Free Learning Resources
PW Books
Notes (Class 10-12)
PW Study Materials
Notes (Class 6-9)
Ncert Solutions
Govt Exams
Class 6th to 12th Online Courses
Govt Job Exams Courses
UPSC Coaching
Defence Exam Coaching
Gate Exam Coaching
Other Exams
Know about Physics Wallah
Physics Wallah is an Indian edtech platform that provides accessible & comprehensive learning experiences to students from Class 6th to postgraduate level. We also provide extensive NCERT solutions, sample paper, NEET, JEE Mains, BITSAT previous year papers & more such resources to students. Physics Wallah also caters to over 3.5 million registered students and over 78 lakh+ Youtube subscribers with 4.8 rating on its app.
We Stand Out because
We provide students with intensive courses with India’s qualified & experienced faculties & mentors. PW strives to make the learning experience comprehensive and accessible for students of all sections of society. We believe in empowering every single student who couldn't dream of a good career in engineering and medical field earlier.
Our Key Focus Areas
Physics Wallah's main focus is to make the learning experience as economical as possible for all students. With our affordable courses like Lakshya, Udaan and Arjuna and many others, we have been able to provide a platform for lakhs of aspirants. From providing Chemistry, Maths, Physics formula to giving e-books of eminent authors like RD Sharma, RS Aggarwal and Lakhmir Singh, PW focuses on every single student's need for preparation.
What Makes Us Different
Physics Wallah strives to develop a comprehensive pedagogical structure for students, where they get a state-of-the-art learning experience with study material and resources. Apart from catering students preparing for JEE Mains and NEET, PW also provides study material for each state board like Uttar Pradesh, Bihar, and others
Copyright © 2025 Physicswallah Limited All rights reserved.
Monoclinic unit cell For a monoclinic cell, 
of the number of atoms of N. Therefore, ratio of the number of atoms of M to that of N is M: N 
= 2 : 3 Thus, the formula of the compound is M 2 N 3 .
p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p-type semiconductor. When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated. When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate. 
Question 26. Non-stoichiometric cuprous oxide, Cu 2 O 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 : In the cuprous oxide (Cu 2 O) prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Cu + ions is slightly less than twice the number of O 2− ions. This is because some Cu + ions have been replaced by Cu 2+ ions. Every Cu 2+ ion replaces two Cu + ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor. 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. Solution : Let the number of oxide (O 2− ) ions be x. So, number of octahedral voids = x It is given that two out of every three octahedral holes are occupied by ferric ions. So, number of ferric (Fe 3+ ) ions = 2/3x Therefore, ratio of the number of Fe 3+ ions to the number of O 2− ions, Fe 3+ : O 2− = 2/3x:x 2/3 :1 =2:3 = 2 : 3 Hence, the formula of the ferric oxide is Fe 2 O 3 . Question 28.Classify each of the following as being either a p-type or an n-type semiconductor: (i) Ge doped with In (ii) B doped with Si. Solution : (i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor. (ii) B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be a p-type semiconductor. Question 29. Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell? Solution : For a face-centred unit cell: a = 2 √2r It is given that the atomic radius, r = 0.144 nm So, a = 2 √2 × 0.144 nm = 0.407 nm Hence, length of a side of the cell = 0.407 nm Question 30. In terms of band theory, what is the difference (i) Between a conductor and an insulator (ii) Between a conductor and a semiconductor Solution : (i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band. 
On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band. (ii) In the case of a conductor, the valence band is partially-filled or it overlaps with a higher energy, unoccupied conduction band. So, the electrons can flow easily under an applied electric field. 
On the other hand, the valence band of a semiconductor is filled and there is a small gap between the valence band and the next higher conduction band. Therefore, some electrons can jump from the valence band to the conduction band and conduct electricity. Question 31. Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres Solution : (i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionic solids. In this defect, an equal number of cations and anions are missing to maintain electrical neutrality. It decreases the density of a substance. Significant number of Schottky defects is present in ionic solids. For example, in NaCl, there are approximately 10 6 Schottky pairs per cm 3 at room temperature. Ionic substances containing similar-sized cations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr, etc. 
(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect. 
