The Solid State

  • Question 165
    CBSEENCH12005513

    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

    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 This means that x M2+ have been replaced by M3+ ions.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

    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

    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%.




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