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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.
Prob 13. A metal crystallizes into two cubic system-face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0Å respectively. Calculate the ratio of densities of fcc and bcc.
Sol. fcc unit cell length = 3.5Å
bcc unit cell length = 3.0Å
Density in fcc = n1 × stomic weight / V1 × Avogardo number
Density in bcc = n2 × stomic weight / V2 × Avogardo number
Dfcc / Dbcc = n1 / n2 × V2 / V1
n1 for fcc = 4; Also V1 = a3 = (3.5 × 10–8)3
n2 for bcc = 2; Also V2 = a3 = (3.0 × 10–8)3
Prob 14. Copper crystal has a face centred cubic structure. Atomic radius of copper atom is
128 pm. What is the density of copper metal? Atomic mass of copper is 63.5.
Sol. In face centred cubic arrangement face diagonal is four times the radius of atoms face diagonal = 4×128 = 512 pm
Face diagonal = √2 × edge length
Edge length = 512 / √2 = 362 × 10–10 cm
Volume of the unit cell = (362 × 10-10)3 cm3 = 47.4 × 10-24 cm3
In a face centred cubic unit cell, there are four atoms per unit cell
Mass of unit cell = 4 × 63.5 / 6.023 × 1023 g = 4.22 × 10–22 g
Density = mass of unit cell / volume of unit cell = 4.22 × 10–22 / 47.4 × 10–24 = 8.9 g cm–3
Prob 15. The first order reflections of a beam of X – rays of wavelength of 1.54A° from the (100) face of a crystal of the simple cubic type occurs at an angle 11.29°. Calculate the length of the unit cell.
Sol. Applying Bragg’s equation
2 d Sinθ = nλ
Given θ = 11.29°, λ = 1.54A° = 1.54×10-8 cm
n = 1
d = 1.54 × 10–8 / 2 × Sin 11.29o = 1.54 × 10–8 / 2 × 0.1957 = 3.93 × 10–8 cm
dhkl = a √h2 + k2 + l2 = a
a = 3.93 × 10–8 cm (length of the unit cell)
Prob 16. X- rays of wavelength equal to 0.134 nm give a first order diffraction from the surface of a crystal when the value of θ is 10.5°. Calculate the distance between the planes in the crystal parallel to the surface examined.
Sol. Given λ = 0.134 nm, θ = 10.5°
n = 1
Applying Bragg’s equation
2d Sinθ = nλ
d = nλ / 2Sinθ = 1 × 0.134 / 2 × Sin 10.5o = 0.134 / 2 × 0.1822 = 3.68 A
Prob 17. What is the difference in the semi conductors obtained by doping silicon with Al or
Sol. Silicon doped with Al produces P – type semi conductors i.e. flow is due to creation of positive holes whereas silicon doped with P produces n – type semi conductors i.e. flow is due to extra electrons carrying negative charge.
Prob 18. Non stoichiometric cuprous oxide, Cu2O can be prepared in the laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1 can you account of the fact that this substance is a p – type semiconductor
Sol. The ratio less that 2:1 in Cu2O show that some cuprous (Cu+) ions have been replaced by cupric (Cu+2) ions. To maintain electrical neutrality, every two Cu+ ions will be replaced by one Cu+2 ion. Thereby creating a hole. As conduction will be due to presence of these positive holes, hence it is a p – type semi conductor.
Prob 19. Classify each of the following as being either a p – type or n – type semiconductor.
(i) Ge doped with In
(ii) B doped with Si
Sol. (i) Ge is group 14 element and In is group 13 element. Hence an electron deficient hole is created and therefore, it is p – type.
(ii) B is group 13 elements and Si is group 14 elements, there will be a free electron. Henc, it is n – type
Prob 20. If NaCl is doped with 10-3 mole% SrCl2 what is the concentration of cation vacancies?
Sol. Doping of NaCl with 10-3 mol% SrCl2 means that 100 moles of NaCl are doped with
10-3 mol of SrCl2
∴1 mole of NaCl is doped with SrCl2 = 10–3 / 100 mole = 10–5 mole
As each Sr+2 ion creates one cation vacancy, therefore concentration of cation vacancies
= 10-5 mol / mol of NaCl
=10-5×6.023×1023 = 6.023×1018 mol–1