(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect. 
(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres. These unpaired electrons impart colour to the crystals. For example, when crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres. 
Question 32. 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 cm 3 of aluminium? Solution : (i) For cubic close-packed structure: a = 2 √2r = 2 √2 × 125 pm = 353.55 pm = 354 pm (approximately) (ii) Volume of one unit cell = (354 pm) 3 = 4.4 × 10 7 pm 3 = 4.4 × 10 7 × 10 −30 cm 3 = 4.4 × 10 −23 cm 3 Therefore, number of unit cells in 1.00 cm 3 = 
= 2.27 × 10 22 Question 33. If NaCl is doped with 10 −3 mol % of SrCl 2 , what is the concentration of cation vacancies? Solution : It is given that NaCl is doped with 10 −3 mol% of SrCl 2 . This means that 100 mol of NaCl is doped with 10 −3 mol of SrCl 2 . Therefore, 1 mol of NaCl is doped with 
mol of SrCl 2 = 10 −5 mol of SrCl 2 Cation vacancies produced by one Sr 2+ ion = 1 
Hence, the concentration of cation vacancies created by SrCl 2 is 6.022 × 10 8 per mol of NaCl. Question 34. Explain the following with suitable examples: (i) Ferromagnetism (ii)Paramagnetism (iii)Ferrimagnetism (iv)Antiferromagnetism (v)12-16 and 13-15 group compounds. Solution : (i) Ferromagnetism: The substances that are strongly attracted by a magnetic field are called ferromagnetic substances. Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field. Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium, and CrO 2 . In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains and each domain acts as a tiny magnet. In an un-magnetised piece of a ferromagnetic substance, the domains are randomly-oriented and so, their magnetic moments get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet. 
Schematic alignment of magnetic moments in ferromagnetic substances (ii) Paramagnetism: The substances that are attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances are O 2 , Cu 2t , Fe 3t , and Cr 3t . Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed. To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism. (iii) Ferrimagnetism: The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism. Examples include Fe 3 O 4 (magnetite), ferrites such as MgFe 2 O 4 and ZnFe 2 O 4 . Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances. On heating, these substances become paramagnetic. 
Schematic alignment of magnetic moments in ferrimagnetic substances (iv) Antiferromagnetism: Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositely-oriented. The oppositely-oriented domains cancel out each other’s magnetic moments. 
Schematic alignment of magnetic moments in antiferromagnetic substances (v) 12-16 and 13-15 group compounds: The 12-16 group compounds are prepared by combining group 12 and group 16 elements and the 13-15 group compounds are prepared by combining group 13 and group15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (IrSb), aluminium phosphide (AlP), and gallium arsenide (GaAS) are typical compounds of groups 13-15. GaAs semiconductors have a very fast response time and have revolutionised the designing of semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements. Question 35. Define the term 'amorphous'. Give a few examples of amorphous solids. Solution : Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic. Question 36. What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass? Solution : The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders. Quartz can be converted into glass by heating and then cooling it rapidly. Question 37. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous. (i) Tetra phosphorus decoxide (P 4 O 10 ) (vii) Graphite (ii) Ammonium phosphate (NH4) 3 PO 4 (viii) Brass (iii) SiC (ix) Rb (iv) I 2 (x) LiBr (v) P 4 (xi) Si Solution : Ionic → (ii) Ammonium phosphate (NH 4 ) 3 PO 4 , (x) LiBr Metallic → (viii) Brass, (ix) Rb Molecular → (i) Tetra phosphorus decoxide (P 4 O 10 ), (iv) I 2 , (v) P 4 . Covalent (network) → (iii) SiC, (vii) Graphite, (xi) Si Amorphous → (vi) Plastic Question 38. (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) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number. (ii) The coordination number of atoms (a) in a cubic close-packed structure is 12, and (b) in a body-centred cubic structure is 8 Question 39. 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 : By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined. Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the mass of one atom of the metal and ‘z’ be the number of atoms in the unit cell. Now, density of the unit cell 
