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Chemistry I Chapter 1 The Solid State

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    NCERT Solution For Class 12 Chemistry Chemistry I

    The Solid State Here is the CBSE Chemistry Chapter 1 for Class 12 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 12 Chemistry The Solid State Chapter 1 NCERT Solutions for Class 12 Chemistry The Solid State Chapter 1 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 12 Chemistry.

    Question 1
    CBSEENCH12005349
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    Why are solids rigid?

    Solution
    Solids are rigid because the internuclear distance between the  molecules of a solid are very less and closely packed and their positions are fixed due to the strong forces of attraction between them.
    Question 2
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    Why do solids have a definite volume?

    Solution
    A solid have a definite volume because the inter molecular distance  between its molecules are fixed.
    Question 3
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    Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

    Solution
    Amorphous solids: Polyurethane, teflon, cellophane, polyvinyal chloride, fibre glass.
    Crystalline solids: naphthlene, benzoic acid, potassium nitrate, copper.
    Question 4
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    Why is glass considred a super-cooled liquid?

    Solution
    Glass is amorphous solids and  have a tendency to flow, though very slowly. Therefore, these are called pseudo solids or super cooled liquids .
    Question 5
    CBSEENCH12005353
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    Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show clevage property?

    Solution
    Refractive index of a solid which have the same value along all directions are isotropic in nature. It would not show cleavage property.
    Question 6
    CBSEENCH12005354
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    Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
    Potassium sulphate, tin, benzene, urea, ammonia, water, zinc
    sulphate, graphite, rubidium, argon, silicon carbide.

    Solution

    Molecular solids: water, argon.
    Ionic solids: potassium sulphate, zinc sulphate.
    Metallic solids: tin, rubidium.
    Covalent or network: benzene, urea, ammonia, graphite, silicon carbide.

    Question 8
    CBSEENCH12005356
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    Ionic solids conduct electricity in liquid state but not in solid state. Explain.

    Solution

    Ionic solids do not conduct electric current in solid state because in the solid ionic compound, the ions are held together in fixed positons by strong electrostatic forces and cannot move freely.
    Ionic solids conduct electricity in liquid state because the crystal structure is broken down and ions become free to move and conduct electricity.

    Question 9
    CBSEENCH12005357
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    What type of solids are electrical conductors, malleable and ductile?

    Solution
    Metallic solids are electrical conductors, malleable and ductile.
    Question 10
    CBSEENCH12005358
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    Give the significance of a 'lattice point'.

    Solution

    (i) Each point in a lattice is called lattice point or lattice side.
    (ii) Each point in a crystal lattice represents one constituent particle which may be an atom, a molecule or an ion.
    (iii) Every lattice point has identical surroundings except those which are on the surface or corner of a crystal.
    (iv) If a line is drawn by joining two lattice points, it would pass through a series of similar points at regular intervals.
    (v) Lattice points are joined by straight lines to bring out the geometry of the lattice. These lines do not represent chemical bond.

    Question 11
    CBSEENCH12005359
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    Name the parameters that characterize a unit cell.

    Solution

    Unit cell is the smallest portion of a crystal lattice.
    A unit cell is characterized by:
    (i) its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular.
    (ii) angles between the edges, α (between b and c), β (between a and c),γ (between a, and b). Thus, a unit cell is characterised by six parameters, a, b, c, α, β and γ.

    Question 12
    CBSEENCH12005360
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    Distinguish between Hexagonal and monoclinic unit cells.

    Solution

    Ans. (i)                                        Hexagonal Unit cell                      Monoclinic unit cell

    (1) Axial distances or edge
    lengths                                              a = b ≠ c                                       a ≠ b ≠ c
    (2)     Axial angles                                α = β = 90°                                    α = γ=90° β ≠ 90°

    (3) Examples:                                   Graphite, ZnO, CdS                       Monoclinic Sulphur, PbCrO2, 
                                                                                                                         Na2SO4.10H2O
                                                                                                                          2 4 2

    Question 13
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    Distinguish between Face-centred and end-centred unit cells.

    Solution

    (ii) Face centered unit cell

    End-centered unit cell

    It contains one particle present at the centre of its each face, besides the ones that are at its corners.

    In this unit cell, one constituent particle is present at the centre of any two opposite faces besides the ones present at its corners.
    Question 14
    CBSEENCH12005362
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    Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.

    Solution

    (i) Primitive cubic unit cell has atoms only at its corner. Each atom at a corner is shared between eight adjacent unit cells . four unit cells in the same layer and four unit cells of the upper (or lower) layer Therefore only 1/8th of an atom (molecule or ion) actually belongs to a particular unit cell.
    since each cubic unit cell has 8 atoms on its corners, the total number of atoms in one unit cell is 8×1/8 = 1 atom.
    (ii) The atom at the body centre wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell:
    (i) 8 corners×1/8 per corner atom =8×1/8 = 1 atom 
    (ii) 1 body centre atom =1×1=1
       therefore total number of atom per unit cell =2 atom.

    Question 15
    CBSEENCH12005363
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    What is the two-dimensional coordination number of a molecule in square close packed layer ?

    Solution
    The co-ordination number is 4.
    Question 16
    CBSEENCH12005364
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    A compound forms hexagonal closed-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

    Solution
    Number of closed packed particeles                          =0.5×6.022×1023=3.011×1023  
    therefore number of octhahedral voids                    =   3.011×1023
    and number oftetrahedral voids =2×3.011×1023=6.022×1023
    therefore total number of voids=3.011×1023 +6.022×1023 =9.033×1023.
    Question 17
    CBSEENCH12005365
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    A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3 of tetrahedral voids. What is the formula of the compound?

    Solution
    The number of tetrahedral voids formed is equal to twice the number of atoms of element N and only 1/3rd of these are occupied by the atoms of element M. Hence the ratio of number of atom of M and N is 2 x (1/3) : 1 or 2 : 3. So the formula of the compound is M2 N3.

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    Question 18
    CBSEENCH12005366
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    Which of the following lattices has the highest packing efficiency:
    • simple cubic 

    • body-centred cubic

    • hexagonal close packed lattice
    Solution

    C.

    hexagonal close packed lattice

    Hexagonal close packed lattice has the highest packing efficiency (74%).

    Question 20
    CBSEENCH12005368
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    What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

    Solution

    Non-stoichiometric defects can arise when a solid is heated.
    The crystals with this defect becomes coloured. Alkali halides like NaCl and KCl show this type of defect. When crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The CI– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy an ionic sites. As a result the crystal now has an excess of sodium. They impart yellow colour to the crystals of NaCl.

    Question 21
    CBSEENCH12005369
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    What type of stoichiometric defect is shown by
    (i) ZnS, (ii) AgBr (iii) KCl

    Solution

    (i) Frenkel defect.
    (ii) Schottky defect and Frenkel defect.
    (iii) Schottky defect.

    Question 22
    CBSEENCH12005370
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    Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

    Solution
    If we mix a small amount of molten strontium chloride (SrCl2) with molten sodium chloride (NaCl) and the resulting solution is cooled, in the crystals of NaCl some Na+ ions will get replaced by Sr2+ ions. In order to maintain electron neutrality too Na+ ions are to leave their respective lattice sites. However, one out of them will be occupied by the Sr2+ion while the other will remain unoccupied or it will be vacant. Thus vacancies are introduced in an ionic solid.
    Question 23
    CBSEENCH12005371
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    Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitale example.

    Solution
    When crystals of NaCl are heated in an atmosphre of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The CI– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by loss electron by sodium atom to form Na+ ions. The released electrons diffuse into the crystals and occupy anionic sites. As a result the crystal now has an excess of sodium. They impart yellow colour to the crystals of NaCl.
    Question 25
    CBSEENCH12005373
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    What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic, justify your answer.

    Solution
    Ferromagnetic substances would make better permanent magnets.In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance the domains are randomly oriented and their magnetic moments get cancelled. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field  and a strong magnetic effect is produced. This ordering of domains persist even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.
    Question 26
    CBSEENCH12005374
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    how many  the number of tetrahedral and octahedral voids are present in a cubic close packed structure of A spheres?

    Solution
     In ccp structure of a sphere there is 2A tetrahedral voids and A octahedral voids.
    Question 27
    CBSEENCH12005375
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     What would be the number of tetrahedral and octahedral voids,In a face centred cubic unit cell (i.e., ccp structure) 4 atoms/ions are present?

    Solution
     there is 8 tetrahedral voids and 4 octhahedral  voids in a face centred cubic unit cell.
    Question 28
    CBSEENCH12005376
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    What type of point defect in a crystal of zinc sulphate?

    Solution
    Frenkel defect.
    Question 29
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    A cubic solid is made of two elements Y and Z. Atoms of Z are at the corners of the cube and Y at the body centre. What is the formula of the compound?

    Solution

    Number of Z per unit cell = 1/8 x 8Z = 1 Z.
    Number of X per unit cell = 1 Y.
    Formula of the compound = ZY.

    Question 30
    CBSEENCH12005378
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    MgO has rock salt type structure. What is the co-ordination number of each ion?

    Solution
    The co-ordination number of each ion in MgO is six.
    Question 31
    CBSEENCH12005379
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    Thallium chloride exists in Cs CI type lattice. What are the co-ordination number of TI and Cl– ions?

    Solution
    The co-ordination number of each ion is 8. Thallium chloride is a 8:8 co-ordination lattice.
    Question 32
    CBSEENCH12005380
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    What is the effect of pressure on NaCl type crystals?

    Solution
    High pressure increases the co-ordination number from 6:6 to 8:8.
    Question 33
    CBSEENCH12005381
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    In a crystal of zinc sulphide, zinc occupies tetrahedral voids. What is the coordination number of zinc?

    Solution
    The coordination number of zinc is 4.
    Question 34
    CBSEENCH12005382
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    Why is potassium chloride sometimes violet instead of pure white?

    Solution
    This  is due to the electrons trapped in anionic vacancies or due to F-centres.
    Question 35
    CBSEENCH12005383
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    When high temperature is given to ferromagnetic substance what effect can be obtain ?

    Solution
    As the temperature is raise the ferromagnetic substance starts to become paramagnetic due to randomisation spins.
    Question 36
    CBSEENCH12005384
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    For hcp crystal structure of an element What is the maximum possible coordination number ?

    Solution

     The maximum coordination number is 12 in hcp crystal .

    Question 37
    CBSEENCH12005385
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    How many atoms can be assigned to its unit cell if an element forms (i) a body centred cubic cell and (ii) a face centred cubic cell?

    Solution
    (i) for a body centred cubic cell 2 atoms.        
    (ii)for  face centred cubic cell 4 atoms.
    Question 38
    CBSEENCH12005386
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    How many Ca+ ions occupy second nearest neighbour locations to a central Cs+ ion in caesium chloride crystal?

    Solution
    Co-ordination number of Cs+ ion is 8 in the crystal of caesium chloride .
    Question 39
    CBSEENCH12005387
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    How many Na+ ions occupy second nearest neighbour locations of a Na+ ion in the structure of sodium chloride crystals?

    Solution
     The second nearest Co-ordination number of Na+ ions is 12.

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    Question 40
    CBSEENCH12005388
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    Why FeO is non-stoichiometric with the formula Fe0.95 O explain?

    Solution
    The non-stoichiometry reflect the ease of oxidation of Fe2+ to Fe3+ effectively replacing a small portion of Fe2+ with two thirds their number of Fe3+. Thus for every three 'missing' Fe2+ ions, the crystal contains two Fe3+ ions to balance the charge .   i.e. 3Fe2+ = 2Fe3+ to maintain electrical neutrality
    Question 41
    CBSEENCH12005389
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    Define non-stiochiometry defect in crystals?

    Solution
    Those defect which lead to change in composition of solids are called non-stoichiometric defects.They are most often inorganic compounds, and almost always solids They contain constituting particles in non-stoichiometric ratio, e.g., Fe0.93 O1.0, Ni0.94 O1.0 are examples of non-stochiometric compounds.
    Question 42
    CBSEENCH12005390
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    What is the arrangement of Zn2+ and S2– ions in Zinc blende (ZnS) crystal structure?

    Solution
    Zinc sulfide crystallizes in a Face-Centered Cubic unit cell (FCC) having an edge length of 5.409 Angstroms
    Sulfide anions have FOUR neighbors of opposite charge arranged at vertices of a tetrahedron. Zinc cations also have FOUR neighbors of opposite charge arrange at vertices of a tetrahedron. So the Zn : S coordination ratio is 4 : 4 or 1 : 1.
    Question 43
    CBSEENCH12005391
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    explain the effect of Frenkel structural defect on the electrical conductance of a crystalline solid?

    Solution
    Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site . It creates a vacancy defect at its original site and an interstitial defect at its new location. this vacancy defect increase the Electrical conductance.
    Question 44
    CBSEENCH12005392
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    An ionic compound AB2 possess CaF2 type crystal structure. What the co-ordination number of A2+ and B– ions in crystal of AB2 explain.

    Solution
    CaF is partof bcc inwhich the Ca2+ is located at the body centred F- is located atthe corners of the unit cell thus co-ordination numberof Ca2+ is 8 and co-ordination number of F- is 4. similarly in AB2   The co-ordination number of A2+ is 8, whereas B- ion co-ordination number is 4.
    Question 45
    CBSEENCH12005393
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    What kind  point defect lowers the density of a crystal?

    Solution
    Schottky defect lowers the density of a crystal.
    Question 46
    CBSEENCH12005394
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    What do you mean by ‘point defects' in crystals?

    Solution
    Point defects are caused by missing or misplaced ion or atom in the crystal lattice i.e., deviation from regular arrangement of ions/ particles in the crystal.
    Point defects can be classified into three types : (i) stoichiometric defects (ii) impurity defects and (iii) non-stoichiometric defects.
    Question 47
    CBSEENCH12005395
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    Define dislocation in crystals.

    Solution
    crystal defect. Line defects, or dislocations, are lines along which whole rows of atoms in a solid are arranged anomalously. The resulting irregularity in spacing is most severe along a line called the line of dislocation. Line defects can weaken or strengthen solids.
    Question 48
    CBSEENCH12005396
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    Explain the term ‘dislocations’ in relation to crystals?

    Solution
    Crystal defect. Line defects, or dislocations, are lines along which whole rows of atoms in a solid are arranged anomalously. The resulting irregularity in spacing is most severe along a line called the line of dislocation. Line defects can weaken or strengthen solids.
    Question 49
    CBSEENCH12005397
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    Why does zinc oxide exhibit enhanced electrical conductivity on heating?

    Solution
    Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow.On heating, zinc oxide loses oxygen


    The excess zinc ions are accomodated in interstitial sites with electrons trapped in the neighbourhood. The enhanced conductivity is due to these trapped electrons.
    Question 50
    CBSEENCH12005398
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    In a body centred unit cell, one of the diagonals contains two extra atoms. Find the number of atoms present per unit cell?

    Solution

    In a body centred unit cell diagonal contain two extra atom .so in per unit cell  = 8×18+2=3.

    Question 51
    CBSEENCH12005399
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    An element occurs in the bcc structure. How many atoms are present in its unit cell explain ?

    Solution
    body centre wholly belongs to the unit cell in which it is present. Thus in a body-centered cubic (bcc) unit cell:
    (i) 8 corners × 1/8 per corner atom 8×1/8 = 1 atom
    (ii) 1 body centre atom = 1 × 1 = 1 atom
    ∴ Total number of atoms per unit cell = 2 atoms.
    Question 52
    CBSEENCH12005400
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    What is the number of atoms per unit cell in a body centred cubic structure?

    Solution
    The number of atoms per unit cell in a body centred cubic structure = 2.
    Question 53
    CBSEENCH12005401
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    What are lattice defects imperfection?

    Solution
    lattice defect can be define as combination of Line defects and plane defects together are called lattice defects.
    Question 54
    CBSEENCH12005402
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    what is Frenkel defect.Why is Frenkel defect found in AgCl ?

    Solution
    Frenkel Defect: This defect is shown by ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site . It creates a vacancy defect at its original site and an interstitial defect at its new location.
     Frenkel defect is shown by ionic substance in which there is a large difference in the size of ions so it is due to difference in size of Ag    + and CI–, Ag+ occupies voids.
    Question 55
    CBSEENCH12005403
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    What  type of void are developed in hexagonal close packing?

    Solution
    In hexagonal close packing octahedral void as well as tetrahedral void are produced.
    Question 56
    CBSEENCH12005404
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    What is the co-ordination number of an octahedral void?

    Solution
    The co-ordination number of an octahedral void  is  6 : 6.
    Question 57
    CBSEENCH12005405
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    Give two exampls of the compounds that exhibit Schottky defect.

    Solution
    KCl, AgBr
    Question 58
    CBSEENCH12005406
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    What are interstitials in a crystal?

    Solution
    The atoms or ions which occupy the voids present in a crystal are termed interstitials.
    Question 59
    CBSEENCH12005407
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    What will happen if NaCl crystal is subjected to high pressure?

    Solution
    It will change to CsCI structure i.e.At High pressure the coordination number change from 6:6 to 8:8 .
    Question 60
    CBSEENCH12005408
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    What is unit cell.

    Solution
    Unit cell: It is the smallest repeated unit in a crystal lattice e.g., in sodium chloride and caesium chloride, the unit cell is a cube. It has all the characteristic properties of crystal.
    Question 61
    CBSEENCH12005409
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    For tetrahedral co-ordination, what should be the range of radius ratio (r+/r–) value?

    Solution
    For tetrahedrtal coordination Radius ratio is 0.225-0.414.
    Question 62
    CBSEENCH12005410
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    What type of structure possessed by a unit cell of CsCl.

    Solution
    CsCl has a body centred cubic structure.
    Question 63
    CBSEENCH12005411
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    What is the effect of increasing temperature on the conductivity of semi-conductors?

    Solution
    In case of semiconductors, the gap between the valence band and conduction band is small  Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semiconductors increases with rise in temperature, since more electrons can jump to the conduction band.
    Question 64
    CBSEENCH12005412
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    Define radius ratio. What is the ideal value for a tetrahedral void?

    Solution
    It is the ratio of the radius of the cation to the radius of the anion in an ionic crystal.the radius ratio of tetahedral is in between 0.225-0.732. Its ideal value is 0.225 for tetrahedral void.
    Question 65
    CBSEENCH12005413
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    How in the  crystal of zinc sulphide, zinc occupies tetrahedral voids. What is the coordination number of zinc?

    Solution
    The zinc(II) ions, which are smaller than the sulfide ions, are inserted into tetrahedral holes and push the sulfide ions apart so that no two sulfide ions are in contact with each other. The resulting structure has (4,4)-coordination. Observe that none of the octahedral holes are occupied. 
    The coordination number of zinc isfour.
    Question 66
    CBSEENCH12005414
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    What is  point defects in solids.

