NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State – Free PDF

NCERT Solutions for Class 12 Chemistry Chapter 1 – The Solid State – FREE PDF Download

CoolGyan provides you Free PDF download of NCERT Solutions for Class 12 Chemistry Chapter 1 – The Solid State solved by Expert Teachers as per NCERT (CBSE) Book guidelines. All Chapter 1 – The Solid State Exercise Questions with Solutions to help you to revise complete Syllabus and Score More marks.

Revision Notes Class 12 chemistry
NCERT Solutions Class 12 chemistry

Subtopics for Class 12 Chemistry Chapter 1 – The Solid State

1. General Characteristics of Solid State
2. Amorphous and Crystalline Solids
3. Classification of Crystalline Solids
   1. Molecular Solids
   2. Ionic Solids
   3. Metallic Solids
   4. Covalent or Network Solids
4. Crystal Lattices and Unit Cells
   1. Primitive and Centred Unit Cells
5. Number of Atoms in a Unit Cell
   1. Primitive Cubic Unit Cell
   2. Body-Centred Cubic Unit Cell
   3. Face-Centred Cubic Unit Cell
6. Close-Packed Structures
   1. The formula of a Compound and Number of Voids Filled
7. Packing Efficiency
   1. Packing Efficiency in HCP and CCP Structures
   2. The efficiency of Packing in Body-Centred Cubic Structures
   3. Packing Efficiency in Simple Cubic Lattice
8. Calculations Involving Unit Cell Dimensions
9. Imperfections in Solids
   1. Types of Point Defects
   2. Electrical Properties
   3. Conduction of Electricity in Metals
   4. Conduction of Electricity in Semiconductors
   5. Magnetic Properties

Class 12 Chemistry NCERT Solutions (The Solid State) – Important Questions

Class 12 Chemistry Chapter 1 The Solid State INTEXT Questions

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?
Solution:
Since the solid has the same value of refractive index along all directions, it is isotropic in nature. It is because there is no long range order and the arrangement is irregular along all the directions and hence, amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead, it would break into pieces with irregular surfaces.

Question 6.
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.
Solution:
Ionic solids : Potassium sulphate, zinc sulphide
Covalent solids : Graphite, silicon carbide
Molecular solids : Benzene, urea, ammonia, water, argon
Metallic solids : Rubidium, tin

Question 7.
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?
Solution:
Covalent or network solid

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Solution:
Ionic solids conduct electricity in molten state since in the molten state, ionic solids dissociate to give free ions. However, in the solid state, since the ions are not free to move about but remain held together by strong electrostatic forces of attraction, they do not conduct electricity.

Question 9.
What type of solids are electrical conductors, malleable and ductile?
Solution:
Metallic solids conduct electricity in solid state and are malleable and ductile.

Question 10.
Give the significance of a ‘lattice point’.
Solution:
Constituent particles of a crystalline solid are arranged in a definite fashion in a three dimensional crystal lattice. Each position which is occupied by the atoms, ions or molecules in the crystal lattice is called a lattice point.

Question 11.
Name the parameters that characterise a unit cell.
Solution:
The size and shape of a unit cell is determined by the lengths of the edges of the unit cell (a, b and c) which may or may not be mutually perpendicular and by the angles α, β and y between the edges b and c, c and a and a and b respectively.

Question 12.
Distinguish between
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Solution:
(i) (a) Hexagonal unit cell: a = b ≠ c
α = β = 90°, y = 120°
Examples are graphite, ZnO.

(b) Monoclinic unit cell : a ≠ b ≠ c
α = y = 90°, β ≠ 90°
Examples are Na₂SO₄.10H₂O, monoclinic S.

(ii) (a) Face-centred unit cell : Lattice points are at the corners and centre of each face.
No. of atoms per unit cell = 8 ×  + 6 ×  = 4

(b) End-centred unit cell : Lattice points are at the corners and at the centres of two end faces.
No. of atoms per unit cell = 8 ×  + 2 ×  = 2

Question 13.
Explain how much portion of an atom located at (i) corner and (ii) body-centre of a cubic unit cell is part of its neighbouring unit cell.
Solution:

1. A point lying at the corner of a unit cell is shared equally by eight unit cells and therefore, only one-eighth (1/8) of each such point belongs to the given unit cell.
2. A body-centred point belongs entirely to one unit cell since it is not shared by any other unit cell.

Question 14.
What is the two dimensional coordination number of a molecule in square close-packed layer?
Solution:
In such an arrangement, each sphere is in contact with four of its neighbours. Thus, the two dimensional coordination number is four.

