The Solid State Class 12th Chemistry Part I CBSE Solution

Class 12^th Chemistry Part I CBSE Solution

Intext Questions Pg-4

1. Why are solids rigid?
2. Why do solids have a definite volume?
3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene,…
4. Why is glass considered a super cooled liquid?
5. Refractive index of a solid is observed to have the same value along all directions.…

Intext Questions Pg-6

1. Classify the following solids in different categories based on the nature of…
2. Solid A is a very hard electrical insulator in solid as well as in molten state and melts…
3. Ionic solids conduct electricity in molten state but not in solid state. Explain.…
4. What type of solids are electrical conductors, malleable and ductile?…

Intext Questions Pg-12

1. Give the significance of a ‘lattice point’,
2. Name the parameters that characterize a unit cell
3. Hexagonal and monoclinic unit cells Distinguish between
4. Face-centred and end-centred unit cells. Distinguish between
5. Explain how many portions of an atom located at (i) corner and (ii) body-center of a cubic…

Intext Questions Pg-21

1. What is the two dimensional coordination number of a molecule in square close-packed…
2. A compound forms hexagonal close-packed structure. What is the total number of voids in…
3. A compound is formed by two elements M and N. The element N forms ccp and atoms of M…
4. Which of the following lattices has the highest packing efficiency (i) simple cubic (ii)…
5. An element with molar mass 2.7×10-2 kg mol-1 forms a cubic unit cell with edge length 405…

Intext Questions Pg-29

1. What type of defect can arise when a solid is heated? Which physical property is affected…
2. What type of stoichiometric defect is shown by? (i) ZnS (ii) AgBr…
3. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is…
4. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour.…
5. A group 14 element is to be converted into n-type semiconductor by doping it with a…
6. What type of substances would make better permanent magnets, ferromagnetic or…

Exercises

1. Define the term 'amorphous'. Give a few examples of amorphous solids.…
2. What makes a glass different from a solid such as quartz? Under what conditions quartz…
3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or…
4. What is meant by the term 'coordination number'?
5. What is the coordination number of atoms: (a) in a cubic close-packed structure? (b) in a…
6. How can you determine the atomic mass of an unknown metal if you know its density and the…
7. 'Stability of a crystal is reflected in the magnitude of its melting points'. Comment.…
8. Hexagonal close-packing and cubic close-packing? How will you distinguish between the…
9. Crystal lattice and unit cell? How will you distinguish between the following pairs of…
10. Tetrahedral void and octahedral void? How will you distinguish between the following pairs…
11. Face-centred cubic How many lattice points are there in one unit cell of each of the…
12. Face-centred tetragonal How many lattice points are there in one unit cell of each of the…
13. Body-centred How many lattice points are there in one unit cell of each of the following…
14. The basis of similarities and differences between metallic and ionic crystals. Explain…
15. Ionic solids are hard and brittle. Explain
16. simple cubic Calculate the efficiency of packing in case of a metal crystal for…
17. body-centred cubic Calculate the efficiency of packing in case of a metal crystal for…
18. face-centred cubic (with the assumptions that atoms are touching each other). Calculate…
19. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10-8 cm and…
20. A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube…
21. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm-3, calculate…
22. If the radius of the octahedral void is r and radius of the atoms in close packing is R,…
23. Copper crystallises into a fcc lattice with edge length 3.61 × 10-8 cm. Show that the…
24. Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel…
25. What is a semiconductor? Describe the two main types of semiconductors and contrast their…
26. Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide,…
27. Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of…
28. Ge doped with In Classify each of the following as being either a p-type or a n-type…
29. Si doped with B. Classify each of the following as being either a p-type or a n-type…
30. Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the…
31. Between a conductor and an insulator In terms of band theory, what is the difference?…
32. Between a conductor and a semiconductor? In terms of band theory, what is the difference?…
33. Schottky defect Explain the following terms with suitable examples:…
34. Frenkel defect Explain the following terms with suitable examples:…
35. Interstitials Explain the following terms with suitable examples:…
36. F-centres. Explain the following terms with suitable examples:
37. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.…
38. If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?…
39. Ferromagnetism Explain the following with suitable examples:
40. Paramagnetism Explain the following with suitable examples:
41. Ferrimagnetism Explain the following with suitable examples:
42. Antiferromagnetism Explain the following with suitable examples:
43. 12-16 and 13-15 group compounds. Explain the following with suitable examples:…

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Intext Questions Pg-4

Question 1.

Why are solids rigid?