    Solution
    Point defects can be define as  the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance.
     Point defects can be classified into three types : (i) stoichiometric defects (ii) impurity defects and (iii) non-stoichiometric defects.
    Question 67
    CBSEENCH12005415
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    What type of compounds exhibit Schottky defect?

    Solution
    schottky is basically vacancy defect. it can be shown by ionic substance in which cation and anion are of almost similar sizes. for example NaCl, KCl etc.
    Question 68
    CBSEENCH12005416
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    The unit cell of a substance has cation A+ at the corners of the unit cell and anions B- the centre. What is the simplest formula of the the substance?

    Solution
    Number of cations A+ at the corners of the unit cell is
    =18×8=1.
    Number of anions B–at the centre of unit cell is=1
    Therefore, simplest formula of the substance = AB.
    Question 69
    CBSEENCH12005417
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    What is the co-ordination number of each ion in Na2O.

    Solution
    The coordination number of Na in Na2O is 4.Na+ ion is surrounded by four oxide ion and O2- ion is surrounded by eight Na+ ions.
     Co-ordination number of Na    + = 4.
    Co-ordination number of O2– = 8.
    Question 70
    CBSEENCH12005418
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    Why is an ionic crystal of NaCl a nonconductor explain?

    Solution
    ionic bonds are the electrostatic force of attraction between oppositely charged ion. The oppositely charged ion are arranged in a regular way to form giant ionic lattices. Ionic crystal of NaCl is a non-conductor of electricity because it has its electron in fully filled lower energy state. There are no free electrons in crystal lattice.
    Question 71
    CBSEENCH12005419
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    What causes the conduction of electricity by semi-conductors?

    Solution
     The conduction of electricity in semi- conductors is because of the  Electrons and holes.
    Question 72
    CBSEENCH12005420
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    What may be the difference between phosphorus doped and gallium doped semiconductors?

    Solution
    Phosphorus doped semiconductors produced n- type semiconductor while when silicon doped with gallium, it produces p- type semiconductor.
    Question 73
    CBSEENCH12005421
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    What are non-stoichiometric compounds?

    Solution
    Non-stoichiometric compounds are chemical compounds, almost always solid inorganic compounds, having elemental composition whose proportions cannot be represented by integers; most often, in such materials, some small percentage of atoms are missing or too many atoms are packed into an otherwise perfect lattice work. exaple FeO, ZnO etc.
    Question 74
    CBSEENCH12005422
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    In which compound  Schottky and Frenkel defects are present together?

    Solution
    Silver bromide (AgBr) has both Schottky and Frenkel defects together.
    Question 75
    CBSEENCH12005423
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    What is the percentage of free space in a body centred cubic crystal?

    Solution
    In body centred cubic crystal packing effciency is 68% thus free space in body centred cubic crystal is free space effciency = 100- 68%
                                  = 32%
    Question 76
    CBSEENCH12005424
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    Give one example of anti ferroelectric solid and one example of ferroelectric solid.

    Solution
    Antiferro electric solid-PbZrO3 (Lead Zirconate).
    Ferro electric solid-BaTiO3 (Barium Titanate).
    Question 77
    CBSEENCH12005425
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    What happens when CdCl2 is added to AgCl?

    Solution
    When AgCl is doped with CdCl2 a cation vacancy defect is created i.e. An Ag+ ion from the lattice is absent its position due to presence of adjacent Cd++ ion.
    Question 78
    CBSEENCH12005426
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    Name the solid substance in which cations occupy all the tetrahedral voids.

    Solution
     Na2O is the solid in which cation occupy all the tetrahedral voids. in Na2O, Na+ ion is surrounded by four oxide ion and oxide ion surrounded by the eight sodium ion.
    Question 79
    CBSEENCH12005427
    Wired Faculty App

    What is the effect of Frenkel defect on electrical conductivity of the solid?

    Solution
    Frenkel defect tends to increase the dielectric constant of the crystal. Compounds having such defect conduct electricity to a small extent. When electric field is applied, an ion moves from its lattice site to occupy a hole, it creates a new hole. In this way, a hole moves from one end to the other. Thus, it conducts electricity across the crystal. Due to the presence of holes, stability (or the lattice energy) of the crystal decreases.

    Sponsor Area

    Question 80
    CBSEENCH12005428
    Wired Faculty App

    Mention one property which is caused due to the presence of F-centre in a solid.

    Solution
    If F-centres are introduced into a transparent crystal lattice then they will induce colour into the materiale for example NaCl heated in sodium vapours the crystal becomes coloured.
    Question 81
    CBSEENCH12005429
    Wired Faculty App

    Name a salt that can be added to AgCl so as to produce cation vacancies.

    Solution
    CdCl2 can be mixed in AgCl to create cation vacancies, where Cd2+ will occupy some of the sites of Ag+ ions.
    Question 82
    CBSEENCH12005430
    Wired Faculty App

    How many lattice points are there in one unit cell of each of the following lattice?
    (a) face centred cubic
    (b) face centred tetragonal
    (c) body centred cubic
    Solution

    A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube.each atom located at the face-centre is shared between two adjacent unit cells and only 1/2 of each atom belongs to a unit cell

    Thus, in a face-centred cubic
    (fcc) unit cell:
    (i) 8 corners atoms × 1/8 atom per unit cell = 8×1/8 = 1 atom
    (ii) 6 face-centred atoms ×1/2 atom per unit cell = 6 ×1/2 = 3 atoms
    ∴ Total number of atoms per unit cell = 4 atoms

    (a) z = 4,
    (b) z = 4,

    body centre wholly belongs to the unit cell in which it is present. Thus
    in a body-centered cubic (bcc) unit cell:
    (i) 8 corners × 1/8 per corner atom= 8×1/8 = 1 atom
    (ii) 1 body centre atom = 1 × 1 = 1 atom
    ∴ Total number of atoms per unit cell =1+1 = 2

    (c) z = 2.

    Question 83
    CBSEENCH12005431
    Wired Faculty App

    Find out the number of atoms per unit cell in a face-centred cubic structure having only single atoms at its lattice points.

    Solution

    A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube.

    It can be seen that each atom located at the face-centre is shared between two adjacent
    unit cells and only1/2 of each atom belongs to a unit cell.
    The number of atoms present at corners per unit cell= 8 corner atoms x 1/8 atoms per unit cell = 1

    The number of atoms present at faces per unit cell
    = 6 atoms at the faces x 1/2 atoms per unit cell
    = 6 x 1/2 = 3 atoms

    Therefore, total number of atoms per unit cell = 4 atoms.

    Question 84
    CBSEENCH12005432
    Wired Faculty App

    What makes alkali metal halides sometimes coloured, which are otherwise colourless?

    Solution
    Metal excess defect is resposible for colour in alkylhalides.

    When crystals of NaCl are 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 electron by sodium atoms to form Na+ ions. The released electrons diffuse into the crystal and occupy anionic sites and impart colour.

    Question 85
    CBSEENCH12005433
    Wired Faculty App

    What is the difference between ccp and fcc lattice?

    Solution
    These are two different names for the same lattice.
    so there is no difference between ccp and fcc.
    Question 86
    CBSEENCH12005434
    Wired Faculty App

    What percent of space is utilised by spheres in (i) simple cubic, (ii) face-centered cubic, (iii) body-centered cubic structure?

    Solution
    (i)Packing efficiency in simple cubic lattice = 52.4%,
    (ii)packing efficiency in face-centred  cubic lattice =  74%,
    (iii) packing efficiency in body centred cubic lattice =  68%.
    Question 87
    CBSEENCH12005435
    Wired Faculty App

    If three elements A, B and C crystallize in a cubic solid lattic with A atom at the corner, B atoms at cube centre and C atoms at edges then what would be the formula of the solid?

    Solution
    Atom A is shared by 8 corners 1
    atoms of C is shared by 4 unit cells
    Atom B is present at centre of the unit cell
    Hence, effective number of atoms of A per unit cell = 8× 1/8 = 1
    Effective number of atoms of C per unit cell = 12/4 = 3
    Effective number of atoms of B per unit cell = 1 Hence, the formula of the compound is ABC3
    Question 88
    CBSEENCH12005436
    Wired Faculty App

    Out of Schottky and Frenkel defect which increases the dielectric constant of crystals?

    Solution
    Frenkel defect tends to increase the dielectric constant of the crystalt. It is because of the presence of the ions in interstitial sites.
    Question 89
    CBSEENCH12005437
    Wired Faculty App

    What are extrinsic semi-conductor?

    Solution
    An extrinsic semiconductor is an improved intrinsic semiconductor with a small amount of impurities added by a process, known as doping, which alters the electrical properties of the semiconductor and improves its conductivity. Introducing impurities into the semiconductor materials (doping process) can control their conductivity.
    Question 90
    CBSEENCH12005438
    Wired Faculty App

    What is Photovoltaic.

    Solution
    Photovoltaics (PV) is a method of generating electrical power by converting solar radiation into direct current electricity using semiconductors that exhibit the photovoltaic effect. Example-amorphous silica.
    Question 91
    CBSEENCH12005439
    Wired Faculty App

    Although pure silicon is an insulator, then how does it behave as a semiconductor on heating.

    Solution
    On heating some covalent bonds among silicon atoms break on heating and electrons become free to move under applied field, hence silicon behaves like semi-conductor at high temperature.
    Question 92
    CBSEENCH12005440
    Wired Faculty App

    What is meant by tetrahedral void in a close-packed structure?

    Solution
     It is the empty space between three spheres in one layer and one sphere in the layer above it or one sphere in the layer below it is called a tetrahedral void
    .
    Question 93
    CBSEENCH12005441
    Wired Faculty App

    What is meant by octahedral void in a close-packed structure?

    Solution
    If a triangular void pointing up in one close-packed layer is covered by a triangular void pointing down in the adjacent layer, then a void surrounded by six spheres results Such a void is called an octahedral void since the six spheres surrounding it lie at the corners of a regular octahedron.

                        octahedral void
    Question 94
    CBSEENCH12005442
    Wired Faculty App

    How many vertex, edges, faces and centres are there in a cubic unit cell?

    Solution
     In a cubic unit cell,there are 8 vertex, 12 edges, 6 faces and one body centre.
    Question 95
    CBSEENCH12005443
    Wired Faculty App

    Agl crystallises in cubic close packed ZnS structure. What fraction of tetrahedral sites are occupied by Ag+ ions?

    Solution
    In AgI, if there are nI- ions, there will be nAg+ ions. As I- ions form the lattice, number of tetrahedral voids = 2n.
    As there are nAg+     ions to occupy these voids, therefore fraction of tetrahedral voids occupied by Ag+ ions = n/2n = ½ = 50%.
    Question 96
    CBSEENCH12005444
    Wired Faculty App

    Write the formula of a compound which is made of two elements A and B. Atom A forms close-packed lattice while atom B occupies all tetrahedral voids.

    Solution
    For every atom in a close packed structure, there are two tetrahedral voids.

    Let the number of close packed spheres be N, then:
    The number of octahedral voids generated = N
    The number of tetrahedral voids generated = 2N

    Therefore formula of the compound is AB2
    Question 97
    CBSEENCH12005445
    Wired Faculty App

    What is possible formula of a compound which is made of two elements P and Q and forms fcc structure. Atoms P occupy corner positions and Q occupy face positions of cube?

    Solution
    In a fcc unit cell, there are 8 corners and 6 faces. Eight corners contribute 8 x 1/8 = 1 atom
    six faces contribute 6 x 1/2 = 3 atoms.
    So the formula of the compound is PQ    3.
    Question 98
    CBSEENCH12005446
    Wired Faculty App

    What is the general structure of ionic solids?

    Solution
    In ionic solids, generally anions form close-packed structure and cations occupy interstitial sites. example NaCl, KCl. These are ionic solids
    Question 99
    CBSEENCH12005447
    Wired Faculty App

    How many octahedral voids are there in 1 mole of a compound having cubic closed packed structure? 

    Solution
    There are 1 mole of octhahedral voids in every one mole ofthe compound cotaining ccp structure.
    since number of particles present in 1 mole compound =6.022 x 1023 
    we know that number of octahedral voids=number of atoms in structure.
    Question 100
    CBSEENCH12005448
    Wired Faculty App

    What is the two-dimentional coordination number of a molecule in square closed packed layer ?
    Solution
    The coordination number is 4.
    Question 101
    CBSEENCH12005449
    Wired Faculty App

    Why is potassium chloride sometimes violet instead of pure white.

    Solution
    when a crystal of KCl is heated in an atomsphere of potassiuim vapour the Katoms lose electron to form K+ ions. The released electron diffuse into the crystal and occupy anionic site (F- centre)
    These electron impart violet colour to the KCl crystal.
    Question 102
    CBSEENCH12005450
    Wired Faculty App

    What is the coordination number of each type of ions in a rock-salt type crystal structure?

    Solution
    Rock salt or NaCl has face centred cubic structure ,here each ion (cation/anion) is surrounded by 6 ions of opposite charges, thus coordination number is 6.
    Question 103
    CBSEENCH12005451
    Wired Faculty App

    What is the total number of atoms per unit cell in a face-centred cubic (fcc) structure?

    Solution

    A face-centred cubic (fcc) unit cell contains atoms at all the corners and at the centre of all the faces of the cube. It can be seen that each atom located at the face-centre is shared between two adjacent
    unit cells and only 1/2 of each atom belongs to a unit cell.

    Thus, in a face-centred cubic
    (fcc) unit cell:
    (i) 8 corners atoms × 1/8 atom per unit cell=8×1/8
      = 1 atom
    (ii) 6 face-centred atoms ×
    1/2 atom per unit cell = 6 ×1/2 = 3 atoms
    ∴ Total number of atoms per unit cell = 4 atoms

    Question 104
    CBSEENCH12005452
    Wired Faculty App

    What type of substances exhibits anti-ferromagnetism?

    Solution
    A substance in which magnetic moments align in such a way that the net magnetic moment is zero, exhibits anti-ferromagnetism. exampe MnO, FeO, NiO etc.
    Question 105
    CBSEENCH12005453
    Wired Faculty App

    Define the term ‘amorphous’. Give a few examples of amorphous solids. 

    Solution

    Solids having constituent particles with irregular shapes and short range order are called amorphous solids. Amorphous solids are isotropic in nature and melt over a range of temperature. Thus, amorphous solids are also referred as pseudo solids or super cooled liquids.

    Amorphous solids do not have definite heat of fusion.

    Amorphous solids give irregular surfaces, when cut with sharp tool.

    Glass, rubber, plastic, etc. are some examples of amorphous solid.

    Question 106
    CBSEENCH12005454
    Wired Faculty App

    What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass? 

    Solution

    It is the arrangement of constituent particles of glass which makes it different from quartz. The constituent particles of glass have short range order while quartz has constituent particles in long range order and short range order both.

    By heating and cooling rapidly quartz can be converted into glass.

    Property

    Quartz

    Glass

    1. Structure

    2. Melting point

    Quartz is crystalline and it has long range order.

    Quartz has a sharp melting point. It changes into viscous liquid at 1983 K.

    Amorphous silica does not have a regular structure of long range order.

    Glass does not have a sharp melting point. On heating it softens and melts over a wide range of temperature.

     

    If SiO2 is melted and the melt is cooled very, rapidly it forms a glass.
    Question 107
    CBSEENCH12005455
    Wired Faculty App

    Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:

    (i) Tetra phosphorous decoxide (P4O10) 

    (ii) Ammonium phosphate (NH4)3PO4

    (iii) SiC

    (iv) I2 

    (v) P4

    (vi) Plastic

    (vii) Graphite

    (viii) Brass

    (ix) Rb

    (x) LiBr

    (xi) Si 

     

    Solution

     

    Solution:

    (i) Tetra phosphorous decoxide (P4O10) - Molecular

    (ii) Ammonium phosphate (NH4)3PO4 – Ionic

    (iii) SiC - Covalent (network)

    (iv) I2 - Molecular

    (v) P4 - Molecular

    (vi) Plastic - Amorphous

    (vii) Graphite – Covalent (network)

    (viii) Brass - Metallic

    (ix) Rb - Metallic

    (x) LiBr - Ionic

    (xi) Si – Covalent (network)

     

    Question 108
    CBSEENCH12005456
    Wired Faculty App

    What is meant by the term ‘coordination number’?

    (ii) What is the coordination number of atoms:

    (a) in a cubic close-packed structure?

    (b) in a body-centred cubic structure?
    Solution
    Co-ordination number : The total number of nearest neighbour atoms of a particular atom in a crystal lattice is known as its co-ordination number.


    Example: In hexagonal close-packing (hcp) arrangement, each particle is surrounded by 12 equidistant spheres, i.e., 6 spheres in contact with in the same plane and 3 each in adjacent layers-one immediately above and the other immediately below. This, co-ordination number in hcp packing is 12.

    (ii) The coordination number of atoms

    (a) In a cubic close-packed structure is 12

    (b) In a body-centered cubic structure is 8

     
    Question 110
    CBSEENCH12005458
    Wired Faculty App

    How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

    Solution

    Solution:

    The atomic mass of an unknown metal can be determined by knowing its density and the dimension of unit cell.

     
    Let ‘a’ be the edge length of a cubic unit cell and it contains ‘Z’ atoms of a substance of atomic mass ‘M’.

    From the above expression for density, d we can calculate the atomic mass, M of metal as other parameters are known.
    Question 111
    CBSEENCH12005459
    Wired Faculty App

    'Stability of a crystal is reflected in the magnitude of its melting points. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the inter molecular forces between these molecules?
    Solution

    Solution:

    Stability of a crystal is reflected in the magnitude of its melting points because higher the melting point, greater is the intermolecular force and greater the intermolecular force greater is the stability. And hence, a substance with higher melting point would be more stable.

    The melting points of the given substances are as follows:

    Solid water - 273 K

    Ethyl alcohol – 158.8 K

    Diethyl ether – 156.85 K

    Methane – 89.34 K

    As we can see the melting point of solid water is highest and melting point of methane is lowest among the given substance. This says that intermolecular force in solid water is strongest and the intermolecular force in methane is weakest.

    Question 115
    CBSEENCH12005463
    Wired Faculty App

    How many lattice points are there in one unit cell of each of the following lattice?
    (i) face centered cubic
    (ii) face centered tetragonal
    (iii) body centered.

    Solution

    Solution:

    (i) One unit cell of a face-centered cubic has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

    (ii) One unit cell of face-centered tetragonal has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

    (iii) One unit cell of body centered has 8 lattice points are corners and 1 lattice points at faces, total 9 lattice points.

    Question 116
    CBSEENCH12005464
    Wired Faculty App

    Explain the basis of similarities and differences between metallic and ionic crystals.
    Solution
    The similarities and differences between metallic and ionic crystals are described as below:

    Characteristics

    Ionic crystals

    Metallic crystals

    1.

