Solid State: Answer the following questions

Chemistry : Solid State

Answer the following questions:

1.        Define unit cell.

• A basic repeating structural unit of a crystalline solid is called a unit cell.

• A unit cell is characterised by the three edge lengths (a, b, c) or lattice constants and the angle between the edges α, β and γ.

2.     Give any three characteristics of ionic crystals.

1. Ionic solids have high melting point.

2. These solids do not conduct electricity, because the ions are fixed in their lattice positions.

3. They conduct electricity in molten state (or) when dissolved in water because, the ions are free to move in the molten state or solution.

4. They are hard as only strong external force can change the relative positions of ions 

3.     Differentiate crystalline solids and amorphous solids.

Crystalline solids

• Long range orderly arrangement of constituents.

• Definite shape

• They are anisotropic in nature

• They are true solids

• Definite heat of fusion

• They have sharp melting point.

• Examples: NaCl, diamond.

Amorphous solids

• Short range, random arrangement of constituents.

• Irregular shape

• They are isotropic

• They are considered as pseudo solids (or) super cooled liquids

• Heat of fusion is not definite

• Gradually soften over a range of temperature and so can be moulded.

• Examples: Rubber, plastics, glass.

4.     Classify the following solids

a. P4 b. Brass c. diamond d. NaCl e. Iodine

a) P₄ - Covalent solid

b) Brass - Metallic solid

c) Diamond - Covalent solid

d) NaCl - Ionic solid

e) Iodine - Molecular solid

5.     Explain briefly seven types of unit cell.

6.     Distinguish between hexagonal close packing and cubic close packing.

Hexagonal close Packing

• The spheres can be arranged so as to fit into the depression in such a way that the third layer placed directly over a first layer

• abab arrangement is known as the hexagonal close packed (hcp) arrangement

• The tetrahedral voids of the second layer are covered by the spheres of the third layer

• The coordination number is 12

Cubic close packing

• The third layer may be placed over the second layer in such a way that all the spheres of the third layer fit in octahedral voids

• abcabcabc... arrangement is called cubic close packed (ccp) structure.

• The octahedral voids of the second layer are covered by the spheres of the third layer

• The coordination number is 12

7.     Distinguish tetrahedral and octahedral voids.

Tetrahedral voids

• During the crystal layer arrangement wherever a sphere of the second layer is above the void of the first layer a tetrahedral void is formed. These voids are called tetrahedral voids

• The void is surrounded by 4 spheres

• Radius of an tetrahedral voids r / R = 0.225

Octahedral voids

• The triangular voids in the second layer are above the triangular voids in the first layer, and these are not overlap.

• The void is surrounded by 6 sphere

• Radius of a octahedral void r / R = 0.414

8.     What are point defects?

If the deviation occurs from the perfect crystal is due to missing atoms, displaced atoms or extra atoms, the imperfection is named as a point defect.

Such defects arise due to imperfect packing during the original crystallisation or they may arise from thermal vibrations of atoms at elevated temperatures.

Example: Schottky defect and Frenkel defect.

9.     Explain Schottky defect.

• Schottky defect arises due to the missing of equal number of cations and anions from the crystal lattice.

• This defect does not change the stoichiometry of the crystal.

• Ionic solids in which the cation and anion are of almost of similar size show schottky defect. Example: NaCl.

• Schottky defects in a crystal, lowers its density.

• For example, the theoretical density of vanadium monoxide (VO) is 6.5 g cm⁻³, but the actual experimental density is 5.6 g cm ⁻³.

10. Write short note on metal excess and metal deficiency defect with an example.

Metal excess defect

• Metal excess defect arises due to the presence of more number of metal ions as compared to anions.

• The electrical neutrality of the crystal can be maintained by the presence of anionic vacancies equal to the excess metal ions (or) by the presence of extra cation and electron present in interstitial position.

 • Example: When NaCl crystals are heated in the presence of sodium vapour, Na⁺ ions are formed and are deposited on the surface of the crystal.

• Chloride ions (CƖ⁻) diffuse to the surface from the lattice point and combines with Na⁺ ion. The electron lost by the sodium vapour diffuse into the crystal lattice and occupies the vacancy created by the CƖ ⁻ ions.

