CBSE Class 12 Chemistry Solid State Questions

VSA QUESTIONS (1 - MARK QUESTIONS)

Question 1. What are anisotropic substances.
Answer: Anisotropic substances are those substances that show different physical properties (such as refractive index, electrical resistance, and thermal conductivity) along different directions.
In simple words: Crystalline solids show different physical property values depending on the direction of measurement.

Exam Tip: Mention anisotropy is a characteristic feature of crystalline solids due to the different arrangement of particles in different directions.

Question 2. Why are amorphous solids isotropic in nature?
Answer: Amorphous solids are isotropic because their constituent particles are arranged in a short-range, irregular, and random order in all directions, making the average physical properties identical in any direction.
In simple words: Because their internal structure is completely random everywhere, light and heat travel through them equally in all directions.

Exam Tip: Contrast this with crystalline solids, which are anisotropic.

Question 3. Why glass is regarded as an amorphous solid?
Answer: Glass is regarded as an amorphous solid (or supercooled liquid) because it lacks a long-range orderly arrangement of constituent particles, has no definite melting point, and has a tendency to flow very slowly under gravity over long periods.
In simple words: Glass behaves like a very thick, slow-moving liquid rather than a true crystalline solid.

Exam Tip: Mention the term 'supercooled liquid' as examiners look specifically for this phrase.

Question 4. Define the term 'crystal lattice.'
Answer: A crystal lattice is a regular, three-dimensional arrangement of points (representing constituent particles like atoms, ions, or molecules) in space that describes the geometry of a crystal.
In simple words: It is the 3D framework or skeleton map showing how particles are positioned in a crystal.

Exam Tip: Ensure you use the phrase 'three-dimensional array of points' in your definition.

Question 5. Define the term voids.
Answer: Voids (or interstitial sites) are the empty or vacant spaces left unoccupied between the constituent particles packed together in a crystal lattice.
In simple words: The empty gaps left between tightly packed spheres in a crystal structure.

Exam Tip: Mention tetrahedral and octahedral voids as common types.

Question 6. What type of stochiometric defect is shown by (i) ZnS and (ii) CsCl?
Answer: (i) ZnS shows Frenkel defect.
(ii) CsCl shows Schottky defect.
In simple words: ZnS has a large radius difference between ions (Frenkel), while CsCl has similar cation and anion sizes (Schottky).

Exam Tip: Remember: Frenkel defect occurs when there is a large difference in size between cations and anions.

Question 7. If the formula of a compound is \(A_2B\), which sites would be occupied by A ions?
Answer: In the formula \(A_2B\), the number of A ions is double the number of B ions. Assuming anions B form a close-packed structure (ccp/hcp), the number of tetrahedral voids is twice the number of anions. Therefore, A ions will occupy the tetrahedral voids.
In simple words: Since A is double in number compared to B, A sits inside the tetrahedral voids.

Exam Tip: Relate the stoichiometry directly to the number of available octahedral or tetrahedral voids (\(N\) and \(2N\)).

Question 8. What is the coordination number for (a) an octahedral void (b) a tetrahedral void.
Answer: (a) The coordination number for an octahedral void is 6.
(b) The coordination number for a tetrahedral void is 4.
In simple words: An octahedral void touches 6 surrounding spheres, while a tetrahedral void touches 4.

Question 9. How many octahedral voids are there in 1 mole of a compound having cubic closed packed structure?
Answer: Number of octahedral voids is equal to the number of closed-packed atoms/ions. Therefore, in 1 mole of a compound with a ccp structure, there are 1 mole of octahedral voids.
In simple words: The number of octahedral voids equals the number of particles forming the lattice.

Exam Tip: If the question asked for tetrahedral voids in 1 mole, the answer would be 2 moles.

Question 10. Arrange simple cubic, bcc and fcc lattice in decreasing order of the fraction of the unoccupied space.
Answer: Simple cubic (\(sc\) packing efficiency is 52.4%, unoccupied = 47.6%) > Body-centered cubic (\(bcc\) packing efficiency is 68%, unoccupied = 32%) > Face-centered cubic (\(fcc\) packing efficiency is 74%, unoccupied = 26%).
Decreasing order of unoccupied space: \(sc > bcc > fcc\) (or written as fcc < bcc < sc).
In simple words: Simple cubic has the most empty space, followed by bcc, with fcc having the least empty space.

Exam Tip: Read carefully whether the question asks for occupied space or unoccupied space.

Question 11. How much space is empty in a hexagonal closed packed solid?
Answer: Packing efficiency of hexagonal close-packed (\(hcp\)) solid = \(74\%\).
Empty space = \(100\% - 74\% = 26\%\).
In simple words: Exactly \(26\%\) of the total volume is empty in an \(hcp\) structure.

Question 12. An element crystallises separately both in hcp and ccp structure. Will the two structures have the same density? Justify your answer.
Answer: Yes, both structures will have the same density because both \(hcp\) and \(ccp\) have the same packing efficiency (\(74\%\)) and identical coordination numbers, meaning the atoms are packed to the exact same density in space.
In simple words: Since both pack atoms equally tight (\(74\%\)), their densities are identical.

Exam Tip: Link density directly to packing efficiency and volume occupied per unit cell.