    Constituent particles

    cations (positive ions) and anions (negative ions)

    Positive ions and mobile electrons.

    2.

    Binding forces

    Strong electrostatic forces (electro valent bonding)

    Electrical attraction between positive ions and mobile electrons (metallic bonding)

    3.

    Physical properties

    Hard and brittle; high m.p. and b.p. good conductors of heat and electricity in molten state and aqueous solution, high heat of fusion.

    Hard but malleable, high m.p., very good conductors of heat and electricity in solid and molten state, moderate heats of fusion.

     
    Question 117
    CBSEENCH12005465
    Wired Faculty App

    Explain Ionic solids are hard and brittle. 
    Solution
     In ionic solids, constituent particles are held together with strong electrostatic force of attraction along with their fixed position. The fixed position of ions and strong electrostatic force of attraction make ionic solids hard and brittle.
    Question 118
    CBSEENCH12005466
    Wired Faculty App

    Give reasons for ZnO becomes yellow on heating.

    Solution

    Metal excess defect due to the presence of extra cations at interstitial sites: Zinc oxide is white in colour at room temperature. On heating it loses oxygen and turns yellow
    Question 119
    CBSEENCH12005467
    Wired Faculty App

    Give reasons for Solids with F-centres are paramagnetic?

    Solution
    f centres are the anionic vacancies occupied by unpaired electron. since the presence of unpaired electrons shows paramagnetic.

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    Question 120
    CBSEENCH12005468
    Wired Faculty App

    Give reasons for Cation vacancies in some crystals make them good catalysts.

    Solution
    The catalytic activity dependsvery much on the surface area per unit mass of the sample.
    Transition metal corresponds show metal deficiency due to absence of metal ion from its lattice. The charge is balanced by an adjacent ion having higher positive charge which ultimately increases chemisorption.
    Question 121
    CBSEENCH12005469
    Wired Faculty App

    Give reasons for Non-stoichiometric sodium chloride is a yellow solid.

    Solution

    When crystals of NaCl are 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 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 byunpaired electrons are called F-centre. They impart 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.

    Question 122
    CBSEENCH12005470
    Wired Faculty App

    What is a semi-conductor? Describe the two main types of semi-conductors and contrast their conduction mechanism.

    Solution

    Solution:

    Semiconductor: - Solids having intermediate range of conductivity, i.e. from 10–6 to 104 ohm–1 m–1 are called semiconductors. Semiconductors are of following two types:

    (i) n – type of semiconductors

    (ii) p – type of semiconductors

    (i) n – type semiconductors – Semiconductors formed after doping with electron rich impurities to increase their conductivity are called n-type of semiconductors.

    Example –

    Silicon and germanium, each has four valence electrons as they belong to 14th group of periodic table. Arsenic and phosphorous belong to 15th group of periodic table and they have valence electrons equal to 5. When silicon or germanium is doped with phosphorous or arsenic, four electrons of phosphorous or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon or germanium.



    n-type semi-conductor.

    Since the electrical conductivity of silicon or phosphorous is increased because of negatively charged particle (electron), thus this is known as n-type of semiconductor.


    (ii) p – type of semiconductor - Semiconductors formed by the doping with electron deficient impurities; to increase their conductivity; are called p-type semiconductors. In p - type of semiconductors, conductivity increase because of formation of electron holes.

    Example - Electrical conductivity of silicon or germanium is doped with elements, such as Boron, Aluminium or Gallium having valence electrons equal to 3. Three valence electrons present in these elements make covalent bonds with three electrons present in valence shell out of four of silicon or germanium leaving one electron delocalized. The place from where one electron is missing is called electron hole or electron vacancy.

    When the silicon or germanium is placed under electrical field, electron from neighbouring atom fill the electron hole, but in doing so another electron hole is created at the place of movement of electron. In the influence of electrical filed electron moves toward positively charge plate through electron hole as appearing the electron hole as positively charged and are moving towards negatively charged plate.




     p-type semi-conductor.

    Question 123
    CBSEENCH12005471
    Wired Faculty App

    A cubic solid 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? What are the coordination numbers of P and Q?

    Solution
    Formula of a compound is same as the formula of unit cell. An atom at the corner of cube contributes only l/8th to the unit cell and there are 8 corners in a cube.
    ∴  No. of atoms of Q in the unit cell
                      =8×18 = 1.
    An atom at centre of cube belongs only to this unit cell and there is only one body centre in the unit cell.
    ∴   No. of atoms of P in the unit cell = 1 x 1 = 1
    Thus, the formula of compound is PQ or QP.
    For body centred cubic unit cell, the coordination number is 8:8.
    ∴ Co-ordination number P = 8 and also coordination number of Q = 8.
    Question 124
    CBSEENCH12005472
    Wired Faculty App

    What is the difference between Schottky defects and Frenkel defects?

    Solution

    Schottky defects

    Frenkel defects

    1. It arises due to vacancy at a cation site and at a nearby anion site in an ionic solid.

    2. Overall density of the crystal decreases.

    3. It arises in ionic solids having high co-ordination number.

    4.Example: NaCl, KCl, CsCl etc.

    1. It arises when a cation leaves its lattice site and enters in an interstitial site in an ionic solid.

    2. There is no change in the overall density.

    3. It arieses in ionic solids having large anions and small cations.

     
    Example: ZnS, AgCl, AgI etc
    Question 125
    CBSEENCH12005473
    Wired Faculty App

    If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

    Solution
    An octahedral void is shown in Fig. Though an octahedral void is surrounded by six spheres, only four are shown. The spheres present above and below the void are not shown. Let us assume that the edge length of the unit cell is a cm and raius of octahedral void is r and the radius of sphere is R.

    In the given figure, let an sphere having centre ‘O’ is fitted in the octahedral void.
    As given, radius of the sphere fitted in the octahedral void = r

    And radius of the atoms in close packing = R

    Question 126
    CBSEENCH12005474
    Wired Faculty App

    Pure silicon is an insulator. Silicon doped with phosphorus is a semi-conductor. Silicon doped with gallium is also a semiconductor. What is the difference between the two doped silicon semi-conductor?

    Solution

    Phosphorus has one excess valence electron (compared with Si) after forming the four covalent bonds normally with silicon. This excess electron gives rise to electronic conduction. That is why silicon becomes semi-conductor on doping with phosphorus. It is called n-type semi-conductor.

    Gallium has only three valence electrons. It creates an electron deficient bond or a hole when it is doped with silicon. Such holes can move in the crystal giving rise to electrical conductivity. Thus silicon doped with gallium is also semi-conductor due to movement of holes. It is called-p-type semiconductor.

    Question 127
    CBSEENCH12005475
    Wired Faculty App

    What are paramagnetic and ferromagnetic substances. Account for the paramagnetic character of transition metal compounds. How does the paramagnetic character of the bivalent ions of the first transition metal series varies from Ti (Z = 22) to Cu (Z = 29)?

    Solution

    Paramagnetic substances are attracted by the magnetic field. This is due to the presence of unpaired electrons in the atoms/ions/molecules of the paramagnetic substance. Paramagnetism is a temporary effect.

    Ferromagnetic substances are strongly attracted by the magnetic field. They show magnetism even when the magnetic field is removed. This is due to a spontaneous alignment of the magnetic dipole in the same direction.

    Transition metal compounds contains one or more unpaired electrons. So, these compounds show para magnetism. The paramagnetism of the bivalent ions of the first transition metal series first increases in going from Ti2+ to Mn2+ (number of unpaired electrons increases) and then decreases upto Cu2+ (because the number of unpaired electrons decreases).
    Transition metal compounds contains one or more unpaired electrons. So, these compounds show para magnetism. The paramagnetism of the bivalent ions of the first transition metal series first increases in going from Ti2+ to Mn2+ (number of unpaired electrons increases) and then decreases upto Cu2+ (because the number of unpaired electrons decreases).

    Question 128
    CBSEENCH12005476
    Wired Faculty App

    Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2 :1. Can you account for the fact that this substance is a p-type semi-conductor ?

    Solution

    In the sample of cuprous oxide, Cu2O prepared in laboratory, copper to oxygen ratio is less than 2 : 1 and this deficiency of copper causes the metal deficiency defects.
    In these defects the positive ions are less in number as compared to anions. These defects are caused in two ways:
    (i) By Cation Vacancies : Some Cu+ may be missing from lattice sites and their positive charges are balanced by presence of extra charge on adjacent cations i.e., some Cu2+ are present and some cation vacancies are present.

    (ii) By presence of extra anions at interstitial sites: The type of defect can arise where some oxide ion, O2– are present at lattice sites and their charge is balanced by neighbouring cations in higher oxidation states i.e., some Cu2+ ions are present instead of Cu2+ ions.

    Metal deficiency defect due cation vacancy.

    Metal deficiency defect due to presence of extra anions.

    Question 129
    CBSEENCH12005477
    Wired Faculty App

    Ferric oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three tetrahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

    Solution

    Let us Suppose the number of oxide (O2–) ions = N.
    Number of octahedral void = number of anions
    So that the number of octahedral voids = N
    We have Given that
    Two out of every three octahedral holes are occupied by ferric ions.
    So that the number of ferric (Fe3+) ions = 2N/3
    The ratio of the number of Fe3+ ions to the number of O2− ions,
    Fe3+ : O2− = 2N/3 : N
    Multiply 3 and divide by N we get
    Fe3+ : O2− = 2 : 3
    Hence, the formula of the ferric oxide is Fe2O3.

    Question 130
    CBSEENCH12005478
    Wired Faculty App

    Classify each of the following as being either a p-type or an n-type semi conducor:
    (i) Ge doped with In
    (ii) B doped with Si.

    Solution

    (i) Ge is an element of 14th group (like Si) and has configuration 4s24p2. It has been doped with In, a 13th group element having 5s25p1 configuration i.e., element of 14th group has been doped with 13th group element. All three valence electrons of impurity atom (In) gets bonded with three out of four eectrons of Ge and one electron of Ge remains unbonded. Conductivity is due to unbonded electron of insulator, Ge. Therefore, it is a p-type semi-conductor.

    (ii) Boron, B is an element of 13th group and has 2s22p1 configuration. It is doped with Si, an element of 14th group having 3s23p2 configuration. All three electrons of boron gets bonded with 3 out of 4 electrons of Si and 4th electron of impurity atom (i.e., Si) is responsible for conductivity. Thus it is a n-type semiconductor.

    Question 131
    CBSEENCH12005479
    Wired Faculty App

    In terms of band theory, what is the difference between a conductor and an insulator?
    Solution

    Solution:

    Molecular orbirals of metals are formed by atomic orbitals. These orbitals are so close to each other as they form band or valence band.

    (i) Difference between conductor and insulator - In conductors there is no energy gap between the valence band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

    While in insulators there is large energy gap between the valence band and electrons cannot jump to it i.e. large energy gap prevents the flow of electricity.

    (ii)Difference between conductors and semiconductor - In conductors there is no energy gap between the valence band and conduction band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

    While in semi conductors, there is small energy gap between valence bond and conduction band. The small gap between band facilitates some electrons to jump to the conduction band by acquiring extra energy.

    Question 132
    CBSEENCH12005480
    Wired Faculty App

    In terms of band theory, what is the difference between a conductor and a semi-conductor?
    Solution

    Semiconductor: In instrinsic semiconductors, the energy gap, Eg, is relatively small (≈ = 1 eV). At absolute zero, conduction band is empty and valence band is full. Therefore, the solid behaves as an insulator at very low temperatures. However, at room temperature, some electrons from the top of valence band acquires enough thermal energy to jump into the conduction band. They leave behind a number of holes. Both electrons and holes movement contribute to conduction.
    In extrinsic semi-conductor of n-type, the donor energy level is close to the bottom of conduction band.
    In p-type the acceptor energy level is close to the top of valence band.

    Question 133
    CBSEENCH12005481
    Wired Faculty App

    Explain the following terms with suitable examples:
    'Schottky defect'

    Solution

    Schottky defects : When cations and anions both are missing from regular sites, the defect is called Schottky Defect. In Schottky Defects, the number of missing cations is equal to the number of missing anions in order to maintain the electrical neutrality of the ionic compound.

    Schottky Defect is type of simple vacancy defect and shown by ionic solids having cations and anions; almost similar in size, such as NaCl, KCl, CsCl, etc. AgBr shows both types of defects, i.e. Schottky and Frenkel Defects.

    Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.

    .
    Question 134
    CBSEENCH12005482
    Wired Faculty App

    Explain the following term with suitable examples:
    'Frenkel defect'

    Solution

    It is a type of vacancy defect. In ionic compounds, some of the ions (usually smaller in size) get dislocated from their original site and create defect. This defect is known as Frenkel Defects. Since this defect arises because of dislocation of ions, thus it is also known as Dislocation Defects. As there are a number of cations and anions (which remain equal even because of defect); the density of the substance does not increase or decrease.

    Ionic compounds; having large difference in the size between their cations and anions; show Frenkel Defects, such as ZnS, AgCl, AgBr, AgI, etc. These compounds have smaller size of cations compared to anions.

    Question 135
    CBSEENCH12005483
    Wired Faculty App

    Explain the following term with suitable examples:
    'Interstitials'.

    Solution
    Interstitials – Sometime in the formation of lattice structure some of the atoms or ions occupy vacant interstitial site, and are known as interstitials. These interstitials are generally small size non-metals, such as H, B, C, etc. Defect arises because of interstitials is called interstitial defect.
    Question 136
    CBSEENCH12005484
    Wired Faculty App

    Explain the following term with suitable examples:
    'F-centres'.

    Solution
    F-centres – This is type of defect and called metal excess defect. These type of defects seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F-centre and responsible for showing colour by the compound. example sodium chloride impart yellow colour beacuse of F centre .
    Question 137
    CBSEENCH12005485
    Wired Faculty App

    Examine the illustration of a portion of the defective crystal given below and answer the following questions:
    (i) What are these types of vacancy defects called?
    (ii) How is the density of a crystal affected by these defects?
    (iii) Name one ionic compound which can show this type of defect in the crystalline state.
    (iv) How is the stoichiometry of the compound effected?
    Solution

    (i) These type of vacancy defects are called Schottky defects.
    (ii) This defect decreases the density of the crystal.
    (iii) NaCl shows this type of defect in the crystalline state.
    (iv) This is the point defect which does not disturb stoichiometry of the solid.

    Question 138
    CBSEENCH12005486
    Wired Faculty App

    Schottky defects generate an equal number of cation and anion vacancies while doping produces only cation vacancies and not anion vacancies. Why?

    Solution
    Schottky defect also exists in pair to maintain electroneutrality. Therefore, it generates equal number of cation and anion vacancies. In case of doping of NaCl by CdCl2, CI– ions enter Cl– ions site in NaCl while Cd2+ displace Na+. In doping so, one positive charge becomes extra. Therefore one Na+ is displaced so as to maintain electro-neutrality. That is doing only cation vacancies exist.
    Question 139
    CBSEENCH12005487
    Wired Faculty App

    The ions of NaF and MgO all have the same number of electrons, and the internuclear distance are about the same (235 pm and 215 pm). Why then are the melting points of NaF and MgO so different. (992°C and 2642°C)?

    Solution
    The crystals of NaF and MgO are formed by Na+ and F- in NaF and Mg2+ and O2–ions in MgO respectively arranged in cubic closed structures. There are strong electrovalent bond forces (strong coulombic forces, attraction between ions) between Na+and F– in NaF and Mg2+ and O2– in MgO.

    But the magnitude of these coulombic forces of attraction is much higher in MgO as compared to that in NaF. (Mg    2+ is divalent while Na+is monovalent, similarly O2– and F–).

    The electrostatic forces of attraction between Mg    2+and O2– is almost 4 times as compared to Na+ and F– ions. Melting point of ionic solids is almost the index of inter-ionic attraction in crystal lattice as lot of energy is required to break these forces/overcome these forces before the substance melts.

    It changes into paramagnetic at hight temperature due to randomization of spins.

                (b)        The ions in MgO carry two unit charges. In NaCl only one unit charge. Hence electrostatic forces of attraction in MgO are stronger.

    Question 140
    CBSEENCH12005488
    Wired Faculty App

    How would you account for the following:
    Frenkel defects are not found in alkali metal halides.

    Solution
    The frenkel defect is favoured by a large diffrance in size between the positive and negtive ion the metal cation are generally smaller than the anion which occupy the vacant lattice site or hole.
    Frenkel defects are not found in alkali metal halides because the ions cannot get into interstitial positions due to their large sizes.
    Question 141
    CBSEENCH12005489
    Wired Faculty App

    How would you account for the following:
    Schottky defects lower the density of related solids.

    Solution
     It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal.

    Thus there is equal number of positive and negtive ion leave correct lattice point and go outside the lattice creating  a pair of vacancy. So there are less number of ions thannbefore whichresult in the decrease in its density.
    Question 142
    CBSEENCH12005490
    Wired Faculty App

    How would you account for the following:
    Impurity doped silicon is a semiconductor.

    Solution
    Silicon forms four covalent bonds thus two types of impurities can be added to silicon.
    for example : boron can be added or phosphorus can be added.
    In case of boron : boron form three bonds with silicon, it will result in an electron deficient bond andwill create a hole. These holes can move through crystal like positive charge giving rise electrical conductivity.
    In case of phosphorus: when silicon forms four bonds with phosphorus one electron of phosphorous atom will remain unbonded due to which it become delocalizedn and contributes to electrical conductivity.
    Question 143
    CBSEENCH12005491
    Wired Faculty App

    Silver crystallises in an fcc lattice. The edge length of its unit cell is 4.077 x 10–8 cm and its density is 10.5 g cm-3. Calculate on this basis the atomic mass of silver. (NA = 6.02 x 1023 mol–1).

    Solution

    We have,
    Edge of length of cell a = 4.07x10–8cm
    Density p = 10.5 g /cm3
    Number of atoms in unit cell of fcc lattice = 4
    Avogadro number NA = 6.022x1023
    By using formula,
    Density space equals straight p space equals fraction numerator ZM over denominator straight a cubed straight N subscript straight A end fraction straight M space equals space fraction numerator Pa cubed straight N subscript straight A over denominator straight Z end fraction Putting space the space value space in space the space above space equation space we space get comma fraction numerator 10.5 space straight x left parenthesis 4.07 right parenthesis cubed space straight x 6.022 space space straight x 10 to the power of 23 over denominator 4 end fraction solving space the space equation space we space get comma straight M equals space 107.09 straight g space Mol to the power of negative 1 end exponent

    Question 144
    CBSEENCH12005492
    Wired Faculty App

    Niobium crystallises in body centered cubic structure. If density is 8.55 g cm–3, calculate atomic radius of niobium using its atomic mass 93 u.