• Such anionic vacancies which are occupied by unpaired electrons are called F centers.

• The formula of NaCl which contains excess Na⁺ ions can be written as Na_1+xCl.

Metal deficiency defect

• Metal deficiency defect arises due to the presence of less number of cations than the anions.

• This defect is observed in a crystal in which, the cations have variable oxidation states.

• Example: FeO crystal, some of the Fe²⁺ ions are missing from the crystal lattice.

• To maintain the electrical neutrality, twice the number of other Fe²⁺ ions in the crystal is oxidized to Fe³⁺ ions.

• In such cases, overall number of Fe²⁺ and Fe³⁺ ions is less than the O²⁻ ions. It was experimentally found that the general formula of ferrous oxide is Fe_xO, where x ranges from 0.93 to 0.98.

11. Calculate the number of atoms in a fcc unit cell.

• Identical atoms present at each corner as well as in the centre of each face.

• The atom in the face centre is shared by two unit cells.

• Each corner atom shared by 8 unit cells and makes 1/8 contribution to the unit cell.

Number of atoms in a fcc unit cell = (N_c / 8) + (N_f / 2)

= (8 / 8) + (6 / 2)

= 1+3 = 4

12. Explain AAAA and ABABA and ABCABC type of three dimensional packing with the help of neat diagram.

(i) AAA... type

(i) Linear arrangement of spheres in one direction is repeated in two dimension

(ii) All spheres of different rows align vertically as well as horizontally in identical pattern

 All the rows are identical and the packing is called AAA type. In this arrangement each sphere is in contact with four of its neighbours.

(ii) ABAB.. Type

In this type, the second row spheres are arranged in such a way that they fit in the depression of the first row.

The second row is denoted as B type. The third row is arranged similar to the first row A, and the fourth one is arranged similar to second one. i.e., the pattern is repeated as ABAB....

Coordination number of ABAB arrangement is 6.

(iii) ABCABC type

Formation of first layer

The first layer is formed by arranging two dimensional ABAB arrangements i.e. the spheres of second row fit into the depression of first row.

Formation of second layer

In the first layer (A) there are two types of voids (or holes) and they are designated as x and y. The second layer (B) can be formed by placing the spheres either on the depression (voids / holes) x (or) on y. Consider the formation of second layer by placing the spheres on the depression (y).

The voids (y) in the first layer (A) are partially covered by the spheres of layer (B), now such a void in (A) is called a octahedral void. This constitutes six spheres - three in the lower layer (A) and three in the upper layer (B). When the centers of these six spheres are joined, an octahedron is formed. Simultaneously new tetrahedral voids (or holes) are also created by three spheres in second layer (B) and one sphere of first layer (A)

Formation of third layer

The third layer may be placed over the second layer in such a way that all the spheres of the third layer fit in octahedral voids. This arrangement of the third layer is different from other two layers (A) and (B), and hence, the third layer is designated (C). If the stacking of layers is continued in abcabcabc... pattern, then the arrangement is called cubic close packed (ccp) structure.

13. Why ionic crystals are hard and brittle?

The structural units of an ionic crystal are cations and anions. They are bound together by strong electrostatic attractive forces. Cations are surrounded by as many anions as possible and vice versa to maximize the attractive force.

14. Calculate the percentage efficiency of packing in case of body centered cubic crystal.

Packing efficiency:

The spheres are touching along the leading diagonal of the cube

Radius of the sphere is given as

 r = √3a  /  4

∴ Volume of the sphere with radius 'r'

= 4/3 πr³

= 4/3 π  (√3/4 a)³

= √3/16  πa³              ……………..(1)

Number of spheres belong to a unit cell in bcc arrangement is equal to two and hence the total volume of all spheres

= 2 × (√3πa³ / 16) = √3πa³ / 8

Formula:

Packing fraction = [ (√3πa³ / N) / (a³) ] × 100

= [ √3π / 8  ] × 100

= √3π × 12.5

= 1.732 × 3,14 × 12.5

= 68%

Packing efficiency = 68%

Void = 32%

15. What is the two dimensional coordination number of a molecule in square close packed layer?

In the square close packed structure, each sphere is in contact with four of its neighbours. The two dimensional coordination number is 4.

