Crystalline Solid State Power Revision From Basics to Advanced For NEET, JEE, IIT-JAM, GATE & CSIR-NET

Chapter – The Crystalline Solid State
Easy Notes for NEET · JEE · IIT-JAM · GATE · CSIR-NET · TGT · PGT · BITSAT

So here's the thing — when you take a handful of salt or a tiny piece of quartz, what you're really holding is billions of atoms sitting in a perfectly repeating pattern. That pattern — and why it exists — is what this whole chapter is about. Don't worry, we're going to walk through it in a way that actually makes sense, not just throw formulas at you.

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7.1 — What Even Is a Crystal?

A crystalline solid has atoms, ions or molecules packed in a neat, repeating 3D arrangement. The smallest chunk of this repeating pattern that still carries all the geometry of the crystal is called the unit cell. Think of it like a single tile that makes up an entire mosaic floor — the tile is the unit cell.

There are 14 possible unit cells (called Bravais lattices) grouped into 7 crystal systems. For competitive exams, you'll mostly deal with cubic, hexagonal, and tetragonal.

Exam Trick: Atoms shared between unit cells are counted as fractions:

• Corner atom → shared by 8 cells → counts as 1/8
• Edge atom → shared by 4 cells → counts as 1/4
• Face atom → shared by 2 cells → counts as 1/2
• Body center → fully inside → counts as 1

Simple Cubic (SC)

Just 8 atoms at the corners, each contributing 1/8. So total atoms = 8 × 1/8 = 1 atom per unit cell. Coordination number (CN) = 6. Packing efficiency = only 52.4% — honestly not very efficient, very few real materials use this.

Body-Centered Cubic (BCC)

Corner atoms (8 × 1/8 = 1) + 1 in the body center = 2 atoms per unit cell. CN = 8. The side of the BCC unit cell = 2.31r (where r = atomic radius). Many metals like Na, K, Cr, W, Fe (at room temp) use this structure.

Face-Centered Cubic (FCC)

Corner atoms (8 × 1/8 = 1) + face atoms (6 × 1/2 = 3) = 4 atoms per unit cell. CN = 12. This is also called cubic close packing (CCP). Packing efficiency = 74% — same as hexagonal close packing.

Solved Example 7.1 — Counting Atoms in FCC

A face-centered cubic unit cell has:
— 8 corner atoms × 1/8 = 1 atom
— 6 face atoms × 1/2 = 3 atoms
Total = 4 atoms

If asked about BCC: 8 × 1/8 + 1 = 2 atoms

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7.2 — Close-Packed Structures (HCP and CCP)

When you throw identical balls into a tray, they naturally settle into the most efficient arrangement — each ball touching 6 others in the same layer. This is a close-packed layer.

Stack these layers and you get two options:

• ABA pattern → Hexagonal Close Packing (HCP) — third layer sits exactly above the first. Example: Mg, Zn, Co, Ti.
• ABC pattern → Cubic Close Packing (CCP) = FCC — third layer is offset from both first and second. Example: Cu, Ag, Au, Ni, Al.

In both HCP and CCP, each atom touches 12 neighbors (CN = 12), and the packing efficiency is 74%. This is the theoretical maximum for packing identical spheres.

Holes in Close-Packed Structures:
For every atom in a close-packed structure, there are:
— 2 tetrahedral holes (CN = 4)
— 1 octahedral hole (CN = 6)

Tetrahedral holes can fit ions of radius 0.225r
Octahedral holes can fit ions of radius 0.414r

Diamond Structure

Carbon in diamond sits in a FCC lattice + 4 extra atoms inside alternate mini-cubes. Every carbon is tetrahedrally bonded to 4 others (CN = 4). Same structure is seen in Si, Ge, and grey Sn (α-tin). It's extremely strong because every bond is a full covalent C–C bond.

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7.3 — Structures of Binary Ionic Compounds

When two types of ions pack together, the smaller cation usually sits inside the holes left by the larger anion lattice. Which hole it picks depends on the relative sizes — and this is where the radius ratio becomes important.

