Chemistry · Volume 1 · Chapter 6

Samacheer Class 12 Chemistry - Solid State

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Q.1 Graphite and diamond are a) Covalent and molecular crystals b) ionic and covalent crystals c) both covalent crystals d) both molecular crystals
Answer: c

Graphite and diamond are allotropes of carbon in which atoms are held by extended covalent bonding (network solids). Hence both are covalent crystals.

Q.2 An ionic compound AxBy crystallizes in fcc type crystal structure with B ions at the centre of each face and A ion occupying corners of the cube. The correct formula of AxBy is a) AB b) AB3 c) A3B d) A8B6
Answer: b

Corners: 8 corners × 1/8 = 1 A per unit cell. Faces: 6 faces × 1/2 = 3 B per unit cell. Formula = A1B3 = AB3.

Q.3 The ratio of close packed atoms to tetrahedral hole in cubic packing is a) 1:1 b) 1:2 c) 2:1 d) 1:4
Answer: b

In close packing (fcc/ccp) number of tetrahedral holes = 2N for N close-packed atoms. So ratio atoms : tetrahedral holes = 1 : 2.

Q.4 Solid CO2 is an example of a) Covalent solid b) metallic solid c) molecular solid d) ionic solid
Answer: c

Solid CO2 (dry ice) consists of discrete CO2 molecules held by weak van der Waals forces — a molecular solid.

Q.5 Assertion: Monoclinic sulphur is an example of monoclinic crystal system. Reason: For a monoclinic system, a ≠ b ≠ c and α = γ = 90° ≠ β. Options: a) Both assertion and reason are true and reason is the correct explanation of assertion. b) Both assertion and reason are true but reason is not the correct explanation of assertion. c) Assertion is true but reason is false. d) Both assertion and reason are false.
Answer: a

Monoclinic sulphur is an example of the monoclinic crystal system. The monoclinic unit cell has unequal edges (a ≠ b ≠ c) with two angles 90° and one different (α = γ = 90°, β ≠ 90°). The reason correctly explains the assertion.

Q.6 In calcium fluoride, having the fluorite structure the coordination number of Ca2+ ion and F- ion are (NEET) a) 4 and 2 b) 6 and 6 c) 8 and 4 d) 4 and 8
Answer: c

In fluorite (CaF2) Ca2+ is surrounded by 8 F- (cubic coordination) and each F- is tetrahedrally coordinated to 4 Ca2+. So CN(Ca2+) = 8, CN(F-) = 4.

Q.7 The number of unit cells in 8 g of an element X (atomic mass 40) which crystallizes in bcc pattern is (NA is the Avogadro number) a) 6.023 × 10^23 b) 6.023 × 10^22 c) 60.23 × 10^23 d) (garbled)
Answer: b

Moles of X = 8/40 = 0.2 mol ⇒ number of atoms = 0.2 N_A. In bcc each unit cell has 2 atoms, so unit cells = (0.2 N_A)/2 = 0.1 N_A = 0.1×6.023×10^23 = 6.023×10^22.

Q.8 In a solid atom M occupies ccp (fcc) lattice and 1/2 of tetrahedral voids are occupied by atom N. Find the formula of solid formed by M and N. a) MN b) M3N c) MN3 d) M3N2
Answer: a

In fcc (ccp) there are 4 M atoms per unit cell and 8 tetrahedral holes. If 1/2 of tetrahedral holes are occupied, N = 8×1/2 = 4. Thus M:N = 4:4 ⇒ formula MN. (Question OCR showed fraction; interpreted as 1/2.)

Q.9 The ionic radii of A+ and B- are 0.98 Å and 1.81 Å. The coordination number of each ion in AB is a) 8 b) 2 c) 6 d) 4
Answer: c

Radius ratio r+/r- = 0.98/1.81 ≈ 0.541. A radius ratio in the range 0.414–0.732 corresponds to octahedral coordination (CN = 6).

Q.10 CsCl has bcc arrangement; its unit cell edge length is 400 pm. Its interatomic distance is a) 400 pm b) 800 pm c) (garbled) d) (garbled)
Answer: d

In CsCl structure the distance between Cs and Cl (corner to body centre) = (√3/2) a. So interatomic distance = (√3/2)×400 pm ≈ 0.866×400 = 346.4 pm. (Option d corresponds to the factor √3/2).

