The Solid State Chemistry MCQ With Answers

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1. If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.023 x 1023 mol-1

  • 1. 6.023 x 1015 mol-1             
  • 2. 6.023 x 1016 mol-1
  • 3. 6.023 x 1017 mol-1             
  • 4. 6.023 x 1014 mol-1      

2. The phenomenon in which polar crystals on heating produce electricity is called:-
1. pyro-electricity
2. piezo-electricity
3. ferro-electricity
4. ferri-electricity 

3. Metals have conductivity of the order of (ohm-1 cm-1):-

  • 1. 1012
  • 2. 108
  • 3. 102
  • 4. 10-6

4. Total volume of atoms present in a face centred cubic unit cell of a metal is ( r is atomic radius):

  • 1. 20/3 πr3
  • 2.  24/3 πr3
  • 3.  12/3 πr3
  • 4.  16/3 πr3

5. Lithium borohydride (LiBH4), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are:- a = 6.81 Å , b = 4.43 Å ,c = 7.17 Å. If the molar mass of LiBH4 is 21.76 g mol-1.The density of the crystal is:-
1. 0.668 gcm-3
2. 0.585 gcm2
3. 1.23 gcm-3
4. None of these

6. A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73Å . The edge length of the cell is:

  • 1. 200pm
  • 2. √3/√2 pm
  • 3. 142.2 pm
  • 4. √2pm

7. Which is covalent solid?

  • Fe2O3
  • Diamod
  • Graphite
  • All of these

8. The anions (A) form hexagonal closest packing and atoms (C) occupy only 2/3 of octahedral voids in it. then the general formula of the compound is-

  • 1. CA
  • 2. A2
  • 3. C2A3
  • 4. C3A2

9. The mass of a unit cell of CsCl corresponds to:-

  • 1. 8Cs+ and Cl
  • 2. 1Cs+ and 6Cl
  • 3. 1Cs+ and 1Cl
  • 4. 4Cs+ and Cl

10. Graphite in an example of-

  • 1. Ionic solid
  • 2. Covalent solid
  • 3. Vander waal’s crystal
  • 4. Metallic crystal

11. Which arrangement of electrons leads to anti-ferromagnetism?

  1. ↑↑↑↑
  2. ↑↓↑↓
  3. Both (1) and (2)
  4. None of these

12. The intermetallic compound LiAg has a cubic crystalline structure in which each Li atom has 8 nearest neighbour silver atoms and vice-versa. What is the type of unit cell?

  1. Body centred cubic
  2. Face centred cubic
  3. Simple cubic for either Li atoms alone or Ag atoms alone
  4. None of the above

13. The arrangement ABC ABC.. is referred to as.

  1. octahedral close packing
  2. hexagonal clase packing
  3. tetrahedral close packing
  4. cubic close packing

14. How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00g? [ Atomic masses Na=23, Cl = 35.5]

  • 1. 2.57 x 1021
  • 2. 5.14 x 1021
  • 3. 1.28 x 1021
  • 4. 1.71 x 1021

15. The ratio of cations to anion in a closed pack tetrahedral is:

  1. 0.714
  2. 0.225
  3. 0.02
  4. none of these

Radius Ratio Rule is not much of a big topic in the chapter “The Solid State”. However, it plays a very important role in the determination of a stable structure in an ionic crystal. It also helps in the determination of the arrangement of the ions in the crystal structure.

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

  1. X3Y4
  2. X4Y3
  3. X2Y3
  4. X2Y

17. A spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in ccp layers. The normal spinel has 1/8th of the tetrahedral void occupied  by one type of metal and one half of the octahedral voids occupied by another type of metal ions. Such a spinel is formed by Zn2+, Al3+ and O2- with Zn2+ in tetrahedral void. Give the simplest formula of the spinel.

  • 1. ZnAl2O4
  • 2. ZnAl2O3
  • 3. ZnAlO
  • 4. None of these

  1. (3) Doping of NaCl with 10-4 mol % of SrCl2 means, 100 moles of NaCl are doped with 10-4  mol of SrCl2.

    ... 1 mol of NaCl is doped with

                          SrCl2 = 10-4/100 = 10-6 mole

     As each Sr2+ ion introduces one cation vacancy.

    ... Concentration of cation vacancies

                                 = 10-6 mol/mol of NaCl

                                 = 10-6 x 6.023 x 1023 mol-1

                                 = 6.023 x 1017 mol-1

  2. It is the defination of pyro-electricity.

    Piezoelectricity is the electric charge that accumulates in certain solid materials (such as crystals, certain ceramics, and biological matter such as bone, DNA and various proteins) in response to applied mechanical stress.

    Ferroelectricity is a characteristic of certain materials that have a spontaneous electric polarization that can be reversed by the application of an external electric field. All ferroelectrics are pyroelectric, with the additional property that their natural electrical polarization is reversible.

  3. 2. The conductance order of metals is 106 to 108 ohm-1 cm-1.
  4. 4. Volume of atoms in cell = 4/3 πr3 x n (n =4 for fcc)

    =  4/3 x πr3 x 4 = 16/3 πr3

  5. ρ=ZM/NV=4×(21.76 g mol1)/(6.023×1023 mol1)(6.81×4.43×7.17×1024 cm3)=0.668 g cm3
  6. 1. ratom = /4; Also closest approach in bcc lattice is 1/2 of body diagonal, i.e, 2r=1.73Å

    or a = 1.73×2/ =1.996 Å=199.6pm

  7. 4
  8. 3. Hexagonal close packing contains 6 atoms per unit cell and hence the number of octahedral voids per unit cell is 6. 
    Hence, number of A atoms per unit cell=6 and number of C atoms per unit cell=6 x 2/3=4
    The formula of ionic compound is given as simplest formula and hence formula is C2A3
  9. CsCl=B.C.C Structureat corners, Cl=8×1/8=1       at   centre, Cs+=1×1=1
  10. 2
  11. (2) Anti-ferromagnetic possess complementary dipoles alignment giving net dipole moment equal to zero.
  12. (a) The bcc structure has co-ordination no. of eight. other scc= 6, fcc=12, ccp=12, hcp=12 
  13. ccp
  14. 1
  15. Coordination numberGeometryρ = rcation/ranion
    2linear0 – 0.155
    3triangular0.155 – 0.225
    4tetrahedral0.225 – 0.414
    4square planar0.414 – 0.732
    6octahedral0.414 – 0.732
    8cubic0.732 – 1.0
    12cuboctahedral1.0
  16. 2. In a ccp array number of tetrahedral voids is twice the number of atoms.

    x:y::1:(2/30 x 2=3:4

  17. 1. Let  the no. of tetrahedral voids = a

    No. of octahedral voids = a/2

    1/8 th tetrahedral voids is occupied by Zn2+

    No. of Zn2+ =a/8

    1/2 of octahedral voids occupied by Al3+

    No. of Al3+ = a/4

    Ratio of Zn2+: Al3+

                 1     :   2

    For 1 mole Zn2+, Al3+=2

    Total +ve charge = 2+3×2=8  therefore, No. of O2- =4

    The simplest formula will be  ZnAl2O

  18.  

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