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>> Solved Problems Part I
Board Type Questions
Prob 1. How can you convert NaCl structure into CsCl structure and vice-versa?
Sol. NaCl structure can be converted into CsCl structure by application of pressure while reverse can be done by heating to 760 K.
Prob 2. AgI crystallizes in cubic close packed ZnS structure. What fraction of tetrahedral sites are occupied by Ag+ ions?
Sol. In the face-centred unit cell, there are eight tetrahedral voids, of these, half are occupied by silver cations.
Prob 3. What is Frenkel defect?
Sol. When some ions (usually cations) are missing from the lattice sites and they occupy the interstitial sites so that electrical neutrality as well as stoichiometry is maintained, it is called Frenkel defect.
Prob 4. What type of crystal defect is produced when sodium chloride is doped with MgCl2?
Sol. It is called impurity defect. A cations vacancy is produced. A substitutional solid solution is formed (because 2Na+ ions are replaced by one Mg++ ion). This defect is also known as metal deficiency defect.
Prob 5. A compound AB2 possesses the CaF2 type crystal structure. Write the co-ordination number of A++ and B- ions in its crystals.
Sol. Co-ordination number of A = 8
Co-ordination number of B = 4
IIT Level Questions
Prob 6. A solid between A and B has the following arrangement of atoms
(i) Atoms A are arranged in ccp array
(ii) Atoms B occupy all the octahedral voids and half the tetrahedral voids. What is the formula of the compound?
Sol. In a close packing, the number of octahedral voids is equal to the number of atoms and the number of tetrahedral voids is twice the number of atoms
Since all the octahedral voids and half the tetrahedral voids are filled there will be one atom of B in tetrahedral void and one atom in octahedral void corresponding to each A. Thus, there will be two atoms of B corresponding to each A.
Hence, formula of the solid is AB2
Prob 7. In corundum, oxide ions are arranged in hcp array and the aluminium ions occupy two thirds of octahedral voids. What is the formula of corundum?
Sol. In ccp or hcp packing there is one octahedral void corresponding to each atom constituting the close packing. In corundum only 2/3rd of the octahedral voids are occupied. It means corresponding to each oxide are 2/3 aluminium ions. The whole number ratio of oxide and aluminium ion in corundum is therefore 3:2 Hence formula of corundum is Al2O3
Prob 8. Calculate the ratio of the alkali metal bromides on the basis of the data given below and predict the form of the crystal structure in each case. Ionic radii (in pm) are given below
Li+ = 74, Na+ = 102, K+ = 138
Rb+ = 148, Cs+ = 170, Br- = 195
Sol. The ratio of cation to that of anion i.e. r+/r– gives the clue for crystal structure
Li+ / Br– = 74 / 195 = 0.379 (tetrahedral)
Na+ / Br– = 102 / 195 = 0.523 (octahedral)
K+ / Br– = 138 / 195 = 0.708 (octahedral)
Rb+ / Br– = 148 / 195 = 0.759 (Body centered)
Cs+ / Br– = 170 / 195 = 0.872 (Body centered)
Prob 9. In the close packed cation in an AB type solid have a radius of 75 pm, what would be the maximum and minimum sizes of the anions filling the voids?
Sol. For close packed AB type solid
r+ / r– = 0.414 – 0.7342
∴ Minimum value of r– = r+ / 0.732 = 75 / 0.732 = 102.5 pm
Maximum value r– = r+ / 0.414 = 75 / 0.414 = 181.2 pm
Prob 10. NH4Cl crystallizes in a body centered cubic lattice, with a unit cell distance of 387 pm. Calculate (a) the distance between the oppositely charged ions in the lattice, and (b) the radius of the NH4+ ion if the radius of the Cl- ion is 181 pm.
Sol. (a) In a body centered cubic lattice oppositely charged ions touch each other along the cross - diagonal of the cube. Hence, we can write,
2r+ 2r– = √3a, r+ + r– = √3a / 2 = √3 / 2 (387 pm) = 335.15 pm
(b) Now, since r– = 181 pm
wehaver+ = (335.15 – 181) pm = 154.15 pm.
Prob 11. Copper has the fcc crystal structure. Assuming an atomic radius of 130pm for copper atom (Cu = 63.54):
(a) What is the length of unit cell of Cu?
(b) What is the volume of the unit cell?
(c) How many atoms belong to the unit cell?
(d) Find the density of Cu.
Sol. As we know
p = n × Mm / NA × a3,
(a) for fcc structure
4r = √2 a
a = 2√2r
=2√2 × 130 pm = 367.64 pm
(b) volume of unit cell = a3 = (367.64 × 10–10 cm)3
= 4.968 × 10–23 cm3
(c) n = 4
(d) p = 4 × 63.54 / 6.023 × 1023 × (3.67 × 10–8 cm3)3 = 8.54 gm / cm3
Prob 12. The density of CaO is 3.35 gm/cm3. The oxide crystallises in one of the cubic systems with an edge length of 4.80 Å. How many Ca++ ions and O–2 ions belong to each unit cell, and which type of cubic system is present?
Sol. From equation
r(density) = 3.35 gm/cm3
a = 4.80 Å
Mm of CaO = (40 + 16) gm = 56 gm CaO
Q r = where n = no. of molecules per unit cell
p = n × Mm / a3 × NA
∴ n = 3.35 × (4.8 ×10–8)3 × 6.023 × 1023 / 56 = 3.98
or n ≈ 4
So, 4-molecules of CaO are present in 1 unit cell
So, no. of Ca+ + ion = 4
No. of O– – ion = 4
So, cubic system is fcc type.