Q.11 A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion will be (choices garbled)
Answer: c

In NaCl (rocksalt) the anions form fcc: a = 2√2 r_–. Also along edge a/2 = r_+ + r_–. Combining: √2 r_– = r_+ + r_– ⇒ r_+ = (√2 − 1)r_– ⇒ r_– = r_+/(√2 − 1). For r_+ = 100 pm, r_– ≈ 100/0.4142 ≈ 241.4 pm (option corresponds to 100/0.414).

Q.12 The vacant space in bcc lattice unit cell is a) 48% b) 23% c) 32% d) 26%
Answer: c

Packing efficiency of bcc = 68.0% occupied, so vacant space = 100% − 68% = 32%.

Q.13 The radius of an atom is 300 pm, if it crystallizes in a face centered cubic lattice, the length of the edge of the unit cell is a) 488.5 pm b) 848.5 pm c) 884.5 pm d) 484.5 pm
Answer: b

For fcc, a = 2√2 r. Thus a = 2√2×300 pm = 600×1.4142 ≈ 848.5 pm.

Q.14 The fraction of total volume occupied by the atoms in a simple cubic is a) π/4 b) π (garbled) c) π (garbled) d) π/6
Answer: d

In simple cubic r = a/2. Fraction occupied = (volume of one atom)/(cell volume) = (4/3)πr^3 / a^3 = (4/3)π (a/2)^3 / a^3 = π/6.

Q.15 The yellow colour in NaCl crystal is due to a) excitation of electrons in F centers b) reflection of light from Cl- ion on the surface c) refraction of light from Na+ ion d) all of the above
Answer: a

Colouring in NaCl (yellow) arises from F-centres: electrons trapped in anion vacancies absorb visible light by electronic excitation.

Q.16 If 'a' stands for the edge length of the cubic system; sc, bcc, and fcc. Then the ratio of radii of spheres in these systems will be respectively (choices garbled)
Answer: c

r_sc = a/2; r_bcc = √3 a/4; r_fcc = √2 a/4. Hence r_sc : r_bcc : r_fcc = 1/2 : √3/4 : √2/4 = 1 : √3/2 : √2/2 (or multiplying by 4/a gives 2 : √3 : √2).

Q.17 If 'a' is the length of the side of the cube, the distance between the body centered atom and one corner atom in the cube will be (choices garbled)
Answer: b

Corner at (0,0,0) and body centre at (a/2,a/2,a/2). Distance = √[(a/2)^2+(a/2)^2+(a/2)^2] = (√3/2) a.

Q.18 Potassium has a bcc structure with nearest neighbour distance 4.52 Å. Its atomic weight is 39. Its density will be a) 915 kg m-3 b) 2142 kg m-3 c) 452 kg m-3 d) 390 kg m-3
Answer: a

Nearest neighbour distance d = 4.52×10^-10 m; for bcc a = 2d/√3 = 5.219×10^-10 m. Volume per cell = a^3 ≈ 1.421×10^-28 m^3. Mass per cell = 2×39/ N_A ≈ 1.295×10^-25 kg. Density = mass/volume ≈ 1.295×10^-25 /1.421×10^-28 ≈ 9.12×10^2 kg m^-3 ≈ 915 kg m^-3.

Q.19 Schottky defect in a crystal is observed when a) unequal number of anions and cations are missing from the lattice b) equal number of cations and anions are missing from the lattice c) an ion leaves its normal site and occupies an interstitial site d) no ion is missing from its lattice.
Answer: b

Schottky defect: vacancies of both cations and anions (equal numbers) are missing from lattice sites, preserving electrical neutrality.

Q.20 The cation leaves its normal position in the crystal and moves to some interstitial position, the defect in the crystal is known as a) Schottky defect b) F center c) Frenkel defect d) non-stoichiometric defect
Answer: c

Frenkel defect: a cation (usually smaller) moves from its lattice site to an interstitial site, creating a vacancy–interstitial pair.

Q.21 Assertion: due to Frenkel defect, density of the crystalline solid decreases. Reason: in Frenkel defect cation and anion leaves the crystal.
Answer: d

Frenkel defect: a cation is displaced from its lattice site to an interstitial site; no ion leaves the crystal and stoichiometry is unchanged. Because no ions are lost, density does not decrease appreciably. Thus both the assertion and the reason are false.