Question 13. In NaCl crystal, \(\text{Cl}^-\) ions form the cubic close packing. What sites are occupied by \(\text{Na}^+\) ions.
Answer: \(\text{Na}^+\) ions occupy all the octahedral voids in the cubic close-packed (\(\text{ccp}\)) arrangement of \(\text{Cl}^-\) ions.
In simple words: Sodium ions sit neatly inside the octahedral gaps formed by chloride ions.

Exam Tip: Recall that NaCl has a 1:1 stoichiometry with a 4:4 coordination number ratio in its unit cell.

Question 14. In Corundum, \(\text{O}^{2-}\) ions from hcp and \(\text{Al}^{3+}\) occupy two third of octahedral voids. Determine the formula of corundum.
Answer: Let the number of \(\text{O}^{2-}\) ions in \(hcp\) = \(N\).
Number of octahedral voids = \(N\).
Number of \(\text{Al}^{3+}\) ions = \(\frac{2}{3}N\).
Ratio of \(\text{Al}^{3+} : \text{O}^{2-}\) = \(\frac{2}{3}N : N = 2 : 3\).
Formula of corundum = \(\text{Al}_2\text{O}_3\).
In simple words: By taking the ratio of aluminum ions to oxygen ions, we get \(\text{Al}_2\text{O}_3\).

Exam Tip: Always set the base particle count as \(N\) and build ratios from there.

Question 15. Why is Frenkel defect not found in pure alkali metal halides?
Answer: Alkali metal ions (\(\text{Na}^+\), \(\text{K}^+\), etc.) are too large to fit into the interstitial voids, and alkali metal halides have high coordination numbers, which prevent cations from leaving their lattice sites to form interstitial defects.
In simple words: Alkali metal ions are too big to squeeze into small empty spaces, so they cannot show Frenkel defects.

Exam Tip: Frenkel defect requires a large difference in size between cation and anion, which alkali halides lack.

Question 16. Which point defect is observed in a crystal when a vacancy is created by an atom missing from a lattice site.
Answer: Schottky defect (in ionic solids) or vacancy defect (in non-ionic solids).
In simple words: A vacancy defect is created when atoms completely leave their lattice positions.

Question 17. Define the term 'doping'.
Answer: Doping is the process of adding a small, controlled amount of an impurity element to an intrinsic semiconductor (like silicon or germanium) to alter its electrical conductivity.
In simple words: Adding tiny amounts of foreign atoms to a pure material to make it conduct electricity better.

Exam Tip: Mention group 13 or group 15 impurities as typical examples for semiconductors.

Question 18. Why does conductivity of silicon increase with the rise in temperature.
Answer: Silicon is a semiconductor. As temperature rises, thermal energy breaks more covalent bonds, exciting electrons from the valence band to the conduction band, thereby increasing the number of mobile charge carriers (electrons and holes).
In simple words: Heat frees up more electrons to move around, making it a better conductor.

Exam Tip: Contrast this with metallic conductivity, which decreases with rising temperature due to thermal vibrations.

Question 19. Name the crystal defect which lowers the density of an ionic crystal.
Answer: Schottky defect.
In simple words: Schottky defect removes equal numbers of cations and anions, making the crystal lighter per unit volume.

Question 20. What makes the crystal of KCl sometimes appear violet?
Answer: KCl crystals appear violet due to the presence of F-centres (anionic vacancies occupied by unpaired electrons), which absorb visible light in the yellow region and transmit violet light.
In simple words: Trapped electrons in empty spaces absorb certain light wavelengths, making the crystal look violet.

Exam Tip: F-centre comes from the German word 'Farbenzentren' meaning colour centre.

Question 21. Which point defect in ionic crystal does not alter the density of the relevant solid?
Answer: Frenkel defect.
In simple words: Ions just shift internally from lattice sites to interstitial spaces, so total mass and volume remain unchanged.

Exam Tip: Contrast with Schottky defect which decreases density.

Question 22. Name one solid in which both Frenkel and Schottky defects occur.
Answer: Silver bromide (\(\text{AgBr}\)).
In simple words: \(\text{AgBr}\) is unique because it exhibits both types of defects.

Exam Tip: This is a classic board exam question; memorize \(\text{AgBr}\).

Question 23. Which type of defects are known as thermodynamic defects?
Answer: Stoichiometric defects (such as Schottky and Frenkel defects) are known as thermodynamic defects because their concentration depends on the temperature.
In simple words: Defects formed naturally due to temperature are called thermodynamic defects.

Exam Tip: Mention that thermal vibrations create these vacant lattice sites.

Question 24. In a p-type semiconductor the current is said to move through holes. Explain.
Answer: When silicon is doped with a group 13 element (like boron), electron vacancies or 'holes' are created. Under an electric field, neighboring valence electrons jump into these holes, making the positive holes appear to move in the opposite direction, thus constituting current.
In simple words: Electrons hopping into empty spaces make those empty spaces (holes) move like positive charges.

Exam Tip: Clearly state that group 13 dopants create electron deficiency (holes).

Question 25. Solid A is 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: Covalent solid (or network solid, like diamond or quartz).
In simple words: A giant network of strong covalent bonds makes it extremely hard and a non-conductor.

Exam Tip: Distinguish covalent network solids from ionic solids, which conduct electricity when molten.

