question 27.<\/strong><br \/>\nOn heating a crystal of KC1 in potassium vapour, the crystal starts exhibiting a violet colour. What is this due to? (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nThe Cl ions diffuse to the surface and combine j with atoms which get ionized by losing electrons. ! These electrons are trapped in anions vacancies j and act as F-centre which imparts violet colour to the crystal.<\/p>\n<p><strong>Question 28.<\/strong><br \/>\nWhich type of ionic substances show Schottky defect in solids? (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nHighly ionic compounds with high coordination rjuniber and small difference in size of cations and anions show schottky defect.<\/p>\n<p><strong>Question 29.<\/strong><br \/>\nHow many atoms per unit cell (z) are present in bcc unit cell? (Comptt. Delhi) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nNumber of atoms in a unit cell of a body centred cubic structure :<br \/>\nContribution by 8 atoms on the corners<br \/>\n= \\(\\frac{1}{8}\\) \u00d7 8 = 1<br \/>\nContribution by the atom presents within the body = 1<br \/>\n\u2234 Total number of atoms present in the unit cell = 1 + 1 = 2 atoms<\/p>\n<p style=\"text-align: center;\"><strong>Click Here: <a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-syllabus\/class-12\/\">NCERT Syllabus for Class 12<\/a><\/strong><\/p>\n<p><strong>Question 30.<\/strong><br \/>\nWhat type of stoichiometric defect is shown by NaCl? (Comptt. Delhi) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSchottky defect is shown by NaCl.<\/p>\n<p><strong>Question 31.<\/strong><br \/>\nWrite a distinguishing feature between a metallic solid and an ionic solid. (Comptt. Delhi) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nThe electrical conductivity in metallic substances is due to free electrons while in ionic substances the conductivity is due to presence of ions.<\/p>\n<p><strong>Question 32.<\/strong><br \/>\nWhy are crystalline solids anisotropic? (Comptt. All India) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nCrystalline solids show different values of their some properties like electrical conductivity, refractive index, thermal expansion etc. in different directions.<\/p>\n<p><strong>Question 33.<\/strong><br \/>\nWhat is meant by \u2018antiferromagnetism\u2019? (Comptt. All India) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nAntiferromagnetism : These substances possess zero net magnetic moment because of presence of equal number of electrons with opposite spins.<\/p>\n<p><strong>Question 34.<\/strong><br \/>\nWrite a distinguishing feature of a metallic solid compared to an ionic solid. (Comptt. All India) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nMetallic solid conducts electricity in solid state but ionic solids do so only in molten state or in solution or metals conduct electricity through electrons and ionic substances through ions. Metallic solids are malleable and ductile while ionic solids are hard and brittle.<\/p>\n<p><strong>Question 35.<\/strong><br \/>\nWhat is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1\/3rd of tetrahedral voids? (Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFormula is X<sub>2<\/sub>Y<sub>3<\/sub>.<\/p>\n<p><strong>Question 36.<\/strong><br \/>\nWhat is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 2\/3rd of tetrahedral voids? (All India) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nY atoms are N (No. of tetrahedral voids are 2N), No. of tetrahedral voids occupied by X are<br \/>\n\\(\\frac{2}{3}\\) \u00d7 2N = \\(\\frac{4 \\mathrm{N}}{3}\\)<br \/>\nX : Y = 4N : 3N<br \/>\nFormula : X<sub>4<\/sub>Y<sub>3<\/sub><\/p>\n<p><strong>Question 37.<\/strong><br \/>\nWhat is the no. of atoms per unit cell (z) in a body-centred cubic structure? (Comptt. Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nContribution by the atoms present at eight comers = 8 \u00d7 \\(\\frac{1}{8}\\) = 1<br \/>\nContribution by the atoms present at centre = 1<br \/>\nTotal number of atoms present in unit cell = 1 + 1 = 2<\/p>\n<p><strong>Question 38.<\/strong><br \/>\nWhat type of stoichiometric defect is shown by AgCl? (Comptt. All India) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nAgCl shows Frenkel defect.<\/p>\n<p><strong>Question 39.<\/strong><br \/>\nWhat type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction? (Delhi) 2016<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFerromagnetism is shown by a substance if magnetic moments of domains are arranged in same direction.<\/p>\n<p><strong>Question 40.<\/strong><br \/>\nGive an example each of a molecular solid and an ionic solid. (All India) 2016<\/p>\n<p><strong>Answer:<\/strong><br \/>\nMolecular solid \u2192 Iodine (I<sub>2<\/sub>)<br \/>\nIonic solid \u2192 Sodium chloride (NaCl)<\/p>\n<p><strong>Question 41.<\/strong><br \/>\nA metallic element crystallises into a lattice having a pattern of AB AB \u2026 and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? (Comptt Delhi) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nTetrahedral void is formed in AB AB \u2026 pattern. The hexagonal close packing (hep) is formed in this arrangement.<\/p>\n<p><strong>Question 42.<\/strong><br \/>\nA metallic element crystallises into a lattice having a ABC ABC \u2026 pattern and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement? (Comptt. Delhi) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nOctahedral voids are formed in ABC ABC \u2026 pattern. The cubic close packing (ccp) is formed in this arrangement.<\/p>\n<p><strong>Question 43.<\/strong><br \/>\nWhat type of Stoichiometric defect is shown by AgCl? (Comptt. Delhi) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFrenkel defect.<\/p>\n<p><strong>Question 44.<\/strong><br \/>\nWhat type of stoichiometric defect is shown by NaCl? (Comptt. All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSchottky defect is shown by NaCl.<\/p>\n<p><strong>Question 45.<\/strong><br \/>\nWhich ionic compound shows both Frenkel and Schottky defects? (Comptt. All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSilver bromide (AgBr) shows both Schottky and Frenkel defect.