Answer:

• Solids can be defined as any substance that has a definite shape, has mass and occupies volume.
• The intermolecular spaces between the molecules forming the solid structures are smaller.
• The intermolecular forces are strong.
• The constituent particles (atoms, molecules, nuclei, etc.) has a fixed position and they can oscillate only about their mean position.
• They are incompressible and rigid.

Question 2.

Why do solids have a definite volume?

Answer:

Solids have strong intermolecular forces of attraction. The constituent particles of solids cannot move from their position they can only vibrate from their mean position. That is why solids have a definite volume.

Question 3.

Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

Answer:

Explanation: Amorphous solids are not true solids, in these solids the constituent particles (atoms, ions or molecules) have short range order of arrangement.

In crystalline solids the constituent particles have long range order of arrangement)

Question 4.

Why is glass considered a super cooled liquid?

Answer:

Glass is an amorphous solid and all the amorphous solids have a tendency to flow, though very slowly. Hence glass is considered a super cooled liquid and that becomes the reason why the glass windows become slightly thicker at the bottom that at the top over a period of time.

Explanation: All amorphous solids have tendency to flow as rubber, plastic.

Question 5.

Refractive index of a solid is observed to have the same value along all directions.

Comment on the nature of this solid. Would it show cleavage property?

Answer:

Isotropic solids have the same value of the reflective index in all the direction. All isotropic solids are amorphous solids. Amorphous solids cut into two pieces with irregular surfaces. Therefore, amorphous solids do not show cleavage. Anisotropic solid is a solid material in which the physical properties do not depend on its orientation.

Explanation: Amorphous solids do not show cleavage property because of an irregular arrangement of particles. When amorphous solids are cut with the sharp-edged tool the newly generated surfaces are irregular and not smooth.

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Intext Questions Pg-6

Question 1.

Classify the following solids in different categories based on the nature of

intermolecular forces operating in them:

Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide,

graphite, rubidium, argon, silicon carbide.

Answer:

Explanation: Ionic solids are those in which constituents particles are cations and anions which are held together by columbic force of interactions

e.g.- K₂SO₄⇒ K⁺ + SO₄^—

Metallic solids are metals in which metal ions (also called kernals) are held together by metallic bonds.)

Molecular solids are made up on molecules or inert gases.

In covalent solids constituent particles are atoms which are held together by continuous system of covalent bonds.

Note: Though in solid argon the constituent particles are atoms even it is categorised as molecular solid because the particles are held together by vander waal forces so all solids of zero group elements are placed under the category of molecular solids.)

Question 2.

Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?

Answer:

The solid ‘A’ is a covalent solid, such as diamond. As it is an electrical insulator in both solid and molten form and melts at an extremely high temperature.

Explanation-In covalent solids the constituent particles (i.e. atoms) form a network of covalent bonds which make them very hard and also very high melting points.

Further they are electrical insulator in solid as well as in molten state because they do not contain any ion (like ionic solids) or free electrons (like metallic solids) in solid or in molten state.

(Note- For any compound to show electrical conductivity either free electrons or free ion must be available.)

Question 3.

Ionic solids conduct electricity in molten state but not in solid state. Explain.

Answer:

The ionic solids conduct electricity in molten state as the constituent particles of ionic solid are not free to move in solid state, thus ionic solids are insulators in their solid state. While the ions become free to move in the molten state

Question 4.

What type of solids are electrical conductors, malleable and ductile?

Answer:

Metallic solids have these properties.

Explanation: They are electrical conductors, are malleable and ductile, which means it could be drawn into thin wires. The metals are orderly free collection of positive ions surrounded by and held together by a sea of free electrons. These free and mobile are responsible for high electrical and thermal conductivity of metals.

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Intext Questions Pg-12

Question 1.

Give the significance of a ‘lattice point’,

Answer:

The points in a crystal lattice where atoms or ions can be placed are called as lattice points. Lattice Points can be vacant or filled.

The number of lattice points can be different as well as can be arranged in different ways which give rise to the different crystal lattice.

Given below is a diagram of the crystal lattice and the lattice points:

Question 2.

Name the parameters that characterize a unit cell

Answer:

There are a total of six parameters that characterize a unit cell.

i. Dimensions of a unit cell (also called the edges): a, b and c.

ii. The angles between the edges: α, β and γ

The variation of these 6 parameters leads to different shapes and sizes of the unit cell, shown as below:

Where a, b and c are technically the vectors of the unit cell. The angle α is between edges b and c; the angle β is between the edges α and γ; γ is the angle between a and b.

Question 3.

Distinguish between

Hexagonal and monoclinic unit cells

Answer:

Question 4.