⇒ d =zm / a 3----------------------------(1) [Since mass of the unit cell = Number of atoms in the unit cell × mass of one atom] [Volume of the unit cell = (Edge length of the cubic unit cell) 3 ] From equation (i), we have: m = da 3 /z -----------------------------------------(ii) Now, mass of one atom of metal (m) 
Therefore, 
If the edge lengths are different (say a, b and c), then equation (ii) becomes: M = d(abc)NAz (iv) From equations (iii) and (iv), we can determine the atomic mass of the unknown metal. Question 40. '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 : Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point. The melting points of the given substances are: Solid water → 273 K Ethyl alcohol → 158.8 K Diethyl ether → 156.85 K Methane → 89.34 K Now, on observing the values of the melting points, it can be said that among the given substances, the intermolecular force in solid water is the strongest and that in methane is the weakest. Question 41. How will you distinguish between the following pairs of terms: (i) Hexagonal close-packing and cubic close-packing? (ii) Crystal lattice and unit cell? (iii) Tetrahedral void and octahedral void? Solution : (a) Cubic close packing: When a third layer is placed over the second layer in a manner that the octahedral voids are covered by the spheres, a layer different from the first (A) and second (B) is obtained. If we continue packing in this manner we get the cubic close packing. Hexagonal close packing: When the third layer is placed over the second layer in a way that the tetrahedral voids are covered by the spheres, a 3D close packing is produced where spheres in each third or alternate layers are vertically aligned. If we continue packing in this order we get hexagonal close packing. (b) Unit cell: It is the smallest 3D dimensional portion of a complete space lattice, which when repeated over and over again in different directions from the crystal lattice. Crystal lattice : it is a regular orientation of particles of a crystal in a 3D space. (c) Octahedral void : it is a void surrounded by 6 spheres. Tetrahedral void : it is a void surrounded by 4 spheres. 
Question 42. Find the packing efficiency of a metal crystal for : (a) simple cubic (b) body-centered cubic (c) face-centred cubic (with the assumptions that atoms are touching each other). Solution : (a) Simple cubic: In a simple cubic lattice, particles are present only at the corners and they touch each other along the edge. 
Let ,the edge length = a Radius of each particle = r. Thus, a = 2r Volume of spheres = πr 3 (4/3) Volume of a cubic unit cell = a 3 = (2r) 3 = 8r 3 We know that the number of particles per unit cell is 1. Therefore, Packing efficiency = Volume of one particle/Volume of cubic unit cell = [ πr 3 (4/3) ] / 8r 3 = 0.524 or 52.4 % (b) Body-centred cubic: 
From ∆FED, we have: b 2 = 2a 2 b = ( 2a ) 1/2 Again, from ∆AFD, we have : c 2 = a 2 + b 2 => c 2 = a 2 + 2a 2 c 2 = 3a 2 => c = (3a) 1/2 Let the radius of the atom = r. Length of the body diagonal, c = 4r => (3a) 1/2 = 4r => a = 4r/ (3) 1/2 or , r = [ a (3) 1/2 ]/ 4 Volume of the cube, a 3 = [4r/ (3) 1/2 ] 3 A BCC lattice has 2 atoms. So, volume of the occupied cubic lattice = 2πr 3 (4/3) = πr 3 ( 8/3) Therefore, packing efficiency = [ πr 3 ( 8/3) ]/ [ { 4r/(3) 1/2 } 3 ] = 0.68 or 68% (iii) Face-centred cubic: 
Let the edge length of the unit cell = a let the radius of each sphere = r Thus, AC = 4r From the right angled triangle ABC, we have : AC = ( a 2 + a 2 ) 1/2 = a(2) 1/2 
Therefore, 4r = a(2) 1/2 => a = 4r/( 2) 1/2 Thus, Volume of unit cell =a 3 = { 4r/( 2) 1/2 } 3 a 3 = 64r 3 /2(2) 1/2 = 32r 3 / (2) 1/2 No. of unit cell in FCC = 4 Volume of four spheres = 4 × πr 3 (4/3) Thus, packing efficiency = [πr 3 (16/3) ] / [32r 3 / (2) 1/2 ] = 0.74 or 74 % Question 43. Explain (i) The basis of similarities and differences between metallic and ionic crystals. (ii) Ionic solids are hard and brittle. Solution : (i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points. The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity. (ii) The constituent particles of ionic crystals are ions. These ions are held together in three-dimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. This is the reason why ionic crystals are hard and brittle. Question 44. Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic (ii) body-centred cubic (iii) face-centred cubic (with the assumptions that atoms are touching each other). Solution : (i) Simple cubic In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge. 