<\/p>\n<h4><span class=\"ez-toc-section\" id=\"The_Solid_State_Class_12_Important_Questions_Short_Answer_Type_-_I_SA_-_1\"><\/span>The Solid State Class 12 Important Questions Short Answer Type \u2013 I (SA \u2013 1)<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>Question 46.<\/strong><br \/>\nExplain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of unit cell of its crystal. (All India) 2011<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSuppose edge of the unit cell = a pm<br \/>\nNumber of atoms present per unit cell = Z<br \/>\n\u2234 Volume of unit cell = (a pm)<sup>3<\/sup><br \/>\n= (a \u00d7 10<sup>-10<\/sup>cm)<sup>3<\/sup> = a<sup>3<\/sup> \u00d7 10<sup>-30<\/sup> cm<sup>3<\/sup><br \/>\nDensity of unit cell = \\(\\frac{\\text { Mass of unit cell }}{\\text { Volume of unit cell }}\\) \u2026\u2026\u2026\u2026\u2026\u2026 (i)<br \/>\nMass of unit cell = Number of atoms in the unit cell \u00d7 mass of each atom<br \/>\n= Z \u00d7 m<br \/>\nMass of each atom = \\(\\frac{\\text { Atomic mass }}{\\text { Avogadro\u2019s no. }}=\\frac{M}{N_{0}}\\)<br \/>\nSubstituting these values in equation (i), we get<br \/>\nDensity of unit cell = \\(\\frac{Z \\times M}{a^{3} \\times 10^{-30} \\times N_{0}}\\)<br \/>\nIf a is in cm, d = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{0}}\\) g\/cm<sup>3<\/sup><br \/>\n\u2234 Molar mass can be calculated as<br \/>\nM = \\(\\frac{d \\times a^{3} \\times N_{0}}{Z}\\)<\/p>\n<p><strong>Question 47.<\/strong><br \/>\nCalculate the packing efficiency of a metal crystal for a simple cubic lattice. (All India) 2011<\/p>\n<p><strong>Answer:<\/strong><br \/>\nPercentage efficiency of packing of simple cubic lattice = 52.4%.<\/p>\n<p><strong>Question 48.<\/strong><br \/>\nDefine the following terms in relation to crystalline solids :<br \/>\n(i) Unit cell (ii) Coordination number Give one example in each case. (All India) 2011<\/p>\n<p><strong>Answer:<\/strong><br \/>\n<strong>(i) Unit cell :<\/strong> The smallest three dimensional portion of a complete space lattice which when repeated over and again in different directions produces the complete space lattice is called the unit cell.<br \/>\nExample: Cubic unit cell, Hexagonal unit cell etc.<br \/>\n<strong>(ii) Coordination number :<\/strong> The number of nearest spheres with which a particular sphere is in contact is called co-ordination number.<br \/>\nExample : Co-ordination number of hexagonal (hep) structures is 12.<\/p>\n<p><strong>Question 49.<\/strong><br \/>\nThe unit cell of an element of atomic mass 108 u and density 10.5 g cm-3 is a cube with edge length, 409 pm. Find the type of unit cell of the crystal. [Given : Avogadro\u2019s constant = 6.023 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>] (Comptt. Delhi) 2012<br \/>\nAnswer:<br \/>\nSo it forms cubic- closed packed (ccp) lattice or fee structure.<\/p>\n<p><strong>Question 50.<\/strong><br \/>\nExplain the following terms with suitable examples : Ferromagnetism and Ferrimagnetism (Comptt. Delhi) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFerromagnetic solids : The solids which are strongly attracted by external magnetic field and do not lose their magnetism when the external field is removed are called ferromagnetic solids. The property, thus exhibited, is termed as ferromagnetism.<br \/>\nExample: Fe, Co and Ni show ferromagnetism at room temperature.<\/p>\n<p>Ferrimagnetic solids : The solids which are expected to show large magnetism due to the presence of unpaired electrons but in fact have small net magnetic moment, are called ferrimagnetic solids.<br \/>\nExample : Fe<sub>3<\/sub>O<sub>4<\/sub> and ferrites.<\/p>\n<h4><span class=\"ez-toc-section\" id=\"More_Resources_for_Class_12\"><\/span>More Resources for Class 12<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<ul>\n<li><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-12\/maths\/\"><strong>NCERT Solutions for Class 12 Maths<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-12\/physics\/\"><strong>NCERT Solutions for Class 12 Physics<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-12\/chemistry\/\"><strong>NCERT Solutions for Class 12 Chemistry<\/strong><\/a><\/li>\n<li><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-solutions\/class-12\/biology\/\"><strong>NCERT Solutions for Class 12 Biology<\/strong><\/a><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-books\/\">NCERT Books<\/a><\/strong><\/li>\n<li><strong><a href=\"https:\/\/infinitylearn.com\/surge\/study-materials\/ncert-books\/class-12\">NCERT Books for Class 12<\/a><\/strong><\/li>\n<\/ul>\n<p><strong>Question 51.<\/strong><br \/>\nAn element X crystallizes in f.c.c structure. 208 g of it has 4.2832 \u00d7 10<sup>24<\/sup> atoms. Calculate the edge of the unit cell, if density of X is 7.2 g cm<sup>-3<\/sup>. (Comptt. Delhi) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nZ = 4(fcc) A = 7,2 g\/cm<sup>3<\/sup> a = ?<br \/>\n4.2832 \u00d7 10<sup>24<\/sup> atoms have mass = 208 g<br \/>\n6.022 \u00d7 10<sup>23<\/sup> atoms have mass<br \/>\n= \\(\\frac{208}{4.2832 \\times 10^{24}}\\) \u00d7 6.022 \u00d7 10<sup>23<\/sup> = 29.24 (at. mass)<br \/>\na<sup>3<\/sup> = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{d \\times \\mathrm{N}_{\\mathrm{A}}}=\\frac{4 \\times 29.24}{7.2 \\times 6.022 \\times 10^{23}}\\)<br \/>\n= 269.6 x 10<sup>-24<\/sup> cm<sup>3<\/sup><br \/>\n\u2234 a = 6.46 x 10<sup>-8<\/sup> cm = 6.46 \u00c5<\/p>\n<p><strong>Question 52.<\/strong><br \/>\nWhat is a semiconductor? Describe the two main types of semiconductors. (Comptt. Delhi) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSemiconductor : The solid materials whose electrical conductivity lies between those of the typical metallic conductors and insulators are termed as semiconductors. The semiconductors possess conductivity in the range of 10<sup>2<\/sup> to 10<sup>-9<\/sup> ohm<sup>-1<\/sup> cm<sup>-1<\/sup>.<br \/>\nThese are of two types :<br \/>\n(a) n-type semiconductors : Doping of higher group element impurity forms n-type semiconductors, e.g. when \u2018As\u2019 is doped in \u2018Ge\u2019.