    Solution

    We have give that,
    Density (d)=8.55 g Cm-3
    Atomic mass (M) =93u =93 gMol-1
    Atomic radius (r) = ?

    We know that, Avogadro number Na =6.022 x 10 23 mol-1

    since given lattice is bcc

    therefore
    Number of atoms per unit cell (z) =2

    we know that

     

    straight d equals space fraction numerator zM over denominator straight a cubed straight N subscript straight A end fraction 8.55 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space 93 space over denominator straight a cubed space straight x 6.022 space straight x 10 to the power of 23 space mol to the power of negative 1 end exponent end fraction straight a cubed space equals space fraction numerator 2 space straight x 93 over denominator 8.55 space straight x space 6.022 space straight x space 10 to the power of 23 end fraction cm cubed straight a cubed space equals 3.6124 space straight x space 10 to the power of negative 23 end exponent straight a cubed space equals 36.124 space straight x space 10 to the power of negative 24 end exponent straight a equals space 3.3057 space straight x 10 to the power of negative 8 end exponent space cm For space bb space unit space cell space radius space left parenthesis straight r right parenthesis space equals fraction numerator square root of 3 over denominator 4 end fraction straight a straight r equals space fraction numerator 1.732 over denominator 4 end fraction space straight x space 3.3057 space straight x 10 to the power of negative 8 end exponent straight r equals fraction numerator 5.725 over denominator 4 end fraction space straight x space 10 to the power of negative 8 end exponent cm straight r equals space 14.31 space straight x space 10 to the power of negative 9 end exponent space cm straight r equals 14.31 nm

     
    Question 145
    CBSEENCH12005493
    Wired Faculty App

    Copper crystallizes into a fcc lattice with edge length 3.61 x 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.

    Solution

    We have given that
    Length of edge, a = 3.61x10–8 cm
    Atomic mass of cupper, M = 63.55 g/mol
    Avogadro constant, NA = 6.022x1023 per mol
    Number of atoms in unit cell of FCC, Z= 4
    Use formula of density,
    straight p equals fraction numerator ZM over denominator straight a cubed straight N subscript straight A end fraction Putting space the space given space value space in space above space equation straight p space equals fraction numerator 4 space straight x space 63.55 over denominator left parenthesis 3.61 space straight x 10 to the power of negative 8 end exponent right parenthesis cubed space left parenthesis 6.022 space straight x 10 to the power of 23 right parenthesis end fraction straight g over Cm cubed straight p equals space fraction numerator 254.2 over denominator left parenthesis 47.04 space space straight x 10 to the power of negative 24 end exponent right parenthesis space left parenthesis 6.022 space straight x 10 to the power of 23 right parenthesis end fraction straight p equals space fraction numerator 254.2 over denominator 28.318 end fraction straight g over cm cubed straight p equals 8.92 space straight g divided by cm cubed

     



    Question 146
    CBSEENCH12005494
    Wired Faculty App

    Analysis shows that nickel oxide has formula Ni 0.98 O1.00 What fractions of Nickel exist as Ni2+ and Ni3+ ions? 

    Solution

    Formula is Ni0.98O1.00
    So the ration of Ni : O = 98:100
    So if there are 100 atoms of oxygen then 98 atoms of Ni,
    Let number of atoms of Ni+2 = x
    Then number of atoms of Ni+3 = 98–x
    Charge on Ni = charge on O
    So that oxygen has charge –2
    3(98–x) + 2x = 2 (100)
    294 –3x +2x = 200
    –x = – 94
    x = 94
    Percentage of Ni+2 = (atom of Ni+2/total number of atoms of Ni)100
    =100(94/98)x100
    = 96%
    Percentage of Ni+3
    =100 – Ni+2
    =100 – 96
    = 4 %


    Question 147
    CBSEENCH12005495
    Wired Faculty App

    Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
    Solution
    Radius, r = 0.144 nm,
    Unit cell is fcc, edge of unit cell, a = ?
    In face centered unit cell, diagonal of face = 4r = a2
                                    [∵ face diagonal = a2]

    Therefore,       a=4r2 = 2r
                         
      = 2 × 1.4142 × 0.144 nm= 0.4073 nm.

    Length of side of unit cell,
                       
    a = 0.4073 nm.
                       
     
    Question 149
    CBSEENCH12005497
    Wired Faculty App

    If NaCl is doped with 10–3 mol % of SrCl2' what is the concentration of cation vacancies?

    Solution

    We know that two Na+ ions are replaced by each of the Sr2+ ions while SrCl2 is doped with NaCl. But in this case, only one lattice point is occupied by each of the Sr2+ ions and produce one cation vacancy.

    Here 10 – 3 mole of SrCl2 is doped with 100 moles of NaCl Thus, cation vacancies produced by NaCl = 10 – 3 mol Since, 100 moles of NaCl produces cation vacancies after doping = 10 –3 mol

    Therefore, 1 mole of NaCl will produce cation vacancies after doping
    =  10 to the power of negative 3 end exponent over 100 space equals 10 to the power of negative 5 end exponent space mol
    therefore, total cationic vacancies
    =10-5 x Avogadro's number
    =10-5  x 6.023 x 1023
    =6.023 x 10-18 vacancies
     

    Question 150
    CBSEENCH12005498
    Wired Faculty App

    The ionic radius of CI– ion is 181 pm. Consider the closest packed structure in which all anions are just touching:
    (i) Calculate the radius of the cation that just fits into the octahedral holes of this lattice of anions. (ii) Calculate the radius of the cation that just fits into the tetrahedral holes of this lattice of anions.

    Solution
    given:
    Radius = 181 pm
    thus,
    Radius of octahedral void = 0.414 r = 0.414 x 181 pm = 74.934 pm
    Cation having radius 74.934 pm will just fit into octahedral voids.

    Radius of tetrahedral void
    = 0.225 r = 0.225 x 181 pm = 40.725 pm
    Cation having radius 40.725 pm will just fit into tetrahedral void.
    Question 151
    CBSEENCH12005499
    Wired Faculty App

    An element occurs in BCC structure with cell edge of 300 pm. The density of the element is 5.2 g cm–3. How many atoms of the element does 200 g of the element contain?

    Solution
    solution:
    we have given that
    z = 2 
    a = 300 pm
    d = 5.2 gm cm    –3
    by using the given formula we obtaine mass of element i.e.           
    Question 152
    CBSEENCH12005500
    Wired Faculty App

    Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell. (Molar mass of FeO = 72 g mol–1, NA = 6.02 x 1023 mol–1).

    Solution

    Solution:
    We  have given
    volume of the unit cell = (5 x 10-8cm)3 = 1.25 x 10-22cm3

    Density of FeO = 4g cm-3

    Density,

    ρ = z×Ma3×NA4= z×72(5×10-8)3×6.02×1023z= 4×1.25×10-22 ×6.02×102372 = 4.18≅ 4

    Each unit cell has four units of FeO. So it has four Fe2+ and four O2– ions.

    Question 153
    CBSEENCH12005501
    Wired Faculty App

    A metallic element exists as a cubic lattice. Each edge of the unit cell is 2.88 A°. The density of the metal is 7.20 g cm–3. How many unit cells will be there in 100 g of the metal?

    Solution

    solution: 
    we have given a= 2.88A0
    Density = 7.20g cm-3
    mass = 100g

    Volume of the unit cell = (2.88 A°)3
                              = (2.88×10-8cm)3= 23.9 × 10-24 cm3

    Volume of 100g of the metal

                     = WeightDensity=1007.20= 13.9 cm3

    No. of unit cell in 13.9 cm3
       
                       = 13.9 cm323.9 × 10-24 cm3= 5.82 × 1023.

    Question 154
    CBSEENCH12005502
    Wired Faculty App

    The unit cell of an element of atomic mass 96, and density 10.3 g cm–3 is a cube with edge length of 314 pm. Find the structure of crystal lattice (simple cubic, F.C.C. or B.C.C.) Avogadro’s constant. NA = 6.023 x 1023 mol–1?

    Solution

    solution:
    we have given

    Density of element,  ρ=10.3 g cm-3
    Cell edge  a= 314pm or 3.14 x 10-10 cm          

         NA =6.023×1023 mol-1

    Atomic mass = 96 g mol-1

                     P = Z×Ma3×NAZ = P×a3×NAM

    = 10.g cm-3×(3.14)3×10-30cm3×6.023×1023 mol-196 g mol-1=2

    The structure of the crystal lattice is B.C.C.

    Question 155
    CBSEENCH12005503
    Wired Faculty App

    An element a crystallises in fcc structure. 200 g of this element has 4.12 x 1024atoms. The density of A is 7.2 g cm-3. Calculate the edge length of the unit cell?

    Solution
    solution:
    we have given that
    mass = 200g
    density = 7.2 g cm-3
    NA = 4.12 x 1024
    Z = 4 
    by using the following equation,
     
    ρ = z× Ma3× NA
                         

    ⇒               a3 = Z×Md×N
                     

             = 4×2007.2×4.12×1024= 80029.664×10-24


                a3 = 26.968×10-24a = 2.997×10-8a = 299.7 pm.
     
    hence the edge length is 299.7pm
    Question 156
    CBSEENCH12005504
    Wired Faculty App

    An element crystallises in BCC structure. The edge length of its unit cell is 288 pm. If the density of crystal is 7.2 g cm–3, what is the atomic mass of the element?

    Solution

    Solution:
    we have given that  
    density, d= 7.2g cm-3
    edge = 288 x 10-10 cm 
    NA = 6.02 x 1023
    for Bcc structure Z= 2

    Atomic mass = ?

     
    d = ZMa3NAM = d×a3×NAZM=7.25×(2.88×10-10)×6.02×10232g/molM= 51.77 g/mol of atoms
            
    Thus, atomic mass of element is =51.77 g/mol

    Question 157
    CBSEENCH12005505
    Wired Faculty App

    Caesium chloride crystallises as a body centred cubic lattice and has a density of 4.0 g cm–3.  Calculate the length of the edge of the unit cell of caesium chloride crystal.
             [Molar Mass of CsCl = 168.5 g mol–1, NA = 6.02 x 1023 mol–1]

    Solution

    Solution;
    we have given that    

       Z =1, ρ = 4 g cm-3,
            M = 168.5 mol-1,NA = 6.02×1023mol-1   ρ = Z×Ma3×N

    or          a3 = Z×Mρ×NA    = 1×168.5g mol-14×6.02×1023    = 168.5×10-2324.08
    log a3 = log 168.5 + log 10-23 - log 24.083 log a = 2.2266 - 23.000-1.38163 log a = -22.1550+2-23 log a = 24+(2.1550)       a = Antilog 8¯.7183 = 5.228×10-8cm         = 5.228×10-8×1010pm = 522.8 pm.

    Question 158
    CBSEENCH12005506
    Wired Faculty App

    The density of chromium is 7.2 g cm–3. If the unit cell is cubic with edge length of 289 pm, determine the type of the unit cell (Atomic mass of Cr = 52 amu).

    Solution

    Solution:
    We have given that 

    Gram atomic mass of Cr(M) = 52.0 g mol-1
       Edge length of unit cell (a) = 289 pm
       Density of unit cell (ρ) = 7.2 g cm-3
           Avogadro's Number (N0) = 6.022×1023 mol-1       

         ρ = Z×Ma3×NA×10-30

    or              Z=ρ×a3×NA×10-30M
       
       Z=(7.2 g cm-3) × (289)3× (6.022×1023 mol-3) × (10-30 cm3)(52.0 g mol-1)=2

    Since the unit cell has 2 atoms, it is body centre in nature.

    Question 159
    CBSEENCH12005507
    Wired Faculty App

    Iron (II) oxide has a cubic structure and each of the unit cell is 5.0 A°. If density of the oxide is 4.0 g cm-3, calculate the number of Fe2+ and O2– ions present in each unit cell.

    Solution

    Volume of the unit cell = (5 A°)
    = (5 x 10–8 cm)3
    = 125 x 10–24 cm3
    = 1.25 x 10–22 cm3
    Density of FeO = 4.0 g cm–3

    Therefore, mass of the unit cell
    = 1.25 x 10–22 cm3 x 4 g cm–3
    = 5 x 10–22g
    Mass of all molecules of FeO
    =726.022×1023= 1.195 × 10-22 g

     Number of FeO molecules/unit cell
    =5×10-22g1.195×10-22g= 4.19≈4

    Thus, there are 4 Fe    2+ ions and 4O2– ions in each unit cell.

    Question 160
    CBSEENCH12005508
    Wired Faculty App

    An element (At. mass 60) have face centred cubic structure has a density of 6.23 g cm–3. What is the edge length of the unit cell?

    Solution
    Solution:
    we have given that
    Atomic mass of element = 60.
    Number of atoms per fcc unit cell = 4
    Density of the element = 6.23 g/cm    3

                                      = 6.231023 g/(pm)3

    Density,         d=N×MNA×a3


    or              a3 = N×Md×NA      =60×4×10306.23×6.02×1023      = 64 × 106 (pm)3


     Therefore, edge of unit cell, a = 400 pm.
    Question 161
    CBSEENCH12005509
    Wired Faculty App

    A metal (at. mass = 50) has a bcc crystal structure. The density of the metal is 5.96 g cm–3. Find the volume of its unit cell?

    Solution
     Solution:
    We have given that
    Atomic mass of the metal = 50g
    bcc unit cell, Z = 2
    Density of metal = 5.96 g/cm    3
    Therefore, volume of the unit,


    a3 = M×ZNA×d     = 50×26.02×1023×5.96    = 2.787 × 10-23 cm3
    Question 162
    CBSEENCH12005510
    Wired Faculty App

    Calculate the density of silver which crystallizes in a face centred cubic lattice with unit cell length 0.4086 nm (At. mass of Ag = 107.88)

    Solution

     Solution:
    We have given that

    Unit cell length,
    a = 0.40806 nm  = 4.086 x 10-10 m

      If fcc lattice the number of atoms per unit cell,
    i.e. Z = 4

            M for Ag = 107.88 g mol-1

                          =1.0788×10-1 kg mol-1

                  NA = 6.023 × 1023


    Density of Ag, d = ZMNA a3

                             = 4×1.0788×10-16.023×1023×(4.086×10-10)3= 1.051 × 104 kg m-3

     Thus the density of silver is 1.051 x 104 kgm-3

    Question 163
    CBSEENCH12005511
    Wired Faculty App

    An element crystallizes in a structure having a fcc unit cell of an edge 200 pm. Calculate its density if 200 g of this element contains 24 x 1023 atoms.

    Solution

    Solution:
    We have given that
    Edge length of the unit cell
    = 200 pm = 200 x 10–10 cm

    Vol. of the unit cell
    = (200 x 10–10 cm)3 = 8 x 10–24 cm3
    Therefore, volume of the substance
    = Vol. of unit cell x Vol. of 1 unit cell.

    Since the element has a fcc unit cell, number of atoms per unit cell = 4.
    Total number of atoms = Atoms/unit cell x number of unit cells.
                                         = 24 x 1023atoms = 4

    atoms/unit cell x No. of unit cells.

    ∴   No. of unit cells = 24×1023 atoms4 atoms/unit cell

                                   = 6×1023 unit cells.

    ∴  Volume of the substance
                       6×1023 unit cell × 8× 10-24 cm3/unit cell = 4.8 cm3


    Now,     Density  = MassVolume               = 200g4.8 cm3 =4.17 g cm-3.



    Question 164
    CBSEENCH12005512
    Wired Faculty App

    An element X with an atomic mass of 60 g/mol has density of 6.23 g/cm–3. If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.

    Solution
    Solution:
    we have given that 
    Density, d = 6.23 g/cm3
                  a = 400 pm
                  M = 60g/mol-1
     
    Volume  = (a3)=(400)3 = (400×10-10 cm)3

    Z=d×a3×NAM

    NAM=6.023×1023 mol-160 g/mol 

    Z = 6.023 g cm-3×(400)3×10-30 cm3×6.023×1023 mol-160 g mol-1    = 6.023×64×6.023600   = 2401.49600=4.

    Hence, the type of cubic unit cell is FCC

    Radius = aa2=40022=2002

                  = 2001.414=141.4 pm. 

    So radius of element is 141.4 pm

    Question 165
    CBSEENCH12005513
    Wired Faculty App

    An atom has fcc crystal whose density is 10 gm–3 and cell edge is 100 pm. How many atoms are present in its 100 g?

    Solution

    Solution:
    We have given that
    Density = 10 gm-3
    Mass = 100g
    edge of unit cell ,a= 4 since it is a Fcc crystal

    we have to find total number of atom, So by following relation we can get the result,

    Total No. of atoms = Z×Ma3.d

    Therefore, number of atoms

    =4×100 g(100×10-12m)3×10 gm-3= 4×1031 atoms
     
    thus the number of atoms is 4 x 1031 

    Question 166
    CBSEENCH12005514
    Wired Faculty App

    Analysis shows that a metal oxide has the empirical formula M0.98 O1.00. Calculate the percentage of M2+ and M3+ ions in this crystal.

    Or

    In an ionic compound the anion (N) form cubic close type of packing. While the cation (M+) ions occupy one third of the tetrahedal voids. Deduce the empirical formula of the compound and the coordination number of (M) ions.

    Solution

    Solution:

    M0.98 O1.00 is non-stoichiometric compound and is a mixture of M2+ and M3+ ions.
    Let x atoms of M3+ are present in the compound. This means that x M2+ have been replaced by M3+ ions.
    ∴  Number of M2+ ions = 0.96 – x.
    For electrical neutrality, positive charge on compound

    = negative charge on compound.
    ∴ 2(0.96 – x) + 3x = 2
    1.92 – 2x + 3x = 2
    or x = 2 – 1.92 = 0.08

    ∴     % of M3+ ions  = 0.080.96×100 = 8.33%

    Thus, M2+ is 92% and M3+ is 8% in given sample having formula, M0.96 O1.00.

    Or

    The number of tetrahedral voids formed is equal

    to twice the number of atoms of element N and only 13rd of these are occupied by the element M. Hence the ratio of number of atoms of M and is 2×13:1  or  2:3.  So, the formula of the compound is M2N3.

    Question 167
    CBSEENCH12005515
    Wired Faculty App

    An element E crystallizes in body centred cubic structure. If the edge length of the cell is 1.469 x 10–10 m and the density is 19.3 g cm–3, calculate the atomic mass of this element. Also calculate the radius of this element.

