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  • Question 1
    5 / -1
    Which primitive unit cell has unequal edge lengths (a ≠ b ≠ c) and all axial angles different from 90°?
    Solution

    Concept:

    Triclinic primitive unit cell has dimensions as, a ≠ b ≠ c and α ≠ β ≠ 90°.

    Among the seven basic or primitive crystalline systems, the triclinic system is most unsymmetrical, the triclinic. In other cases, edge length and axial angles are given as follows:

    Hexagonal: a = b ≠ c and α = β = 90°, γ = 120°

    Monoclinic: a ≠ b ≠ c and α = γ = 90°, β ≠ 90°

    Tetragonal: a ≠ b ≠ c and α = β = γ = 90°

    Additional Information

    • The parameters of all other crystal systems are given below:

    Primitive and non-primitive unit cell 

  • Question 2
    5 / -1
    Solids can be classified as ______ on the basis of the nature of order present in the arrangement of their constituent particles.
    Solution

    The correct answer is crystalline or amorphous.

    • Solids can be classified as crystalline or amorphous based on the nature of order present in the arrangement of their constituent particles.
    •  A crystalline solid usually consists of a large number of small crystals, each of them having a definite characteristic geometrical shape.
      • In a crystal, the arrangement of constituent particles (atoms, molecules, or ions) is ordered.
      • It has long-range order which means that there is a regular pattern of arrangement of particles that repeats itself periodically over the entire crystal.
      • Sodium chloride and quartz are typical examples of crystalline solids.
    • An amorphous solid consists of particles of irregular shape.
      • The arrangement of constituent particles (atoms, molecules, or ions) in such a solid has only short-range order.
      • Such portions are scattered and in between the arrangement is disordered.
      • Quartz is a form of SiO2  (silica). It has tetrahedral SiO4  (silicate) units which are orderly arranged in crystalline quartz. When SiO2  is melted and the melt is cooled, it forms amorphous quartz glass.

    Important Points

    • The distinction between Crystalline and Amorphous Solids:
      PropertyCrystalline SolidAmorphous Solid
      ShapeThe definite characteristic geometrical shape.Irregular shape.
      Melting pointMelt at a sharp and characteristic temperature.Gradually soften over a range of temperatures.
      AnisotropyAnisotropic in nature.Isotropic in nature.
      NatureTrue solids.Pseudo solids or supercooled liquids.
  • Question 3
    5 / -1
    If KCl is doped with 10-4 mol% of SrCl2, then what is the concentration of cation vacancies?
    Solution

    Concept:

    Doping:

    • The process of deliberate addition of desirable impurity in an intrinsic semiconductor is Doping.
    • Semiconductor materials themselves do not have free electrons or space for free electrons.
    • So, some impurities are added to it to make it either negatively charged or positively charged.
    • Doping converts intrinsic semiconductors to extrinsic semiconductor.

    Defects in a crystal:

    • Defects in a crystal are imperfections in the regular arrangement of atoms.
    • These defects are formed during the slow formation of crystals.
    • The defects are classified into Point defects and line defects.
    • The type of point defects are:
    • Frenkel defect:
      • It is a type of point defect in a crystal lattice when an atom or ion leaves its own lattice site vacant and instead of that, it occupies a normally vacant site called interstitial sites.
    • Schottky defect:
      • It is a type of point defect in a crystal lattice that occurs when oppositely charged ions or atoms leave their lattice sites, creating vacancies.

    Stoichiometric defects:

    • Interstitial defect:
      • It is a point defect where an atom occupies an interstitial position between surrounding atoms in normal sites.
    • Vacancy defects:
      • This type of defect occurs when cations are missing from their sites creating vacancies.
    • When we dope KCl with Sr2+, two cations of potassium are replaced by one Strontium because Sr is dipositive whereas K is unipositive.
    • One of the positions of the cation remains vacant, this is called a cation vacancy.

    Calculation:

    Given:

    •  KCl is doped with 10-4 mol% of SrCl2.
    • This signifies that in 100 moles of KCl, 10-4 moles of Sr2+ is present.
    • We know that one Sr2+ ion creates one cationic vacancy.

    Hence, in one mole of KCl, \({10^{-4} \over100}=10^{-6} \) moles of Sr2+ or cationic vacancies are present.

    One mole = 6.022 × 1023 atoms or ions.

    So, 10-6 moles of Sr2+ = 6.022 × 1023 × 10-6 = 6.02 x 1017 ions of  Sr2+.

    Hence, if KCl is doped with 10-4 mol% of SrCl2, then the concentration of cation vacancy is 6.02 x 1017.

  • Question 4
    5 / -1
    Amorphous materials are
    Solution

    Amorphous material:

    • Amorphous materials are in which there is no definite atomic structure and atoms exist in a random pattern just as in a liquid.
    • An amorphous material (AM) has a non-crystalline structure that differs from that of its iso-chemical liquid and does not undergo structural relaxation and the glass transition when heated. Examples are Glass, Gels, plastics, various polymers, wax, thin films.
    • The terms amorphous and non-crystalline are synonymous under this definition. The term glassy has the same structural meaning, but besides it also usually implies that the material exhibits a ‘glass transition’. Although the presence of dynamic disorder in the case of liquids complicates matters, the average atomic structure of liquids can be described in similar ways to that of amorphous solids.
    • Amorphous alloys are a class of metal alloy that, unlike ordinary metals do not have a long-range crystal structure.
    • Amorphous alloys have no grain boundaries.