<br \/>\n(b) p-type semicondctors : Impurity of lower groups forms electron deficient bond in the structure. Electron deficiency develops to p-hole.<\/p>\n<p><strong>Question 53.<\/strong><br \/>\nAccount for the following:<br \/>\n(i) Schottky defects lower the density of related solids.<br \/>\n(ii) Conductivity of silicon increases on doping it with phosphorus. (All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(i) Schottky defect produced due to missing of equal number of cation and anion from lattice as a result of which the density of the lattice solid decreases.<br \/>\n(ii) The conductivity of silicon increases due to negatively charged extra electron of doped pentavalent phosphorus.<\/p>\n<p>Question 54.<br \/>\nAluminium crystallizes in an fee structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal? (All India) 2013<br \/>\nAnswer:<br \/>\nFor fee, Formula : r = \\(\\frac{a}{2 \\sqrt{2}}\\)<br \/>\nGiven: r = 125 pm<br \/>\n\u2234 a = \\(2 \\sqrt{2} r\\) \u00f7 a = \\(2 \\sqrt{2}\\) \u00d7 125<br \/>\n\u21d2 a = 2 \u00d7 1.414 \u00d7 125 = 353.5 pm<\/p>\n<p><strong>Question 55.<\/strong><br \/>\n(a) Why does presence of excess of lithium makes LiCl crystals pink?<br \/>\n(b) A solid with cubic crystal is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? (All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).<br \/>\n(b) As atoms of Q are present at the 8 centres of the cube, therefore, number of atoms of Q in the unit cell = \\(\\frac{1}{8}\\) \u00d7 8 = 1<br \/>\nThe atom P is at the body centre .-. Number of atoms = 1<br \/>\nRatio of atoms P : Q = 1 : 1<br \/>\nHence, the formula of the compound is PQ.<\/p>\n<p><strong>Question 56.<\/strong><br \/>\n(a) What change occurs when AgCl is doped with CdCl<sub>2<\/sub>?<br \/>\n(b) What type of semiconductor is produced when silicon is doped with boron? (All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Impurity defect of ionic solids is produced when AgCl is doped with CdCl2. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.<br \/>\n(b) p-type semi-conductor is obtained when silicon is doped with boron.<\/p>\n<p>Question 57.<br \/>\nIf NaCl is doped with 10<sup>-3<\/sup> mole percent SrCl<sub>2<\/sub>, what will be the concentration of cation vacancies? (N<sub>A<\/sub> = 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>) (Comptt. All India) 2013<br \/>\nAnswer:<br \/>\n10<sup>-3<\/sup> mol percent means 100 moles of NaCl are doped with 10<sup>-3<\/sup> moles of SrCl<sub>2<\/sub><br \/>\n\u2234 1 mole of NaCl is doped with SrCl<sub>2<\/sub><br \/>\n= \\(\\frac{10^{-3}}{100}\\) = 10<sup>-5<\/sup> mole<br \/>\nSince each Sr<sup>2+<\/sup> ion introduces one cation vacancy<br \/>\n\u2234 Concentration of cation vacancies<br \/>\n= 10<sup>-5<\/sup> mol\/mol of NaCl<br \/>\n= 10<sup>-5 <\/sup>\u00d7 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup><br \/>\n= 6.02 \u00d7 10<sup>18<\/sup> mol<sup>-1<\/sup><\/p>\n<p><strong>Question 58.<\/strong><br \/>\nWhat is a semiconductor? Describe the tivo main types of semiconductors and contrast their conduction mechanism. (Comptt. All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nSemiconductor : The solids which have intermediate conductivities between metals and non-metals i,e. between 10<sup>-6<\/sup> to 10<sup>4<\/sup> \u03c0<sup>-1<\/sup> m<sup>-31<\/sup> are called semiconductors.<br \/>\nExample : Germanium and Silicon.<br \/>\nMain types of semiconductors are of two types :<br \/>\n<strong>(i) Intrinsic semiconductor :<\/strong> These are insulators at room temperature and become semiconductors when temperature is raised<br \/>\n<strong>(ii) Extrinsic semiconductor :<\/strong><br \/>\np-type semiconductor<br \/>\nn-type semiconductor<br \/>\nThese are formed by dopping impurity of lower or higher group.<br \/>\nThese are subdivided into two types :<br \/>\n<strong>\u2022 p-type semiconductor :<\/strong> When a silicon crystal is doped with atoms of group-13 elements like B, Al, Ga etc., the atom forms only 3 covalent bonds with the Si atom and 4<sup>th<\/sup> missing electron creates a hole which conducts electricity.<br \/>\n<strong>\u2022 n-type semiconductor :<\/strong> When a silicon crystal is doped with atoms of group-15 elements like P, As etc., then only four of the five valence electrons of each impurity atom, participate in 4 covalent bond formation and 5<sup>th<\/sup> e<sup>\u2013<\/sup> conducts electricity.<\/p>\n<p><strong>Question 59.<\/strong><\/p>\n<p>A compound forms hep structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids? (Comptt. All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nNo. of atoms in the hep = 0.5 \u00d7 6.022 \u00d7 10<sup>23<\/sup><br \/>\n= 3.011 \u00d7 10<sup>23<\/sup><br \/>\nNo. of octahedral voids<br \/>\n= No. of atoms in packing = 3.011 \u00d7 10<sup>23<\/sup><br \/>\nNo. of tetrahedral voids<br \/>\n= 2 \u00d7 No. of atoms in packing<br \/>\n= 2 \u00d7 3.011 \u00d7 10<sup>23<\/sup> = 6.022 \u00d7 10<sup>23<\/sup><br \/>\n\u2234 Total no. of voids<br \/>\n= 3.011 \u00d7 10<sup>23 <\/sup>+ 6.022 \u00d7 10<sup>23<\/sup> = 9.033 \u00d7 10<sup>23<\/sup><\/p>\n<p><strong>Question 60.<\/strong><br \/>\nAn element crystallizes in a structure having fee unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 \u00d7 10<sup>23<\/sup> atoms. (Comptt. All India) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n24 \u00d7 10<sup>23<\/sup> atoms of an element have mass = 200 g<br \/>\n\u2234 6.022 \u00d7 10<sup>23<\/sup> atoms of an element have mass<br \/>\n= \\(\\frac{200}{24 \\times 10^{23}}\\) \u00d7 6.022 \u00d7 10<sup>23<\/sup> = 50.18 g<br \/>\nGiven : a = 200 pm = 200 \u00d7 10<sup>-12<\/sup> cm,<br \/>\nZ = 4 (For fee), M = 50.18 g<\/p>\n<p><strong>Question 61.<\/strong><br \/>\nAn element with density 11.2 g cm<sup>-3<\/sup> forms a f.c.c. lattice with edge length of 4 \u00d7 10<sup>-8<\/sup> cm. Calculate the atomic mass of the element. (Given : N<sub>A<\/sub> = 6.022 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>) (Delhi) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : p = 11.2 g cm<sup>-3<\/sup>, a = 4 \u00d7 10<sup>-8<\/sup> cm<br \/>\nFor fee lattice, Z = 4<br \/>\nUsing formula,<\/p>\n<p><strong>Question 62.