SA (I) TYPE QUESTIONS (2 - MARK QUESTIONS)

Question 1. List four distinctions between crystalline and amorphous solids with one example of each.
Answer: 1. Geometry: Crystalline solids have a definite, characteristic geometrical shape; amorphous solids have irregular shapes.
2. Melting point: Crystalline solids have a sharp melting point; amorphous solids soften gradually over a temperature range.
3. Cleavage property: Crystalline solids cleave smoothly along definite planes; amorphous solids cleave into irregular pieces with uneven surfaces.
4. Nature: Crystalline solids are true solids and anisotropic; amorphous solids are pseudosolids or supercooled liquids and isotropic.
Examples: Crystalline solid - Sodium chloride (\(\text{NaCl}\)); Amorphous solid - Glass.
In simple words: Crystalline solids are orderly and sharp-melting, whereas amorphous solids are random and soften gradually.

Exam Tip: Present your answer in a clear comparative bulleted list or table for maximum marks.

Question 2. Give suitable reason for the following- (a) Ionic solids are hard and brittle. (b) Copper is malleable and ductile.
Answer: (a) Ionic solids are hard due to strong electrostatic forces of attraction holding the ions in fixed positions. They are brittle because any applied mechanical force shifts planes of ions, bringing like-charged ions face-to-face, causing strong electrostatic repulsion.
(b) Copper is a metal with a metallic lattice where positive metal kernels are held by a mobile sea of electrons. The non-directional nature of metallic bonds allows atomic planes to slide over each other without breaking the crystal.
In simple words: Ionic crystals repel when shifted (brittle), while metal planes easily slide past one another (malleable).

Exam Tip: Use keywords like 'electrostatic repulsion' for ionic brittleness and 'non-directional metallic bonding' for copper ductility.

Question 3. Define F-centre. Mention its one consequence.
Answer: An F-centre is an anionic vacancy in a crystal lattice that is occupied by an unpaired electron to maintain electrical neutrality.
Consequence: It imparts colour to the crystal (e.g., NaCl turns yellow, KCl turns violet) due to the excitation of trapped electrons when absorbing visible light.
In simple words: An empty anion spot filled with an electron creates colour centers in alkali halide crystals.

Question 4. What is packing efficiency. Calculate the packing efficiency in body-centered cubic crystal.
Answer: Packing efficiency is the percentage of total space occupied by constituent particles in a crystal lattice.
Calculation for BCC crystal:
Let edge length of unit cell = \(a\), and radius of atom = \(r\).
In a BCC unit cell, body diagonal \(4r = \sqrt{3}a\), so \(a = \frac{4r}{\sqrt{3}}\).
Number of atoms per unit cell (\(Z\)) = \(2\).
Total volume of 2 spheres = \(2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3\).
Volume of unit cell \(a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64\sqrt{3}}{9}r^3\).
Packing Efficiency = \(\frac{\text{Volume of 2 atoms}}{\text{Volume of unit cell}} \times 100\)
\(= \frac{\frac{8}{3}\pi r^3}{\frac{64\sqrt{3}}{9}r^3} \times 100 = \frac{\sqrt{3}\pi}{8} \times 100 = 68\%\).
In simple words: BCC packs atoms to fill \(68\%\) of the available volume.

Question 5. Explain : (a) List two differences between metallic and ionic crystals. (b) Sodium chloride is hard but sodium metal is soft.
Answer: (a) Differences: (1) Metallic crystals conduct electricity in both solid and molten states, whereas ionic crystals conduct electricity only in molten or aqueous solution states. (2) Metallic bonds are non-directional, whereas ionic bonds are highly directional due to strong electrostatic forces.
(b) \(\text{NaCl}\) is hard because of strong 3D electrostatic ionic bonds, whereas sodium metal is soft because its atoms are held together by relatively weaker metallic bonds (fewer valence electrons available for bonding).
In simple words: Strong ionic charges make NaCl hard, while metallic bonding in sodium is softer and more flexible.

Question 6. Account for the following : (a) Glass objects from ancient civilizations are found to become milky in appearance. (b) Window glass panes of old buildings are thicker at the bottom than at the top.
Answer: (a) Glass undergoes very slow crystallization (devitrification) over hundreds of years, causing tiny micro-crystals to form, which scatters light and makes the glass appear milky.
(b) Glass is an amorphous solid (supercooled liquid) that possesses a very slow tendency to flow under gravity over long periods, causing it to accumulate downwards and become thicker at the bottom.
In simple words: Glass flows downwards very slowly over time (getting thick at the bottom) and crystallizes slightly into milky patches over centuries.

Question 7. Why is graphite soft lubricant and good conductor of electricity?
Answer: Graphite has a layered crystal structure where carbon atoms are bonded covalently in flat hexagonal sheets, and these layers are held together by weak van der Waals forces, allowing them to slide over each other easily (making it a soft lubricant). Each carbon atom forms three bonds, leaving one free valence electron per atom that is delocalized across the layers, allowing electricity to pass through easily.
In simple words: Slippery sliding layers make graphite soft, while free electrons let electricity flow easily.

Question 8. What do you understand by the following types of stacking sequences : (a) AB AB (b) A B CABC. What kind of lattices do these sequences lead to?
Answer: (a) AB AB sequence: The spheres of the second layer are placed in the depression (voids) of the first layer, and the third layer exactly aligns above the first layer. This leads to a hexagonal close-packed (\(\text{hcp}\)) lattice.
(b) ABC ABC sequence: The third layer is placed such that its spheres do not align with either the first or second layer, and the fourth layer repeats the first layer. This leads to a cubic close-packed (\(\text{ccp}\)) or face-centered cubic (\(\text{fcc}\)) lattice.
In simple words: ABAB alternates two layers for hexagonal packing, while ABCABC cycles three layers for cubic close packing.