Question 15.
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?
Solution:
Number of atoms in 0.5 mol close packing = 0.5 × 6.022 × 10²³ = 3.011 × 10²³
Number of octahedral voids = Number of atoms in the packing = 3.011 × 10²³
Number of tetrahedral voids = 2 × Number of atoms in the packing = 2 × 3.011 × 10²³ = 6.022 × 10²³
Total number of voids = 3.011 × 10²³ + 6.022 × 10²³ = 9.033 × 10²³

Question 16.
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?
Solution:
Suppose no. of atoms of element N = a
Number of tetrahedral voids = 2a
Number of atoms of element M =  × 2a = 
Ratio of M and N =  ; a = 2 : 3
Formula of the compound = M₂N₃

Question 17.
Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body- centred cubic and (iii) hexagonal close-packed lattice?
Solution:

1. Simple cubic – 52%
2. Body-centred cubic – 68%
3. Hexagonal close-packed – 74%

Hence hcp lattice has the highest packing efficiency.

Question 18.
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?
Solution:

Number of atoms of the element present per unit cell = 4. Hence, the cubic unit cell must be face-centred or cubic dose packed (ccp).

Question 19.
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Solution:
On heating a solid, vacancy defect is produced in the crystal. This is because on heating, some lattice sites become vacant. As a result of this defect, the density of the substance decreases because some atoms or ions leave the crystal completely.

Question 20.
What type of stoichiometric defect is shown by:

1. ZnS
2. AgBr

Solution:

1. ZnS – Frenkel defect
2. AgBr – Both Frenkel and Schottky defect

Question 21.
Explain how vacancies are introduced in an ionic solid when a cation of higher valency is added as an impurity in it.
Solution:
When a cation of higher valency is added as an impurity in the ionic solid, some of the sites of the original cations are occupied by the cations of higher valency. Each cation of higher valency replaces two or more original cations and occupies the site of one original cation and the other site(s) remains vacant.

Question 22.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Solution:
A negative ion may be missing from its lattice site, leaving a hole which is occupied by an electron, thereby maintaining the electrical balance. The electrons thus trapped in the anion vacancies are called E-centres because they are responsible for imparling colour to the crystals.
For example, when NaCl is heated in an atmosphere of Na vapour, the excess of Na atoms deposit on the surface of NaCl crystal. The Cl^– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by loss of electrons by sodium atoms to form Na⁺ ions. The released electrons diffuse into the crystal and occupy anionic sites. As a result the crystal now has an excess of sodium. The anionic sites are occupied by unpaired electrons.

They impart yellow colour to the crystals of NaCl. The colour results by the excitation of these electrons when they absorb energy from the visible light falling on crystals.

Question 23.
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?
Solution:
n-type semicondcutor means conduc-tion due to presence of excess of electrons. Therefore, to convert group 14 element into n-type semicondutor, it should be doped with group 15 element.

Question 24.
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Solution:
Ferromagnetic substances make better permanent magnets. This is because the metal ions of a ferromagnetic substance are grouped into small regions called ‘domains’. Each domain acts as a tiny magnet. These domains are randomly oriented. When a ferromagnetic substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field and a strong magnetic field is produced. Such order of domains persists even when the external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

Class 12 Chemistry Chapter 1 The Solid State  – NCERT Exercises

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Solution:
A solid is said to be amorphous if the constituent particles are not arranged in any regular fashion. They may have only short range order. Amorphous solids are generally obtained when the melts are rapidly cooled, e.g., glass, plastics, amorphous silica, etc.

Question 2.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Solution:
Glass is an amorphous solid in which the constituent particles (Si04 tetrahedra) have only a short range order and there is no long range order. In quartz, the constituent particles (SiO₄ tetrahedra) have both short range as well as long range orders. On melting quartz and then cooling it rapidly, it is converted into glass.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

1. Tetraphosphorus decoxide, P₄O₁₀
2. Ammonium phosphate, (NH₄)₃PO₄
3. SiC
4. l₂
5. P₄
6. Plastic
7. Graphite
8. Brass
9. Rb
10. LiBr
11. Si

Solution:
P₄O₁₀ – molecular, (NH₄)₃PO₄ – ionic, SiC – network (covalent), I₂ – molecular, P₄ – molecular, plastic – amorphous, graphite – covalent, brass – metallic, Rb – metallic, LiBr – ionic, Si – covalent

Question 4.
(i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atoms:
(a) in a cubic close-packed structure?
(b) in a body-centred cubic structure?
Solution:
(i) Coordination number is defined as the number of nearest neighbours in a close packing. In ionic crystals, coordination number of an ion in the crystal is the number of oppositely charged ions surrounding that particular ion.