     

    Properties of Amorphous Solids:

    Amorphous solids are sometimes described as supercooled liquids because their molecules are arranged randomly somewhat as in a liquid state.

    1. Lack of long-range order: Amorphous Solid does not have a long-range order of arrangement of their constituent particles. However, they may possess small regions of orderly arrangement. These crystalline parts of an otherwise amorphous solid are known as crystallites.
    2. No sharp melting point: An amorphous solid does not have a sharp melting point but melts over a range of temperatures. For example, glass on heating first softens and then melts over a temperature range. Glass, therefore, can be molded or blown into various shapes. Amorphous solid does not possess the characteristic heat of fusion.
    3. Conversion into a crystalline form: Amorphous solid, when heated and then cooled slowly by annealing, becomes crystalline at some temperature. That is why glass objects of ancient times look milky because of some crystallization having taken place.

    Additional Information

    Difference between Crystalline and Amorphous Solid :

    Properties

    Crystalline solid

    Amorphous solid

    Structure

    The constituent particles, atoms, ions, or molecules are arranged in regular and definite three-dimensional patterns. For example, sodium chloride, diamond, sugar, etc.

    The constituent particles are arranged in irregular three-dimensional patterns.

    Cutting with a knife

    Gives clean, sharp cleavage.

    Unclean cleavage.

    Compressibility

    Rigid and incompressible

    Usually rigid and cannot be compressed to any appreciable extent.

    Melting point

    They have a sharp and definite melting point.

    The melting point is not definite. Melt over a range of temperatures.

    Heat of fusion

    Definite

    Not definite.

    Physical properties

    These are anisotropic in that their physical properties are not identical in all directions.

    These are isotropic, that is their physical properties are identical in all directions.

  • Question 5
    5 / -1
    Element 'B' forms ccp structure and 'A' occupies half of the octahedral voids, while oxygen atoms occupy all the tetrahedral voids. The structure of bimetallic oxide is:
    Solution

    Concept:

    Atom A:

    The number of atoms in octahedral void = 4 atoms

    Half of the octahedral voids = \(\frac{4}{2} = 2\) atoms

    Number of A atoms = 2 atoms

    Atom B:

    Number of atoms in CCP lattice structure = 4 atoms

    Number of B atoms = 4 atoms

    Atom Oxygen:

    Number of atoms in all tetrahedral voids = 8

    Number of oxygen atoms = 8 atoms

    Thus, the compound formed is: A2B4O8 ⇒ AB2O4
  • Question 6
    5 / -1
    Name the crystal system in which none of the axial lengths are equal but all the axial angles are 900.
    Solution

    Concept:

    • Crystals are formed by the symmetrical arrangement of constituent atoms.
    • The atoms are arranged in repeating units and the basic unit is called a unit cell.
    • There are three types of unit cells depending on the arrangement of atoms.
    • A unit cell tells us about the number of atoms and their nature of the arrangement.
    • Multiple unit cells put together in an ordered manner gives us Crystal lattice.

    Lattice constant - Wikipedia

    • The type of crystal a system will form depends upon the intrinsic nature of the constituents.
    • There are basically seven crystal systems depending on the symmetrical parameters like bond angles, lengths, shape.

    Explanation:

    Lattice parameters:

    • A unit cell is characterized by three non-collinear axes 'a', 'b', 'c' which defines the axial lengths of the cells.
    • The interfacial angles between the cells are called axial angles.
    • The angle between axes c and b is 'α', between axes c and a is 'β' and between axes a and b is 'γ'.

    The 7 types of crystal structures :

    • Monoclinic
    • Triclinic
    • Orthorhombic
    • Rhombohedral
    • Hexagonal 
    • Cubic
    • Tetragonal

    The crystal system in which none of the axial lengths is equal but all the angles are is Orthorhombic.

    Orthorhombic @ Chemistry Dictionary & Glossary

    Additional Information

    • The parameters of all other crystal systems are given below:

    Primitive and non-primitive unit cell 

     

  • Question 7
    5 / -1
    NaCl crystals appear yellow due to:
    Solution

    Concept:

    Stoichiometric Defects:

    • The Stoichiometric defects are those in which the imperfections are such that the ratio between the cations and anions remains the same as represented by the chemical formula.
    • Types of Stoichiometric defects are:-Schottky defect and Frenkel defect.

    Non-Stoichiometric Defect:

    • The Non- Stoichiometric defects are those in which the imperfections are such that the ratio between the cations and anions differs from the ideal chemical formula.
    • Here, the balance of + and - charges are maintained either by having extra electrons or +ve charge.
    • Its types include:- Metal excess (i. due to anion vacancy and ii. due to interstitial cations) and Metal Deficiency.