    Solution
    Solution:
    We have given that 
    Z= 2 since it is bcc structure
    edge length, a = 1.469 x 10-10
    Density, d= 19.3 g cm-3
    Mass = ?

    d=2×3a3×NA19.3 = 2×M(1.469×10-8)3 × 6.023 × 1023or M = 19.3×3.17×10-34×6.023×10232         = 19.3×3.17×6.023×102          = 18.4 g mol-1.

    mass of element is 18.4g mol-1


    (ii) Cell edge,

             a=1.469×10-10m

    Body diagonal = 4R

    Then 4R = 39
                 = 3×1.469×10-10m
    or     R = 3×1.469×10-104m    = 0.64 × 10-10m    = 6.4 × 10-11m.

    Radius of element is 6.4 x 10-11m.
    Question 168
    CBSEENCH12005516
    Wired Faculty App

    Calculate the efficiency of packing in case of a metal crystal for
    (i) simple cubic
    (ii) body- centred cubic
    (iii) face - centred cubic . (With the assumptions that atoms are touching each other).

    Solution

    Solution:

    (i) Simple Cubic: A simple cubic unit cell has one sphere (or atom) per unit cell. If r is the radius of the sphere, then volume occupied by one sphere present in unit cell = 43πr3
               
    Edge length of unit cell (a) = r + r = 2r
    Volume of cubic (a3) = (2r)3 = 8r3
    Volume of occupied by sphere  = 43πr3
    Percentage volume occupied = percentage of efficiency of packing

    = Volume of sphereVolume of cube ×100= 43πr38r3×100= 16×3.143×100 = 52.4%

    For simple cubic metal crystal the efficiency of packing = 52.4%.

    (b) Body centred cubic: From the figure it is clear that the atom at the centre will be in touch with other two atoms diagonally arranged and shown with solid boundaries.

    In ∆EFD         b2 = a2+a2                              = 2a2                        b = 2aNow in ∆AFD                        c2 = a2+ b2                            = a2+2a2 = 3a2                        c = 3a

    The length of the body diagonal c is equal to 4r where r is the radius of the sphere (atom). But c = 4r, as all the three spheres along the diagonal touch each other

    ∴              3a = 4r

    or                   a=4r3

    or                    r=34a.



    Fig. BCC unit cell. In this sort of structure total number of atoms is two and their volume is 2×43πr3.

    Volume of the cube, a    3 will be equal to
    43r or a3 = 43r3.

    Therefore, Percentage of efficiency Volume occupied by four - spheres

    = in the unit cellTotal volume of the unit cell×100%= 2×43πr3×100(4/3r)3%= (8/3) πr3 × 100[64/(33)r3]% = 68%


    (c) Face centred cubic: A face centred cubic cell (fcc) contains four spheres (or atoms) per unit volume occupied by one sphere of radius

    r = 4/3 πr3

    Volume occupied by four spheres present in the unit cell

    r = 4/3 πr3 x 4 = 16/3 πr3





    Fig.  Face centred cubic unit cell. Figure indicates that spheres placed at the corners touches a face centred sphere. Length of the face diagonal
    = r + 2r + r = 4r





    Hence, for face centred cubic, efficiency of packing = 74%.
    Question 169
    CBSEENCH12005517
    Wired Faculty App

    Explain the following with suitable examples:
    Ferromagnetism.

    Solution
    Ferromagnetism:–
    Ferromagnetism: few substances like iron, cobalt, nickel, gadolinium and CrO2 are strongly attracted by a magnetic field. Such substances are known as ferromagnetic substances. These substances can be permanently magnetised with the help of strong electrostatic field.  The metal ions of ferromagnetic substances are grouped together into small regions called domains in solid state. So that each domain acts as a tiny magnet. The domains of  un magnetized piece of a ferromagnetic are randomly oriented so that their magnetic moments get cancelled out. When such substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field  and a strong magnetic effect is produced. This ordering of domains remains same  even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.


    Fig.  Schematic alignments of magnetic moments in ferromagnetism.
    Question 170
    CBSEENCH12005518
    Wired Faculty App

    Explain the following with suitable examples:
    Paramagnetism

    Solution
    paramagnetism:

    few substances like O2, Cu2+, Fe3+, Cr3+  are weekly attracted by a magnetic field.
    These substabces are magnetised in a magnetic field in the same direction. When we remove magnetic field ,they lose their magnetism. paramagnetism is takes place due to presence of one or more unpaired electrons. These unpaired electron are attracted by the magnetic field.
    Question 171
    CBSEENCH12005519
    Wired Faculty App

    Explain the following with suitable examples:
    Ferrimagnetism

    Solution
    Ferrimagnetism:

    This effect is observed when the magnetic moments of the domains in the substance are aligned in parallel and anti–parallel directions in unequal numbers. Ferrimagnetic substances are weakly attracted by magnetic field as compared to ferromagnetic substances. Magnetite like Fe    3O4  and ferrites like MgFe2O4 and ZnFe2O4 are examples 
    Question 172
    CBSEENCH12005520
    Wired Faculty App

    Explain the following with suitable examples:
    Piezoelectric effect (or pressure electricity)

    Solution

    Piezoelectric effect (or pressure electricity):
    The word piezoelectricitymeans electricity resulting from pressure.
    Insulators do not conduct electricity be the electrons present in them are held tightly to the individual atoms or ions and do not move. However, when an electric field is applied polarisation takes place and newly formed dipoles may align themselves in an ordered manner so that such crystals have a net moment.
    When mechanical stress is applied on a polar crystals so as to deform them, electricity produced due to displacement of ions. This is known as piezoelectric effect and electric so this produced is known as Piezo electricity or pressure electricity. Example are : titanium, barium and lead, lead zirconate (PbZrO3), ammonium dihydrogen phosphate and quartz.

    Question 173
    CBSEENCH12005521
    Wired Faculty App

    Explain the following with suitable examples:
    Antifluorite structure 

    Solution
    Antifluorite structure: 
    The antifluorite structure is the opposite arrangment with anion in FCC array with coordination number 8 and cation in the tetrahedral holes with coordination number 4. example of the antifluorite structure are K2O, LiO, Na2O and K2S.
    Question 174
    CBSEENCH12005522
    Wired Faculty App

    Give reasons for the following:
    Molecular solids are generally soft and easily compressible.

    Solution
    A molecular soild consist of small, non-polar covalent molecule and is held together by landon dispersion forces. thus the force of attraction between the molecule is less de to this reason molecular soilds are generally soft and easily compressible.
    Question 175
    CBSEENCH12005523
    Wired Faculty App

    Give reasons for the following:
    The energy required to vapourise one mole of copper is smaller than that of energy required to vapourise one mole of diamond.

    Solution
    The bonding have great influence on the any substance. Copper is a metallic solid having metallic bonds while diamond is a covalent solid having covalent bonds. Metallic bonds are weaker than covalent bonds thus lesser amount of energy is required to break metallic bonds than covalent bonds.
    Question 176
    CBSEENCH12005524
    Wired Faculty App

    Give reasons for the following:
    All metals have metallic bonds but some metals are soft and having low melting points and some are hard and having high melting points.

    Solution
    The strength of metallic bond depends upon the numbr of valence electrons of the metal atom and the magnitude of forces binding these electrons to the nuclei. Thus the strength of metallic bond in different metals may be weak or strong, hence their nature (soft or hard) and their melting points may be different.
    Question 177
    CBSEENCH12005525
    Wired Faculty App

    Give reasons for the following:
    Covalent crystals like diamond and silicon carbide are quite hard and difficult to break.

    Solution
    In covalent crystals, the shared pair of electrons give rise to very strong forces between them, therefore covalent solids like diamond and silicon carbide are quite hard and difficult to break and show very high melting points.
    Question 178
    CBSEENCH12005526
    Wired Faculty App

    Give reasons for the following:
    Ionic solids, inspite of being made up of ions, do not conduct electricity.

    Solution

    Solid ionic compounds do not conduct electricity because the ions (charged particles) are locked into a rigid lattice or array. The ions cannot move out of the lattice, so the solid cannot conduct electricity.

    When molten, the ions are free to move out of the lattice structure.

     

    • Cations (positive ions) move towards the negative electrode (cathode) 
      M+ + e -----> M

       

    • Anions (negative ions) move towards the positive electrode (anode) 
      X- -----> X + e

    When an ionic solid is dissolved in water to form an aqueous solution, the ions are released from the lattice structure and are free to move so the solution conducts electricity just like the molten (liquid) ionic compound.

    Question 185
    CBSEENCH12005533
    Wired Faculty App

    Electron trapped in anion vacancies are known as__________.

    Solution

    F-centre

    Question 187
    CBSEENCH12005535
    Wired Faculty App

    __________ is the most unsymmetrical primitive unit cell.

    Solution
    Triclinic
    Question 190
    CBSEENCH12005538
    Wired Faculty App

    A.

    The number of octahedral sites per sphere in fcc is 4.
    Solution
    A. FALSE
    Question 191
    CBSEENCH12005539
    Wired Faculty App

    A.

    AgBr can show both Schottky and Frenkel defect.

    Solution
    A. TRUE
    Question 199
    CBSEENCH12005547
    Wired Faculty App

    The number of atoms in a face-centred cubic unit cell is:

    • 4

    • 6

    • 1

    • 2
    Solution

    A.

    4

    Question 200
    CBSEENCH12005548
    Wired Faculty App

    Schottky defect generally appears in
    • NaCl
    • KCl
    • CsCl 
    • all the above
    Solution

    D.

    all the above
    Question 205
    CBSEENCH12005553
    Wired Faculty App

    CsCl has which type of lattice?
    • SC
    • fcc

    • bcc

    • hcp
    Solution

    C.

    bcc

    Question 208
    CBSEENCH12005556
    Wired Faculty App

    Name the crystal defect which lowers the density of an ionic crystal

    Solution

    Density of crystal is lowered by shottky defect.

    Question 209
    CBSEENCH12005557
    Wired Faculty App

    What makes the crystal of KCl appear sometimes violet?

    Solution
    This is due to F centre, electron become exicted and when they come to ground state they radiate energy in visible reigon.
    Question 210
    CBSEENCH12005558
    Wired Faculty App

    What is the effect of Schottky and Frenkel defects on the density of crystalline solids?

    Solution

    Schottky defect decrease the density solid where as no such defect arise due to frenkel defect.

    Question 211
    CBSEENCH12005559
    Wired Faculty App

    What is the maximum possible coordination number of an atom in an hcp crystal structure of an element?

    Solution

    The maximum possible coordination number of an atom in hcp structure of an element is 12.

    Question 212
    CBSEENCH12005560
    Wired Faculty App

    How many atoms can be assigned to its unit cell if an element forms
    (i) a body centred cubic cell, and (ii) a face centred cu    bic cell?

    Solution

    Body centre wholly belongs to the unit cell in which it is present. Thus
    in a body-centered cubic (bcc) unit cell:
    (i) 8 corners × 1/8per corner atom= 8 x 1/8 = 1 atom
    (ii) 1 body centre atom = 1 × 1 = 1 atom
    ∴ Total number of atoms per unit cell = 2 atoms
    For face centred 

    (i) 8 corners atoms × 1/8 atom per unit cell
    = 8  x1/8 = 1 atom

    (ii) 6 face-centred atoms ×1/2 atom per unit cell = 6 ×1/2 = 3 atoms
    ∴ Total number of atoms per unit cell = 4 atoms

    Question 213
    CBSEENCH12005561
    Wired Faculty App

    Name an element with which germanium can be doped to produce an n-type semiconductor.

    Solution

    arsenic (As) or phosphorus (P).

    Question 214
    CBSEENCH12005562
    Wired Faculty App

    What is the nature of crystal defect produced when sodium chloride is doped with MgCl2?

    Solution

    2Na+ will replaced by Mg2+ ion to maintain electrical neutrality thus ahole created for lattice site for every Mg2+ ion introduce thus the defect produced will schottky defect

    Question 215
    CBSEENCH12005563
    Wired Faculty App

    What happens when a ferromagnetic substance is subjected to high temperature?

    Solution

    When ferromagnetic substance is subjected to high temperature it loses its magnetic property and become paramagnetic in nature this happen due to disorder ness in electronic arrangment.

    Question 216
    CBSEENCH12005564
    Wired Faculty App

    Name a substance which on addition to AgCl causes cation vacancy in it.

    Solution

    Cadium chloride is substance which on addition to AgCl cause cation vacancy in it.

    Question 217
    CBSEENCH12005565
    Wired Faculty App

    What makes alkali metal halides sometimes coloured, which are otherwise colourless?

    Solution

    Alkali metal halides have anionic sites occupied by unpaired electrons. These are called F-centres, and impart colour to the crystals of alkali metal halides. For example, the excess of lithium in LiCl makes it pink.

    Question 218
    CBSEENCH12005566
    Wired Faculty App

    Define the term ‘amorphous’.

    Solution
    Amorphous solids are the solids whose constituent particles have a completely random arrangement. Amorphous solids do not have a sharp melting point and melt over a range of temperature. These solids are isotropic in nature. Amorphous solid have elasticity, electric conductance, reflective index and rate of dissociation are the same throughout all the direction. Amorphous solid always flow very slowly therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. Amorphous solid do not have definite heat of fusion. When we cut a piece of amorphous solid cut with a sharp–edged tool, they cut into two pieces with irregular surfaces. Examples :– glass, rubber, and plastic.
     
     
    Question 219
    CBSEENCH12005567
    Wired Faculty App

    Which point defect lowers the density of a crystal?

    Solution

    Shottkey defect can lower the density of a crystal.

    Question 220
    CBSEENCH12005568
    Wired Faculty App

    What is coordination number of a metal atom in ccp structure?

    Solution

    Coordination number of CCP is 12.

    Question 222
    CBSEENCH12005570
    Wired Faculty App

    In an alloy of gold and cadmium, gold crystallizes in cubic structure occupying the corners only and cadmium fits into the face centre voids. What is quantitative composition of the alloy?

    Solution

    In an alloy of gold and cadmium, gold crystallizes in cubic structure occupying the corners only and cadmium fits into the face centre voids.

    Number of gold atoms     =1/8 x 8 =1Au

    Number of candium atoms =1/2 x 6 =3Cd

    Hence, quantitative composition on = AuCd3

    Question 223
    CBSEENCH12005571
    Wired Faculty App

    What is the difference between glass and quartz, though both contain SiO44– units?

    Solution

    Both naturally-occurring and artificial quartz crystal contains at least ninety-nine percent silicone dioxide, while cut glass crystal only possesses at most eighty percent silicone dioxide. 

    Question 224
    CBSEENCH12005572
    Wired Faculty App

    What is the effect of Frenkel defect on electrical conductivity of the solid?

    Solution

    Compounds having frenkel defect conduct electricity to a small extent. When electric field is applied, an ion moves from its lattice site to occupy a hole, it creates a new hole. In this way, a hole moves from one end to the other. Thus, it conducts electricity across the crystal. Due to the presence of holes, stability (or the lattice energy) of the crystal decreases.

    Question 225
    CBSEENCH12005573
    Wired Faculty App

    What other element may be added to silicon to make electrons available for the conduction of electric current?

    Solution

    When doping a semiconductor, such as the group IV element silicon (Si), with arsenic (As), a pentavalent n-type dopant from group V in the periodic table the dopant behaves as an electron donor.

    Question 226
    CBSEENCH12005574
    Wired Faculty App

    Why does Frenkel defect not change the density of AgCl crystal?

    Solution
    In the frenkel defect the ions are not removed from the crystal to maintain electrical neutrality. so there will be no change in the crystal structure. that is there is no decrease in the no of ions.all the ions are inside the crystal.they are only dislocated.
    Question 227
    CBSEENCH12005575
    Wired Faculty App

    What is the difference between ferromagnetic and ferrimagnetic substances?

    Solution

    In ferromagnetic materials, neighboring dipoles tend to line up in the same direction, whereas in ferrimagnetic materials, neighboring dipoles tend to line up in opposing directions.

    Question 228
    CBSEENCH12005576
    Wired Faculty App

    Which ‘point defect’ lowers the density of ionic crystals?

    Solution

    Schottky Defect lowers the density of ionic crystal.

    Question 229
    CBSEENCH12005577
    Wired Faculty App

    Assign reasons the following:
    (i) Phosphorus doped silicon is a semi-conductor.
    (ii) Schottky defect lowers the density of a solid.
    (iii) Some of the very old glass objects appear slightly milky instead of being transparent.

    Solution

    (i) When silicon is doped with phosphorus four valence electrons of phosphorus are involved in bond formation with the neighbouring silicon atoms, while the fifth valence electron is left free to conduct electricity. This type of conduction which arises due to the availability of excess electrons is called n-type conduction.

    (ii) Schottky defect arises when equal number of cations and anions are missing from their lattice sites. As the number of ions decreases due to this defect In order to maintain electrical neutrality, the number of missing cations and anions are equal, the mass decreases, but the volume remains the same. As a result, the density of the solid decreases.

    (iii) Some of the very old glass objects appear slightly milky instead of being transparentbecause of some crystallisation at that point.

    Question 230
    CBSEENCH12005578
    Wired Faculty App

    What is a semiconductor? Describe two main types of semiconductors, giving examples and their distinctive features.

    Solution

    Semiconductors : These are the solids with conductivities in the intermediate range from 10–6 to 104 ohm–1m–1.

    In case of semiconductors, the gap between the valence band and conduction band is small Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semiconductors increases with rise in temperature, since
    more electrons can jump to the conduction band. Substances like silicon and germanium show this type of behaviour and are called intrinsic semiconductors.
    The conductivity of these intrinsic semiconductors is too low to be of practical use. Their conductivity is increased by adding an appropriate amount of suitable impurity. This process is called doping.

    There are two types of semiconductor
    i) n- type semiconductor
    ii) p- type semiconductor

    Element of group14 of the periodic table, when doped with a group 15 element like P or As, Four out of five electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron is extra and becomes delocalised.

    These delocalised electrons increase the conductivity of doped silicon (or germanium). Here the increase in conductivity is due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor.

    p-type : 

    Silicon or germanium can also be doped with a group 13 element like B, Al or Ga which contains only three valence electrons. The place where the fourth valence electron is missing is called electron hole or electron vacancy. This hole can be act as charge carrier in the lattice. This known as p - type semiconductor.



    Question 231
    CBSEENCH12005579
    Wired Faculty App

    With reference to crystal structure, what is meant by coordination number? 

    Solution

    The coordination number of an atom, ion or molecule is the number of constituent particles which touch that particular atom, atom or molecule.

    Question 232
    CBSEENCH12005580
    Wired Faculty App

    What is the coordination number of atoms
    (i) in a cubic close packed structure?
    (ii) in a body-centred cubic structure?

    Solution

    i) The coordination number of  cubic close packed structure is 12.

    ii) The coordination number of body-centred cubic structure is 8.

    Question 233
    CBSEENCH12005581
    Wired Faculty App

    Calculate the efficiency (percentage of volume occupied and unoccupied) of packing in case of a metal crystal for simple cubic.