<\/strong><br \/>\nExamine the given defective crystal 2014<br \/>\nAnswer the following questions :<br \/>\n(i) What type of stoichiometric defect is shown by the crystal?<br \/>\n(ii) How is the density of the crystal affected by this defect?<br \/>\n(tii) What type of ionic substances show such defect? (Delhi)<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(i) Schottky defect<br \/>\n(ii) Density of the crystal decreases<br \/>\n(iii) NaCl (Ionic solids having approximate equal size of cations and anions)<\/p>\n<p><strong>Question 63.<\/strong><br \/>\nAn element with density 2.8 g cm<sup>-3<\/sup> forms a f.c.c. unit cell with edge length 4 \u00d7 10<sup>-8<\/sup> cm. Calculate the molar mass of the element.<br \/>\n(Given : N<sub>A<\/sub> = 6.022 \u00d7 10<sup>23<\/sup>mol<sup>-1<\/sup>) (All India) 2014<\/p>\n<p><strong>Question 64.<\/strong><br \/>\n(i) What type of non-stoichiometric point<br \/>\ndefect is responsible for the pink colour of LiCl?<br \/>\n(ii) What type of stoichiometric defect is shown by NaCl? (All India) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(i) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres).<br \/>\n(ii) Schottky defect is shown by NaCl.<\/p>\n<p><strong>Question 65.<\/strong><br \/>\nHow will you distinguish between the following pairs of terms :<br \/>\n(i) Tetrahedral and octahedral voids<br \/>\n(ii) Crystal lattice and unit cell (All India) 2014<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td width=\"15\"><\/td>\n<td style=\"text-align: center;\" width=\"121\"><strong>Tetrahedral voids<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"132\"><strong>Octahedral voids<\/strong><\/td>\n<\/tr>\n<tr>\n<td width=\"15\">1.<\/td>\n<td width=\"121\">It is much smaller than the size of spheres in  the packing.<\/td>\n<td width=\"132\">Size is much larger than tetrahedral voids.<\/td>\n<\/tr>\n<tr>\n<td>2.<\/td>\n<td width=\"121\">Each tetrahedral void is surrounded by 4 spheres. Hence, co\u00adordination no. is 4.<\/td>\n<td width=\"132\">Each octahedral void is surrounded by 6 spheres. Hence, its co\u00adordination no. is 6.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>(ii) A regular arrangement of the constituent particles of a crystal in a three dimensional space is called crystal lattice.<br \/>\nThe smallest three dimensional portion of a complete crystal lattice, which when repeated over and again in different directions produces the complete crystal lattice is called the unit cell.<\/p>\n<p>Question 66.<br \/>\n(i) Write the type of magnetism observed when the magnetic moments are oppositively aligned and cancel out each other.<br \/>\n(ii) Which stoichiometric defect does not change \u00bb the density of the crystal? (All India) 2014<br \/>\nAnswer:<\/p>\n<ol>\n<li>Diamagnetism is observed when the magnetic moments are oppositively aligned and cancel out each other.<\/li>\n<li>Frenkel defect does not change the density of the crystal.<\/li>\n<\/ol>\n<p><strong>Question 67.<\/strong><br \/>\n(i) Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers.<br \/>\n(ii) Which stoichiometric defect decreases the density of the crystal? (All India) 2014<br \/>\nAnswer:<\/p>\n<ol>\n<li>Ferrimagnetism is observed.<\/li>\n<li>Schottky defect decreases the density of the crystal.<\/li>\n<\/ol>\n<p>68. Define the following terms: (Comptt. Delhi) (2016)<br \/>\n(i) n-type semiconductor<br \/>\n(ii) Ferrimagnetism<\/p>\n<p><strong>Answer:<\/strong><br \/>\n<strong>(i) n-type semiconductor :<\/strong> When Si\/Ge is doped with group 15 element.<br \/>\n<strong>(ii) Ferrimagnetism :<\/strong> When magnetic domains are aligned in parallel and anti-parallel directions in unequal numbers.<\/p>\n<p>Question 69.<br \/>\nExplain the following terms with suitable examples : (Comptt. All India) (2016)<br \/>\n(i) Frenkel defect (ii) F-centres<br \/>\nAnswer:<br \/>\n<strong>(i) Frenkel defect :<\/strong> The defect in which the smaller ion\/cation is dislocated to a nearby interstitial site.<br \/>\nExample : Silver halides, ZnS.<br \/>\n<strong>(ii) F-centres :<\/strong> The anion vacancy occupied by an electron is called F-centre in Alkali metal halides.<br \/>\nExample : NaCl, KC1, Li Cl.<\/p>\n<p><strong>Question 70.<\/strong><br \/>\nCalculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a face-centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol<sup>-1<\/sup>) (Comptt. All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\n1 mole of Aluminium = 27 g = 6.022 \u00d7 10<sup>23<\/sup><br \/>\nHence, No. of atoms present in 27 g of Al<br \/>\n= \\(\\frac{6.022 \\times 10^{23}}{27}\\)<br \/>\nAs f.c.c. unit cell contains 4 atoms<br \/>\n\u2234 No. of f.c.c. unit cells present<br \/>\n= \\(\\frac{6.022 \\times 10^{23} \\times 8.1}{27 \\times 4}\\)<br \/>\n= 0.45165 \u00d7 10<sup>23<\/sup> = 4.5165 \u00d7 10<sup>22<\/sup><\/p>\n<h3><span class=\"ez-toc-section\" id=\"The_Solid_State_Class_12_Important_Questions_Short_Answer_Type_-_II_SA_II\"><\/span>The Solid State Class 12 Important Questions Short Answer Type \u2013 II [SA II]<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>Question 71.<\/strong><br \/>\nIron has a body centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm<sup>-3<\/sup>. Use this information to calculate Avogadro\u2019s number (At. mass of Fe = 56 g mol<sup>-1<\/sup>). (Delhi &amp; All India) 2009<\/p>\n<p>Answer:<br \/>\nGiven :<br \/>\na = 286.65 pm = 286.65 \u00d7 10<sup>-10<\/sup>,<br \/>\nd = 7.87 g cm<sup>-3<\/sup>, M = 56 g mol<sup>-1<\/sup><br \/>\nZ = 2 N<sub>A<\/sub> = ?<br \/>\n\u2234 Avogadro\u2019s number N<sub>A<\/sub> = 6.022 \u00d7 10<sup>23<\/sup><\/p>\n<p><strong>Question 72.<\/strong><br \/>\nSilver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four comer atoms.) (All India) 2009<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : a = 409 pm r = ?<br \/>\nFor fee unit cell, the formula is a<br \/>\nr = \\(\\frac{a}{2 \\sqrt{2}}\\)<br \/>\nor r = \\(\\frac{409}{2 \\sqrt{2}}=\\frac{409}{2 \\times 1.414}=\\frac{409}{2.828}\\)<br \/>\n\u2234 r = 144.62<br \/>\n\u2234 Radius of an atom of silver = 144.62 pm<\/p>\n<p><strong>Question 73.