NaCl Structure

Cl⁻ ions form a CCP (FCC) lattice. Na⁺ ions sit in all the octahedral holes. Each Na⁺ is surrounded by 6 Cl⁻ and vice versa — so CN = 6 for both. Many alkali halides have this structure. Unit cell contains 4 formula units of NaCl.

CsCl Structure

Cl⁻ ions form a simple cubic lattice. Cs⁺ sits in the body center (not BCC — the center and corner ions are different!). CN = 8 for both. Only a few salts use this: CsCl, CsBr, CsI, TlCl, TlBr, TlI.

Don't confuse this: CsCl is NOT body-centered cubic. BCC requires the same atom at corner and center. In CsCl, Cs⁺ and Cl⁻ are different — so it's called a simple cubic lattice with interpenetrating cubes.

Zinc Blende (ZnS)

S²⁻ ions in a CCP lattice. Zn²⁺ ions occupy half the tetrahedral holes (alternate ones). CN = 4 for both. Same geometry as diamond but with alternating Zn and S atoms.

Wurtzite (ZnS)

Same as zinc blende but S²⁻ is in an HCP lattice instead of CCP. Zn²⁺ still in half the tetrahedral holes. CN = 4 for both. This form appears at higher temperatures.

Fluorite (CaF₂)

Ca²⁺ in a CCP lattice. F⁻ fills all tetrahedral holes (there are 2 per Ca²⁺, and the formula 1:2 works out). CN = 8 for Ca²⁺, CN = 4 for F⁻. Antifluorite is the reverse — anions in CCP, cations in all tetrahedral holes (example: Na₂O, Li₂O, K₂S).

NiAs Structure

As atoms in an HCP lattice. Ni atoms fill all the octahedral holes. Both Ni and As have CN = 6. Common in transition metal chalcogenides and pnictides (e.g., NiS, FeS, CoTe).

Rutile (TiO₂)

Ti⁴⁺ has CN = 6 (octahedral), O²⁻ has CN = 3. TiO₆ octahedra share edges to form chains. This structure is adopted by MgF₂, ZnF₂, MnO₂, SnO₂ as well.

Structure	Anion Packing	Cation Position	CN (cation:anion)	Examples
NaCl	CCP	All octahedral holes	6:6	NaCl, KCl, MgO, FeO
CsCl	Simple cubic	Body centre	8:8	CsCl, CsBr, TlCl
ZnS (zinc blende)	CCP	Half tetrahedral holes	4:4	ZnS, CdS, GaAs
ZnS (wurtzite)	HCP	Half tetrahedral holes	4:4	ZnS (HT), AgI, BeO
CaF₂ (fluorite)	CCP	All tetrahedral holes	8:4	CaF₂, BaF₂, SrCl₂
Na₂O (antifluorite)	CCP	All tetrahedral holes	4:8	Na₂O, K₂S, Li₂O
NiAs	HCP	All octahedral holes	6:6	NiS, FeS, CoSe
TiO₂ (rutile)	Distorted	Octahedral	6:3	TiO₂, MnO₂, SnO₂

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7.4 — The Radius Ratio Rule

The basic idea: a cation has to fit into the hole in the anion lattice without rattling around. If it's too small, the structure collapses; if it's too big, the anions get pushed apart and a higher CN is needed.

r₊/r₋ Range	Predicted CN	Geometry	Example
< 0.155	2	Linear	—
0.155 – 0.225	3	Triangular planar	—
0.225 – 0.414	4	Tetrahedral	ZnS
0.414 – 0.732	6	Octahedral	NaCl, TiO₂
0.732 – 1.000	8	Cubic	CsCl, CaF₂
= 1.000	12	Cubooctahedral	Metals (HCP/CCP)