Q.22 The crystal with a metal deficiency defect is a) NaCl b) FeO c) ZnO d) KCl
Answer: b

FeO exhibits metal deficiency (non-stoichiometry) due to cation vacancies and variable oxidation states (Fe2+/Fe3+), giving a formula like Fe1–xO. NaCl and KCl are stoichiometric; ZnO commonly shows metal excess (oxygen vacancies or interstitial electrons).

Q.23 A two dimensional solid pattern formed by two different atoms X and Y is shown below. The black and white squares represent atoms X and Y respectively. The simplest formula for the compound based on the unit cell from the pattern is a) XY8 b) X4Y9 c) XY2 d) XY4
Answer: c

Counting atoms in the repeating unit cell gives 1 X and 2 Y (corner/sharing contributions accounted). Hence simplest formula = XY2.

Answer the following questionsAnswer the following questions24 questions
Q.1Define unit cell.v

A unit cell is the smallest repeating structural unit of a crystal lattice that, by translation in three dimensions, reproduces the entire crystal. It is defined by the edge lengths a, b, c and interaxial angles α, β, γ and contains the arrangement of atoms for the crystal.

Q.2Give any three characteristics of ionic crystals.v
Solution

Additional features: composed of alternate cations and anions, generally soluble in polar solvents, crystalline and have high lattice energies.

Answer:

1. High melting and boiling points due to strong electrostatic forces. 2. Hard but brittle (cleave along planes). 3. Conduct electricity only in molten state or in solution (ions mobile); poor conductors as solids.

Q.3Differentiate crystalline solids and amorphous solids.v
Solution

Crystalline: atoms/molecules arranged periodically; Amorphous: only short-range order and random arrangement over long distances (e.g., glass, plastics).

Answer:

Crystalline solids: long-range order, definite geometric shape, sharp melting points, anisotropic. Amorphous solids: no long-range order, no definite shape, show glass transition rather than sharp melting, isotropic.

Q.4Classify the following solids a. P4 b. Brass c. diamond d. NaCl e. Iodinev
Solution

Molecular solids: held by van der Waals (P4, I2). Metallic: delocalised electrons (brass). Covalent network: strong covalent bonds throughout (diamond). Ionic: electrostatic attraction between ions (NaCl).

Answer:

a. P4 — molecular solid; b. Brass — metallic (alloy) solid; c. Diamond — covalent (network) solid; d. NaCl — ionic solid; e. Iodine (I2) — molecular solid.

Q.5Explain briefly seven types of unit cell.v
Solution

Brief descriptions: - Cubic: a=b=c, α=β=γ=90°. (e.g. NaCl, fcc, bcc) - Tetragonal: a=b≠c, α=β=γ=90°. - Orthorhombic: a≠b≠c, α=β=γ=90°. - Monoclinic: a≠b≠c, α=γ=90°, β≠90°. - Triclinic: a≠b≠c, α≠β≠γ≠90° (no right angles). - Hexagonal: a=b≠c, α=β=90°, γ=120° (e.g. graphite, hcp metals). - Trigonal/rhombohedral: a=b=c, α=β=γ≠90° (a rhombohedron).

Answer:

The seven crystal systems (types of unit cells): cubic, tetragonal, orthorhombic, monoclinic, triclinic, hexagonal, trigonal (rhombohedral).

Q.6Distinguish between hexagonal close packing and cubic close packing.v
Solution

Other differences: symmetry — hcp is hexagonal lattice, ccp is cubic (face-centered). Examples: Mg (hcp), Cu (fcc).

Answer:

hcp stacking sequence = ABAB...; ccp (fcc) stacking sequence = ABCABC.... Both have packing efficiency 74% and coordination number 12, but unit cells differ: hcp has 6 atoms per conventional cell, ccp (fcc) has 4 atoms per unit cell.

Q.7Distinguish tetrahedral and octahedral voids.v
Solution

Locations: tetrahedral voids lie between a triangular face and opposite atom; octahedral voids lie at edge centers and body centers of close-packed arrays.