<\/strong><br \/>\nThe well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca<sup>2+<\/sup> ions and 8 F<sup>\u2013<\/sup> ions and that Ca<sup>2+<\/sup> ions are arranged in a fee lattice. The F<sup>\u2013<\/sup> ions fill all the tetrahedral holes in the face centred cubic lattice of Ca<sup>2+<\/sup> ions. The edge of the unit cell is 5.46 \u00d7 10<sup>-8<\/sup> cm in length. The density of the solid is 3.18 g cm<sup>-3<\/sup>. Use this information to calculate Avogadro\u2019s number (Molar mass of CaF<sub>2<\/sub> = 78.08 g mol<sup>-1<\/sup>). (Delhi) 2010<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven<br \/>\nEdge of the unit cell (a) = 5.46 \u00d7 10<sup>-8<\/sup> cm<br \/>\nDensity (P) = 3.18 g cm<sup>-3<\/sup><\/p>\n<p><strong>Question 74.<\/strong><br \/>\nThe density of copper metal is 8.95 g cm<sup>-3<\/sup>. If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure?<br \/>\n(Given : At. mass of Cu = 63.54 g mol<sup>-1<\/sup> and N<sub>A<\/sub> = 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>) (Delhi &amp; All India) 2010<\/p>\n<p><strong>Answer:<\/strong><br \/>\nIf copper atom were simple cubic :<br \/>\na = 2 \u00d7 r = 2 \u00d7 127.8 pm = 255.6 pm<br \/>\n= 255.6 pm = 255.6 \u00d7 10<sup>-10<\/sup> cm<br \/>\nZ = 1<br \/>\np = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{\\mathrm{A}}}=\\frac{1 \\times 63.54}{\\left(255.6 \\times 10^{-10}\\right)^{3} \\times\\left(6.02 \\times 10^{23}\\right)}\\)<br \/>\n\u2234 P = 6.34 g cm<sup>-3<\/sup><br \/>\nActual density = 8.95 g cm<sup>-3<\/sup><br \/>\nHence copper atom is not simple cubic.<br \/>\nIf copper atom were body-centred :<br \/>\n\u2234 P = 8.21 g cm<sup>-3<\/sup><br \/>\nHence, copper atom is not body centered<br \/>\nIf copper atom were face-centered<br \/>\na = \\(2 \\sqrt{2}\\)<br \/>\nor a = 2 \u00d7 1.414 \u00d7 127.8 pm<br \/>\n= 361.4 pm = 361.4 \u00d7 10<sup>-10<\/sup> cm<br \/>\nP = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{\\mathrm{A}}}=\\frac{4 \\times 63.54}{\\left(361.4 \\times 10^{-10}\\right)^{3} \\times 6.02 \\times 10^{23}}\\)<br \/>\n\u2234 P = 8.94 cm<sup>-3<\/sup><br \/>\nHence, copper is face-centred cubic.<\/p>\n<p><strong>Question 75.<\/strong><br \/>\nSilver crystallises in face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of silver atom? (All India) 2010<\/p>\n<p><strong>Answer:<\/strong><br \/>\nRefer to Q. 72, Page 9<br \/>\nGiven : a = 409 pm r = ?<br \/>\nFor fee unit cell, the formula is a<br \/>\nr = \\(\\frac{a}{2 \\sqrt{2}}\\)<br \/>\nor r = \\(\\frac{409}{2 \\sqrt{2}}=\\frac{409}{2 \\times 1.414}=\\frac{409}{2.828}\\)<br \/>\n\u2234 r = 144.62<br \/>\n\u2234 Radius of an atom of silver = 144.62 pm<\/p>\n<p><strong>Question 76.<\/strong><br \/>\nSilver crystallizes in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four comer atoms.) (Delhi) 2011<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : a = 400 pm, r = ?<br \/>\nFor fee unit cell : r = \\(\\frac{a}{2 \\sqrt{2}}\\)<br \/>\nor r = \\(\\frac{400}{2 \\times 1.4142}=\\frac{400}{2.828}\\) \u2234 r = 141.4<br \/>\n\u2234 Radius of the silver, r = 141.4 pm<\/p>\n<p><strong>Question 77.<\/strong><br \/>\nThe density of lead is 11.35 g cm<sup>-3<\/sup> and the metal crystallizes with fee unit cell. Estimate the radius of lead atom.<br \/>\n(At. Mass of lead = 207 g mol<sup>-1<\/sup> and N<sub>A<\/sub> = 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>) (Delhi) 2011<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : d = 11.35 g cm<sup>-3<\/sup><br \/>\nAccording to the formula<\/p>\n<p><strong>Question 78.<\/strong><br \/>\nTungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom? (Delhi) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nFor bee, unit cell : radius, r = \\(\\frac{\\sqrt{3} a}{4}\\) (\u2235 a = 316.5 pm)<br \/>\n= \\(\\frac{\\sqrt{3} \\times 316.5 \\mathrm{pm}}{4}=\\frac{1.732 \\times 316.5 \\mathrm{pm}}{4}\\) = 137 pm<\/p>\n<p><strong>Question 79.<\/strong><br \/>\nIron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm<sub>-3<\/sub>. Use this information to calculate Avogadro\u2019s number. (At mass of Fe = 55.845 u) (Delhi) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\n\u2234 Avogadro\u2019s number, N<sub>A<\/sub> = 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup><\/p>\n<p><strong>Question 80.<\/strong><br \/>\nCopper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal.<br \/>\n(Atomic mass of Cu = 63.55 u and Avogadro\u2019s number N<sub>A<\/sub> = 6.02 \u00d7 102<sup>23<\/sup> mol<sup>-1<\/sup>) (All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\nIn fee, lattice, Z = 4 atoms [fee = face centred cubic]\n<p><strong>Question 81.<\/strong><br \/>\nIron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm<sup>-3<\/sup>. Use this information to calculate Avogadro\u2019s number. (Atomic mass of Fe = 56.0 u) (All India) 2012<\/p>\n<p><strong>Question 82.<\/strong><br \/>\n(a) Some of the glass objects recovered from ancient monuments look milky instead of being transparent. Why?<br \/>\n(b) Iron (II) oxide has a cubic structure and each side of the unit cell is 5\u00c5. If density of the oxide is 4 g cm<sup>-3<\/sup>, calculate the number of Fe<sup>2+<\/sup> and O<sup>2-<\/sup> ions present in each unit cell. [Atomic mass : Fe = 56 u, O = 16 u; Avagadro\u2019s number = 6.023 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>]\n(Comptt. All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Some of the glass objects found from ancient monuments look to be milky in appearance because of crystallisation of glass.<br \/>\n(b) Volume of unit cell = a<sup>3<\/sup> = (5 A) = (5 \u00d7 10<sup>-8<\/sup>)<sup>3<\/sup><br \/>\n= 1.25 \u00d7 10<sup>-22<\/sup> cm<br \/>\nDensity of FeO = 4g cm<sup>-3<\/sup><br \/>\nMass of unit cell = Volume \u00d7 Density<br \/>\n= 1.25 \u00d7 10<sup>-22<\/sup> \u00d7 4 g<br \/>\n= 5 \u00d7 10<sup>-22<\/sup> g<br \/>\nMass of FeO molecule per unit cell<br \/>\n= \\(\\frac{5 \\times 10^{-22} g}{1.195 \\times 10^{-22} g}\\) = 4.19 \u2248 4<br \/>\nThus 4Fe<sup>+2<\/sup>, 4O<sup>-2<\/sup> will be present in each unit cell.