Solved Example 7.2 — Radius Ratio for NaCl

r(Na⁺) = 116 pm, r(Cl⁻) = 167 pm (for CN = 6)
r₊/r₋ = 116/167 = 0.695
This falls in the 0.414–0.732 range → predicted CN = 6 ✓ (matches the real NaCl structure)

Solved Example 7.3 — Radius Ratio for ZnS

For CN = 4: r(Zn²⁺) = 74 pm, r(S²⁻) = 170 pm
r₊/r₋ = 74/170 = 0.435 → predicts CN = 6
But experimentally, Zn²⁺ sits in tetrahedral holes (CN = 4) — so the radius ratio prediction is off here. This is because ZnS is significantly covalent, and ions aren't ideal hard spheres. The rule is an approximation only — use it with caution.

GATE/CSIR Tip: The radius ratio rule works well at CN = 8 but is often wrong at CN = 4 (covalent character messes things up). Studies show it's correct only ~2/3 of the time overall. Don't treat it as gospel.

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7.5 — Born–Haber Cycle and Lattice Enthalpy

You can't directly measure how much energy holds an ionic crystal together. So chemists use a clever thermodynamic trick — the Born–Haber cycle — where you break the formation of an ionic compound into separate measurable steps and use Hess's Law.

For LiF, the steps are:

1. Li(s) → Li(g) : sublimation, ΔH = +161 kJ/mol
2. ½F₂(g) → F(g) : bond dissociation, ΔH = +79 kJ/mol
3. Li(g) → Li⁺(g) + e⁻ : ionisation energy, ΔH = +520 kJ/mol
4. F(g) + e⁻ → F⁻(g) : electron affinity, ΔH = –328 kJ/mol
5. Li⁺(g) + F⁻(g) → LiF(s) : lattice enthalpy, ΔH = –1050 kJ/mol

Sum of steps 1–5 = ΔH_f(LiF) = –618 kJ/mol ✓

Madelung Constant (M): When calculating the lattice energy theoretically, you can't just count nearest-neighbor pairs. You need to account for all ion–ion interactions across the whole crystal — that correction factor is M. The Born–Mayer equation is:

U = NMZ₊Z₋/r₀ × (e²/4πε₀) × (1 – ρ/r₀)

where ρ ≈ 30 pm for most simple compounds. Madelung constants: NaCl = 1.748, CsCl = 1.763, CaF₂ = 2.519, ZnS (zinc blende) = 1.638.

Solubility and HSAB

Here's an interesting pattern: salts made of two small ions (like LiF) or two large ions (like CsI) tend to be less soluble than salts with one large + one small ion (like LiI or CsF).

For LiF: huge lattice energy that hydration can't overcome → poorly soluble.
For CsI: both ions large, hydration enthalpies tiny → lattice energy still wins → poorly soluble.
This aligns with Hard-Soft Acid-Base theory: hard–hard (LiF) and soft–soft (CsI) combinations are more stable than mixed ones.

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7.6 — Band Theory and Electronic Structure

When n atoms come together and form molecular orbitals, you get n MOs. For a mole of atoms, that's 6 × 10²³ MOs — so closely spaced they form a continuous band rather than discrete levels.

• Valence band: highest energy band containing electrons
• Conduction band: next empty band above the valence band
• Band gap: energy difference between them

Conductors, Insulators, Semiconductors

Material	Band gap	Behaviour	Examples
Metals/Conductors	None (overlapping bands or half-filled)	High conductivity; decreases with temperature	Cu, Fe, Na
Insulators	Large (>4 eV)	No conductivity	Diamond, NaCl
Intrinsic semiconductors	Small (0.5–3 eV)	Conductivity increases with temperature	Si (1.11 eV), Ge (0.66 eV)

JEE/NEET Trick: For metals, resistance increases with temperature (more vibration = more electron scattering). For semiconductors, resistance decreases with temperature (more electrons promoted to conduction band). This is a very common MCQ point.