Answer:

Tetrahedral void: formed by four atoms surrounding a small sphere; coordination number 4; number of tetrahedral voids = 2 per atom in close packing; radius ratio r/R ≈ 0.225 for fit. Octahedral void: formed by six atoms; coordination number 6; number of octahedral voids = 1 per atom; radius ratio r/R ≈ 0.414.

Q.8What are point defects?v

Point defects are localized imperfections involving one or a few atomic or ionic sites in a crystal (zero-dimensional defects). Types include:

  • Vacancies (missing atoms/ions, e.g., Schottky defect)
  • Interstitials (extra atoms/ions in interstitial sites, e.g., Frenkel defect)
  • Substitutional impurities (foreign atoms replacing host atoms)
  • Interstitial impurities (foreign atoms in interstitial sites)

They affect properties such as density, electrical conductivity and diffusion.

Q.9Explain Schottky defect.v

Schottky defect: paired vacancies of both cations and anions in an ionic solid such that stoichiometry is maintained (e.g., NaCl, KCl). It lowers the density of the solid and increases ionic mobility; its concentration depends on temperature and lattice energy.

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

In metal excess two mechanisms: (i) metal ions in interstitial sites or (ii) anion vacancies with trapped electrons (F-centres). Metal deficiency often arises from cation vacancies compensated by higher oxidation states.

Answer:

Metal excess: occurs when there are extra metal atoms or electrons (e.g. Na excess in NaCl gives F-centres — electrons trapped in anion vacancies causing colour). Metal deficiency: occurs when metal sites are vacant or metals have mixed valency causing vacancies (e.g. FeO is metal deficient: some Fe2+ oxidises to Fe3+ and vacancies form).

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

Contribution: 8 corner atoms × 1/8 = 1; 6 face-centred atoms × 1/2 = 3; total = 1 + 3 = 4 atoms.

Answer:

4 atoms per fcc unit cell.

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

Stacking sequences: - AAAA: every layer identical (no offset). - ABAB: second layer fits into holes of first, third layer repeats first. - ABCABC: third layer occupies a different set of holes so that three distinct layers repeat. (All represent arrangements of close-packed spheres with packing efficiency 74%.)

Answer:

AAAA: each layer lies directly above previous (simple hexagonal stacking). ABAB: alternate layers repeat (hexagonal close packing, hcp). ABCABC: three-layer repeat (cubic close packing, ccp or fcc).

Q.13Why ionic crystals are hard and brittle?v
Solution

Thus rigidity gives hardness but directional shift under stress results in fracture rather than plastic deformation.

Answer:

Hardness: strong electrostatic (ionic) bonds hold ions rigidly in place. Brittleness: when a stress shifts ionic planes, like-charged ions can be brought adjacent causing strong repulsive forces and the crystal cleaves along planes.

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

Derivation: bcc has 2 atoms/cell. Let atomic radius = R. Cell edge a = 4R/√3. Volume occupied by atoms = 2 × (4/3)πR^3. Volume of cell = a^3 = (4R/√3)^3. Efficiency = [2(4/3)πR^3]/(4R/√3)^3 = (π√3)/8 ≈ 68.02%.

Answer:

Packing efficiency (bcc) = (π√3)/8 ≈ 0.6802 = 68.02%.

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

In a square close-packed (square lattice) layer each atom/molecule has four nearest neighbours (one at each side), so 2D coordination number = 4.

Answer:

4

Q.16What is meant by the term "coordination number"? What is the coordination number of atoms in a bcc structure?v
Solution

In bcc a central atom touches eight corner atoms (cube corners), so CN = 8.

Answer:

Coordination number: the number of nearest neighbouring atoms/ions surrounding a given atom/ion in a structure. In bcc the coordination number = 8.

Q.17An element has bcc structure with a cell edge of 288 pm. the density of the element is 7.2 gcm-3. how many atoms are present in 208g of the element.v
Solution

Given: a = 288 pm = 2.88 × 10^-8 cm, a^3 = (2.88 × 10^-8)^3 = 2.3888 × 10^-23 cm^3. Density ρ = 7.2 g cm^-3. Mass of one unit cell = ρ·a^3 = 7.2 × 2.3888×10^-23 = 1.7199×10^-22 g. For bcc, n = 2 atoms/cell so mass per atom = 1.7199×10^-22 / 2 = 8.5995×10^-23 g. Number of atoms in 208 g = 208 / 8.5995×10^-23 ≈ 2.417×10^24 ≈ 2.42×10^24 atoms.