<\/p>\n<p><strong>Question 83.<\/strong><br \/>\n(a) What are intrinsic semi-conductors? Give an example.<br \/>\n(b) What is the distance between Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> ions in NaCl crystal if its density is 2.165 g cm<sup>-3<\/sup>? [Atomic Mass of Na = 23u, Cl = 35.5u; Avogadro\u2019s number = 6.023 \u00d7 10<sup>23<\/sup>] (Comptt. All India) 2012<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Intrinsic semi-conductors : These are insulators at room temperature and become semi-conductors when temperature is raised, Example : silicon and germanium.<br \/>\n(b) Applying the formula<\/p>\n<p><strong>Question 84.<\/strong><br \/>\n(a) What type of semiconductor is obtained when silicon is doped with boron?<br \/>\n(b) What type of magnetism is shown in the following alignment of magnetic moments?<br \/>\n(c) What type of point defect is produced when AgCl is doped with CdCl<sub>2<\/sub>? (Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) p-type semi-conductor is obtained when silicon is doped with boron.<br \/>\n(b) Ferromagnetism is shown when the alignment of magnetic movements will be<br \/>\n(c) Impurity defect of inonic solids is produced when AgCl is doped with CdCl<sub>2<\/sub>. Due to this defect vacancies are created that result in higher electrical conductivity of the solid.<\/p>\n<p><strong>Question 85.<\/strong><br \/>\nAn element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.0 g cm<sup>-3<\/sup>. Also calculate the radius of an atom of this element. (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : For bcc structure, Z = 2<br \/>\nEdge of the unit cell, a = 250 pm<br \/>\nDensity of the element, p = 8.0 g\/ cm<sup>3<\/sup><br \/>\nM = ? r = ?<br \/>\nUsing the formula<\/p>\n<p><strong>Question 86.<\/strong><\/p>\n<p>Iron (II) oxide has a cubic structure and each unit cell has a size of 5 \u00c5. If density of this oxide is 4 g cm<sup>-3<\/sup>, calculate the number of Fe<sup>2+<\/sup> and O<sup>2-<\/sup> ions present in each unit cell.<br \/>\n(Atomic mass of Fe = 56, O = 16, N<sub>A<\/sub> = 6.023 \u00d7 10<sup>23<\/sup> and 1 \u00c5 = 10<sup>-8<\/sup> cm) (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : p = 4g cm<sup>-3<\/sup><br \/>\na = 5\u00c5 = 5 \u00d7 10<sup>-8<\/sup> cm M = 72 g\/\u2019mol, Z = ?<br \/>\nUsing the formula for cubic crystals<br \/>\nThere are four formula units of FeO present per unit cell. Hence it has face-centred cubic lattice where each Fe<sup>2+<\/sup> and O<sup>2-<\/sup> are four in number.<\/p>\n<p><strong>Question 87.<\/strong><br \/>\nNiobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm<sup>-3<\/sup>, calculate atomic radius of niobium, given its atomic mass 93u. (Comptt. Delhi) 2013<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : p = 8.55 g cm<sup>3<\/sup>,<br \/>\nM = 93 g\/mol<sup>-1<\/sup><br \/>\nZ = 2 (For bcc) a = ?<br \/>\nUsing formula<br \/>\na = (36.1)<sup>1\/3<\/sup> \u00d7 10<sup>2<\/sup> \u03c1m<br \/>\n\u2234 a = 3.304 \u00d7 10<sup>2<\/sup> \u03c1m = 330.4 \u03c1m<br \/>\nFor bcc r<sub>1<\/sub> = \\(\\frac{\\sqrt{3}}{4}\\) a = \\(\\frac{\\sqrt{3}}{4}\\) \u00d7 330.4 = 143.1 \u03c1m<\/p>\n<p><strong>Question 88.<\/strong><br \/>\nThe density of copper is 8.95 g cm<sup>-3<\/sup>. It has a face centred cubic structure. What is the radius of copper atom?<br \/>\n(Atomic mass Cu = 63.5 g mol<sup>-1<\/sup>, N<sub>A<\/sub> = 6.02 \u00d7 10<sup>23<\/sup> mol<sup>-1<\/sup>) (Comptt. Delhi) 2014<\/p>\n<p><strong>Question 89.<\/strong><br \/>\nIron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm<sup>-3<\/sup>. Use this information to calculate Avogadro\u2019s number.<br \/>\n(Atomic mass of Fe = 55.84 g mol<sup>-1<\/sup>) (Comptt. Delhi) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven :<br \/>\na = 286.65 pm = 286.65 \u00d7 10<sup>-10<\/sup>,<br \/>\nd = 7.87 g cm<sup>-3<\/sup>, M = 56 g mol<sup>-1<\/sup><br \/>\nZ = 2 N<sub>A<\/sub> =?<br \/>\n\u2234 Avogadro\u2019s number N<sub>A<\/sub> = 6.022 \u00d7 10<sup>23<\/sup><\/p>\n<p><strong>Question 90.<\/strong><br \/>\nIron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm<sup>-3<\/sup>. Use this information to calculate Avogadro\u2019s number.<br \/>\n(Gram atomic mass of Fe = 55.84 g mol<sup>-1<\/sup>). (Comptt. All India) 2014<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven :<br \/>\na = 286.65 pm = 286.65 \u00d7 10<sup>-10<\/sup>,<br \/>\nd = 7.87 g cm<sup>-3<\/sup>, M = 56 g mol<sup>-1<\/sup><br \/>\nZ = 2 N<sub>A<\/sub> = ?<br \/>\n\u2234 Avogadro\u2019s number N<sub>A<\/sub> = 6.022 \u00d7 10<sup>23<\/sup><\/p>\n<p><strong>Question 91.<\/strong><br \/>\nAn element with molar mass 27 g mol<sup>-1<\/sup> forms a cubic unit cell with edge length 4.05 \u00d7 10<sup>-8<\/sup> cm. If its density is 2.7 g cm<sup>-3<\/sup>, what is the nature of the cubic unit cell? (Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\n= 6.022 \u00d7 10<sup>-2<\/sup> \u00d7 66.43 = 4.0004 = 4<br \/>\nIt has Face centred cubic cell\/fee.<\/p>\n<p><strong>Question 92.<\/strong><br \/>\nExamine the given defective crystal :<br \/>\nAnswer the following questions :<br \/>\n(i) Is the above defect stoichiometric or non- stoichiometric?<br \/>\n(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.<br \/>\n(iii) How does this defect affect the density of the crystal? (All India) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(i) It is stoichiometric defect.<br \/>\n(ii) Schottky defect, e.g. NaCl.<br \/>\n(iii) Density of crystal decreases.<\/p>\n<p><strong>Question 93.<\/strong><br \/>\nDefine the following :<br \/>\n(i) Schottky defect<br \/>\n(ii) Frenkel defect<br \/>\n(iii) F-centre (Comptt. Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(i) Schottky defect: If in an ionic crystal of type A B , equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called Schottky defect.<br \/>\n(ii) Frenkel defect : If an ion leaves its site from its lattice site and occupies the interstitial site and maintains electrical neutrality, then it is called Frenkel defect.