    Solution

    Let ‘a’ be the edge length of the unit cell and r be the radius of sphere.

    As sphere are touching each other

    Therefore a = 2r

    No. of spheres per unit cell = 1/8 × 8 = 1 

    Volume of the sphere = 4/3 πr3 

    Volume of the cube = a3= (2r)3 = 8r3

    ∴ Fraction of the space occupied = 1/3πr3 / 8r3 = 0.524

    ∴ % occupied = 52.4 %



    Question 234
    CBSEENCH12005582
    Wired Faculty App

     (a) Determine the type of cubic lattice to which a given crystal belongs if it has edge length of 290 pm and density is 7.80 g cm–3. (Molecular mass.= 56 g mol–1)

    (b) Why does zinc oxide exhibit enhanced electrical conductivity on heating?
    Solution

    we know that

    d =ZMa3NA

    as we have given 
    d= 7.80 g cm–3.
    M=56 g mol–1
    a= 290 pm or a3 =2.43 x10-23
    putting all value in above equation we get,

    Z =a3 x d xNAMZ= 2.43 x10-23 x 7.80 x6.022 x102356 =2.03

    hence it belong to bcc crystal lattice.

    b) Zinc oxide is white in colour at room temperature. On heating it loses electron and turns yellow in colour. 


    ZnO→∆ Zn2+ +12O2 +2e- 

    The electron liberate after heating, can act as charge carrier and thus on heating zinc oxide it exhibit electrical conductivity.

    Question 235
    CBSEENCH12005583
    Wired Faculty App

    The compound CuCl has the ZnS (Cubic) structure. Its density is 3.04 g cm–3. What is the length of the edge of unit cell? (At. mass of Cu = 63.5, CI = 35.5) 

    Solution

    Formula mass of CuCl =63.5 +35.5

                                          =  99.0

    The number of formula units per cell of ZnS is 4. It has face centred cubic structure.

     we have Z  =4density (d) =3.4 g Cm-3N0 (Avogadro's number) =6.023 x1023Mass =99.0 gramTherefore using formula density =Mass x Zunit cell volume(Cm)3 xNavod.no.d =M x Za3 x N0a3 =99 x43.4 x6.023 x1023a3 =1.932 x10-22cm3unit cell length a= (1.932 x10-22)13takinig log both sidelog a =13log  1.932 x10-22       =13(0.2860-22) =13(-21.7140)       = -7.238 =0.762 x10-8taking antilog a =antilog 0.762 x10-8 =5.758 x10-8 cm = 5.758 x 10-8cm




    Question 236
    CBSEENCH12005584
    Wired Faculty App

    Potassium crystallises in a simple cubic unit cell. It has an atomic mass of 209 and its density is 91.5 g m–3. What is the edge length of its unit cell? 

    Solution

    we have given,
    Mass= 209 g
    Number of  atom per unit cell =1 (Simple cubic)
    density =91.5 g m-3
    NA =6.023 x1023
    edge length of unit cell =?
     
    By applying formula


    density = Mass x Number of atom per unit cellvolume of unit cell  x avogadro's numberd= M x Za3 x NA91.5 = 209 x 1a3 x 6.023 x 1023a3 =209 x 191.5 x 6.023 x 1023

    a= 15.59 x10-8 cm

    Question 237
    CBSEENCH12005585
    Wired Faculty App

    An element crystallises in fcc structure with an edge of 200 pm. Calculate density if 200 g of this element contains 24 x 1023 atoms. 

    Solution

    we have 
    edge length =200Pm
    volume of the unit = ( 200 x10-10 cm)3
                             = 8 x10-24 cm-3

    In a FCC unit cell there are four atoms per unit cell.therefore  mass of unit cell  =200 x 424 x1023 =33.3 x10-23gDensity  =mass of unit cell volume of unit cell =33.3 x10-23g8 x10-24cm3  =41.6 g cm-3

    density = 41.6 g cm-3.





    Question 238
    CBSEENCH12005586
    Wired Faculty App

    The nearest neighbour Ag atoms in the silver crystal are 2.87 x 10–10 m apart. What is the density of silver? Silver crystallises in fcc form.

    Solution

    we have given :
    mass of silver = 107.87 g
    Z(FCC)          = 4
    distance between nearest neighbour Ag atoms = 2.87 x10-10

    face diagonal 2 =2.87 x10-10 m2a2 =2.87 x10-10 m a= 2.87 x10-10 x 2a= 2.87 x10-10 x 1.414a= 4.05 x 10-10now using formula d= Z x Ma3 x NAd =4 x 107.87(4.05 x 10)3 x 6.02 x1023d= 10.84 g cm-3

    Question 239
    CBSEENCH12005587
    Wired Faculty App

    The cell edge of a fcc element of atomic mass 108 is 409 pm. Calculate its density.

    Solution

    we have given,.
    mass =108 g
    edge (a) = 409pm  
    a3 = 6.84 x1023 cm3
    Na =6.023 x10-233
    Z =4

    Apply formula

     d= Z xMa3 x NAd = 4 x 1086.84 x 1023 x 6.02 x10-23 = 10.98 g cm-3

    density =  10.98 g cm–3

    Question 240
    CBSEENCH12005588
    Wired Faculty App

    An element ‘A’ of atomic mass 100 having bcc structure has unit cell edge of 400 pm. Calculate the density of ‘A’ and the number of unit cells for 10 g of ‘A’.

    Solution

    length of the unit cell edge = 400Pm =400x10-10 cm
     volume of the unit cell  = ( 400 x10-10cm)3
                                     = 6.4 x 10-23 cm-3

    as the element A forms a body centred  cubic lattice so no of atom per units cell is 2

    z= 2 atoms unit cell

    atomic mass of the element  =100 g/mol
    density of element is given by

    md= M xZa3 xNA d= 100 x26.4 x 10-23 x 6.023 x 10 23 = 51.88 g/cm3volume of 10 g of A = massDensity                             = 10g5.188g cm-3                           = 1.9275 cm3number of unit cell in 1.9275 cm3 volume  =volume of the substance unit cell volume                                                           = 3.0 x1022 unit cell



















    Question 241
    CBSEENCH12005589
    Wired Faculty App

    Aluminium forms fcc cubic crystals. The density of aluminium is 2.7 g cm–3. Calculate the length of the edge of the unit cell of Al.

    Solution

    we have give that aluminium forms Fcc cubic crystal.
    density = 2.7g cm-3
    Mass of aluminium = 27g
    number of unit cell =4 (FCC)
    we have find edge length
    Thus using formula

    d =Z xMa3 x NAputting the value in this equation 2.7 =4 x276.023 x 1023 x a3 a3 = 4 x 27 6.023 x 10 23 x 2.7a =4.05 x 10-8


    Question 242
    CBSEENCH12005590
    Wired Faculty App

    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 a fcc structure.

    Solution

    we know that

    d= ZMa3NA

    as given that,
    density= 2.75 g cm–3
    a = 654 pm or a3 = 2.79 x 10-23
    mass of KBr is 119 gram



    Z =2.75 x6.54 x10-23 x6.022 x1023119 =4

    Question 243
    CBSEENCH12005591
    Wired Faculty App

    Tungsten has a density of 19.35 g cm–3 and the length of the side of the unit cell is 316 pm. The unit cell is a body centred unit cell. How many atoms does 50 grams of the element contain ?

    Solution

    we have given,

    density =19.35 g cm-3
    edge length =316 pm
    Number of unit cell =2 (BCC)
    Mass =?

    Thus,
    Length of edge of the unit cell = 316 pm =316 x 10-10cmvolume of the unit cell = (316 x 10-10cm)3                              = 3.2 x 10-23 cm3/unit cellDensity  =atomic mass x Zunit cell volume  x NAAtomic mass = Density xUnit volume  xNAZ=19.35 x 3.2 x 10-23 x 6.023 x 10 232 =186.5 g/ molAs 186.5 g the element contain NA atoms   = 6.023 x 10 23 atomsso 50g of the element contain NA atoms =6.023 x 1023 x 50186.5=1.614 x 1023 atoms

    Question 244
    CBSEENCH12005592
    Wired Faculty App

    In the cubic crystal of CsCl (d = 3.97 g cm–3), the eight corners are occupied by CI–with a Cs+ at the centre and vice versa. Calculate the distance between the neighbouring Cs+ and CI– ions. What is the radius ratio of the two ions ? (At. mass of Cs = 132.91 and CI = 35.45).

    Solution

    In a unit cell there are one Cs and 1 x 8/8 =1 chlorine  (Cl-) such that one CsCl molecule 

    therefore 
    As we have given 
    density = 3.97 g cm-3
    Mass of CsCl = 168.36g
    Number of unit cell(Z) = 1

    apply formula d= Z xMa3 x NA3.97 = 1 x 168.36a3 x 6.023 x 1023a = 4.13 x 10-8 cma =4.13 A0

    for a cube of side  length 4.13A0 diagonal

                                        =3a i.e.,3 x 4.13 = 7.15A0

    as it is a BCC with  Cs+ at centre radius r+ and Cl- at corner radius r- so, 

    2r+ +2r- =7.15 or r+ +r- =3.57A0

    such that distance between neighbouring Cs+ and Cl- =3.57A0

    now assume two Cl- ion touch with each other so length of unit cell = 2r- =4.13
    r- =2.06A0
    r+ =3.57 -2.06 =1.51

    r+/r-=1.51/2.06 =0.73

    Question 245
    CBSEENCH12005593
    Wired Faculty App

    How the crystalline solids are classified on the basis of the nature of bonding? Give suitable examples and nature of the forces present in different types of solids.

    Solution

    i) 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 which repeats itself periodically over the entire crystal. Sodium chloride and quartz are typical examples of crystalline solids.

    ii) Crystalline solid have a sharp melting point.

    Crystalline solids can be classified on the basis of nature of intermolecular forces operating in them into four categories.
    i) molecular
    ii)ionic
    iii)metallic
    iv)covalent solids.

    i) molecular: Molecules are the constituent particles of molecular solids. These are further sub divided into the following categories:

    a) Non polar Molecular Solids: They comprise of either atoms, for example, argon and helium or the molecules formed by non polar covalent bonds for example H2, Cl2.

    b) Polar Molecular Solids: The molecules in which solids are held together by relatively stronger dipole-dipole interactions.for example HCl, SO2, etc.

    c) Hydrogen Bonded Molecular Solids: The molecules of such solids contain polar covalent bonds between H and F, O or N atoms. Strong hydrogen bonding binds molecules of such solids like H2O (ice).

    ii) Ionic solid : Ions are the constituent particles of ionic solids. In ionic solid cations and anions bound by strong coulombic (electrostatic) forces. for example NaCl, KCl etc.

    iii) metallic solid: These solids contain metal atoms as constituent particles. As metals have a good tendency to lose their valence electron and change in to positively charged metal ions (kernel). These electrons can easily move throughout the whole crystal and form the sea of free electrons. for example iron, calcium etc.

    iv) covalent solids: A wide variety of crystalline solids of non-metals result from the formation of covalent bonds between adjacent atoms throughout the crystal. They are also called giant molecules. for example diamond, silicon carbide, etc.



    Question 246
    CBSEENCH12005594
    Wired Faculty App

    Write two important features, coordination numbr of the ions and number of formula units per unit cell for the following crystals:
    (i) Cesium chloride, (ii) Zinc sulphide, (iii) Calcium fluorite, (iv) Sodium oxide.

    Solution

    i) CsCl is simple cubic cell. Cesium ion is surrounded by eight chloride ion which are also disposed towards the corner of a cube therefore both type ions are in equivalent positions and the stoichiometry is 1:1 . the coordination of CsCl is 8:8.

    ii) Zinc sulfide is a FCC unit cell. The net number of zinc cation per unit cell is four, and the net number of sulfide anions per unit cell is four therefore, the ratio of ZnS ion in the cell is 1:1.

    iii)Calcium fluoride is a FCC unit cell. The net number of Calcium cation per unit cell is four and the net number of fluoride anion is eight. Therefore the ratio of CaF2 ion in cell is 1:2 

    iv) Na2O has the structure opposite to CaF2. In
    this case coordination number of Na+ ions is 4 and that of O2- ion is 8. Thus Na2O has 4:8 coordination.



    Question 247
    CBSEENCH12005595
    Wired Faculty App

    Explain the reasons why:
    (a) Frenkel defect is not found in pure alkali halides?
    (b) Zinc oxide appears yellow on heating?
    (c) Solid containing F-centres are para magnetic?
    (d) Uncharged atoms or molecules never crystallize in a simple cubic structure?
    (e) A given element will have the same density in both a hexagonal close-packed structure and a cubic close-packed structure?

    Solution

    a) Frenkel defect is not found in the pure alkali metal halides because this defect is due to vacancy of ion ,which is shifted in interstitial spaces and because the ions are too large and cannot get into interstitial sites.

    b) ZnO becomes yellow on heating due to frankel defect, when ZnO is heated, it loses oxygen reversibly and turns yellow in colour.

    Zno + heat -->Zn2+ +1/O2 +2e-

    The Zn2+ ions formed the vacant voids in the crystal to form non- stoichiometric solid. The released electrons are trapped in its neighborhood. It is therefore non-stoichiometric ZnO is yellow and shows increased conductivity.

    c) The free electrons trapped in the anion vacancies are called as F-centers. Solids containing F-centres are paramagnetic because the electrons occupying the vacant sites are unpaired.

    d) Uncharged molecules and atoms are packed more efficiently in closed-packet structures. Hence theses uncharged molecules and atoms do not crystallize in simple cubic structures.

    e) In both the structure, the fraction of the total volume occupied is 0.74. The two structure have the same coordination number of 12.



    Question 248
    CBSEENCH12010037
    Wired Faculty App

    What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 1/3rd of tetrahedral voids?
    Solution

    Number of tetrahedral voids formed = 2 x Number of atoms of element Y

    Number of atoms of element Y in the ccp unit cell = 4

    Number of tetrahedral voids formed = 2 x 4 = 8


    Number of tetrahedral voids occupied by atoms of X = 1/3x8

    Ratio of the numbers of atoms of X and Y = 8/3: 4 =2: 3


    Hence, the formula of the compound is X2Y3.

    Question 249
    CBSEENCH12010071
    Wired Faculty App

    An element with molar mass 27 g mol-1  forms a cubic unit cell with edge length 4.05 x 10-8 cm . If its density is 2.7 g cm-3 , what is the nature of the cubic unit cell?
    Solution

    Molar mass of the given element, M = 27 g mol-1 = 0.027 kg mol-1

    Edge length, a = 4.05 x 10-8 cm = 4.05 x 10-10 m

    Density, d = 2.7 g cm-3 = 2.7 x 103 kg m-3

     Applying the relation,

      straight d space equals fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction

    Where, Z is the number of atoms in the unit cell and NA is the Avogadro number. Thus,

     

    straight Z equals space fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A space over denominator straight m end fraction fraction numerator 2.7 space straight x space 10 cubed space straight x space left parenthesis 4.05 space straight x space 10 to the power of negative 10 end exponent right parenthesis cubed space straight x space 6.022 space straight x space 10 to the power of 23 over denominator 0.027 end fraction space equals 4

    Since the number of atoms in the unit cell is four, the given cubic unit cell has a face-centred cubic (fcc) or cubic-closed packed (ccp) structure.

    Question 250
    CBSEENCH12010094
    Wired Faculty App

    An element with density 11.2 g cm-3  forms a f.c.c. lattice with edge length of 4 x10-8

    Calculate the atomic mass of the element. (Given:  NA = 6.022x 10-23 (mol-1) 
    Solution

    Density, d = 11.2 g cm-3

    Edge length, a = 4x10-8 cm

    Avogadro number, NA = 6.022x1023 mol-1

    Number of atoms present per unit cell, Z (fcc) = 4


    We know for a crystal system,

      straight d equals space fraction numerator straight z space straight x space straight m over denominator straight a cubed space straight x space straight N subscript straight A end fraction straight m space equals fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction We space get comma straight M space equals fraction numerator 11.2 space straight x space 64 space straight x space 10 to the power of negative 24 end exponent space straight x space 6.022 straight x space 10 to the power of 23 over denominator 4 end fraction equals space 107.91 space straight g

       

    Thus, the atomic mass of the element is 107.91 g.

     

    Question 251
    CBSEENCH12010095
    Wired Faculty App

    Examine the given defective crystal:
     

    Answer the following questions:


    (i) What type of stoichiometric defect is shown by the crystal?

    (ii) How is the density of the crystal affected by this defect?

     (iii) What type of ionic substances shows such defect? 
    Solution

    (i) Schottky defect is shown by the mentioned crystal, as an equal number of cations and anions are missing in the crystal lattice.

    (ii) This defect leads to decrease in density, as an equal number of the cations and anions are missing from the crystal lattice. A number of such defects in ionic solids are quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature.

    (iii) This kind of defect is shown by that ionic substance in which the cations and anions are of almost similar sizes. 


    Examples:  NaCl, KCl and CsCl.  

    Question 252
    CBSEENCH12010137
    Wired Faculty App

    How many atoms constitute one unit cell of a face-centered cubic crystal? 
    Solution

    Number of atoms in one face centred cubic unit cell can be determined:-

     

    (i) 8 corners atoms × 1/8 per corner atom =   1 over 8 x 8= 1 atom

    (ii) 6 face-centered atoms × 1/ 2 atom per unit cell =  1 half x 6= 3 atoms

     

    ∴ Total number of atoms per unit cell = 4 atoms

    Question 253
    CBSEENCH12010159
    Wired Faculty App

    (a) What type of semiconductor is obtained when silicon is doped with boron?

    (b) What type of magnetism is shown in the following alignment of magnetic moments?

    bold upwards arrow bold space bold space bold space bold upwards arrow bold space bold upwards arrow bold space bold upwards arrow bold space bold upwards arrow

    (c) What type of point defect is produced when AgCl is doped with CdCl2?
    Solution

    (a) When silicon is doped with boron, a p-type semiconductor is obtained.

     (b) The magnetism shown in the alignment of magnetic moments is ferromagnetism.

    Ferromagnetism is the basic mechanism by which certain materials (such as iron) form permanent magnets, or are attracted to magnets.

     (c) Impurity defect is produced when AgCl is doped with CdCl2.

    Question 254
    CBSEENCH12010183
    Wired Faculty App

    What is meant by ‘doping’ in a semiconductor?
    Solution

    Doping is the process of increasing the conductivities of the intrinsic semiconductors by adding suitable impurity.