(ii) (a) Coordination number of atoms in a cubic close-packed structure is 12.
(b) Coordination number of atoms in a body-centred cubic structure is 8.

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
Solution:

Question 6.
‘Stability of a crystal is reflected in the magnitude of its melting point’. 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?
Solution:
The melting points of some compounds are given below :
Water = 273 K, Ethyl alcohol = 155.7 K, Diethyl ether = 156.8 K, Methane = 90.5 K

Higher the melting point, stronger are the forces holding the constituent particles together and hence greater is the stability.

The intermolecular forces in water and ethyl alcohol are mainly hydrogen bonding. Higher melting point of water as compared to alcohol shows that hydrogen bonding in ethyl alcohol molecules is not as strong as in water molecules. Diethyl ether is a polar molecule. The intermolecular forces present in them are dipole-dipole attraction. Methane is a non-polar molecule. The only forces present in them are the weak van der Waals forces (London dispersion forces).

Question 7.
Howwill you distinguish between the following pairs of terms?

1. Hexagonal close-packing and cubic close¬packing
2. Crystal lattice and unit cells
3. Tetrahedral void and octahedral void

Solution:
(i) Hexagonal close-packing (hep) : The first layer is formed utilizing maximum space, thus wasting minimum space. In every second row, the particles occupy the depressions (also called voids) between the particles of the row (fig.). In the third row, the particles are vertically aligned with those in the first row giving AB AB AB …….. arrangement. This structure has hexagonal symmetry and is known as hexagonal close-packing (hep) structure. This packing is more efficient and leaves small space which is unoccupied by spheres. In two dimension central sphere is in contact with six other spheres. Only 26% space is free. In three dimension, the coordination number is 12. A single unit cell has 4 atoms.

leaves third layer not resembling with either first or second layer, but fourth layer is similar to first, fifth layer to second, sixth to third
and so on giving pattern ABC ABC ABC ……. This arrangement has cubic symmetry and is known as cubic close-packed (cep) arrangement. This is also called face centred cubic (fee).

(ii) Crystal lattice : A regular arrangement of the constituent particles (i.e., atoms, ions or molecules) of a crystal in three dimensional space is called crystal lattice or space lattice.

Unit cells : The smallest three dimensional portion of a complete space lattice which when repeated over and over again in different directions produces the complete space lattice is called the unit cell.

(iii) The empty spaces left between closed packed spheres are called voids or holes.

(a) Octahedral voids : This void is surrounded by six spheres and formed by a combination of two triangular voids of the first and second layer. There is one octahedral void per atom in a crystal. The radius ratio

(b) Tetrahedral voids : These voids are surrounded by four spheres which lie at the vertices of a regular tetrahedron. There are two tetrahedral voids per atom in a crystal and the radius ratio () is 0.225

Question 8.
How many lattice points are there in one unit cell of each of the following lattice?

1. Face – centred cubic
2. Face – centred tetragonal
3. Body – centred

Solution:
Lattice points in face-centred cubic or face-centred tetragonal = 8 (at corners) + 6 (at face-centres) = 14
Lattice points in body-centred cube = 8 (at corners) + 1 (at body-centre) = 9

Question 9.
Explain:

1. The basis of similarities and differences between metallic and ionic crystals.
2. Ionic solids are hard and brittle.

Solution:
(i) Similarities : Both ionic and metallic crystals have electrostatic forces of attraction. In ionic crystals, these are between the oppositely charged ions. In metals, these are among the valence electrons and the kernels. That is why both have non-directional bonds.

Differences : (a) In ionic crystals, the ions are not free to move. Hence, they cannot conduct electricity in the solid state. They can do so only in the molten state or in aqueous solution. In metals, the valence electrons are free to move. Hence, they can conduct electricity in the solid state.

(b) Ionic bond is strong due to electrostatic forces of attraction. Metallic bond may be weak or strong depending upon the number of valence electrons and the size of the kernels.

(ii) Ionic crystals are hard because they have strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because ionic bond is non-directional.