<br \/>\n(iii) F-centre : The centres which are created by trapping of electrons in anionic vacancies<br \/>\nand which are responsible for imparting colour to the crystals are called F-centres. (F = Fabre)<\/p>\n<p><strong>Question 94.<\/strong><br \/>\nSilver crystallises in fee lattice. If edge length of the unit cell is 4.077 \u00d7 10<sup>-8<\/sup> cm, then calculate the radius of silver atom. (Comptt. All India) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : a = 4.077 \u00d7 10<sup>-8<\/sup> cm r = ? for fee lattice<br \/>\nUsing formula,<br \/>\nRadius (r) = \\(\\frac{a}{2 \\sqrt{2}}\\)<br \/>\nr = \\(\\frac{4.077 \\times 10^{-8}}{2 \\times 1.414} \\mathrm{cm}=\\frac{4.077 \\times 10^{-8}}{2.828}\\)<br \/>\n\u2234 r = 1.441 \u00d7 10<sup>-8<\/sup> cm<\/p>\n<p><strong>Question 95.<\/strong><br \/>\nAn element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 \u00d7 10<sup>24<\/sup> atoms. (Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven: a = 250 pm = 250 \u00d7 10<sup>-10<\/sup> cm<br \/>\nz = 4 (for fee)<br \/>\nM = ? d = ?<br \/>\nUsing formula : d = \\(\\frac{z \\times M}{a^{3} N_{A}}\\)<br \/>\n\u2235 2 \u00d7 10<sup>24<\/sup> atoms of an element have mass = 300g<br \/>\n\u2234 6.022 \u00d7 10<sup>23<\/sup> atoms of an element have mass<br \/>\n= \\(\\frac{300 \\times 6.022 \\times 10^{23}}{2 \\times 10^{24}}\\) = 90.33 g<br \/>\nNow M = 90.33 g<\/p>\n<p><strong>Question 96.<\/strong><br \/>\nAn element crystallizes in a b.c.c. lattice with cell edge of 500pm. The density of the element is 7.5g cm<sup>-3<\/sup>. How many atoms are present in 300 g of the element? (All India) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven: For b.c.c. structure, z = 2<br \/>\nEdge of the unit cell, a = 500 pm = 500 \u00d7 10<sup>10<\/sup> cm<br \/>\nDensity d = 7.5 g cm<sup>-3<\/sup><br \/>\nUsing the formula,<br \/>\n\u2235 282.28 g of the element contains = 6.022 \u00d7 10<sup>23<\/sup> atoms<br \/>\n\u2234 300 g of the element contains<br \/>\n= \\(\\frac{6.022 \\times 10^{23}}{282.28}\\) \u00d7 300<br \/>\n= \\(\\frac{1806.6}{282.28} \\times 10^{23}\\) \u00d7 10<sup>23<\/sup> = 6.40 \u00d7 10<sup>23<\/sup> atoms<\/p>\n<p><strong>Question 97.<\/strong><\/p>\n<p>If NaCl is doped with 10<sup>-3<\/sup> mol % of SrCl<sub>2<\/sub>, what is the concentration of cation vacancies? (Comptt. Delhi) 2015<\/p>\n<p><strong>Answer:<\/strong><br \/>\nConcentration of SrCl<sub>2<\/sub> = 10<sup>-3<\/sup> mol% = 10<sup>-3<\/sup>\/100 mol = 10<sup>-5<\/sup> mol<br \/>\n1 mol of NaCl on doping procuces = 6.022 \u00d7 10<sup>23<\/sup> cation vacancies<br \/>\nTherefore, 10<sup>-5<\/sup> mol of NaCl on doping produces = 6.022 \u00d7 10<sup>23<\/sup> \u00d7 10<sup>-5<\/sup> = 6.022 \u00d7 10<sup>18<\/sup> cation vacancies<\/p>\n<p><strong>Question 98.<\/strong><br \/>\nSilver crystallises in f.c.c. lattice. If edge length of the cell is 4.07 \u00d7 10<sup>-8<\/sup> cm and density is 10.5 g cm<sup>-3<\/sup>, calculate the atomic mass of silver.  (Comptt. All India) 2015<\/p>\n<p><strong>Question 99.<\/strong><br \/>\n(a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon<br \/>\n(b) ZnO turns yellow on heating. Why?<br \/>\n(c) What is meant by groups 12-16 compounds? Give an example. (All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Silicon carbide is a covalent or network solid while Argon is a non-polar molecular solid.<br \/>\n(b) ZnO shows metal excess defect due to presence of extra cations, i.e., Zn<sup>2+<\/sup> ions in interstitial sites which on heating changes into yellow due to loss of oxygen.<br \/>\n\\(\\mathrm{ZnO} \\stackrel{\\Delta}{\\longrightarrow} \\mathrm{Zn}^{2+}+\\frac{1}{2} \\mathrm{O}_{2}+2 \\mathrm{e}^{-}\\)<br \/>\n(c) Group 12-16 compounds are imperfect covalent compounds in which the ionic character depends on the electronegativities of the two elements, e.g., ZnS, CdS, etc.<\/p>\n<p><strong>Question 100.<\/strong><br \/>\n(a) Based on the nature of intermolecular forces, classify the following solids: Benzene, Silver<br \/>\n(b) AgCl shows Frenkel defect while NaCl does not. Give reason.<br \/>\n(c) What type of semiconductor is formed when Ge is doped with Al? (All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Benzene \u2014 Molecular solid (non-polar)<br \/>\nSilver \u2014 Metallic solid<br \/>\n(b) Due to intermediate radius of AgCl, the size of Ag<sup>+<\/sup> is smaller than larger Na<sup>+<\/sup> ion of NaCl so it can easily occupy interstitial spaces and shows Frenkel defect.<br \/>\n(c) p-type semiconductor is formed when Ge is doped with Al.<\/p>\n<p><strong>Question 101.<\/strong><br \/>\n(a) Based on the nature of intermolecular forces, classify the following solids: Sodium sulphate, Hydrogen<br \/>\n(b) What happens when CdCl<sub>2<\/sub> is doped with AgCl?<br \/>\n(c) Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances? (All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\n(a) Sodium sulphate \u2014 Ionic solid<br \/>\nHydrogen \u2014 Molecular solid (non-polar)<br \/>\n(ib) Cd<sup>2+<\/sup> ion is dipositive and therefore addition of one Cd<sup>2+<\/sup> ion results in the loss of two Ag<sup>+<\/sup> ions from the lattice. But out of 2 holes obtained, one is occupied by Cd<sup>2+<\/sup> ion and one left empty. Hence, addition of CdCl<sub>2<\/sub> results in an impurity defect with cation vacancy.<br \/>\n(c) In ferrimagnetism, domains \/magnetic moments are aligned in opposite direction in unequal numbers while in antiferromagnetic substances, the domains align in opposite direction in equal numbers so they cancel magnetic moments completely, i.e., net magnetism is zero.<\/p>\n<p><strong>Question 102.<\/strong><br \/>\nAn element crystallises in b.c.c. lattice with cell edge of 400 pm. Calculate its density if 500 g of this element contains 2.5 \u00d7 10<sup>24<\/sup> atoms. (Comptt. Delhi) 2017<\/p>\n<p>Answer:<br \/>\nGiven : a = 400 pm = 400 \u00d7 10<sup>-10<\/sup> cm<br \/>\nZ = 2 (for bcc) M = ? d = ?<br \/>\nUsing formula, d = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{\\mathrm{A}}}\\)<br \/>\n\u2235 2.5 \u00d7 10<sup>24<\/sup> atoms of an element have mass = 500 g<br \/>\n\u2234 6.022 \u00d7 20<sup>23<\/sup>atoms of an element have mass<br \/>\n= \\(\\frac{240.88}{6.4 \\times 6.022}=\\frac{240.88}{38.5408}\\)<br \/>\n\u2234 d = 6.25 g cm<sup>-3<\/sup><\/p>\n<p><strong>Question 103.