16. Experiment shows that Nickel oxide has the formula Ni0.96 O1.00 . What fraction of Nickel exists as of Ni2+ and Ni3+ ions?

Formula is Ni_0.96O₁.₀₀

So the ratio of Ni : O = 96:100

So if there are 100 atoms of oxygen then 96 atoms of Ni

Let number of atoms of Ni²⁺ = x

Then number of atoms of Ni³⁺ = 96−x

Charge on Ni = charge on Oxygen

So that oxygen has charge = −2

3(96−x) + 2x = 2 (100)

288 −3x + 2x = 200

−x = −88

x = 88

Therefore, number of Ni²⁺ions = 88

 And, number of Ni³⁺ions = 96 − 88 = 8

 Fraction of nickel that exists as Ni²⁺

= 88/96 = 0.916

Fraction of nickel that exists as Ni³⁺

= 8/96 = 0.083

Percentage of nickel that exists as Ni²⁺

 = 0.916 × 100 = 91.6%

Percentage of nickel that exists as Ni³⁺

= 0.083 × 100 = 8.33%

17. What is meant by the term “coordination number”? What is the coordination number of atoms in a bcc structure?

• The number of nearest neighbours that surrounding a particle in a crystal is called the coordination number of that particle.

• In bcc structure the entire corner atom touches the atom that occupies the body centre. Hence, each atom is surrounded by eight nearest neighbors and coordination number is 8.

18. An element has bcc structure with a cell edge of 288 pm. the density of the element is 7.2 gcm-3. how many atoms are present in 208g of the element.

Given:

Cell edge = 208pm ; Density = 7.2 g cm⁻¹; n= 2

Solution:

volume of 208g of element 

mass / density = ( 208g ) / (7.2g cm^-3 )

= 28.88 cm³

The number of unit cell = [ 28.88 cm³ ] / [ 2.39 ×10^-23] cm³ / unitcell

BCC Contains 2 atoms per unit cell

Total number of atoms Unit cell = 2 × 12.08 × l0²³

Number of atoms present = 24.16 × 10²³ atoms

19. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. calculate the edge length of unit cell.

Answer:

Given:

Radius = 125 pm; Edge length = ?

Formula: 4r = √2a

Solution:

4r = √2a

a =4r / √2 = 4×125 / √2 = 500/√2

Edge length = 353.5 pm

20. if NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration of cation vacancy?

Given:

Concentration of SrCl₂ = 10⁻² mol%

Solution:

= 10⁻² / 100 moles of SrCl₂

= 10⁻⁴ mol of SrCl₂

Concentration is in percentage so that take total 100 mol of solution

Number of moles of NaCl = 100 − moles of SrCl₂

Moles of SrCl₂ is very negligible as compare to total moles

Number of moles of NaCl = 100

1 mol of NaCl is doped with = 10⁻²/100 moles of SrCl₂

= 10⁻⁴ mol of SrCl₂

Cation vacancies per mole of NaCl =10⁻⁴ mol

1 mol = 6.022 × l0²³ particles

Cation vacancies per mole of NaCl

 = 10⁻⁴ × 6.022 × l0²³ = 6.02 × l0¹⁹

21. KF crystallizes in fcc structure like sodium chloride. calculate the distance between K+ and F− in KF.( given : density of KF is 2.48 g cm−3 )

Given:

n = 4

Density = 2.48gcm⁻³

 M = 58.08g / mol⁻¹

 N_A = 6.023 × 10²³

Formula:

ρ = nM / N_Aa³

a ³ = nM / ρN_A

Solution:

a³ = [ 4 × 58.8 ] / [ 248 × 6.023 × 10²³ ]

= [ 232.32 / 1.5×10^-24] cm³

a = 5.37 × l0⁻¹⁰ cm = 537 pm

Edge length (a) = 2r

r = a / 2

537 / 2 = 268 pm

22. An atom crystallizes in fcc crystal lattice and has a density of 10 gcm−3 with unit cell edge length of 100pm. calculate the number of atoms present in 1 g of crystal.