Doped Semiconductors

Pure Si is intrinsic. Add a tiny bit of something else and it becomes extrinsic (doped):

• n-type: dope Si with P or As (Group 15, 5 valence electrons). Extra electrons sit just below the conduction band. Conductivity comes from electron movement.
• p-type: dope Si with B or Al (Group 13, 3 valence electrons). Creates electron holes just above the valence band. Conductivity comes from apparent hole movement.

p–n Junctions and Applications

Putting p-type and n-type layers together creates a p–n junction. This is the heart of all modern electronics.

• Diode: forward bias → current flows. Reverse bias → current blocked.
• Solar cell / Photovoltaic cell: light excites electrons across the gap → current flows through external circuit. Used in calculators, solar panels, emergency lighting.
• LED: forward-biased junction where electrons recombine with holes and release energy as light. GaAs → infrared; GaP → red/green; GaN → blue/UV.
• Laser diode: LED with precise dimensions — photon reflection between edges causes stimulated emission. Red laser pointers work this way.

Quantum Dots

When semiconductor particles get smaller than ~10 nm, they stop behaving like bulk materials and start showing quantised energy levels. These are quantum dots. Key properties:

• Smaller particle → larger band gap → higher energy (shorter wavelength) emission
• Larger particle → smaller band gap → lower energy (longer wavelength) emission
• CdS, InP, InAs, ZnSe are commonly studied quantum dot materials
• Applications: biosensors, drug delivery tracking, solar cells, displays, LED lighting

Blinking Problem in Quantum Dots: CdSe QDs sometimes stop emitting light temporarily even while being excited. Two reasons: (1) Auger electron emission (nonradiative decay) and (2) surface electron traps from dangling orbitals. CdSe/ZnSe core-shell gradient structures fix the blinking problem.

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7.7 — Superconductivity

Below a critical temperature (T_c), some metals suddenly lose all electrical resistance. This is superconductivity, discovered by Kamerlingh Onnes in 1911 for mercury at 4.2 K.

BCS Theory (Cooper Pairs)

Bardeen, Cooper, and Schrieffer explained it in 1957. As an electron moves through the lattice, it slightly attracts nearby positive ions, creating a momentary region of higher positive charge density. A second electron (with opposite spin) is attracted to this. The two form a Cooper pair. The lattice vibrations help them move without scattering — zero resistance. Above T_c, thermal motion breaks these pairs apart.

Meissner Effect

Type I superconductors expel all magnetic flux below T_c. This is the Meissner effect — and it's why a magnet can float above a cooled superconductor. Type II superconductors allow partial flux penetration above a lower critical field, making levitation demonstrations more stable (the trapped field lines resist sideways motion).

High-Temperature Superconductors

The real excitement started in 1986 when Bednorz and Müller found La₂CuO₄ doped with Ba superconducted above 30 K. Then in 1987, YBa₂Cu₃O₇ ("1-2-3" compound) was found to have T_c = 93 K — crucially above liquid nitrogen's boiling point (77 K). Current record: HgBa₂Ca₂Cu₃O₈₋δ under pressure at 164 K.

GATE/CSIR Note: High-T_c superconductors are all ceramic cuprates with copper-oxide planes. They're brittle and can't be drawn into wire easily. The structure of YBa₂Cu₃O₇ has square-pyramidal and square-planar Cu coordination environments — removing oxygen gives YBa₂Cu₃O₆ which is NOT superconducting.

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7.8 — Imperfections in Crystals

Real crystals are never perfect. Here are the main defect types you need to know:

Point Defects

• Vacancy (Schottky defect): an atom is simply missing from its lattice site. For ionic crystals, both cation and anion vacancies appear in pairs to maintain charge neutrality (Schottky pair).
• Frenkel defect: an ion moves from its normal site into an interstitial position. Common in AgCl, ZnS, AgBr. The crystal density doesn't change (nothing is actually lost).
• Self-interstitial: an atom moves into a hole not normally occupied. More distorting than a vacancy.
• Substitutional impurity: a foreign atom replaces a host atom (e.g., Ni replacing Cu).