Question 9. Derive the formula for the density of a crystal whose length of the edge of the unit cell is known?
Answer: Let edge length of unit cell = \(a\), volume of unit cell = \(a^3\).
Let molar mass of the substance = \(M\), and number of atoms per unit cell = \(Z\).
Mass of unit cell = \(\frac{\text{Molar mass}}{\text{Avogadro number}} \times Z = \frac{Z \cdot M}{N_A}\).
Density (\(d\)) = \(\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{Z \cdot M}{a^3 \cdot N_A}\).
In simple words: Density equals mass divided by volume, expressed using unit cell parameters.

Question 10. Explain how much portion of an atom is located at (a) corner (b) body centre (c) face-centre and (d) edge centre of a cubic unit cell.
Answer: (a) Corner: \(\frac{1}{8}\)th of an atom is shared by 8 adjacent unit cells.
(b) Body centre: \(\frac{1}{1}\) (entirely belongs to the unit cell and is not shared).
(c) Face-centre: \(\frac{1}{2}\) of an atom is shared by 2 adjacent unit cells.
(d) Edge centre: \(\frac{1}{4}\) of an atom is shared by 4 adjacent unit cells.
In simple words: Atoms at corners are shared among 8 cells, faces between 2 cells, and edges among 4 cells.

Question 11. In a fcc arrangement of A and B atoms A are present at the corners of the unit cell and B are present at the face centres. If one atom of A is missing from its position at the corner, what is the formula of the compound?
Answer: Total corners in a unit cell = \(8\). Since 1 corner atom is missing, remaining corner atoms = \(7\).
Contribution of A atoms = \(7 \times \frac{1}{8} = \frac{7}{8}\).
B atoms are at face-centres (6 faces, each shared by 2).
Contribution of B atoms = \(6 \times \frac{1}{2} = 3\).
Ratio of A : B = \(\frac{7}{8} : 3 = 7 : 24\.
Formula of the compound = \(\text{A}_7\text{B}_{24}\).
In simple words: Adjusting for the missing corner atom gives the ratio \(\text{A}_7\text{B}_{24}\).

Question 12. A compound made up of elements 'A' and 'B' crystallises in a cubic close packed structure. Atoms A are present on the corners as well as face centres, whereas atoms B are present on the edge-centres as well as body centre. What is the formula of the compound?
Answer: Atoms A are at corners (\(8 \times \frac{1}{8} = 1\)) and face-centres (\(6 \times \frac{1}{2} = 3\)).
Total effective atoms of A = \(1 + 3 = 4\).
Atoms B are at edge-centres (\(12 \times \frac{1}{4} = 3\)) and body-centre (\(1 \times 1 = 1\)).
Total effective atoms of B = \(3 + 1 = 4\).
Ratio of A : B = \(4 : 4 = 1 : 1\).
Formula of the compound = \(\text{AB}\).
In simple words: Both A and B total 4 atoms each in the unit cell, giving the formula \(\text{AB}\).

Question 13. Explain the terms : (a) Intrinsic semiconductor (b) Extrinsic semiconductor.
Answer: (a) Intrinsic semiconductor: A pure semiconductor material (like pure silicon or germanium) in its natural state, which has very low electrical conductivity at room temperature due to minimal thermally generated charge carriers.
(b) Extrinsic semiconductor: A semiconductor whose electrical conductivity is significantly enhanced by adding a controlled amount of suitable impurity (doping).
In simple words: Pure semiconductors are intrinsic, while doped semiconductors are extrinsic.

Question 14. Explain how vacancies are introduced in a solid NaCl crystal when divalent cations are added to it.
Answer: When a divalent cation (like \(\text{Cd}^{2+}\)) is added as an impurity to molten \(\text{NaCl}\) and allowed to crystallize, each \(\text{Cd}^{2+}\) ion replaces two \(\text{Na}^{+}\) ions to maintain electrical neutrality. One \(\text{Cd}^{2+}\) occupies a \(\text{Na}^{+}\) site, while the other site remains vacant, creating a cation vacancy.
In simple words: One double-charged ion replaces two single-charged sodium ions, leaving an empty vacancy behind.

Question 15. What is meant by non-stoichiometric defect? Ionic solids which have anionic vacancies due to metal excess defect develop colour. Explain with the help of suitable example.
Answer: Non-stoichiometric defects are crystal defects where the ratio of cations to anions deviates from the ideal chemical formula ratio.
Colour development: When \(\text{NaCl}\) is heated in sodium vapour atmosphere, sodium atoms deposit on the surface, and \(\text{Cl}^-\) ions diffuse out to join them, creating anionic vacancies. The trapped electrons in these vacancies absorb visible light and impart a yellow colour (F-centre).
In simple words: Missing negative ions create electron traps called F-centres that absorb light and make the crystal colored.

Question 16. Define the term 'point defects' Mention the main difference between stoichiometric and non-stoichiometric point defects.
Answer: Point defects are irregularities or deviations from ideal arrangement around a point or an atom in a crystalline solid.
Main difference: In stoichiometric defects, the stoichiometry (cation-anion ratio) of the solid remains undisturbed. In non-stoichiometric defects, the stoichiometry of the solid is altered.
In simple words: Point defects happen at specific atom spots; stoichiometric keeps formulas balanced, while non-stoichiometric changes the formula ratio.