    Question 256
    CBSEENCH12010204
    Wired Faculty App

    Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm-3. Use this information to calculate Avogadro's number (At. Mass of Fe = 55.845 u)
    Solution

    a = 286.65 pm

    a= 286.65 x 10-10cm

    Density ( ) = 7.874 g cm-3
    At mass of Fe = 55.845 u

    Z = 2 (For body centred cubic unit cell)

    Avogadro number (N0) =?
    straight rho space equals space fraction numerator straight Z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript 0 end fraction 7.875 space straight g space cm to the power of negative 3 end exponent space equals fraction numerator 2 space straight x space left parenthesis 55.845 straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent space cm right parenthesis space space straight x space left parenthesis 7.874 cm to the power of negative 3 end exponent right parenthesis space straight x space straight N subscript 0 end fraction straight N subscript 0 space equals space fraction numerator 2 space straight x space left parenthesis 55.845 space straight g space mol to the power of negative 1 end exponent right parenthesis over denominator left parenthesis 286.65 space straight x space 10 to the power of negative 10 end exponent cm right parenthesis cubed space straight x space left parenthesis 7.874 space cm to the power of negative 3 end exponent right parenthesis end fraction space equals space 6.022 space straight x space 10 to the power of 23 space mol to the power of negative 1 end exponent

    Question 257
    CBSEENCH12010221
    Wired Faculty App

    ‘Crystalline solids are anisotropic in nature’. What does this statement mean?
    Solution

    All crystalline solids are not anisotropic. Those crystalline solids which are anisotropic have their atoms arranged and spaced in a different manner in three different planes (X, Y and Z). Therefore, the physical properties of crystalline solids such as electrical resistance or refractive index show different values when measured along different directions in the same crystals.

    Example NaCl, Quartz, Ice, HCl, Iron, etc.

     

    Question 259
    CBSEENCH12010277
    Wired Faculty App

    Define the following terms:

    (i) Primitive unit cells

    (ii) Schottky defect

    (iii) Ferromagnetism
    Solution

    (i) Primitive unit cells are defined as the unit cells in which the constituent particles are present only at the corner positions.

    (ii) Schottky defect: Schottky defect is basically a vacancy defect shown ionic solids. In this defect, an equal no cation and anion are missing to maintain electrical neutrality. It decreases the density of a substance. An Ionic substance containing the similar size of cations and anion show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc.

    (iii) Ferromagnetism is defined as the phenomenon in which substances, such as iron, cobalt and nickel, are strongly attracted by a magnetic field. Such substances are called ferromagnetic substances.

    Question 260
    CBSEENCH12010297
    Wired Faculty App

    (a) Why does the presence of an excess of lithium make LiCl crystals pink?

    (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-center. What is the formula of the compound? 
    Solution

    (a) When crystals of LiCl is heated in presence of excess of lithium, Cl- ions from crystal diffuse on the surface and combine with ionized Li to form LiCl. The released unpaired electrons occupy the anionic sites known as F-centers. The pink colour results by excitation of these electrons when they absorb energy from visible light falling on them.

    (b) It is given that the atoms of Q are present at the corners of the cube. Therefore, number of atoms of Q in one unit cell =

    Also, it is also given that the atoms of P are present at the body-center.

    Therefore, number of atoms of P in one unit cell equals 1 over 8 straight x 8 = 1

    This means that the ratio of the number of P atoms to the number of Q atoms, P: Q = 1:1

    Hence, the formula of the compound is PQ.

    Question 261
    CBSEENCH12010304
    Wired Faculty App

    Aluminum crystallizes in an fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal? 
    Solution

    Aluminum crystallizes in FCC structure atomic radius (r) = 125 pm

    Length of the side of the unit cell in FCC structure,   straight a equals fraction numerator 4 over denominator square root of 2 end fraction straight r space equals space 2 square root of 2 straight r end root

     Therefore            a= 2 x 1.414 x 125

    a = 353.5 pm

    Question 262
    CBSEENCH12010324
    Wired Faculty App

    What are n-types semiconductors?
    Solution

    The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. These are generated when the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As.

    Question 264
    CBSEENCH12010344
    Wired Faculty App

    Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. The density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u)
    Solution

    a = 286.65 pm

    = 286.65 x 10-10 cm

    Density (Ï) = 7.874 g cm-3

    At mass of Fe = 56.0 u

    Z = 2 (For body centered cubic unit cell)

    Avogadro number (N0) =?
    Error converting from MathML to accessible text.

    Question 265
    CBSEENCH12010359
    Wired Faculty App

    What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction?
    Solution

    When the magnetic moments of domains are arranged in the same direction then, the substance shows ferromagnetism.

    Question 267
    CBSEENCH12010386
    Wired Faculty App

    Given an example each of molecular solid and an ionic solid.
    Solution

    (i)  Molecular Solid   - I2

    (ii)  Ionic Solid  - NaCl

    Question 269
    CBSEENCH12010420
    Wired Faculty App

    Calculate the packing efficiency of a metal crystal for a simple cubic lattice.
    Solution

    In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

    In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.

    Let the edge length of the cube be ‘a’ and the radius of each particle be r.

    So, we can write:

    a = 2r

    Now, volume of the cubic unit cell = a3 = (2r)3 = 8r3

    We know that the number of particles per unit cell is 1.

    Therefore, volume of the occupied unit cell =  4 over 3 pi r cubed
    Hence comma space packing space efficiency space equals space fraction numerator Volume space of space one space particle over denominator Volume space of space cubic space unit space cell end fraction space straight x space 100 percent sign space equals space fraction numerator begin display style 4 over 3 end style πr cubed over denominator 8 straight r cubed end fraction space straight x space 100 percent sign equals 1 over 6 straight pi space straight x space 100 percent sign equals 1 over 6 space straight x 22 over 7 space straight x 100 percent sign space equals 52.4 percent sign

    Question 270
    CBSEENCH12010421
    Wired Faculty App

    Explain how you can determine the atomic mass of an unknown metal if you know its mass density and the dimensions of a unit cell of its crystal.
    Solution

    By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.

    Let ‘a’ be the edge length of a unit cell of a crystal,  ‘d’’  be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z ’ be the number of atoms in the unit cell.
    Now comma space density space of space the space unit space cell space equals space fraction numerator Mass space of space the space unit space cell over denominator Volume space of space the space unit space cell end fraction rightwards double arrow space straight d space equals zm over straight a cubed space space.. left parenthesis straight i right parenthesis left square bracket Since space mass space of space the space unit space cell space equals space number space of space atoms space in space the space unit space cell space straight x space Atomic right square bracket left square bracket Volume space of space the space unit space cell equals space left parenthesis edge space length space of space the space unit space cell right parenthesis cubed right square bracket From space equation space left parenthesis straight i right parenthesis comma space we space have colon space rightwards double arrow straight d equals space da cubed over straight z space space left parenthesis ii right parenthesis thin space Now comma space mass space of space the space metal space left parenthesis straight m right parenthesis space equals space fraction numerator Atomic space mass over denominator Avogadro apostrophe straight s space number space left parenthesis straight N subscript straight a right parenthesis end fraction therefore comma space straight M space equals space fraction numerator da cubed straight N subscript straight a over denominator straight z end fraction space space left parenthesis iiii right parenthesis If space the space edge space length space are space different space left parenthesis say space straight a comma space straight b space and space straight c right parenthesis comma space space then space equation space left parenthesis ii right parenthesis space becomes colon rightwards double arrow space straight m equals space fraction numerator straight d left parenthesis abc right parenthesis straight N subscript straight A over denominator straight z end fraction space space space.. space left parenthesis iv right parenthesis space from space equation space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space can space determine space the space atomic space mass space of space the space unknown space metal.

    Question 271
    CBSEENCH12010450
    Wired Faculty App

    An element with density 2.8 g cm−3 forms of the f.c.c. unit cell with edge length 4 X10−8 cm. Calculate the molar mass of the element.
    (Given: NA = 6.022 X 1023 mol −1)
    Solution

    Edge length, a = 4 x 10-8 cm

    Density, d = 2.8 g cm-3

    As the lattice is fcc type, the number of atoms per unit cell, z is 4.

    Avogadro's number, NA = 6.022 x 1023 mol-1

    Molar mass can be calculated with the help of given relation:
    straight d space equals space fraction numerator straight Z space straight x space straight M over denominator straight N subscript straight A space straight x space straight a cubed end fraction straight m equals space fraction numerator straight N subscript straight A space space straight x space straight a cubed space straight x space straight d over denominator straight Z end fraction fraction numerator 2.8 space straight x space left parenthesis 4 space straight x space 10 to the power of negative 8 end exponent right parenthesis cubed space straight x space 6.022 space straight x space 10 to the power of 23 over denominator 4 end fraction space equals space 26.98 space straight g space mol to the power of negative 1 end exponent approximately equal to space 27 space straight g space mol to the power of negative 1 end exponent

    Question 272
    CBSEENCH12010451
    Wired Faculty App

    (i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl?
    (ii) What type of stoichiometric defect is shown by NaCl?

    OR
    How will you distinguish between the following pairs of terms:
    (i) Tetrahedral and octahedral voids
    (ii) Crystal lattice and unit cell.
    Solution

    i) The pink colour of LiCl is because of the metal excess defect caused by the anionic vacancies.
    ii) NaCl shows Schottky defect. In the Schottky defect, an equal number of cations and anion are missing from their regular sites.

    Or
    (i) Tetrahedral void surrounded by four spheres is called a tetrahedral void.

    Octahedral void surrounded by six spheres is called an octahedral void.

    (ii) Crystal lattice: A crystal lattice is a diagrammatic representation of the constituent particles such as atoms, ions and molecules of a crystal in a regular three-dimensional arrangement.
    Unit cell: A unit cell is the smallest three-dimensional portion of a crystal lattice. When it is repeated in different directions, it generates the entire crystal lattice.

    Question 273
    CBSEENCH12010528
    Wired Faculty App

    An element has atomic mass 93 g mol–1 and density 11.5 g cm–3. If the edge
    the length of its unit cell is 300 pm, identify the type of unit cell.
    Solution

    Given,
    M= 93 g mol-1
    d= 11.5 g cm-3
    a = 300 pm = 300 x 10-10 cm = 3 x 10-8
    We know that,
    straight d space equals space fraction numerator straight Z space straight x space straight M over denominator straight N subscript straight A space straight x space straight a cubed end fraction straight Z space equals space fraction numerator straight d space straight x space straight N subscript straight A space straight x space straight a cubed over denominator straight M end fraction space equals fraction numerator 11.5 space straight x 6.022 space straight x space 10 to the power of 23 space left parenthesis space 3 straight x 10 to the power of negative 8 end exponent right parenthesis cubed over denominator 93 end fraction space equals 2.01
    The number of atoms present in given unit cells is coming nearly equal to 2. Hence, the given unit cell is body centred cubic unit cell (BCC).

    Question 274
    CBSEENCH12010529
    Wired Faculty App

    Write any two differences between amorphous solids and crystalline solids.
    Solution
    Property Crystalline Solids Amorphous solids
    Arrangement of Constituent particles Regular Irregular
    Isotropy Regular Cleavage Irregular Cleavage
    Examples Diamond, Quartz, Copper Sulphate, Sodium Chloride etc Glass, Rubber, Wood, Amorphous Silica
    Question 275
    CBSEENCH12010530
    Wired Faculty App

    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–1)
    Solution

    Given
    Mass of Al =8.1
    Atomic mass of Al = 27 g mol-1
    Number of atoms = Number of moles x 6.022 x1023 space equals fraction numerator 8.1 over denominator 27 end fraction space straight x space 6.022 space straight x space 10 to the power of 23 space equals space 0.3 space straight x 6.022 space straight x space 10 to the power of 23 space equals space 1.8066 space straight x space 10 to the power of 23
    As one fcc unit has 4 atoms
    Hence number of unit cells having 1.8066 x1023 atoms
    fraction numerator 1.8066 space straight x space 10 to the power of 23 over denominator 4 end fraction space equals space 4.51 space straight x 10 to the power of 22 space unit space cell

    Question 276
    CBSEENCH12010531
    Wired Faculty App

    Give reasons :
    In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel
    defect.
    Solution

    NaCl, cation has large size thus it can not fit into voids so thus not show Frankel defect.

    Question 277
    CBSEENCH12010532
    Wired Faculty App

    Give reasons :
    Silicon on doping with Phosphorus forms the n-type semiconductor.
    Solution
    P has 5  outer electron which is one more than the silicon atoms. Four outer electrons combine with silicon atom, while the fifth electron is free to move and serves as the charge carrier and forms n type semiconductor. 
    Question 278
    CBSEENCH12010533
    Wired Faculty App

    Give reason:
    Ferrimagnetic substances show better magnetism than antiferromagnetic
    substances.
    Solution

    In the presence of magnetic field, the magnetic moments of domains of an antiferromagnetic substances are ordered in such a way half of the magnetic moments are aligned in one direction while the remaining half is in opposite direction. As a result of it, a net magnetic moment will be zero whereas in case of ferrimagnetic substances the magnetic moments of the domains of ferrimagnetic substances are aligned in parallel and antiparallel directions in unequal numbers. As a result, there will be the small value of magnetic moment for ferrimagnetic substances.

    Question 279
    CBSEENCH12010548
    Wired Faculty App

    Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.
    Solution

    The non-stoichiometry reflect the ease of oxidation of Fe2+ to Fe3+ effectively replacing a small portion of Fe2+ with two thirds their number of Fe3+. Thus for every three 'missing' Fe2+ ions, the crystal contains two Fe3+ ions to balance the charge.   i.e. 3Fe2+ = 2Fe3+ to maintain electrical neutrality.

    Question 280
    CBSEENCH12010558
    Wired Faculty App

    An element ‘X’ (At. mass = 40 g mol–1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol–1 )
    Solution

    We know,

    d = Z x MNA x a3For FCC Z is 4d = 4 x 406.022 x 1023 x (400 x 10-10)3 = 4.1514 gm cm-3Volume  of 4 gm X is 44.1514 cm3 = 0.96 cm3Volume of 1 unit cell is = (400 x 10-10)3 cm3 = 64 x 10-24 cm3Number  of unit cell  = 0.9664 x 10-24 = 1.5  x 1022

    Question 281
    CBSEENCH12010612
    Wired Faculty App

    Which of the following compounds is metallic and ferromagnetic?

    • VO2

    • MnO2

    • TiO2

    • CrO2
    Solution

    D.

    CrO2

    Only three elements iron (Fe) cobalt (Co) and Nickel (Ni) show ferromagnetism at room temperature. CrO2 is also a metallic and ferromagnetic compound which  is used to make magnetic tapes for cassette recorders

    Question 282
    CBSEENCH12010629
    Wired Faculty App

    Sodium metal crystallises in a body centred cubic lattice with a unit cell edge of 4.29 Å The radius of sodium atom is approximate:

    • 1.86 Å

    • 3.22Å

    • 5.72 Å

    • 0.93 Å
    Solution

    A.

    1.86 Å

    According to the figure
    (AC)2 = (AB)2 + (BC)2
    (AC)2 = a2 + a2 = 2a2
    (AD)2 = (AC)2+ (DC)2
    (4r)2 = 2a2 + a2

    16r2 = 3a2
    straight r space equals space fraction numerator square root of 3 over denominator 4 end fraction straight a
    Now, When Na metal crystallises in bcc unit cell with unit cell edge,
    a = 4.29Å 
    We have the formula for radius,
    i.e,  straight r space equals space fraction numerator square root of 3 over denominator 4 end fraction straight x 4.29 straight Å space =1.86Å 
    Question 283
    CBSEENCH12010653
    Wired Faculty App

    CsCl crystallises in body centred cubic lattice. If 'a' its edge length, then which of the following expression is correct?

    • straight r subscript Cs to the power of plus space plus straight r subscript Cl to the power of minus end subscript space equals space 3 straight a
    • straight r subscript Cs to the power of plus space plus straight r subscript Cl to the power of minus end subscript space equals space fraction numerator 3 straight a over denominator 2 end fraction
    • straight r subscript Cs to the power of plus space plus straight r subscript Cl to the power of minus end subscript space equals space fraction numerator square root of 3 over denominator 2 end fraction straight a
    • straight r subscript Cs to the power of plus space plus straight r subscript Cl to the power of minus end subscript space equals space square root of 3 straight a end root

    Solution

    C.

    straight r subscript Cs to the power of plus space plus straight r subscript Cl to the power of minus end subscript space equals space fraction numerator square root of 3 over denominator 2 end fraction straight a

    In CsCl, Cl- lie at corners of simple cube and Cs+ at the body centre, Hence, along the body diagonal, Cs+ and Cl- touch each other so
    FD space equals space straight b equals space square root of straight a squared plus straight a squared end root space equals space square root of 2 straight a In space increment AFD comma straight c squared space equals space straight a squared space plus straight b squared space equals straight a squared space plus space left parenthesis square root of 2 straight a right parenthesis squared space space equals straight a squared space plus 2 straight a squared straight c squared space equals space 3 straight a squared straight c equals square root of 3 straight a end root As space increment AFD space is space an space equilibrium space triangles comma therefore space square root of 3 straight a space equals space 4 straight r rightwards double arrow straight r space equals space fraction numerator square root of 3 straight a over denominator 4 end fraction Hence comma space straight r subscript cs to the power of plus end subscript space plus straight r subscript Cl space equals space 2 straight r space equals space 2 space straight x fraction numerator square root of 3 straight a over denominator 4 end fraction space equals space fraction numerator square root of 3 straight a over denominator 2 end fraction straight a space

    Question 284
    CBSEENCH12010693
    Wired Faculty App

    Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be

    • 75 pm

    • 300 pm

    • 240 pm

    • 152 pm
    Solution

    D.

    152 pm

    In body centred cubic structure,
    Edge length = a = 351 pm, radius= r=?
    straight r space equals space fraction numerator square root of 3 straight a end root over denominator 4 end fraction space equals fraction numerator square root of 3 space straight x space 351 over denominator 4 end fraction space equals space 152 space pm

    Question 286
    CBSEENCH12010743
    Wired Faculty App

    The edge length of a face centred cubic cell of an ionic substance is 508 pm. If the radius of the cation is 110 pm, the radius of the anion is 

    • 144 pm

    • 288 pm

    • 398 pm

    • 618 pm
    Solution

    A.

    144 pm

    For FCC lattice (assuming cation in octahedral void and anion in FCC)
    a = 508 pm
    (r+ + r-) = a/2 = 508/2 = 254 pm
    (r+ + r-) = 254 pm
    r- = 254-110 = 144 pm

    Question 287
    CBSEENCH12010745
    Wired Faculty App

    Percentages of free space in cubic close packed structure and in body centred packed structure are respectively

    • 48% and 26%

    • 30% and 26%

    • 26% and 32%

    • 32% and 48%
    Solution

    C.