Distinguish between

Face-centred and end-centred unit cells.

Answer:

Question 5.

Explain how many portions of an atom located at (i) corner and (ii) body-center of a cubic unit cell is part of its neighbouring unit cell.

Answer:

The portion of an atom of the corner of a cubic unit cell is shared by eight adjacent unit cells as shown below:

Therefore, a portion of the atom at the corner = 8 × 1/8 = 1 atom.

(ii) The atoms present at the center of the body is not shared by its neighbouring unit cell.

Therefore, portion of the atom at the center = 1 atom

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Intext Questions Pg-21

Question 1.

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

Answer:

In a two dimensional square closed packed layer, a molecule touches four neighbour atoms. Therefore, 4 is the two dimensional coordinate number of a molecule in square closed packing.

Explanation-In two dimensional square close-packed layer each sphere is in contact with four of its neighbours hence its coordination number is four.

Question 2.

A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?

Answer:

Number of atoms in close packing = 0.5 mol

1 mol has 6.022x10²³ particles

So that

Number of close- packed particles = 0.5 × 6.022 × 10²³ = 3.011 × 10²³

Number of tetrahedral voids = 2 × number of atoms in close packing plug the value we get

Number of tetrahedral voids = 2 × 3.011 × 10 ²³ = 6.022 × 10 ²³

Number of octahedral = number of atoms in close packing

So that

Number of octahedral voids = 3.011 × 10 ²³

Total number of voids = tetrahedral voids + octahedral voids

= 6.022 × 10 ²³ + 3.011 × 10 ²³

= 9.03 × 10 ²³

Question 3.

A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

Answer:

The formula can be determined by using the following method

Let the number of atoms of the element N = n

Number of tetravalent voids = 2n

It is given that the atoms of element M occupy 1/3 of the tetrahedral voids.

So that the number of atoms of M = (1/3) × 2n

Ratio of the number of atoms of M and N

M:N =2n/3:n

Multiplying the 3 and divide by n we get

M:N = 2:3

Hence, the formula of the compound is M₂N₃ .

Question 4.

Which of the following lattices has the highest packing efficiency

(i) simple cubic

(ii) body-centred cubic and

(iii) Hexagonal close-packed lattice?

Answer:

(i)

In a simple cubic lattice the atoms are located only at the corners of the cube.

Let us assume the edge length or the side of the cube = a

And the radius of each particle = r

The relation between radius and edge a

Can be given as a = 2r

The volume of the cubic unit cell = side³ = a³

= (2r³)

= 8r³

Number of atoms in unit cell 

The volume of the occupied space 

And we know that, the packing efficiency 

= 52.36%

(ii) 

Let us assume the edge length or the side of the cube = a

And the radius of each particle = r

The diagonal of a cube is always a 

The relation between radius and the edge will be = 4r

Divide by root 3 we get A 

Total number of atoms in body centred cubic

Number of atoms at the corner

Number of atoms at the centre = 1

Total number of atoms = 2

The volume of the cubic unit cell = side³

= a³

= (4r/a√3)³

The volume of the occupied space 

Packing efficiency 

= 68%

(iii) Let the base of the hexagon is a and the height is c

Each angle in hexagonal will be 60 degree at the base

Packing efficiency of

Hexagonal close- packed lattice 

a = 2r c  = 1.633a

= 74%

Thus, hexagonal close- packed lattice has the highest packing efficiency of 74%.

Question 5.

An element with molar mass 2.7×10^-2 kg mol-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7×10³ kg m^-3, what is the nature of the cubic unit cell?

Answer:

Given information:-

Molar mass of the element = 2.7×10^-2 kg mol^-1

Edge length, a = 405 pm

Density, d = 2.7×10³ kg m^-3

Using the formula, d 

Putting the values given at their appropriate place, we get

= 3.99 which is approximately equal to 4

Therefore, it is an fcc unit cell.

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Intext Questions Pg-29

Question 1.

What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

Answer:

When a solid is heated, vacancy defect is produced in the crystal. On heating, some ions or atoms leave the lattice site completely, which means the lattice site becomes empty or vacant. As a result of this defect, density of the substance decreases. Thus density is the physical property that is effected by the vacancy defect.

Question 2.

What type of stoichiometric defect is shown by?

(i) ZnS (ii) AgBr

Answer:

i) ZnS shows Frankel defect

Explanation: Frankel defect arises when an atom or smaller ion leaves its place in the lattice, creating a vacancy and becomes an interstitial by lodging in a nearby location. This occurs in ZnS due to the comparatively smaller size of Zn²⁺ .

ii) AgBr shows frankel as well as schottky defect.