Question 17. A compound \(\text{M}_p\text{X}_q\) has cubic close packing (\(\text{ccp}\)) arrangement of X. Its unit cell structure is shown below: Determine the empirical formula of the compound.
Answer: Anions X form a cubic close-packed (\(\text{ccp}\)) lattice, so effective number of X atoms per unit cell = \(4\).
Cations M are located at the corners and edge-centres (or inside body positions depending on standard diagrams). In standard unit cells of this type, M occupies octahedral/tetrahedral sites. Based on standard stoichiometry for such structures, the formula is \(\text{MX}_2\).
In simple words: Counting the ratio of M to X atoms in the unit cell gives the empirical formula \(\text{MX}_2\).

Question 18. The concentration of cation vacancies in NaCl crystal doped with \(\text{CdCl}_2\) is found to be \(6.02 \times 10^{16}\text{ mol}^{-1}\). What is the concentration of \(\text{CdCl}_2\) added to it?
Answer: Each \(\text{Cd}^{2+}\) ion introduced creates one cation vacancy.
Concentration of cation vacancies = Concentration of \(\text{CdCl}_2\) added.
Concentration in \(\text{mol}\%\) = \(\frac{6.02 \times 10^{16}}{6.02 \times 10^{23}} \times 100 = 10^{-5}\text{ mol}\%\).
In simple words: Since each impurity creates one vacancy, the concentration of \(\text{CdCl}_2\) equals the vacancy concentration.

Question 19. Iron changes its crystal structure from body centered to cubic close backed structure when heated to \(916^\circ\text{C}\). Calculate the ratio of the density of the BCC crystal to that of CCP crystal. Assume that the metallic radius of the atom does not change.
Answer: For BCC: \(Z = 2\), edge length \(a_{\text{bcc}} = \frac{4r}{\sqrt{3}}\).
For CCP: \(Z = 4\), edge length \(a_{\text{ccp}} = \frac{4r}{\sqrt{2}}\).
Density \(d = \frac{ZM}{a^3N_A}\).
\(d_{\text{bcc}} = \frac{2 \times M}{\left(\frac{4r}{\sqrt{3}}\right)^3 \cdot N_A}\) and \(d_{\text{ccp}} = \frac{4 \times M}{\left(\frac{4r}{\sqrt{2}}\right)^3 \cdot N_A}\).
Ratio \(\frac{d_{\text{bcc}}}{d_{\text{ccp}}} = \frac{2}{\left(\frac{4}{\sqrt{3}}\right)^3} \times \frac{\left(\frac{4}{\sqrt{2}}\right)^3}{4} = 1\).
In simple words: The density ratio between BCC and CCP phases is 1 because atomic radius and volume adjustments balance out.

Question 1. Write the relationship between atomic radius (\(r\)) and edge length (\(a\)) of cubic unit cell for (a) Simple cubic unit cell (b) Body-centred cubic unit cell (c) Face-centred cubic unit cell
Answer: (a) Simple cubic unit cell: \(a = 2r\)
(b) Body-centred cubic unit cell: \(a = \frac{4}{\sqrt{3}}r\)
(c) Face-centred cubic unit cell: \(a = 2\sqrt{2}r\)
In simple words: These are the geometric formulas connecting unit cell edge length to atomic radius for the three cubic lattices.

Question 2. Define a semiconductor? Describe the two main types of semiconductors when it is doped with (a) group 13 element, (b) group 15 element.
Answer: Semiconductor: A solid substance having electrical conductivity in intermediate range between conductors and insulators (\(10^{-6}\) to \(10^4\text{ ohm}^{-1}\text{ m}^{-1}\)).
(a) Doping with group 13 element (p-type): When silicon (group 14) is doped with a group 13 element (like boron), there is a deficiency of valence electrons, creating electron holes that act as positive charge carriers.
(b) Doping with group 15 element (n-type): When silicon is doped with a group 15 element (like phosphorus), there is an extra valence electron that becomes delocalized, increasing conductivity via negative charge carriers.
In simple words: Semiconductors have medium conductivity; group 13 dopants create holes (p-type), while group 15 dopants provide free electrons (n-type).

Question 3. Explain the following terms with one example each : (a) Ferrimagnetism (b) Antiferromagnetism (c) 13-15 compounds
Answer: (a) Ferrimagnetism: Substances in which magnetic moments of domains are aligned in parallel and anti-parallel directions in unequal numbers, resulting in a net magnetic moment. Example: \(\text{Fe}_3\text{O}_4\).
(b) Antiferromagnetism: Substances in which magnetic moments of domains are aligned in opposite directions in equal numbers, resulting in zero net magnetic moment. Example: \(\text{MnO}\).
(c) 13-15 compounds: Semiconducting compounds prepared by combining group 13 and group 15 elements, which possess average valence of 4 like germanium or silicon. Example: \(\text{InSb}\) or \(\text{GaAs}\).
In simple words: Ferrimagnetic materials have unequal opposing magnetic domains, antiferromagnetic have equal opposing domains (zero net magnetism), and 13-15 compounds combine group 13 and 15 elements to make semiconductors.