    26% and 32%

    Packing fraction of ccp is 74%
    % free space in ccp = 26%
    Packing fraction of bcc is 68%
    %free space in bcc = 32%

    Question 288
    CBSEENCH12010755
    Wired Faculty App

    A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is'a', the closest approach between two atoms in metallic crystal will be

    • 2a

    • 2 square root of 2 straight a
    • square root of 2 straight a end root
    • fraction numerator straight a over denominator square root of 2 end fraction

    Solution

    D.

    fraction numerator straight a over denominator square root of 2 end fraction

    In FCC unit cell atoms are in constant along face diagonal
    So, √2a = 4R
    therefore, the closest distance
    left parenthesis 2 straight R right parenthesis space equals space fraction numerator square root of 2 straight a over denominator 2 end fraction space equals space fraction numerator straight a over denominator square root of 2 end fraction

    Question 290
    CBSEENCH12010808
    Wired Faculty App

    In a compound atoms of element Y from ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be 

    • X4Y3

    • X2Y3

    • X2Y

    • X3Y4
    Solution

    A.

    X4Y3

    No. of atoms of Y = 4
    No. of atoms of X = 2 x (8/3) 
    Formula of compound will be X4Y3

    Question 294
    CBSEENCH12010896
    Wired Faculty App

    Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point.

    • LiCl

    • NaCl

    • KCl

    • RbCl
    Solution

    B.

    NaCl

    Although lattice energy of LiCl higher than NaCl but LiCl is covalent in nature and NaCl ionic thereafter, the melting point decreases as we move NaCl because the lattice energy decreases as a size of alkali metal atom increases (lattice energy ∝ to melting point of alkali metal halide)

    Question 295
    CBSEENCH12010937
    Wired Faculty App

    What type of crystal defect is indicated in the diagram below?

    • Frenkel defect

    • Frenkel and Schottky defects

    • Interstitial defect

    • Schottky defect
    Solution

    D.

    Schottky defect

    When the equal number of cations and anions are missing. Then it is the case of Schottky defect.

    Question 296
    CBSEENCH12010994
    Wired Faculty App

    Which type of ‘defect’ has the presence of cations in the interstitial sites?

    • Metal deficiency defect

    • Schottky defect

    • Vacancy defect

    • Frenkel defect
    Solution

    D.

    Frenkel defect

    In Frenkel defect, smaller ion displaces from its actual lattice site into the interstitial sites.

    Question 297
    CBSEENCH12011045
    Wired Faculty App

    Lithium has a bcc structure. Its density is 530 kg m-3 and its atomic mass is 6.94 g mol-1. Calculate the edge length of a unit cell of lithium metal. (NA = 6.02 x 1023 mol-1).

    • 352pm

    • 527pm

    • 264pm

    • 154pm
    Solution

    A.

    352pm

    Given, Li has a bcc structure
    Density (ρ) =530 kg-m-3
    atomic mass (M) = 6.94 g mol-1
    Avogadro's number of atoms per unit cell in bcc (Z) = 2.
    ∴ we have the formula for density.
    straight rho space equals space fraction numerator straight z space straight x space straight M over denominator straight N subscript straight A space straight x space straight a cubed end fraction where space straight a space equals space edge space length space of space straight a space unit space cell straight a space equals space 3 square root of fraction numerator straight Z space xM over denominator straight rho space straight x space straight N subscript straight A end fraction end root straight a space equals space 3 square root of fraction numerator 2 space straight x 6.94 space straight g space mol to the power of negative 1 end exponent over denominator 0.53 space straight g space straight x space 6.02 space straight x space 10 to the power of 23 space mol to the power of negative 1 end exponent end fraction end root straight a space equals space space 3 square root of 4.35 space straight x space 10 to the power of negative 23 end exponent space cm to the power of negative 3 end exponent end root space equals space 3.52 space straight x space 10 to the power of negative 8 end exponent straight a space equals space 3.52 space pm

    Question 298
    CBSEENCH12011071
    Wired Faculty App

    A given metal crystallises out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom.

    • 40 pm

    • 127 pm

    • 80 pm

    • 108 pm
    Solution

    B.

    127 pm

    Given, edge length = 361 pm
    Four metal atoms in one unit cell
    i.e effective number in unit cell (z) = 4 (given)
    therefore,
    It is a FCC structure
    Face diagonal  = 4r

    square root of 2 straight a end root space equals space 4 straight r space equals space straight r equals space fraction numerator square root of 2 space end root space straight x space 361 over denominator 4 end fraction space equals space 127 space pm

    Question 299
    CBSEENCH12011091
    Wired Faculty App

    If a is the length of the side of a cube, the distance between the body centred atom and one corner atom in  the cube will be

    • fraction numerator 2 over denominator square root of 3 end fraction a
    • fraction numerator 4 over denominator square root of 3 end fraction a
    • fraction numerator square root of 3 over denominator 4 end fraction a
    • fraction numerator square root of 3 over denominator 2 end fraction a

    Solution

    D.

    fraction numerator square root of 3 over denominator 2 end fraction a

    In bcc, 2 atoms are present. One atom lies the centre of the cube while other lies at the corner of the cube. Hence, the distance between the body centred and one corner atom is half of the body diagonal, i.e., fraction numerator square root of 3 over denominator 2 end fraction a.

    Question 300
    CBSEENCH12011118
    Wired Faculty App

    The number of carbon atoms per unit cell of diamond unit cell is

    • 4

    • 8

    • 6

    • 1
    Solution

    B.

    8

    The diamond lattice contains an fcc Bravias point lattice which has two identical atoms per lattice point.
    The diamond lattice contains 4 lattice points per unit cell but contains 8 atoms per unit cell.

    Question 301
    CBSEENCH12011120
    Wired Faculty App

    A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm-3. The molar mass of the metal is (NA Avogadro's constant= 6.02 x 1023 mol-1)

    • 40 g mol-1

    • 30 g mol-1

    • 27 g mol-1

    • 20 g mol-1
    Solution

    C.

    27 g mol-1

    Given, cell is fcc, So Z =4
    Edge length, a = 404 pm = 4.04 x 10-8 cm
    Density of metal, d = 2.72 g cm-3
    NA = 6.02 x 1023 mol-1
    Molar mass ofg the metal, M =?
    We know that


    density, d=  fraction numerator straight z space straight x space straight M over denominator straight a cubed space straight x space straight N subscript straight A end fraction therefore space straight M equals space fraction numerator straight d space straight x space straight a cubed space straight x space straight N subscript straight A over denominator straight Z end fraction space equals fraction numerator 2.72 space xc space left parenthesis 4.04 space straight x space 10 to the power of negative 8 end exponent right parenthesis cubed space straight x space 6.02 space straight x space 10 to the power of 23 over denominator 4 end fraction space equals 27 space straight g space mol to the power of negative 1 end exponent
    Question 302
    CBSEENCH12011133
    Wired Faculty App

    Which of the following statements about the interstitial compounds is incorrect?

    • They retain metallic conductivity

    • They are chemically reactive

    • They are much harder than the pure metal

    • They have higher melting points than the pure metal
    Solution

    B.

    They are chemically reactive

    Interstitial compounds are obtained when small atoms like H, B, C, resemebleN etc.fit into the lattice etc. fit into the lattice of other elements. These retain metallic conductivity resemble their parent metal in chemical properties (reactivity)but differ in physical properties like hardness, melting point etc.

    Question 304
    CBSEENCH12011154
    Wired Faculty App

    The number of octahedral voids (s) per atom present in a cubic close-packed structure is 

    • 1

    • 3

    • 2

    • 4
    Solution

    A.

    1

    A number of octahedral voids = a number of atoms in the close-packed structure. Since number of atoms= 1
    So, number of octahedral voids = 1

    Question 305
    CBSEENCH12011171
    Wired Faculty App

    A structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

    • ABO2

    • A2BO2

    • A2B3O4

    • AB2O2
    Solution

    D.

    AB2O2

    According to ccp,
    Number of O2- ions = 4
    So, tetrahedral void = 8
    and octahedral void = 4
    Since A ions occupied 1/4th of the tetrahedral void.
    Therefore,
    Number of A ions = 1/4 x 8 = 2
    Again, B ions occupied all octahedral void.
    Therefore, Number of B ions = 4 
    A: B:O = 2:4:4
     = 1:2:2
    Structure of oxide= AB2O2

    Question 306
    CBSEENCH12011217
    Wired Faculty App

    The correct statement regarding defects in the crystalline solid is 

    • Schottky defects have no effect on the density of crystalline solid

    • Frenkel defect decreases the density of crystalline solids

    • Frenkel defect is a dislocation defect

    • Frenkel defect is found in halides of alkaline metals
    Solution

    C.

    Frenkel defect is a dislocation defect

    In Frenkel defect, ions in solid dislocate from their position. Hence, Frenkel defect is a dislocation defect. 

    Question 307
    CBSEENCH12011227
    Wired Faculty App

    The vacant space in bcc lattice cell is 

    • 26%

    • 48%

    • 23%

    • 32%
    Solution

    D.

    32%

    Packing efficiency in bcc lattice = 68%
    Vacant space in bcc lattice = 100-68 = 32%

    Question 309
    CBSEENCH12011267
    Wired Faculty App

    AB crystallises in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositely charged ion in the lattice is

    • 335 pm

    • 250 pm

    • 200 pm

    • 300 pm
    Solution

    A.

    335 pm

    For body centred cubic (bcc) lattice, distance between two oppositely charged ions,
    straight d space equals space fraction numerator square root of 3 straight a end root over denominator 2 end fraction space equals space fraction numerator square root of 3 space end root space straight x space 387 over denominator 2 end fraction space pm equals space 355.15 space pm space

    Question 310
    CBSEENCH12011304
    Wired Faculty App

    Copper crystallises in a face-centered cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm?

    • 128

    • 157

    • 181

    • 108
    Solution

    A.

    128

    In case of the face-centred cubic lattice,
    radius = 2a4
    therefore, Radius of the copper atom (fcc lattice)
    =2 x 3614 = 128 pm

    Question 311
    CBSEENCH12011310
    Wired Faculty App

    Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell lithium is 351 pm, the atomic radius of the lithium will be 

    • 240.8 pm

    • 151.8 pm

    • 75.5

    • 300.5 pm
    Solution

    B.

    151.8 pm

    In case of body centred cubic (bcc) crystal,
    straight a square root of 3 space equals space 4 straight r
    Hence, atomic radius of lithium,
    straight r equals fraction numerator space straight a-th root of 3 over denominator 4 end fraction equals space fraction numerator 351 space straight x space 1.732 over denominator 4 end fraction equals space 151.1 space 98 space pm

    Question 312
    CBSEENCH12011329
    Wired Faculty App

    Volume occipied by one molecule of water (density = 1 g cm-3) is

    • 9.0 x 10-23

    • 6.023 x 10-23 cm3

    • 3.0 x 10-23 cm3

    • 5.5 x 10-23 cm3
    Solution

    C.

    3.0 x 10-23 cm3

    6.02 x 1023 molecules of water = 1 mol
                                                      = 18 g
    Therefore, Mass of one molecule of water
      fraction numerator 18 over denominator 6.023 space straight x space 10 to the power of 23 space straight g end fraction straight d equals space straight m over straight V therefore comma space straight V space equals space straight m over straight d space equals space fraction numerator 18 over denominator 6.023 space straight x space 10 to the power of 23 space straight x 1 end fraction space equals space 3 space straight x space 10 to the power of negative 23 space end exponent space cm cubed

    Question 313
    CBSEENCH12011333
    Wired Faculty App

    Which of the following statements is not correct? 

    • The fraction of the total volume occupied by the atoms in a primitive cell is 0.48

    • molecules solids are generally volatile

    • the number of carbon atoms in a unit cell

    • The number of Bravais latices in which a crystal can be categorised is 14
    Solution

    A.

    The fraction of the total volume occupied by the atoms in a primitive cell is 0.48

    D.

    The number of Bravais latices in which a crystal can be categorised is 14

    Packing fraction of lattice structure or fraction of total volume occupied

    equals space fraction numerator Volume space occupied space by space atoms space in space straight a space unit space cell over denominator volume space of space the space unit space cell end fraction Packing space fraction space of space simple space cube space equals space fraction numerator straight z space straight x begin display style 4 over 3 end style space πr cubed over denominator straight a cubed end fraction space equals fraction numerator 1 space straight x space begin display style 4 over 3 end style space πr squared over denominator left parenthesis 2 straight r right parenthesis cubed end fraction therefore comma space For space primitive space cell space straight a space space equals space 2 straight r volume space occupied space space equals space 52 percent sign

    Question 314
    CBSEENCH12011334
    Wired Faculty App

    If 'a' stands for the edge length of the cubic system:simple cubic , body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be repsectively,

    • 1 half a semicolon space fraction numerator square root of 3 over denominator 4 end fraction a colon space fraction numerator 1 over denominator square root of 2 end fraction a
    • 1 half a space colon space square root of 3 space a space colon thin space fraction numerator 1 over denominator square root of 2 end fraction straight a
    • 1 half a space colon thin space fraction numerator square root of 3 over denominator 2 end fraction straight a space colon space fraction numerator square root of 2 over denominator 2 end fraction straight a
    • 1 straight a space colon space square root of 3 straight a end root colon space square root of 2 straight a

    Solution

    A.

    1 half a semicolon space fraction numerator square root of 3 over denominator 4 end fraction a colon space fraction numerator 1 over denominator square root of 2 end fraction a

    For simple cubic, 
    a = 2r
    r = a/2
    For body centred cubic,
    straight a space equals space fraction numerator 4 straight r over denominator square root of 3 end fraction straight r space equals space fraction numerator square root of 3 straight a end root over denominator 4 end fraction For space face space centred space cubic comma space straight a space equals space 2 square root of 2 straight r straight r space equals space fraction numerator straight a over denominator 2 square root of 2 end fraction
    Hence, the ratio of radii in simple cubic, body centred cubic and face centred cubic is 
    1 half a semicolon space fraction numerator square root of 3 over denominator 4 end fraction a colon space fraction numerator 1 over denominator square root of 2 end fraction a

    Question 315
    CBSEENCH12011337
    Wired Faculty App

    Percentage of free space in a body centred cubic unit cell is 

    • 30%

    • 32%

    • 34%

    • 28%
    Solution

    B.

    32%

    Packing fraction

    space equals space fraction numerator Volume space occupied space by space atoms space in space straight a space unit space cell over denominator Volume space of space the space unit space cell end fraction For space body space centred space cube comma space Packing space fraction space equals space fraction numerator 2 space straight x begin display style 4 over 3 end style πr cubed over denominator open parentheses begin display style fraction numerator 4 straight r over denominator square root of 3 end fraction end style close parentheses squared end fraction space equals space fraction numerator square root of 3 space straight pi over denominator 8 end fraction space equals space 0.68 where comma space edge space length space straight a space equals space fraction numerator 4 straight r over denominator square root of 3 end fraction therefore comma space volume space space ccupiedf space equals space 68 and space volume space vacant space space equals space 32 percent sign

    Question 316
    CBSEENCH12011338
    Wired Faculty App

    With which one of the following elements silicon should be doped so as to give p - type of semiconductor?

    • Germanium

    • Arsenic

    • Selenium 

    • Boron 
    Solution

    D.

    Boron 

    The n- type semiconductors are obtained when Si or Ge are doped with elements of group 15, eg, Arenic (As), while p-type semiconductors are obtained when Si or GGe are doped with traces of elements of group 13, ie indium (In), Boron (B).

    Question 317
    CBSEENCH12011368
    Wired Faculty App

    The fraction of totoal volume occupied by the atoms present in a simple cube is: 

    • π/6

    • fraction numerator straight pi over denominator 3 square root of 2 end fraction
    • fraction numerator straight pi over denominator 4 square root of 2 end fraction

    • π /4
    Solution

    A.

    π/6

    For simple cubic
    space Radius space left parenthesis straight r right parenthesis space equals space straight a over 2 Volume space of space the space atom space space equals space 4 over 3 straight pi open parentheses straight a over 2 close parentheses cubed Packing space fraction space equals space open parentheses fraction numerator 4 over 3 straight pi open parentheses straight a over 2 close parentheses cubed over denominator straight a cubed end fraction close parentheses space equals space straight pi over 6

    Question 318
    CBSEENCH12011383
    Wired Faculty App

    If NaCl is doped with 10-4 mol% of SrCl2 the concentration of cation vacancies will be 
    (NA = 6.02 x 1023 mol-1):

    • 6.02 x 1015 mol-1

    • 6.02 x 1016 mol-1

    • 6.02 x 1017

    • 6.02 x 1014
    Solution

    A.

    6.02 x 1015 mol-1

    If NaCl is doped with 10-4 mol% of SrCl2, 
    2 Na+ ions doped by Sr2+ NA = 6.02 x 1023
    The concentration of cation vacancies
     =  6.02 x 1023 x10-8
    = 6.02 x 1015 mol-1

    Question 320
    CBSEENCH12011396
    Wired Faculty App

    The appearance of colour is solid alkali metal halides is generally due to:

    • F- centres

    • Schottky defect

    • Frenkel defect

    • interstitial positions
    Solution

    A.

    F- centres

    The appearance of colour in solid alkali metal halides is generally due to F-centres.

    Question 321
    CBSEENCH12011412
    Wired Faculty App

    Which is the incorrect statement?

    • FeO0.98 has non stoichiometric metal deficiency defect

    • Density decreases in case of crystals with Schottky's defect

    • NaCl(s) is insulator, silicon is semiconductor,silver is conductor, quartz is piezo electric crystal

    • Frenkel defect is favoured in those ionic compounds in which sizes of cation and anionsare almost equal
    Solution

    A.

    FeO0.98 has non stoichiometric metal deficiency defect

    D.

    Frenkel defect is favoured in those ionic compounds in which sizes of cation and anionsare almost equal

    Frenkel defect occurs in those ionic compounds in which size of cation and anion is largely different.
    The non-stoichiometric ferrous oxide is Fe0.93–0.96O1.00 and it is due to metal deficiency defect.

    Question 322
    CBSEENCH12011471
    Wired Faculty App

    Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

    • 32

    • 4332

    • 1/2

    • 3342
    Solution

    D.

    3342

    For bcc lattice: Z = 2, a = 4r3

    For fcc lattice: Z = 4, a = 22r

    ∴d250 Cd900oC  = ZMNAa3bccZMNAa3fcc = 2422r4r33 = 3342

    Question 323
    CBSEENCH12011503
    Wired Faculty App

    A solid AB has a NaCl structure. If the radius of cation A+ is 170 pm, then the maximum possible radius of the anion B- is

    • 397.4 pm

    • 347.9 pm

    • 210.9 pm

    • 410.6 pm
    Solution

    D.

    410.6 pm

    For NaCl type structure,rA+rB- = 0.414 -0.737 

    For maximum possible radium of B-

    rA+rB- = 0.414, 170rB- = 0.414rB-  = 1700.414 = 410.6 pm

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