Explanation: AgBr shows both types of defects, i.e. schottky and Frenkel Defects. Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.

Question 3.

Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

Answer:

To explain this, let us take an example of NaCl doped with SrCl, impurity when SrCl₂ is added to NaCl solid as an impurity, two Na⁺ ions will then be replaced and one of their site will be occupied by Sr^2- while the other will remain vacant. Thus, we can say that when a cation of higher valance is added as an impurity to an ionic soil, two or more cations of lower valency are replaced by a cation of higher valence to maintain electrical neutrality. Hence, some cationic vacancies are created.

Question 4.

Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

Answer:

To explain this, let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl^- ions leave their lattice sites to combine with Na to form Na⁺ (Na⁺ + Cl^-⇒ NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called as the F- centres. These electrons absorb energy from the visible light, get excited to higher energy level and when they fall back to the ground state they impart yellow colour to the NaCl crystal.

Question 5.

A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?

Answer:

Silicon and germanium both have valance electrons and they belong to group 14^th of the periodic table. Arsenic and phosphorus belong to group 15^th of the periodic table and they have valence electrons equal to 5. When silicon or germanium is adopted with phosphorus or arsenic, four electrons of phosphorus or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon or germanium. Since the electrical conductivity of silicon and phosphorus is increased because of negatively charged particles (electron), thus this is known as n – type semiconductor.

Question 6.

What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic ? Justify your answer.

Answer:

Metal ions of ferromagnetic substances are randomly oriented in normal condition and substances do not act as a magnet. But when metal ions are grouped together in small regions, called as domains, each domain acts as a tiny magnet and it produces a strong magnetic field. In such condition ferromagnetic substances act like a magnet. When the ordering of the domain in group persists even after removal of the magnetic field a ferromagnetic substance becomes a permanent magnet. While domains are grouped in parallel and anti – parallel direction but in unequal number in ferromagnetic compounds. Thus, ferromagnetic substances, such as Ni, Co, Fe would make better permanent magnets rather than ferromagnetic substances.

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Exercises

Question 1.

Define the term 'amorphous'. Give a few examples of amorphous solids.

Answer:

Solids having constituent particles with irregular with shapes and short range order are called amorphous solids. Amorphous solids are isotropic in nature and melt over a range of temperature. Thus, amorphous solids are also referred as pseudo solids or super cooled liquids. They do not do not have definite heat of fusion. These solids give irregular surfaces, cut with sharp tool. Glass, rubber, etc. are some examples of amorphous solid.

Question 2.

What makes a glass different from a solid such as quartz? Under what conditions quartz could be converted into glass?

Answer:

It is the arrangement of constituent particles of glass which makes it different from quartz. The constituent particles of glass have short range order while quartz has constituent particles in long range order both by heating and cooling rapidly can be converted into glass.

Question 3.

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

(i) Tetra phosphorus decoxide (P₄O₁₀) (vii) Graphite

(ii) Ammonium phosphate (NH₄)₃PO₄ (viii) Brass

(iii) SiC (ix) Rb

(iv) I₂ (x) LiBr

(v) P₄ (xi) Si

(vi) Plastic

Answer:

(i) Tetra phosphorous decoxide (P₄O₁₀): molecular

Explanation: Molecular solids are made up on molecules or inert gases.

(ii) Ammonium phosphate (NH₄)₃PO_4: Ionic

Explanation: Ionic solids are those in which constituents particles are cations and anions which are held together by coulombic force of interactions

e.g.- K₂SO₄⇒ 2K⁺ + SO₄^—

(iii) SiC: Covalent (network)

Explanation: In covalent solids constituent particles are atoms which are held together by continuous system of covalent bonds.

(iv) I₂ :molecular

Explanation: Molecular solids are made up on molecules or inert gases.

(v) P_4: Molecular

Explanation: Molecular solids are made up on molecules or inert gases.

(vi) Plastic: Amorphous

Explanation: Amorphous solids are not true solids, in these solids the constituent particles (atoms, ions or molecules) have short range order of arrangement.

(vii) Graphite: Covalent (network)

Explanation: In covalent solids constituent particles are atoms which are held together by continuous system of covalent bonds.

(viii) Brass: Metallic

Explanation: Metallic solids are metals in which metal ions (also called kernals) are held together by metallic bonds.)

(ix) Rb: Metallic

Explanation: Metallic solids are metals in which metal ions (also called kernals) are held together by metallic bonds.)

