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Solid State Test - 55

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Solid State Test - 55
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  • Question 1
    1 / -0
    The volume of the hexagonal prism is ( radius of every sphere is $$r$$)

    Solution
    Height of a unit cell: $$\mathrm{\sqrt{\cfrac{2}{3}}\times 4r}$$

    Base area of Hexagonal Prism : $$\mathrm{6 \times \sqrt{\cfrac{3}{4}} \times 2r^2}$$

    Volume of Hexagonal Prism =$$ Base \ area \times Height \ of \ the \ Unit \ cell$$ = $$\mathrm{6 \times \sqrt{\cfrac{3}{4}} \times 2r^2 \times \sqrt{\cfrac{2}{3}}\times 4r}$$ = $$24 \sqrt2 \times r^3$$.

    Hence, Option "A" is the correct answer.
  • Question 2
    1 / -0
    The cell content $$(Z)$$ of the unit cell of a h.c.p. space lattice is 

    Solution
    The cell content $$(Z)$$ of the unit cell of a h.c.p. space lattice is $$2$$.
    Hence, Option "A" is the correct answer.
  • Question 3
    1 / -0

    Directions For Questions

    When an atom or an ion is missing from its normal lattice site, a lattice vacancy (Schottky defect) is created. In stoichiometric crystals, a vacancy of one ion has to be accompanied by the vacancy of the oppositely charged ion in order to maintain electrical neutrality.
    In a Frenkel defect an ion leaves its position in the lattice and occupies an interstitial void. This is the Frenkel defect commonly found along with the Schottky defects and interstitials. In pure alkali halides, Frenkel defects are not found since the ions cannot get into the interstitial sites. Frenkel defects are found in silver halides because of the small size of the $$Ag^+$$ ion. Unlike Schottky defects, Frenkel defects do not change the density of the solids. In certain ionic solids (e.g. $$AgBr$$) both Schottky and Frenkel defects occur.
    The defects discussed above do not disturb the stoichiometry of the crystalline material. There is large variety of non-stoichiomertic inorganic solids which contain an excess or deficiency of one of the elements. Such solids showing deviations from the ideal stoichiometric composition form an important group of solids. For example in the vanadium oxide, $$VO_x, \ x $$ can be anywhere between $$0.6$$ and $$1.3$$. There are solids which are difficult to prepare in the stoichiometric composition. Thus, the ideal composition in compounds such as $$FeO$$ is difficult to obtain (normally we get a composition of $$Fe_{0.95} \ O$$ but it may range from $$Fe_{0.93} \ O$$ to $$Fe_{0.96} \ O$$). Non-stoichiometric behaviour is most commonly found fro transition metal compounds though is also known for also lanthanoids and actinoids.
    Zinc oxide loses oxygen reversibly at high temperatures and turns yellow in colour. The excess metal is accommodated interstitially, giving rise to electrons trapped in the neighbourhood the enhanced electrical conductivity of the non-stoichiometric $$ZnO$$ arises from these electrons.
    Anion vacancies in alkali halides are produced by heating the alkali halide crystals in an atmosphere of the alkali metal vapour. When the metal atoms deposit on the surface they diffuse into the crystal and after ionisation the alkali metal ion occupies cationic vacancy whereas electron occupies anionic vacancy. Electrons trapped in anion vacancies are referred to as F-centres (from Farbe the German word for colour) that gives rise to interesting colour in alkali halides. Thus, the excess of potassium in $$KCl$$ makes the crystal appear violet and the excess of lithium in $$LiCl$$ makes it pink.

    ...view full instructions

    In this crystal of $$Fe_{0.93}O$$, the percentage of $$Fe(III)$$ will be:
    Solution
    $$3x + (0.93 - x) \times  2 =2$$

    $$x= 0.14$$

    $$\%$$ of $$Fe$$ as $$Fe (III) = \dfrac{0.14}{0.93} \times 100 =15\% $$
  • Question 4
    1 / -0
    In hexagonal close packing of sphere in three dimensions:
    Solution


    $$HCP = AB\ AB\ AB$$......................pattern repeat
     
    For calculating voids between two layers $$A$$ and $$B$$.

    Total tetrahedral voids $$= 12$$ (represented by dots)out of which $$8$$ are completely inside but rest are shared by other unit cells.

    Total octahedral voids $$=6$$ (represented by cross). All are completely inside.

  • Question 5
    1 / -0
    How many atoms ( spheres ) are contained within the hexagonal prism shown below in h.c.p. space lattice?

    Solution
    No. of atoms in a h.c.p cell = $$12 \times \cfrac{1}{6} + 2\times \cfrac{1}{2} + 3 \times 1$$

    $$\implies$$No. of atoms in a h.c.p cell = $$6$$

    Hence, Option "B" is the correct answer.
  • Question 6
    1 / -0
    In melting lattice, structure of solid
    Solution
    in melting lattice , struture of solid  $$changes$$ Because of the repeating geometric structure of the crystal, all the bonds between the particles have equal strength. A crystalline solid will have a distinct melting point, because applying heat will break all the bonds at the same time.

  • Question 7
    1 / -0
    Bravais lattices are of:
    Solution
    $$14$$ kinds of Bravais lattices (space lattices) are possible in a crystal.
     Hence  C is correct option
  • Question 8
    1 / -0
    Crystalline solids are:
    Solution
    Sugar is a crystalline solid while glass, rubber and plastic are amorphous solids.
  • Question 9
    1 / -0
    The correct statement in the following is
    Solution
    A crystal system is hexagonal if its unit cell having $$a = b \neq c$$ axial ratio and $$\alpha = \beta = 90^{\circ}, \gamma = 120^{\circ}$$ axial angles.
  • Question 10
    1 / -0
    Under which category, iodine crystals are placed among the following?
    Solution
    Iodine crystals are molecular crystals, in which constituent particles are molecules having interparticle forces are Vander Waals forces.
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