Let the edge length of the cube be ‘a’ and the radius of each particle be r. So, we can write: a = 2r Now, volume of the cubic unit cell = a 3 = (2r) 3 = 8r 3 We know that the number of particles per unit cell is 1. Therefore, volume of the occupied unit cell 
Hence, packing efficiency 
(ii) Body-centred cubic 
It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged. From ΔFED, we have: 
Again, from ΔAFD, we have: 
Let the radius of the atom be r. Length of the body diagonal, c = 4π 
or, 
Volume of the cube, 
A body-centred cubic lattice contains 2 atoms. So, volume of the occupied cubic lattice 
(iii) Face-centred cubic Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b. 
From ΔABC, we have: 
Let r be the radius of the atom. Now, from the figure, it can be observed that: 
Now, volume of the cube, 
We know that the number of atoms per unit cell is 4. So, volume of the occupied unit cell 
= 74% Question 45. 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. Solution : It is given that the edge length, a = 4.077 × 10 −8 cm Density, d = 10.5 g cm −3 As the lattice is fcc type, the number of atoms per unit cell, z = 4 We also know that, N A = 6.022 × 10 23 mol −1 Using the relation: 
= 107.13 gmol −1 Therefore, atomic mass of silver = 107.13 u Question 46. 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 : It is given that the atoms of Q are present at the corners of the cube. Therefore, number of atoms of Q in one unit cell 
It is also given that the atoms of P are present at the body-centre. Therefore, number of atoms of P in one unit cell = 1 This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1 Hence, the formula of the compound is PQ. The coordination number of both P and Q is 8. Question 47. 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 : It is given that the density of niobium, d = 8.55 g cm −3 Atomic mass, M = 93 gmol −1 As the lattice is bcc type, the number of atoms per unit cell, z = 2 We also know that, N A = 6.022 × 10 23 mol −1 Applying the relation: 
= 3.612 × 10 −23 cm 3 So, a = 3.306 × 10 −8 cm For body-centred cubic unit cell: 
= 1.432 × 10 −8 cm = 14.32 × 10 −9 cm = 14.32 nm Question 48. If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R. Solution : 
A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled ∠POQ = 90 0 Now, applying Pythagoras theorem, we can write: 
Question 49. 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 . Solution : Edge length, a = 3.61 × 10 −8 cm As the lattice is fcc type, the number of atoms per unit cell, z = 4 Atomic mass, M = 63.5 g mol −1 We also know that, NA = 6.022 × 10 23 mol −1 Applying the relation: 
= 8.97 g cm −3 The measured value of density is given as 8.92 g cm −3 . Hence, the calculated density 8.97 g cm −3 is in agreement with its measured value. Question 50. Analysis shows that nickel oxide has the formula Ni 0.98 O 1.00 . What fractions of nickel exist as Ni 2+ and Ni 3+ ions? Solution : The formula of nickel oxide is Ni 0.98 O 1.00 . Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100 Now, total charge on 100 O 2− ions = 100 × (−2) = −200 Let the number of Ni 2+ ions be x. So, the number of Ni 3+ ions is 98 − x. Now, total charge on Ni 2+ ions = x(+2) = +2x And, total charge on Ni 3+ ions = (98 − x)(+3) = 294 − 3x Since, the compound is neutral, we can write: 2x + (294 − 3x) + (−200) = 0 ⇒ −x + 94 = 0 ⇒ x = 94 Therefore, number of Ni 2+ ions = 94 And, number of Ni 3+ ions = 98 − 94 = 4 Hence, fraction of nickel that exists as Ni 2+ = 94 / 98 = 0.959 And, fraction of nickel that exists as Ni 3+ = 4/98 = 0.041 Alternatively, fraction of nickel that exists as Ni 3+ = 1 − 0.959 = 0.041