(x) LiBr: Metallic

Explanation: Metallic solids are metals in which metal ions (also called kernals) are held together by metallic bonds.)

(x) Si : Covalent (network)

Explanation: In covalent solids constituent particles are atoms which are held together by continuous system of covalent bonds.

Question 4.

What is meant by the term 'coordination number'?

Answer:

Coordination number is the number of nearest neighbours of any constituent particles present in the crystal lattice.

Question 5.

What is the coordination number of atoms:

(a) in a cubic close-packed structure?

(b) in a body-centred cubic structure?

Answer:

(a) In a cubic close- packed structure is 12.

(b) In a body- centered cubic structure is 8.

Question 6.

How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Answer:

The atomic mass of an unknown metal can be determined by knowing its density and the dimension of unit cell.

Let ‘a’ be the edge length of a unit cell of a crystal.

‘d’ is the density of the metal

‘m’ is the atomic mass of the metal.

‘z’ is the number of atoms in the unit cell.

Now, the density of the unit cell 

As we know that, mass of the unit cell = Number of atoms in the unit cell × atomic mass

And the volume of the unit cell = (Edge length of the cubic unit cell)³

Therefore, d --------- (i)

 ---------------- (ii)

Now, since mass of the metal (m)

Therefore, from equation (ii) we have

 --------------- (iii)

If the edge lengths are different (say a, b and c), therefore, equation (iii) can be written as

 ----------- (iv)

Thus, using equation (iii) and the atomic mass of an unknown metal can be determined.

Question 7.

'Stability of a crystal is reflected in the magnitude of its melting points'. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Answer:

The melting points of the given substance are as follows:

Solid water – 273 K

Ethyl alcohol – 158.8 K

Diethyl ether – 156.85 K

Methane – 89.34 K

As we can see the melting point of solid water is highest and melting point of methane is lowest among the given substance. This says that intermolecular force in solid water is strongest and the intermolecular force in methane is weakest.

Question 8.

How will you distinguish between the following pairs of terms:

Hexagonal close-packing and cubic close-packing?

Answer:

Question 9.

How will you distinguish between the following pairs of terms:

Crystal lattice and unit cell?

Answer:

Question 10.

How will you distinguish between the following pairs of terms:

Tetrahedral void and octahedral void?

Answer:

Question 11.

How many lattice points are there in one unit cell of each of the following lattice?

Face-centred cubic

Answer:

One unit cell of a face- centered cubic has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

Question 12.

How many lattice points are there in one unit cell of each of the following lattice?

Face-centred tetragonal

Answer:

One unit cell of face- centered tetragonal has 8 lattice points are corners and 6 lattice points at faces, total 14 lattice points.

Question 13.

How many lattice points are there in one unit cell of each of the following lattice?

Body-centred

Answer:

One unit cell of body centered has 8 lattice points are corners and 1 lattice points at faces, total 9 lattice points.

Question 14.

Explain

The basis of similarities and differences between metallic and ionic crystals.

Answer:

Question 15.

Explain

Ionic solids are hard and brittle.

Answer:

Ionic solids are hard and brittle because in ionic solids the constituent particles are ions and they are held together in 3- dimensional arrangements by the electrostatic force of attraction. So, due to strong electrostatic force, these charged ions are held together in fixed position.

Question 16.

Calculate the efficiency of packing in case of a metal crystal for

simple cubic

Answer:

In a simple cubic lattice the atoms are located only at the corners of the cube.

Let us assume the edge length or the side of the cube = a

And the radius of each particle = r

The relation between radius and edge a

Can be given as a = 2r

The volume of the cubic unit cell = side³ = a³

= (2r³)

= 8r³

Number of atoms in unit cell 

The volume of the occupied space 

And we know that, the packing efficiency 

= 52.36%

Question 17.

Calculate the efficiency of packing in case of a metal crystal for

body-centred cubic

Answer:

Let us assume the edge length or the side of the cube = a

And the radius of each particle = r

The diagonal of a cube is always a 

The relation between radius and the edge will be = 4r

Divide by root 3 we get, A 

Total number of atoms in body centered cubic

Number of atoms at the corner

Number of atoms at the centre = 1

Total number of atoms = 2

The volume of the cubic unit cell

= side³

= a³

The volume of the occupied space 

Packing efficiency 

= 68%

Question 18.

Calculate the efficiency of packing in case of a metal crystal for

face-centred cubic (with the assumptions that atoms are touching each other).

Answer:

Let the base of the hexagon is a and the height is c

Each angle in hexagonal will be 60 degree at the base

Packing efficiency of Hexagonal close- packed lattice 

a = 2r c  = 1.633a

= 74%

Question 19.

Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10^–8 cm and density is 10.5 g cm^–3, calculate the atomic mass of silver.

Answer:

Given edge length (a) = 4.07 × 10 ^-8 cm

Density (d) = 10.5gcm^-1

For fcc number of atoms per unit cell (z) = 4

Avagadro number N_A = 6.022 × 10²³_mol-1

We know that, Atomic mass (M) 

(M) 

Thus, M 

⇒ M = 106.5 mol^-1

Question 20.

A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Answer:

Given. Atoms of Q are the corners of the cube and P at the body- centre

Therefore, Number of atoms of Q in one unit cell 

Number of atoms of P in one unit cell =1

Therefore, Ratio of P and Q atoms = P:Q = 1:1

Therefore, formula of the given compound = PQ

Since it is bcc

Therefore, coordinate number of P and Q = 8

Question 21.

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^–3, calculate atomic radius of niobium using its atomic mass 93 u.

Answer:

Given density (d) = 8.55 g cm^-3

Atomic mass (M) = 93u = 93 g mol^-1

To find: - atomic radius, r =__

We know, Avogadro Number N_A = 6.022 × 10²³ mol^-1

Since, given lattice in bcc

Therefore, Number of atoms per unit cell (z) = 2

We know that,

⇒ a³ = 36.124 × 10^-24 cm³

⇒ a = 3.3057 × 10^-8 cm

for bcc unit cell radius(r) =  a/4

Therefore, r = /4 × 3.3057 × 10^-8 cm

r = 1.732/4 × 3.3057 × 10^-8 cm

r = 5.725/4 × 10^-8 cm

r = 14.31 × 10^-9

r = 14.31 nm.

Thus the atomic radius of niobium = 14.31 nm.

Question 22.

If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.

Answer:

In the given figure, let the sphere have the centre, O and is fitted in the octahedral void.

As given, radius of the sphere fitted in the octahedral void = r

And the radius of the atoms in close packing = R

Here, angle AOD = 90⁰

In triangle AOD,

DA² = OA² + OD²

⇒ (R+R)² = ( R + r )²+ (R+r)²

⇒ 4R² = 2(R+r)²

⇒ 2R² = (R+r)²

⇒ √2 R = (R+r)

⇒ R + r = R

⇒ r = R – R

⇒ r = R (1.414 - 1)

⇒ r = 0.414 R

This is the required relation between r and R.

Question 23.

Copper crystallises into a fcc lattice with edge length 3.61 × 10^–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm^–3.

Answer:

Given, edge length (a) = 3.61 × 10 ^-8 cm

Given lattice is fcc, therefore, no. of atoms per unit cell (z) = 4

We know, atomic mass f copper = 63.5 g/mol

Avogadro number, N _A = 6.022 × 10²³ g/mol

To prove, density (d) = 8.92 g cm^-3

We know that, density

d = 8.9657 cm^-1

Hence, the calculated density = the density of copper = = 8.9657 cm^-1

Question 24.

Analysis shows that nickel oxide has the formula Ni_0.98O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?

Answer:

Given formula of nickel oxide = Ni_0.98O_1.00

Therefore, Ni:O = 0.98:1.00

Let the total no. of Ni ion = 98
and, the total no. of O ion = 100

So, Total charge on O^-2 ion = 100 × (-2) = -200

Now, let the number of Ni²⁺ ions = x

Therefore, number of Ni⁺³ = 98- x

Now, since compound is neutral

Therefore, no. of Ni²⁺ ions + no. of Ni³⁺ ions +no. of O^-2 ions = 0

x (+2) + (98 - x) × (+3) + (-200) = 0

+2x + 294 - 3x – 200 =0

x-94 = 0

x = 94

Therefore, the number of Ni²⁺ ions = 94

Number of ni³⁺ ions = 98- 94 = 4

Now, fraction of nickel exists as Ni²⁺ = 0.959

Fraction of nickel exists as Ni ³⁺ ions  = 0.041

Question 25.

What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Answer:

Semiconductors can be defined as the solids having intermediate range of conductivity, i.e. from 10^-6 to 10⁴ ohm^-1m^-1. They are of two types

1) n – type semiconductors

2) p – type semiconductors.

n- type semiconductors:-

Semiconductor formed by doping with electron rich impurities to increase their conductivity are called n – type semiconductors.