    The classification of defects in solids is given below:

    Explanation:

    Metal excess defect due to anionic vacancies:

    • Alkali halide like NaCl and KCl show this type of defect.
    • When crystals of NaCl is heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal.
    • The Cl ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl.
    • This happens by loss of an electron by sodium atoms to form Na+ ions.
    • The released electrons diffuse into the crystal and occupy anionic sites.

    • As a result, the crystal now has an excess of sodium.
    • The anionic sites occupied by unpaired electrons are called F-centres.
    • They impart a yellow colour to the crystals of NaCl.
    • The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals.
    • Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).

    Hence, NaCl crystals appear yellow due to F - centres.

    Additional Information

     Schottky Defect:

    • In order to maintain electrical neutrality, the number of missing cations and anions are equal.
    • It decreases the density of the substance.
    • The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes.
    • For example, NaCl, KCl, CsCl and AgBr.

    Frenkel Defect:

    • The smaller ion (usually cation) is dislocated from its normal site to an interstitial site.
    • It creates a vacancy defect at its the original site and an interstitial defect at its new location.
    • Frenkel defect is also called a dislocation defect.
    • It does not change the density of the solid.
    • Frenkel defect is shown by the ionic substance in which there is a large difference in the size of ions.
    • For example, ZnS, AgCl, AgBr and AgI due to small size of Zn2+ and Ag+ ions.

    Interstitial Defect:

    • When some constituent particles (atoms or molecules) occupy an interstitial site, the crystal is said to have an interstitial defect
    • This defect increases the density of the substance.
    • Ionic solids must always maintain electrical neutrality. 
  • Question 8
    5 / -1
    A compound of formula A2 B3 has the hcp lattice. Which atom forms the hcp lattice and what fraction of tetrahedral voids is occupied by the other atoms?
    Solution

    Concept:

    The total effective number of atoms in hcp unit lattice = Number of octahedral voids in hcp = 6

    ∴ Number of tetrahedral voids (TV) in hcp 

    = 2 × Number of atoms in hcp lattice 

    = 2 × 6 = 12

    As the formula of the lattice is A2B3

    Suppose, A B

    \(\left( {\frac{1}{3} \times TV} \right)\) (hcp)

    \(⇒ {\rm{\;}}\frac{1}{3} \times 12\) (6)

    \(⇒ \frac{2}{3}\) 1

    ⇒ 2, 3

    So, \(A = \frac{1}{3}\) tetrahedral voids, B = hcp lattice.
  • Question 9
    5 / -1
    An element has a face-centred cubic (fcc) structure with a cell edge of a. The distance between the centers of two nearest tetrahedral voids in the lattice is:
    Solution

    Concept:

    The distance between the centers of two nearest tetrahedral voids in the lattice and also the minimum distance between two tetrahedral voids is \(\frac{a}{2}\)..

    In fcc (face centered cubic), tetrahedral voids are located on the body diagonal at a distance of \(\frac{{\sqrt {3a} }}{4}\) from the corner. Together they form a smaller cube of edge length\(\frac{a}{2}\).

    Therefore, distance between centres of two nearest tetrahedral voids in the lattice is also \(\frac{a}{2}\).

    Face-centered cubic (fcc) refers to a crystal structure consisting of an atom at each cube corner and an atom in the center of each cube face. It is a close-packed plane in which on each face of the cube atoms are assumed to touch along face diagonals.

  • Question 10
    5 / -1
    In a solid, oxide ions are arranged in CCP. 2/3 of the tetrahedral voids are occupied by M cations and 1/3 of the octahedral voids are occupied by N cations. What is the formula of the compound?
    Solution

    Concept:

    Cubic closed Pack or Face Centered Cubic lattice:

    • In this crystal structure, the atoms are located at the corners of a unit cell.
    • There are also atoms at each face of the cube.​
    • The corner atoms are shared via 8 cubes, so the contribution from each corner is 1/8.
    • Total number of atoms on the corner of a cube = 8

    So,

    \({\rm{total\;contribution\;from\;corners\;is\;}} = {\rm{\;}}\frac{1}{8} × 8 = 1\)

    • Contributions from the faces of the cube = 3
    • The total number of atoms in an FCC structure is four.
    • There is twice the number of tetrahedral voids as there are atoms in a lattice.
    • The number of octahedral voids is the same as that of atoms per unit cell. 

    Calculation:

    Given:

    • The oxides are present in CCP, so the number of oxide ions is 4.
    • The total number of octahedral voids = 8
    • 2/3 of the tetrahedral voids are occupied by M cations.
    • The number of M cations =

    2/3 × 8 = 16/3

    • 1/3 of the octahedral voids are occupied by N cations.
    • The number of N cations =

    1/3 × 4 = 4/3

    The formula of the compound thus becomes M\(16\over3\)N\(4\over 3\)Owhich is equal to M16N4O12

    Simplifying, we can write M16N4O12 as M4NO3.

    Hence, the formula of the compound is M4NO3.

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