Answer:

2.42 × 10^24 atoms (approximately).

Q.18Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm. calculate the edge length of unit cell.v
Solution

In fcc (ccp) the face diagonal = 4r and also = a√2. Hence a = 4r/√2 = 2√2·r. With r = 125 pm,

\(a = 2\sqrt{2}\times125\,\text{pm} = 250\times1.4142\,\text{pm} \approx 353.6\,\text{pm}.\)

Answer:

a = 2√2 r = 353.6 pm

Q.19If NaCl is doped with 10-2 mol percentage of strontium chloride, what is the concentration of cation vacancy?v
Solution

10^-2 mol% = 10^-2% = 0.01% = 10^-4 (fraction). Doping with SrCl2 introduces Sr^{2+} ions; each Sr^{2+} in a Na^+ site requires one Na^+ vacancy for charge neutrality. Thus vacancy concentration = mole fraction of SrCl2 = 10^-4 (i.e. 0.01%).

Answer:

1.0×10^-4 (fraction) = 0.01% vacancies

Q.20KF crystallizes in fcc structure like sodium chloride. calculate the distance between K + and F - in KF.( given : density of KF is 2 48 . g cm -)v
Solution

For NaCl type (fcc) unit cell: 4 formula units per cell. M(KF)=39.10+19.00=58.10 g·mol^{-1}. Density ρ=2.48 g·cm^{-3}.

Cell volume V = 4M/(N_A·ρ).

V = 4×58.10/(6.022×10^{23}×2.48) ≈ 1.557×10^{-22} cm^{3}.

a = V^{1/3} ≈ 5.37×10^{-8} cm = 537 pm.

Nearest‑neighbour (K^+–F^-) distance in NaCl structure = a/2 ≈ 537/2 ≈ 268.5 pm ≈ 269 pm.

Answer:

K–F distance ≈ 269 pm

Q.21An atom crystallizes in fcc crystal lattice and has a density of 10 gcm - with unit cell edge length of 100pm. calculate the number of atoms present in 1 g of crystal.v
Solution

a = 100 pm = 100×10^{-12} m = 1×10^{-8} cm. Volume of unit cell V = a^3 = (1×10^{-8} cm)^3 = 1×10^{-24} cm^3.

Mass of one unit cell = ρV = 10 g·cm^{-3}×1×10^{-24} cm^3 = 1×10^{-23} g.

In fcc, atoms per cell = 4, so mass per atom = 1×10^{-23} g /4 = 2.5×10^{-24} g.

Number of atoms in 1 g = 1 g / (2.5×10^{-24} g) = 4.0×10^{23} atoms.

Answer:

4.0×10^{23} atoms

Q.22Atoms X and Y form bcc crystalline structure. Atom X is present at the corners of the cube and Y is at the centre of the cube. What is the formula of the compound?v
Solution

In bcc: 8 corner atoms ×1/8 = 1 atom X per cell; 1 centre atom Y per cell. Formula per cell = X_1Y_1 → XY.

Answer:

XY

Q.23Sodium metal crystallizes in bcc structure with the edge length of the unit cell 4 3 . × -cm. calculate the radius of sodium atom.v
Solution

Interpreting edge length a = 4.3×10^{-8} cm (typical value). In bcc the body diagonal = a√3 = 4r. Thus r = (√3/4)·a.

r = (1.732/4)×4.3×10^{-8} cm ≈ 0.433×4.3×10^{-8} cm ≈ 1.86×10^{-8} cm = 1.86×10^{-10} m = 186 pm.

Answer:

r ≈ 1.86×10^{-8} cm = 186 pm

Q.24Write a note on Frenkel defect.v
Solution

A Frenkel defect occurs when a smaller ion (usually a cation) leaves its lattice site and occupies an interstitial site, creating a vacancy–interstitial pair. It does not change overall stoichiometry or the number of ions, so mass is unchanged; density is nearly unchanged (small local volume change). Common in ionic crystals with large size difference between ions (e.g. AgCl, ZnS). Frenkel defects increase ionic conductivity (mobile interstitial ions) and create lattice disorder.

Answer:

Frenkel defect: cation vacancy–interstitial pair; stoichiometry unchanged.