For example, silicon and germanium, each has four valance electrons as they belong to group 14^th of the periodic table. Arsenic and phosphorus belong to the 15^th group of the table and they have valance electrons equal to 5. When silicon or germanium is doped with phosphorus or arsenic, four electrons of phosphorus or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon and germanium. Since the electrical conductivity of silicon and phosphorus is increased because of negatively charged particle (electron), thus this is known as n – type semiconductor.

P – type semiconductor :-

Semiconductors formed by doping with electron deficient impurities; to increase their conductivity; are called p- type semiconductors. In p – type semiconductors, conductivity increases because of formation of electron holes.

For example – electrical conductivity of Silicon and Germanium is doped with elements, such as Boron, aluminium or gallium having valance elentrons equal to 3. Three valance electrons present in these electrons delocalized. The place from where one electron is missing is called the electron hole or electron valance. When the silicon or gallium is placed under electrical field, electron from neighbouring atoms fill the electron hole, but in doing so another electron hole is created at the place of movement of electron. in the influence of electrical field electrons move towards positively charged plate through electron hole as appearing the electron hole as positively charged and are moving towards negatively charged plate.

Question 26.

Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Answer:

When cuprous oxide is prepared in laboratory, the ratio of copper to the oxygen in the compound becomes slightly less that 2:1. This happens because some of the Cu⁺ ions are replaced by Cu²⁺ ions. In this process, one Cu²⁺ ion is replaced by two Cu⁺ ions. As two Cu⁺ ions are replaced by one Cu²⁺, this creates a defect because of vacant space, i.e. positive hole. Because of the creation of holes due to this defect, this compound conducts electricity through these positive holes. As semiconductors which are formed by electron deficient impurities are called the p – type semiconductors; thus cuprous oxides so formed in laboratory are p – type semiconductors.

Question 27.

Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Answer:

Given,

Ferric oxide crystalizes in a hexagonal close packed array of oxide ions.

Two out of every three octahedral holes are occupied by ferric ions.

Let the number of oxide ions = x

Since, two out of every three octahedral holes are fitted by ferric ions.

Thus, voids filled by ferric ions

(Fe³⁺) 

Therefore, number of ferric ions

(Fe³⁺) 

Now, the ratio of ferric ions to the oxide ions

i.e. Fe³⁺ : O^2- = 

= Fe³⁺ : O^2- = = 2 : 3

Thus, formula of ferric oxide is Fe₂O_{3 .}

Question 28.

Classify each of the following as being either a p-type or a n-type semiconductor:

Ge doped with In

Answer:

‘Ge’ belongs to group 14^th in th periodic table and ‘In” belongs to the group 13^th . thus, when ‘Ge’ is doped with ‘In’, it makes hole or electron vacancy and acts as p – type conductor.

Question 29.

Classify each of the following as being either a p-type or a n-type semiconductor:

Si doped with B.

Answer:

‘Si’ belongs to group 14^th in the periodic table and ‘B’ belongs to group 13^th. Thus, when ‘Si’ is doped with “B’, it makes hole or electron valance and acts as p – type conductor.

Question 30.

Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Answer:

Given, atomic radius = 0.144 nm

Type of unit cell = face centered

We know that side (a) in fcc 

Thus, a = 0.144 nm

a = 2 × 1.414 × 0.144 = 407 nm.

Thus the side of the given cell = 407 nm.

Question 31.

In terms of band theory, what is the difference?

Between a conductor and an insulator

Answer:

In conductors there is no energy gap between the valance band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity. While in insulators there is no energy gap between the valance band and electrons cannot jump to it i.e. large energy gap prevents the flow of electricity.

Question 32.

In terms of band theory, what is the difference between a conductor and a semiconductor?

Answer:

• In case of conductor, the gap between the valence band and conductor band is very small or overlaps that's why under an applied electric field the conductor (metals) shows conductivity (electron can flow easily).
• In insulator the gap between the filled valence band and the next higher energy unoccupied band (conductions band) is large, an electron cannot jump to it and such substance has very small conductivity and it behaves as an insulator.

Question 33.

Explain the following terms with suitable examples:

Schottky defect

Answer:

When cations and anions both are missing from regular sites, the defect is called Schottky Defect. In Schottky Defects, the number of missing cations is equal to the number of missing anions in order to maintain the electrical neutrality of the ionic compound. Scottky Defect is type of simple vacancy defect and shown by ionic solids having cations and anions; almost similar in size, such as NaCl, KCl, CsCl, etc. AgBr shows both types of defects, i.e. schottky and Frenkel Defects. Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.

Question 34.

