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 This means that x M2+ have been replaced by M3+ ions.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