Question 10.
Calculate the efficiency of packing in case of a metal crystal for

1. simple cubic
2. face-centred cubic (with the assumptions that atoms are touching each other).
3. body-centred cubic

Solution:

Question 11.
Silver crystallises in fee lattice. If edge length of thecell is 4.07 × 10^-8 cm anddensity is 10.5 cm^-3, calculate the atomic mass of silver.
Solution:

Question 12.
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 ?
Solution:
As atoms Q are present at the eight corners of the cube, therefore number of atoms of Q in the unit cell =  × 8 = 1
As atoms P are present at the body-centre, therefore, number of atoms of P in the unit cell = 1
Hence, the formula of the compound is PQ. Coordination number of each of P and Q = 8

Question 13.
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.
Solution:
Given d = 8.55 g cm^-3, M = 93 g mol^-1, Z = 2 (for bcc), N_A = 6.022 × 10²³, r = ?
Using formula

Question 14.
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.
Solution:
R and r are the radii of the octahedral site and atoms respectively, then from Pythagoras theorem we get

AC² = AB² + BC²
(2 R)² = (R + r)² + (R + r)²
or, R = R + r
or, (-1) R = r
∴ r = 0.414 R

Question 15.
Copper crystallises into a fee 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.
Solution:

Question 16.
Analysis shows that nickel oxide has the formula Ni_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni²⁺ = x
Then Ni present as Ni³⁺ = 98 – x
Total charge on x Ni²⁺ and (98 – x) Ni³⁺ should
be equal to charge on 100 O^2- ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94
∴ Fraction of Ni present as Ni²⁺ =  × 100 = 96%
Fraction of Ni present as Ni³⁺ =  × 100 = 4%

Question 17.
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Solution:
Those solids which have intermediate conductivities ranging from 10^-6 to 10⁴ ohm^-1 m^-1 are classified as semiconductors. As the temperature rises, there is a rise in conductivity value because electrons from the valence band jump to conduction band.

(i) n-type semiconductor : When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor : When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-tvpe semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.

Question 18.
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?
Solution:
The ratio less than 2 : 1 in Cu₂0 shows cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. For maintaining electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal dose- packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Solution:
Suppose the number of oxide ions (O^2-) in the packing = 90
∴ Number of octahedral voids = 90
As 2/3^rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present =  × 90 = 60
∴ Ratio of Fe³⁺ : O^2- = 60 : 90 = 2 : 3
Hence, the formula of ferric oxide is Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or n-type semiconductor :

1. Ge doped with In
2. B doped with Si.

Solution:

1. Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
2. B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
Solution:
Given r = 0.144 nm, a = ?
For fcc, a = 2 r = 2 × 1.414 × 0.144 nm = 0.407 nm

Question 22.
In terms of band theory, what is the difference

1. between a conductor and an insulator
2. between a conductor and a semiconductor?

Solution:
In most of the solids and in many insulating solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic solids, it is the ions that are responsible for the conducting behaviour due to their movement.

(i) In metals, conductivity strongly depends upon the number of valence electrons available in an atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other, as to form a band. If this band is partially filled or it overlaps with the higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal behaves as a conductor.

If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator.

(ii) If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump to the conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. Conductors have no forbidden band.

Question 23.
Explain the following terms with suitable examples :

1. Schottky defect
2. Frenkel defect
3. Interstitial defect
4. F-centres.

Solution:
(i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the crystal lattice due to the absence of equal number of cations and anions from their lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect : This defect arises when some of the ions in the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.

(iii) Interstitial defect : When some constituent particles (atoms or molecules) occupy an interstitial site of the crystal, it is said to have interstitial defect. Due to this defect the density of the substance increases.

(iv) F-Centres : These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.

1. What is the length of the side of the unit cell?
2. How many unit cells are there in 1.00 cm³ of aluminium?

Solution:

Question 25.
If NaCI is doped with 10^-3 mol % SrCl₂, what is the concentration of cation vacancies ?
Solution:
Let moles of NaCI = 100
∴ Moles of SrCl₂ doped = 10^-3
Each Sr²⁺ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.
∴ Moles of cation vacancy in 100 moles NaCI = 10^-3
Moles of cation vacancy in one mole
NaCI = 10^-3 × 10^-2 = 10^-5
∴ Number of cation vacancies
= 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸ mol^-1

Question 26.
Explain the following with suitable example:

1. Ferromagnetism
2. Paramagnetism
3. Ferrimagnetism
4. Antiferromagnetism
5. 12-16 and 13-15 group compounds.

Solution:
(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO₂ show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO₂, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are weakly attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. They are magnetised in the same direction as that of the applied field. This property is shown by those substances whose atoms, ions or molecules contain unpaired electrons, e.g., O₂, Cu²⁺, Fe³⁺, etc. These substances, however, lose their magnetism in the absence of the magnetic field.

(iii) Ferrimagnetic substances : Substances which are expected to possess large magnetism on the basis of the unpaired electrons but actually have small net magnetic moment are called ferrimagnetic substances, e.g., Fe₃O₄, ferrites of the formula M²⁺Fe₂O₄ where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic moment.

(iv) Antiferromagnetic substances : Substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is due to the presence of equal number of magnetic moments in the opposite directions

(v) 13-15 group compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc.

12-16 group compounds : Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character.