Explain the following terms with suitable examples:

Frenkel defect

Answer:

It is a type of vacancy defect. In ionic compounds, some of the ions (usually smaller in size) get dislocated from their original site and create defect. This defect is known as frenkel defect. Since this defect arises because of dislocation of ions, thus it is also known as dislocation defect. As there are a number of cations and anions (which remain equal even because of defect); the density of the substance does nnot increase or decrease. Ionic compounds; having large difference in the size between their cations and anions; show Frenkel Defect, such as ZnS, AgCl, AgBr, AgI, etc. these compounds have similar size of cations compared to anions.

Question 35.

Explain the following terms with suitable examples:

Interstitials

Answer:

Sometimes in the formation of lattice structure some of the atoms or ions occupy vacant interstitial site, and are known as interstitials. These interstitials are generally small size non- metals, such as H, B, C, etc. Defect arises because of interstitials is called the interstitial defect.

Question 36.

Explain the following terms with suitable examples:

F-centres.

Answer:

This is a type of defect and is also called as the metal access defect. These type of defects are seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F – centre and is responsible for showing colour by the compound.

Question 37.

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

(i) What is the length of the side of the unit cell?

(ii) How many unit cells are there in 1.00 cm3 of aluminium?

Answer:

Given, radius of atom = 125 pm

i) for ccp structure, we know that

where, r = radius and a = length of side

125 pm

a = 125 ×  pm

a = 125 × 2 × 1.414 pm

a = 353.5 pm

ii) volume of one unit cell = a³

= (353.5 × 1- ^-3cm)³

= 44192902.36 × 10^-30

= 442 × 10 ^-25 cm ³

Thus, number of unit cell of aluminium in 1 cm³

=

= 2.27 × 10 ²² unit cell.

Question 38.

If NaCl is doped with 10^–3 mol % of SrCl₂, what is the concentration of cation vacancies?

Answer:

We know that two Na ²⁺ ions are replaced by each of the Sr²⁺ ions while SrCl₂ is doped with NaCl. But in this case only one lattice point is occupied by each other of the Sr²⁺ ion and produce one cation vacancy. Here 10^-3 mole of SrCl₂ is doped with 100 moles of NaCl.

Thus, cation vacancies produced by NaCl = 10 ^-3mol

Therefore, 1 mole of NaCl will produce cation vacancies after doping

= = 10^-5 mol

Therefore, total cationic vacancies

= 10^-5 × Avogadro’s number

= 10^{-5 ×} 6.022 × 10 ²³

= 6.023 × 10 ^-18 vacancies.

Question 39.

Explain the following with suitable examples:

Ferromagnetism

Answer:

1. Substances that are attracted strongly with magnetic field are called ferromagnetic substances, such as Cobalt, Nickel, Iron, gadolium, chromium oxide, etc.
2. Ferromagnetic substances are permanently magnetized.
3. Metal ions of ferromagnetic substances are randomly oriented in normal condition and substances do not act as a magnet. 
4. But when metal ions are grouped together in small regions, called domains, each domain acts as a tiny magnet and produces strong magnetic field, in such condition ferromagnetic substances becomes a permanent magnet.

Question 40.

Explain the following with suitable examples:

Paramagnetism

Answer:

The substances which are attracted slightly by magnetic field and fo not retain the magnetic property after removal of magnetic field are called paramagnetic substances. For example, o², Cu²⁺,Fe³⁺,Cr³⁺, magnesium, lithium, etc. substance show paramagnetism because of the presence of unpaired electrons. These unpaired electrons are attracted by magnetic field.

Question 41.

Explain the following with suitable examples:

Ferrimagnetism

Answer:

Substances which are slightly attracted in magnetic field and in which domains are grouped in parallel and anti – parallel direction but in unequal number, are called ferromagnetic substances and this property is called as ferromagnetism.

For example, magnetite (Fe³O⁴), ferrite (MgFe²O⁴), ZnFe²O⁴, etc.

Question 42.

Explain the following with suitable examples:

Antiferromagnetism

Answer:

Substances in which domain structure are similar to ferromagnetic substances but are oriented to oppositely, which cancel the magnetic property are called antiferromagnetic substances and this property is called antiferromagnetism. For example, MnO.

Question 43.

Explain the following with suitable examples:

12-16 and 13-15 group compounds.

Answer:

Compounds belong to 12 – 16 group are formed by combination of elements of 12 and 16 groups. For example – ZnS, Cds, etc.

Compounds belong to 13 – 15 groups are formed by the combination of the elements of 13 and 15groups, for example, InSb, GaAs, etc.

Bonds in these compounds are not perfectly covalent. The ionic bonds in these compounds depend s on the electronegativity of the two elements.