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Solid State Test - 19

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Solid State Test - 19
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  • Question 1
    1 / -0

    CsCI crystallises in a cube that has CP at each corner and Cs+ at the centre of the unit cell. If  rCs = 169 pm and rcl- =181 pm, then edge length of the cube is 

    Solution

    (d) CsCI has bcc structure.
    Closest dista nce between Cs+ and Cl- is the sum of radii of Cs+ and CI- = 169 + 181 = 350 pm

    Cubic diagonal = √3 a

    Closest distance is one-half of the cubic diagonal

  • Question 2
    1 / -0

    A mineral having the formula AB2 crystallises in the cubic closest - packed lattice with the A-atoms occupying the lattice points. Fraction of the tetrahedral sites occupied by 6-atoms is

    Solution

    (d) ABcrystallises as cubic close packed (ccp).

    Thus, A2+ is surrounded by eight B- and B- is surrounded by four A2+ .

    Thus, coordinatio n number of A2+ = B

    coordination number of B- = 4

    Number of atoms in tetrahedral in small cube of unit cell of atom B = 4

    Thus, fraction occupied = 4/4 = 1 (100%)

  • Question 3
    1 / -0

    Silver metal crystallises in a cubic closest packed arrangement with edge length 407 pm. Thus, radius of the silver atom is

    Solution

    (c)

    AB = a, AC = a

    BC2 = AB2 + AC2 =2a2

    Radius =r

    then BC = CD + DE + EB = r + 2r + r = 4r 

    ∴ 16r2 = 2a2

  • Question 4
    1 / -0

    What is the formula of a magnetic oxide of cobalt used in recording tapes that crystallises with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closest packed array of oxide ions?

    Solution

    (c)

    Tetrahedral holes = 8, Octahedral holes = 4 Cobalt atoms = 8/8 =1

    Oxide ions = 4/2 =2

    Thus, formula is CoO2

  • Question 5
    1 / -0

    The total number of tetrahedral voids in the face-centred unit cell is

    Solution

    (b) Unit cell of fee is divided into eight small cubes. Each cube has four atoms at alternate positions. When joined to each other, they make a regular tetrahedron.
    Thus, number of tetrahedral void in small cube = 1 and in eight cubes, voids = 8.

    Note  If number of atoms in a cube = N, then, number of octahedral voids = N and number of tetrahedral voids = 2N

  • Question 6
    1 / -0

    Packing efficiency (%) of different types of unit cells is given. Select the correct packing efficiency.

    Solution

    (a) Edge length = a, Volume of one face = a3 Volume of six faces of a cube = 6a3 = V Radius = r

    fee can be hexagonal closest packed (hep)

    or cubic closest packed (ccp)

    z = number of atoms in unit cell.

  • Question 7
    1 / -0

    Coordination numbers in a square close packed structure and hexagonal close packed structure respectively, are

    Solution

    (c) Close packing in two-dimensions can be Type I A A A ...... ty p e The second row may be ^ placed in contact with the first one such that the spheres of A the second row are exactly . above those of the first row.
    The spheres of the two rows A are aligned horizontally as well as vertically. In this arrangement, each sphere is in contact with four of its neighbours.
    Thus, its coordination number = 4 and is called square close packing in two-dimensions.
    Atom A (•) is in contact with four others.

    Type II ABAB... type The second row may be placed above the first in a staggered manner such that its spheres fit in the depressions of the first row.
    First row arrangement: A type Second row arrangement: 8 type Third row arrangement: A type and so on There is less free space and packing is more efficient than the square close packing. Each sphere is in contact with six of its neighbours and coordination number = 6 in two-dimensions. It is called two dimensional hexagonal close-packing. Voids formed in this case are triangular in shape. If the apex of the triangles downward in the first row, it is upward in the second row.

     

  • Question 8
    1 / -0

    The length of the unit cell edge of a bcc lattice metal is 352 pm. Thus, volume of atoms in one mole of the metal is

    Solution

     (a)

    In a bcc structure, the atoms tou ch each other along the body diagonal.

    Radius of the atom = r

    AB2 = 8C2 + AC2 
    BC2 = BD2 + CD2 = a2 + a2 = 2a2 

    Volume of one atom = 

    Volume of N0 atoms in one mole of the metal

    = 6.02 x 1023 x 1.48 x 10-23 cm3 = 8.93 cm3

  • Question 9
    1 / -0

    A compound Mp Xq has cubic close packing arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is

     

    Solution

    (b)

    Structure is cubic close packing.
    Number of atoms of M = (from edge centre) +1 (from body centre) =2

    Number of atoms of X =   (from corners of the cube) +  (from face centre)  =4

    Thus, M : X = 2 : 4 = 1 : 2 Hence, the empirical formula of the compound is MX2.

  • Question 10
    1 / -0

    Radii of A+ and that of X- and Y- have been given as 

    A+= 1.00 pm

    X- = 1.00 pm

    Y- = 2.00 pm

    Thus, ratio of volumes of A X and A Y unit cells is 

    Solution

    (a) Ratio of r+ /r_ will decid e the type of unit cell 

    For AX    hence, AX has bcc structure.

    Thus, octahedral or square planar structure.

  • Question 11
    1 / -0

    In the structure of NaCI given below, ratio rNa+/rcl is 

    Solution

     

  • Question 12
    1 / -0

    Matching List Type
    Choices for the correct combination of elements from Column I and Column II are given as options (a), (b), (c) and (d), out of which one is correct.
     

        

    Solution

    (i)

    Body-centred cubic Every cation (Cs+) is coordinated to eight anions (Br- ) and every anion (Br- ) is coordinated to eight cations (Cs+).
    Thus, (i) -> (s)

    (ii) Face-centred cubic Every cation (Na+) is coordinated to six anions (Cl- ) and every anion (Cl- ) is coordinated to six cations (Na+).
    Thus, (ii) -> (q)

    (iii) Face-centred cubic (tetrahedral void) Every cation (Zn2+) is coordinated to four anions (S2-) and vice-versa.
    Thus, (iii) —> (p)

    (iv) Face-centred cubic (antifluorite) EveryCa2+ is coordinated to eight fluoride and every F- is coordinated to fourCa2+ (to balance charge).

    4Ca2+ + 8F-
    Thus, (iv) —> (r)

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