Line Defects (Dislocations)

• Edge dislocation: an extra half-plane of atoms is inserted. The local structure is compressed on one side and stretched on the other.
• Screw dislocation: part of a layer is shifted by one cell dimension. These are growth sites for crystals and form helical paths.

Work hardening: When you hammer a metal, dislocations cluster together, making the remaining structure more uniform and resistant to deformation. Heating the metal redistributes dislocations and restores flexibility (annealing).

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7.9 — Silicates (for IIT-JAM, CSIR-NET, GATE)

The Earth's crust is mostly oxygen + silicon, combined as silicates. The fundamental unit is the SiO₄ tetrahedron. These tetrahedra share corners (never edges or faces) to build increasingly complex structures.

Class	Formula	Structure	Example
Orthosilicate (Nesosilicate)	[SiO₄]⁴⁻	Isolated tetrahedra	Forsterite (Mg₂SiO₄)
Pyrosilicate (Sorosilicate)	[Si₂O₇]⁶⁻	Two tetrahedra sharing one O	Thortveitite
Ring (Cyclosilicate)	[Si₃O₉]⁶⁻, [Si₆O₁₈]¹²⁻	Closed rings	Beryl (emerald)
Single chain (Inosilicate)	[SiO₃]²⁻	Endless chain, 2 corners shared	Pyroxenes
Double chain	[Si₄O₁₁]⁶⁻	Two chains linked	Amphiboles, tremolite
Sheet (Phyllosilicate)	[Si₂O₅]²⁻	3 corners shared, 2D sheet	Talc, mica, kaolinite
Framework (Tectosilicate)	[SiO₂]	All 4 corners shared	Quartz, zeolites

Al³⁺ can replace Si⁴⁺ in any of these structures, but an extra cation must be added to balance the charge (e.g., K⁺ in feldspars, Na⁺ in sodalite).

Zeolites

These are aluminosilicate frameworks with large internal cavities and tunnels. Key points:

• Can absorb water, organic molecules — used as molecular sieves
• Ion exchange capability — water softening (remove Ca²⁺, Mg²⁺)
• Used as heterogeneous catalysts in petroleum cracking
• Methanol-to-hydrocarbons conversion depends on zeolite pore size

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Solved MCQ 1

Q: In a face-centered cubic (FCC) crystal, the number of atoms per unit cell is:

(A) 1   (B) 2   (C) 4   (D) 6

Answer: (C) 4
8 corners × 1/8 = 1; 6 faces × 1/2 = 3; Total = 4.

Solved MCQ 2

Q: In NaCl structure, Cl⁻ ions form a CCP lattice. Na⁺ ions occupy:

(A) All tetrahedral holes   (B) Half tetrahedral holes   (C) All octahedral holes   (D) Half octahedral holes

Answer: (C) All octahedral holes
There is 1 octahedral hole per Cl⁻, and 1 Na⁺ per Cl⁻ — perfect match.

Solved MCQ 3

Q: The Madelung constant for CsCl is:

(A) 1.638   (B) 1.748   (C) 1.763   (D) 2.519

Answer: (C) 1.763

Solved MCQ 4

Q: Which statement about n-type semiconductors is correct?

(A) They are doped with Group 13 elements
(B) Conduction is through positive holes
(C) They are doped with Group 15 elements
(D) They have fewer electrons than the host lattice

Answer: (C) — n-type dopants (P, As, Sb) have 5 valence electrons and donate extra electrons to the conduction band.

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Practice Questions (25–30 Questions)

Based on NEET · JEE · IIT-JAM · CSIR-NET · GATE · TGT · PGT patterns

Q1. The coordination number of Na⁺ in NaCl crystal structure is:

(A) 4    (B) 6    (C) 8    (D) 12

Answer: (B) 6

Q2. Which of the following has the highest packing efficiency?