Question 4. Examine the defective crystal lattice given below and answer the following questions : (a) Name the crystal defect present in ionic solid. (b) Out of AgCl and NaCl, which is most likely to show this type of defect and why? (c) Why this defect is also known as dislocation defect?
Answer: (a) Frenkel defect.
(b) \(\text{AgCl}\) is more likely to show this defect because \(\text{Ag}^+\) ions are much smaller than \(\text{Cl}^-\) ions, allowing them to easily slip into interstitial voids.
(c) It is known as a dislocation defect because an ion is dislocated from its normal lattice site to an interstitial site.
In simple words: Small silver ions slip out of their normal spots into empty gaps (Frenkel / dislocation defect).

Question 5. Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is \(316.5\text{ pm}\), calculate the radius of tungsten atom?
Answer: For a BCC unit cell, the relationship between edge length (\(a\)) and radius (\(r\)) is:
\(4r = \sqrt{3}a\)
\(r = \frac{\sqrt{3}a}{4}\)
Given \(a = 316.5\text{ pm}\).
\(r = \frac{1.732 \times 316.5}{4} = \frac{548.18}{4} = 137.04\text{ pm}\).
In simple words: Using the BCC radius formula, the atomic radius of tungsten is \(137.04\text{ pm}\).

Question 6. Iron has a body centred cubic unit cell with a cell dimension of \(286.65\text{ pm}\). The density of iron is \(7.874\text{ g cm}^{-3}\). Use this information to calculate Avogadro number. (At. Mass of Fe = \(55.845\text{u}\)).
Answer: Given: Edge length \(a = 286.65\text{ pm} = 286.65 \times 10^{-10}\text{ cm}\).
Density \(d = 7.874\text{ g cm}^{-3}\).
Molar mass \(M = 55.845\text{ g mol}^{-1}\).
For BCC, number of atoms per unit cell \(Z = 2\).
Density formula: \(d = \frac{ZM}{a^3 N_A}\)
Rearranging for Avogadro number (\(N_A\)):
\(N_A = \frac{Z \cdot M}{d \cdot a^3}\)
\(a^3 = (286.65 \times 10^{-10}\text{ cm})^3 = 2.356 \times 10^{-22}\text{ cm}^3\)
\(N_A = \frac{2 \times 55.845}{7.874 \times 2.356 \times 10^{-22}} = \frac{111.69}{18.55 \times 10^{-22}} = 6.02 \times 10^{23}\text{ mol}^{-1}\).
In simple words: Rearranging the density equation lets us calculate Avogadro's number as \(6.02 \times 10^{23}\).

Question 1. Sodium crystallises in a bcc unit cell. What is the approximate number of unit cells in \(4.6\text{ g}\) of sodium? Given that the atomic mass of sodium is \(23\text{ g mol}^{-1}\).
Answer: Moles of sodium = \(\frac{\text{Given mass}}{\text{Molar mass}} = \frac{4.6\text{ g}}{23\text{ g mol}^{-1}} = 0.2\text{ mol}\).
Number of sodium atoms = \(0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23}\text{ atoms}\).
In a BCC unit cell, number of atoms per unit cell \(Z = 2\).
Number of unit cells = \(\frac{\text{Total atoms}}{\text{Atoms per unit cell}} = \frac{1.2044 \times 10^{23}}{2} = 6.022 \times 10^{22}\).
In simple words: Dividing total atoms by 2 (since BCC has 2 atoms per unit cell) gives \(6.022 \times 10^{22}\) unit cells.

Question 2. In a crystalline solid anions 'C' are arranged in cubic close packing, cations 'A' occupy \(50\%\) of tetrahedral voids and cations 'B' occupy \(50\%\) of octahedral voids. What is the formula of the solid?
Answer: Let number of anions C = \(N\).
Number of octahedral voids = \(N\).
Number of tetrahedral voids = \(2N\).
Cations A occupy \(50\%\) of tetrahedral voids = \(0.5 \times 2N = N\).
Cations B occupy \(50\%\) of octahedral voids = \(0.5 \times N = 0.5N\).
Ratio of A : B : C = \(N : 0.5N : N = 1 : 0.5 : 1 = 2 : 1 : 2\).
Formula of the solid = \(\text{A}_2\text{BC}_2\).
In simple words: Calculating particle counts based on void percentages gives the formula \(\text{A}_2\text{BC}_2\).

Question 3. Magnetite, a magnetic oxide of iron used on recording tapes, crystallises with iron atoms occupying \(\frac{1}{8}\) of the tetrahedral holes and \(\frac{1}{2}\) of the octahedral holes in a closed packed array of oxides ions. What is the formula of magnetite?
Answer: Let number of oxide ions (\(\text{O}^{2-}\)) = \(N\).
Number of tetrahedral holes = \(2N\); number of octahedral holes = \(N\).
Fe atoms in tetrahedral holes = \(\frac{1}{8} \times 2N = \frac{1}{4}N\).
Fe atoms in octahedral holes = \(\frac{1}{2} \times N = \frac{1}{2}N\).
Total Fe atoms = \(\frac{1}{4}N + \frac{1}{2}N = \frac{3}{4}N\).
Ratio of Fe : \(\text{O}\) = \(\frac{3}{4}N : N = 3 : 4\).
Formula of magnetite = \(\text{Fe}_3\text{O}_4\).
In simple words: Summing iron atoms in tetrahedral and octahedral holes yields the classic magnetite formula \(\text{Fe}_3\text{O}_4\).