<\/strong><br \/>\nAn element crystallises in fee lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 \u00d7 10<sup>24<\/sup> atoms. (Comptt. Delhi) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : a = 400 pm = 400 \u00d7 10<sup>-10<\/sup> cm<br \/>\nZ = 4 (for fee), M = ?, d = ?<br \/>\nUsing formula, d = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{\\mathrm{A}}}\\)<br \/>\n\u2235 2.5 \u00d7 10<sup>24<\/sup> atoms of an element have mass = 250 g<br \/>\n\u2234 6.022 \u00d7 10<sup>23<\/sup> atoms of an element have mass<br \/>\n= \\(\\frac{250 \\times 6.022 \\times 10^{23}}{2.5 \\times 10^{24}}\\)<br \/>\n\u2234 M = 60.22 g<br \/>\nSubstituting all values in formula :<br \/>\nd = \\(\\frac{4 \\times 60.22}{\\left(400 \\times 10^{-10}\\right)^{3} \\times 6.022 \\times 10^{23}}=\\frac{240.88}{38.5408}\\)<br \/>\n\u2234 d = 6.25 g cm<sup>-3<\/sup><\/p>\n<p><strong>Question 104.<\/strong><br \/>\nAn element crystallises in bcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contains 2.5 \u00d7 10<sup>24<\/sup> atoms.<br \/>\n(Comptt. Delhi) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : a = 400 pm = 400 \u00d7 10<sup>-10<\/sup> cm<br \/>\nZ = 2 (for bcc), M = ?, d = ?<br \/>\nUsing formula, d = \\(\\frac{\\mathrm{Z} \\times \\mathrm{M}}{a^{3} \\times \\mathrm{N}_{\\mathrm{A}}}\\)<br \/>\n\u2235 2.5 \u00d7 10<sup>24<\/sup> atoms of an element have mass = 250g<br \/>\n\u2234 6.022 \u00d7 10<sup>23<\/sup> atoms of an element have mass<br \/>\n= \\(\\frac{250 \\times 6.022 \\times 10^{23}}{2.5 \\times 10^{24}}\\)<br \/>\n\u2234 M = 60.22 g<br \/>\nSubstituting all values in formula : .<br \/>\nd = \\(\\frac{2 \\times 60.22}{\\left(400 \\times 10^{-10}\\right)^{3} \\times 6.022 \\times 10^{23}}=\\frac{120.44}{38.5408}\\)<br \/>\n\u2234 d = 3.125 g cm<sup>-3<\/sup><\/p>\n<p><strong>Question 105.<\/strong><br \/>\nAn element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is 7.2 g\/cm<sup>3<\/sup>. (Comptt. All India) 2017<\/p>\n<p><strong>Answer:<\/strong><br \/>\nGiven : Cell edge, a = 288 pm = 288 \u00d7 10<sup>-10<\/sup> cm<br \/>\nDensity, d = 7.2 g\/cm<sup>3<\/sup><br \/>\nFor bcc formula, units per cell Z = 2, M = ?<br \/>\nUsing formula and substituting values,<\/p>\n<h4><span class=\"ez-toc-section\" id=\"The_Solid_State_Class_12_Important_Questions_Long_Answer_Type_LA\"><\/span>The Solid State Class 12 Important Questions Long Answer Type (LA)<span class=\"ez-toc-section-end\"><\/span><\/h4>\n<p><strong>Question 106.<\/strong><br \/>\n(a) An element has an atomic mass 93 g mol<sup>-1<\/sup> and density 11.5 g cm<sup>-3<\/sup>. If the edge length of its unit cell is 300 pm, identify the type of unit cell.<br \/>\n(b) Write any two differences between amorphous solids and crystalline solids. (Delhi) (2017)<\/p>\n<p><strong>Answer:<\/strong><\/p>\n<p>(a) Given:<br \/>\nM = 93 g mol<sup>-1<\/sup>; \u03c1 = 11.5 g cm<sup>-3<\/sup>;<br \/>\na = 300 pm = 300 \u00d7 10<sup>-10<\/sup> cm = 3 \u00d7 10<sup>-8<\/sup> cm<br \/>\nUsing formula,<br \/>\nZ = \\(\\frac{\\rho \\times a^{3} \\times N_{A}}{M}\\)<br \/>\n= \\(\\frac{11.5 \\times\\left(3 \\times 10^{-8}\\right)^{3} \\times 6.022 \\times 10^{23}}{93}\\)<br \/>\n= 2.01 (approx.)<br \/>\nAs the number of atoms present in given unit cells are coming nearly equal to 2, hence the given units cell is body centered cubic unit cell (BCC).<\/p>\n<table border=\"2\">\n<tbody>\n<tr>\n<td style=\"text-align: center;\" width=\"312\"><strong>Amorphous solids<\/strong><\/td>\n<td style=\"text-align: center;\" width=\"312\"><strong>Crystalline solids<\/strong><\/td>\n<\/tr>\n<tr>\n<td width=\"312\">(i) They are  isotropic, i.e., they will show same value of physical perties in directions.<\/td>\n<td width=\"312\">They are aniso\u00adtropic, i.e.,  tropic, i.e., value   of physical properties will be different when measured along different directions.<\/td>\n<\/tr>\n<tr>\n<td width=\"312\">(ii) They have short range order.<\/td>\n<td width=\"312\">(ii) They have long range order.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong>Question 107.<\/strong><br \/>\n(a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = 27 g mol<sup>-1<\/sup>)<br \/>\n(b) Give reasons:<br \/>\n(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.<br \/>\n(ii) Silicon on doping with Phosphorus form n-type semiconductor.<br \/>\n(iii) Ferrimagnetic substances show better magnetism than antiferromagnetic substances. (Delhi) 2017<br \/>\n(a) Given:<br \/>\nMass of Al = 8.1,<br \/>\nAtomic mass of Al = 27 g mol<sup>-1<\/sup><br \/>\nNo. of atoms = \u03b7 \u00d7 6.022 \u00d7 10<sup>23<\/sup><br \/>\n= \\(\\frac{8.1}{27}\\) \u00d7 6.022 \u00d7 10<sup>23<\/sup><br \/>\n= 0.3 \u00d7 6.022 \u00d7 10<sup>23<\/sup><br \/>\n= 1.8066 \u00d7 10<sup>23<\/sup><br \/>\nSince one f.c.c. unit cell has 4 atoms<br \/>\n\u2234 No. of unit cells = \\(\\frac{1.8066 \\times 10^{23}}{4}\\)<br \/>\n= 4.5 \u00d7 10<sup>22<\/sup> unit cells<br \/>\n(b) (i) Schottky defect is shown by the ionic solids having very small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. NaCl exhibits Schottky defect because radius of both Na<sup>+<\/sup> and Cl<sup>\u2013<\/sup> have very small difference.<br \/>\n(ii) Phosphorus is pentavalent that is it has 5 valence electrons, an extra electron results in the formation of n-type semi conductors on doping with Silicon. The conductivity is due to presence of extra electrons.<br \/>\n(iii) In antiferromagnetic substances the magnetic moments of domains are half aligned in one direction and remaining half in opposite direction in the presence of magnetic field so magnetic moment will be zero while in ferrimagnetic substances the magnetic moments of domains are aligned in parallel and anti-parallel directions in unequal numbers, hence shows some value of magnetic moment.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Find important questions with answers for Class 12 Chemistry, Chapter 1: Solid State. 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