Answer:

Given:-

Edge length = 100pm

Density = l0g cm⁻³

n = 4

Solution:

Volume of unit cell = (100pm)³

= (100 ×l0⁻¹²m)³

 = (100 × 10⁻¹⁰cm)³

 = l × l0⁻²⁴cm³

Volume of 1 g of crystal = mass / density

= 1g / 10 gcm⁻³ = 0.1 cm³

Number of unit cell = [ 0.1 cm³ ] / [ 1 × 10⁻²⁴cm³/unitcell ]

= l × 10²³ unitcell

Fcc contains 4 atoms

Total number of atoms for 1 g = 4 × 1 × 10²³

= 4 × 10²³ atoms

23. Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?

Each corner shared by 8 unit cells. Eight corners contributes 8/8 = 1

X= 1

Y present at the centre of the cube.

Centre atom shared by one unit cell. The formula of the compound is XY.

24. Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4.3 ×10−8 cm . calculate the radius of sodium atom.

Given:

Na crystallizes in bcc structure

Edge length = 4.3 × 10⁻⁸cm

Solution:

Distance 4r = √3 a

r = √3/4 × 4.3 × 10⁻⁸

= 7.44 / 4 × 10⁻⁸

= 1.86 × 10⁻⁸ cm

Radius of Na atom = 1.86 × 10⁻⁸ cm

 = 1.86 Ǻ

= 186 pm

25.   Write a note on Frenkel defect.

Frenkel defect arises due to the dislocation of ions from its crystal lattice. The missing ion occupies an interstitial position. This defect is shown by ionic solids in which cation and anion differ in size. This defect does not affect the density of the crystal.

For example, In AgBr, small Ag⁺ ion leaves its normal site and occupies an interstitial position

EVALUATE YOURSELF:

1. An element has a face centered cubic unit cell with a length of 352.4 pm along an edge. The density of the element is 8.9 g cm⁻³. How many atoms are present in 100 g of an element?

Given:

Density: 8.9 g cm⁻³

a = 352.4 pm = 3.524 × l0⁻⁸ cm, Mass = 100g, Z = 4 

M = [ a³ρ × N_A ] / n

= [ (3.524 × 10⁻⁸)³ × 8.9 × 6.023 × 10²³ ] / 4

= [ 2345.9 × 10⁻¹ ] / 4 = 586.5 × 10⁻¹ = 58.65 gmol⁻¹

Number of atoms of an element

= ( mass / atomic mass ) × 6.023 × 10²³

Number of atoms in l00g of an element

= [100/58.65] × 6.023 × 10²³ = 10.27 × 10²³ atoms

2. Determine the density of CsCl which crystallizes in a bcc type structure with an edge length 412.1 pm.

Given:

M = 133 + 35.5 = 168.5 g mol⁻¹,

a = 412.1 pm = 4.121 × 10⁻⁸ cm, Z = 1

ρ = ( M × n )/ ( a^{3 ×} N_A ) = [ 168.5 × 1 ] / { (4.121 × 10⁻⁸)³ × 6.023 × 10²³ } = 168.5 / 42.15 = 3.99

Density of CsCl crystal = 4 g cm⁻³

3. A face centered cubic solid of an element (atomic mass 60) has a cube edge of 4A. Calculate its density.

Given:

M = 60 g mol⁻¹ ; a = 4Ǻ = 4 × 10⁻⁸ cm, Z = 4

ρ = [M × n ] / [ a³ × N _A ] = [ 60 × 4] / [ (4 × l0⁻⁸)³ × 6.023 × 10²³ ]

= 240 / 38.54 = 6.227 g cm⁻³

Density of an element = 6.227 g cm⁻³

4. Barium has a body centered cubic unit cell with a length of 508 pm along an edge. What is the density of barium in g cm⁻³?

Given:

M = 137.3 g mol⁻¹, a = 508 pm = 5.08 × 10⁻⁸ cm,

 Z = 2

ρ = [ M × n ] / [ a³ × N_A ]

= [ 137.3 × 2 ] / [ (5.08 × 10⁻⁸)³ × 6.023 × 10²³ ]

= 274.6 / 78. 96 = 3.478 g cm⁻³ = 3.5 g cm⁻³