Solved Problems

Question: 1

It can be seen now that both fcc and Hexagonal Primitive Structure have the same packing fraction.  Moreover this is also the highest packing fraction of all the possible unit cells with one type of atom with empty voids.  Can you explain this?

Solution:

HCP is of the ABAB… type and FCC is of ABCABC… type. In HCP and FCC both, the layers are formed of same atoms of the same type in the same arrangement (each sphere surrounded by six) but the placement of the layers on top of each other is different.

Both will have same packing fraction.

his is the highest packing fraction possible because in both types, maximum number of atoms of the same size are present around the atom (namely six).  

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Question: 2

The radius of a calcium ion is 94 pm and of an oxide ion is 146 pm. Predict the crystal structure of calcium oxide.

Solution:

The ratio  r+/r- = 94/146 = 0.644 . The prediction is an octahedral arrangement of the oxide ions around the calcium. Because the ions have equal but opposite charges, there must also been an octahedral arrangement of calcium ions around oxide ions. Thus we would expect a rock salt (NaCl) structure.

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Question: 3

The unit cell of silver iodide (AgI) has 4 iodine atoms in it. How many silver atoms must be there in the unit cell.

Solution:

The formula, AgI, tells us that the ratio of silver atoms to iodine atoms is 1:1. Hence, if there are four iodine atoms in the unit cell, there must be four silver atoms.

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Question: 4

The co-ordination number of the barium ions, Ba2+, in barium chloride (BaF2) is 8. What must be the co-ordination number of the fluoride ions?

Solution:

The co-ordination number of barium ions tells us that it is surrounded by  eight fluoride ions (charge 8 ´ (-1) = -8). In order to balance out the eight negative charges, we need four barium ions (charge 4 ´ (+2) = +8). Hence, the co-ordination number of the fluoride ions must be 4. You will find that the co-ordination number and the charge on ions always
balance out to neutrality.

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Question: 5

 A solid between A and B has the following arrangement of atoms

• Atoms A are arranged in ccp array

• Atoms B occupy all the octahedral voids and half the tetrahedral voids.

What is the formula of the compound?

Solution

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

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Question: 6

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?

Solution

In ccp or hcp packing there is one octahedral void corresponding to each atom constituting the close packing. In corundum only 2/3^rd 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

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Question: 7

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

Solution  

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

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Question: 8

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?

Solution  

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

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Question: 9

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.

Solution 

(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.

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Question: 10

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.

Solution       

As we know

p = n × M_m / N_A × 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

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Question: 11

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?

Solution        

From equation

r(density) = 3.35 gm/cm³

 a = 4.80 Å

M_m of CaO = (40 + 16) gm = 56 gm CaO

Q r =  where n = no. of molecules per unit cell

p = n × M_m / a³ × N_A          

∴ 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.

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Question: 12

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.     

Solution      

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

D_fcc / D_bcc = 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

= 1.259

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Question: 13

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.

Solution       

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

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Question: 14

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.

Solution       

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 11.29^o = 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)

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Question: 15

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.

Solution 

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 10.5^o = 0.134 / 2 × 0.1822 = 3.68 A 

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Question: 16

What is the difference in the semiconductors obtained by doping silicon with Al or  with P?

Solution 

Silicon doped with Al produces P – type semiconductors 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.

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Question: 17

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

Solution  

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.

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Question: 18

Classify each of the following as being either a p – type or n – type semiconductor.

• Ge doped with In

• B doped with Si

Solution       

• Ge is group 14 element and In is group 13 element. Hence an electron deficient hole is created and therefore, it is p – type.

• B is group 13 elements and Si is group 14 elements, there will be a free electron. Henc, it is n – type

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Question: 19

If NaCl is doped with 10^-3 mole% SrCl₂ what is the concentration of cation vacancies?

Solution   

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

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Question: 20

In LiI Crystal, I– ions form a cubical closest packed arrangement and Li+ ions occupy octahedral holes. What is the relationship between the edge-length of the unit cells and the radii of the I– ions? Calculate the limiting ionic radii of Li+ and I–- if a= 600pm.

Solution

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