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Solid State Test - 7

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Solid State Test - 7
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  • Question 1
    4 / -1

    The edge length of a fcc  is 508 pm. If the radius of cation is 110 pm, the radius of anion is

    Solution

    For fcc,
    2(r+ + r-) = a
    2(110 + r-) = 508
    r- = 508/2 - 110 = 144 pm

  • Question 2
    4 / -1

    A metal X crystallises in a face-centred cubic arrangement with the edge length 862 pm. What is the shortest separation of any two nuclei of the atom ?

    Solution

    For fcc arrangement , distance of nearest neighbour (d) is

    = 609.6 pm

  • Question 3
    4 / -1

    The edge length of sodium chloride unit cell is 564 pm. If the size of Cl- ion is 181 pm. The size of Na+ ion will be

    Solution

    2(r+ + r-) = a
    2(2Na+ + rCl-) = 564
    2Na+  = 564/2 - 181 = 101 pm

  • Question 4
    4 / -1

    If the distance between Na+ and Cl- ions in NaCl crystals is 265 pm, then edge length of the unit cell will be?

    Solution

    In NaCl crystal , Edge length = 2 x distance between Na+ and Cl−
    =2 x 265 = 530 pm

  • Question 5
    4 / -1

    The radius of Na+ is 95pm and that of Cl- is 181 pm. The edge length of unit cell in NaCl would be (pm).

    Solution

    a = (r+ + r-) = 2(95 + 181) = 552 pm

  • Question 6
    4 / -1

    Copper crystallises in fcc with a unit cell length of 361 pm. What is the radius of copper atom?157 pm

    Solution

    For fcc,
    r = √2 / 4 x a = √2 / 4 x 361
    = 
    = 127 pm

  • Question 7
    4 / -1

    Total volume of atoms present in a fcc unit cell of a metal with radius r is

    Solution

    a = 2√2 r
    Volume of the cell = a3 = (2√2 r)3 =  16√2r3
    No. of shperes in fcc = 8 x 1/8 + 6 x 1/2 = 4
    Volume of the four spheres = 4 x 4/3πr3
    = 16πr3

  • Question 8
    4 / -1

    The relation between atomic radius and edge length 'a' of a body centred cubic unit cell:

    Solution

    Distance between nearest neighbours, d = AD/2
    In right angled △ABC,AC2 = AB2 + BC2 
    AC2 = a2 + a2  or AC = √2a
    Now in right angled △ADC,
    AD2 = AC2 + DC2
    ∴ d = √3a / 2
    Radius , r = d/2 = √3/4 a

  • Question 9
    4 / -1

    Edge length of unit cell of Chromium metal is 287 pm with the arrangement. The atomic radius is the order of:

    Solution

    In bcc lattice, r = √3/4
    r = 
    = 124.27 pm

  • Question 10
    4 / -1

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

    Solution

    For simple cube,
    Radius(r) = a / 2 [a = edge length]
    Volume of the atom = 
    Packing fraction = 
    Volume of the sphere (atoms) in an unit cell.
    For simple cubic, Z = 1 atom
    Packing fraction = 

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