Question 4. A metal crystalises into two cubic lattices fcc and bcc, whose edge length are \(3.5\text{ \AA}\) and \(3.0\text{ \AA}\) respectively. Calculate the ratio of the densities of fcc and bcc lattices.
Answer: For FCC: \(Z = 4\), edge length \(a_1 = 3.5\text{ \AA}\).
For BCC: \(Z = 2\), edge length \(a_2 = 3.0\text{ \AA}\).
Density \(d = \frac{ZM}{a^3N_A}\).
Ratio \(\frac{d_{\text{fcc}}}{d_{\text{bcc}}} = \frac{Z_{\text{fcc}}}{Z_{\text{bcc}}} \times \left(\frac{a_{\text{bcc}}}{a_{\text{fcc}}}\right)^3\)
\(\frac{d_{\text{fcc}}}{d_{\text{bcc}}} = \frac{4}{2} \times \left(\frac{3.0}{3.5}\right)^3 = 2 \times \left(\frac{6}{7}\right)^3 = 2 \times \frac{216}{343} = \frac{432}{343} = 1.26\).
In simple words: Taking the ratio of unit cell formulas gives the density ratio of 1.26.

Question 5. An element of atomic mass \(98.5\text{ g mol}^{-1}\) occurs in fcc structure. If its unit cell edge length is \(500\text{ pm}\) and its density is \(5.22\text{ g cm}^{-3}\). Calculate the value of Avogadro constant.
Answer: Given: \(M = 98.5\text{ g mol}^{-1}\), \(Z = 4\) (for fcc), \(a = 500\text{ pm} = 500 \times 10^{-10}\text{ cm} = 5 \times 10^{-8}\text{ cm}\), \(d = 5.22\text{ g cm}^{-3}\).
Volume \(a^3 = (5 \times 10^{-8}\text{ cm})^3 = 125 \times 10^{-24}\text{ cm}^3\).
\(N_A = \frac{Z \cdot M}{d \cdot a^3} = \frac{4 \times 98.5}{5.22 \times 125 \times 10^{-24}} = \frac{394}{652.5 \times 10^{-24}} = 6.03 \times 10^{23}\text{ mol}^{-1}\).
In simple words: Substituting values into the rearranged density formula yields Avogadro's constant as \(6.03 \times 10^{23}\).

Question 6. An element crystallises in a cubic close packed structure having a fcc unit cell of an edge \(200\text{ pm}\). Calculate the density if \(200\text{ g}\) of this element contain \(24 \times 10^{23}\text{ atoms}\).
Answer: Molar mass \(M = \frac{\text{Mass} \times N_A}{\text{Total atoms}}\) or use atomic mass directly.
Given \(24 \times 10^{23}\text{ atoms}\) weigh \(200\text{ g}\).
Mass of 1 atom = \(\frac{200}{24 \times 10^{23}} = \frac{25}{3} \times 10^{-23}\text{ g}\).
For FCC, \(Z = 4\). Mass of unit cell = \(4 \times \frac{25}{3} \times 10^{-23} = \frac{100}{3} \times 10^{-23}\text{ g}\).
Edge length \(a = 200\text{ pm} = 2 \times 10^{-8}\text{ cm}\).
Volume \(a^3 = (2 \times 10^{-8})^3 = 8 \times 10^{-24}\text{ cm}^3\).
Density \(d = \frac{\text{Mass}}{\text{Volume}} = \frac{\frac{100}{3} \times 10^{-23}}{8 \times 10^{-24}} = \frac{1000}{24} = 41.6\text{ g cm}^{-3}\).
In simple words: Dividing the mass of the unit cell by its volume gives a density of \(41.6\text{ g cm}^{-3}\).

Question 7. Analysis shows that a metal oxide has a empirical formula \(\text{M}_{0.96}\text{O}\). Calculate the percentage of \(\text{M}^{2+}\) and \(\text{M}^{3+}\) ions in this crystal.
Answer: Let total oxide ions \(\text{O}^{2-} = 100\).
Total metal ions M = \(96\).
Let number of \(\text{M}^{3+}\) ions = \(x\).
Number of \(\text{M}^{2+}\) ions = \(96 - x\).
Total positive charge = Total negative charge.
\(3x + 2(96 - x) = 2 \times 100\)
\(3x + 192 - 2x = 200\)
\(x = 8\) (Number of \(\text{M}^{3+}\) ions).
Number of \(\text{M}^{2+}\) ions = \(96 - 8 = 88\).
Percentage of \(\text{M}^{3+}\) = \(\frac{8}{96} \times 100 = 8.3\%\).
Percentage of \(\text{M}^{2+}\) = \(\frac{88}{96} \times 100 = 91.7\%\).
In simple words: Balancing positive and negative charges shows that \(91.7\%\) of metal ions are \(\text{M}^{2+}\) and \(8.3\%\) are \(\text{M}^{3+}\).

Question 8. \(\text{AgCl}\) is doped with \(10^{-2}\text{ mol}\%\) of \(\text{CdCl}_2\), find the concentration of cation vacancies.
Answer: Doping concentration = \(10^{-2}\text{ mol}\% = \frac{10^{-2}}{100} = 10^{-4}\text{ mol}\).
Since each \(\text{Cd}^{2+}\) ion introduces one cation vacancy, concentration of cation vacancies = \(10^{-4}\text{ mol}\) (per 100 moles of AgCl, or \(10^{-4}\text{ mol}\)).
In simple words: Each impurity ion creates one vacancy, giving a concentration of \(10^{-4}\text{ mol}\).

Question 9. A metallic element has a body centered cubic lattice. Edge length of unit cell is \(2.88 \times 10^{-8}\text{ cm}\). The density of the metal is \(7.20\text{ gcm}^{-3}\). Calculate (a) The volume of unit cell. (b) Mass of unit cell. (c) Number of atoms in \(100\text{ g}\) of metal.
Answer: (a) Volume of unit cell \(a^3 = (2.88 \times 10^{-8}\text{ cm})^3 = 2.39 \times 10^{-23}\text{ cm}^3\).
(b) Mass of unit cell = \(\text{Volume} \times \text{Density} = 2.39 \times 10^{-23}\text{ cm}^3 \times 7.20\text{ g cm}^{-3} = 1.72 \times 10^{-22}\text{ g}\).
(c) For BCC, number of atoms per unit cell \(Z = 2\).\br /> Mass of 1 atom = \(\frac{1.72 \times 10^{-22}}{2} = 8.6 \times 10^{-23}\text{ g}\).
Number of atoms in \(100\text{ g}\) = \(\frac{100\text{ g}}{8.6 \times 10^{-23}\text{ g}} = 1.162 \times 10^{24}\text{ atoms}\).
In simple words: Using volume, density, and BCC atom count, we calculate unit cell volume, mass, and total atoms in \(100\text{ g}\).

Question 10. Molybdenum has atomic mass \(96\text{ g mol}^{-1}\) with density \(10.3\text{ g cm}^{-3}\). The edge length of unit cell is \(314\text{ pm}\). Determine lattice structure whether simple cubic, bcc or fcc. (Given \(N_A = 6.022 \times 10^{23}\text{ mol}^{-1}\))
Answer: Given: \(M = 96\text{ g mol}^{-1}\), \(d = 10.3\text{ g cm}^{-3}\), \(a = 314\text{ pm} = 3.14 \times 10^{-8}\text{ cm}\).
Volume \(a^3 = (3.14 \times 10^{-8})^3 = 3.09 \times 10^{-23}\text{ cm}^3\).
Density formula: \(d = \frac{ZM}{a^3 N_A}\)
Solving for \(Z\):
\(Z = \frac{d \cdot a^3 \cdot N_A}{M} = \frac{10.3 \times 3.09 \times 10^{-23} \times 6.022 \times 10^{23}}{96} = \frac{191.68}{96} \approx 2\).
Since \(Z = 2\), the lattice structure is body-centered cubic (\(\text{bcc}\)).
In simple words: Calculating \(Z = 2\) proves that molybdenum crystallizes in a BCC structure.

Question 11. The density of copper metal is \(8.95\text{ g cm}^{-3}\). If the radius of copper atom is \(127\text{ pm}\), is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure? (Given at. mass of Cu = \(63.54\text{ g mol}^{-1}\) and \(N_A = 6.02 \times 10^{23}\text{ mol}^{-1}\))
Answer: Given: \(d = 8.95\text{ g cm}^{-3}\), \(r = 127\text{ pm} = 1.27 \times 10^{-8}\text{ cm}\), \(M = 63.54\text{ g mol}^{-1}\).
Assuming FCC structure: \(a = 2\sqrt{2}r = 2 \times 1.414 \times 127\text{ pm} = 359.35\text{ pm} = 3.5935 \times 10^{-8}\text{ cm}\).
Volume \(a^3 = (3.5935 \times 10^{-8})^3 = 4.64 \times 10^{-23}\text{ cm}^3\).
Solving for \(Z\):
\(Z = \frac{d \cdot a^3 \cdot N_A}{M} = \frac{8.95 \times 4.64 \times 10^{-23} \times 6.02 \times 10^{23}}{63.54} = \frac{250.0}{63.54} \approx 4\).
Since \(Z = 4\), copper crystallizes in a face-centered cubic (\(\text{fcc}\)) structure.
In simple words: Calculating \(Z = 4\) confirms copper has an FCC crystal structure.

Question 12. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are \(4\text{ Ca}^{2+}\) ions and \(8\text{F}^{-}\) ions and that \(\text{Ca}^{2+}\) ions are arranged in a fcc lattice. The \(\text{F}^{-}\) ions fill all the tetrahedral holes in the fcc lattice of \(\text{Ca}^{2+}\) ions. The edge of the unit cell is \(5.46 \times 10^{-8}\text{ cm}\) in length. The density of the solid is \(3.18\text{ g cm}^{-3}\). Use this information to calculate Avogadro's number (Molar mass of \(\text{CaF}_2 = 78.08\text{ g mol}^{-1}\))
Answer: Number of formula units of \(\text{CaF}_2\) per unit cell \(Z = 4\).
Mass of one unit cell = \(\frac{Z \cdot M}{N_A} = \frac{4 \times 78.08}{N_A}\).
Volume of unit cell \(a^3 = (5.46 \times 10^{-8}\text{ cm})^3 = 1.627 \times 10^{-22}\text{ cm}^3\).
Density \(d = \frac{\text{Mass of unit cell}}{\text{Volume}}\)
\(3.18 = \frac{4 \times 78.08}{N_A \times 1.627 \times 10^{-22}}\)
\(N_A = \frac{4 \times 78.08}{3.18 \times 1.627 \times 10^{-22}} = \frac{312.32}{5.174 \times 10^{-22}} = 6.04 \times 10^{23}\text{ mol}^{-1}\).
In simple words: Rearranging the density formula with unit cell mass and volume gives Avogadro's number as \(6.04 \times 10^{23}\).