(A) Simple cubic    (B) BCC    (C) FCC    (D) Simple cubic and BCC are equal

Answer: (C) FCC — 74%, same as HCP. BCC = 68%, SC = 52.4%.

Q3. In CaF₂ (fluorite) structure, the coordination numbers of Ca²⁺ and F⁻ respectively are:

(A) 4 and 8    (B) 6 and 6    (C) 8 and 4    (D) 4 and 4

Answer: (C) 8 and 4 — Ca²⁺ is surrounded by 8 F⁻; each F⁻ by 4 Ca²⁺.

Q4. The number of tetrahedral voids per atom in a close-packed structure is:

(A) 1    (B) 2    (C) 3    (D) 4

Answer: (B) 2

Q5. CsCl has CN = 8 for both ions. Which crystal structure does this correspond to?

(A) FCC    (B) BCC    (C) Simple cubic with interpenetrating cubes    (D) HCP

Answer: (C) — CsCl is NOT BCC. The two ion types are different.

Q6. The radius ratio r₊/r₋ for predicting tetrahedral coordination (CN = 4) lies in the range:

(A) 0.155 – 0.225    (B) 0.225 – 0.414    (C) 0.414 – 0.732    (D) 0.732 – 1.00

Answer: (B) 0.225 – 0.414

Q7. Electrical conductivity of a metal:

(A) Increases with temperature    (B) Decreases with temperature    (C) Is independent of temperature    (D) First increases then decreases

Answer: (B) — More vibrations at higher T → more electron scattering → less conductivity.

Q8. The Schottky defect in ionic crystals involves:

(A) An extra atom in the interstitial position    (B) Missing cation and anion pair    (C) Replacement of one ion by another    (D) Extra electron in the lattice

Answer: (B) — Both a cation and an anion are missing to keep charge balance.

Q9. Which defect does NOT change the density of a crystal?

(A) Schottky    (B) Frenkel    (C) Interstitial    (D) Edge dislocation

Answer: (B) Frenkel — the ion just moves within the crystal, no mass is lost.

Q10. GaAs LEDs emit red light. Adding more phosphorus (making GaP) would:

(A) Reduce the band gap and shift to red    (B) Increase the band gap and shift to higher energy (green/UV)    (C) Have no effect    (D) Only change the intensity

Answer: (B) — GaP has a larger band gap (2.27 eV) than GaAs (1.44 eV), so emission shifts to higher energy (shorter wavelength).

Q11. In wurtzite structure, the anion lattice is:

(A) Simple cubic    (B) CCP    (C) BCC    (D) HCP

Answer: (D) HCP

Q12. For quantum dots, as particle size decreases:

(A) Band gap decreases, emission shifts to longer wavelength
(B) Band gap increases, emission shifts to shorter wavelength
(C) Band gap stays constant
(D) Emission wavelength is independent of size

Answer: (B) — Quantum confinement increases energy level spacing in smaller particles.

Q13. The Madelung constant is largest for which crystal structure?

(A) NaCl    (B) CsCl    (C) ZnS    (D) Al₂O₃ (corundum)

Answer: (D) Al₂O₃ — M = 4.040, due to higher ionic charges.

Q14. In the Born–Haber cycle for NaCl, which step has the largest positive ΔH?

(A) Sublimation of Na    (B) Dissociation of Cl₂    (C) Ionisation energy of Na    (D) Electron affinity of Cl

Answer: (C) Ionisation energy of Na (+496 kJ/mol) is the largest endothermic step.

Q15. Diamond is an electrical insulator but a good thermal conductor. Why?

(A) Large band gap prevents electron flow; rigid covalent bonds allow phonon-based heat conduction
(B) Diamond has free electrons for heat but not electricity
(C) Carbon has too few valence electrons
(D) Diamond is a metallic crystal

Answer: (A)

Q16. Which of these is an antifluorite structure compound?

(A) CaF₂    (B) TiO₂    (C) Na₂O    (D) NiAs

Answer: (C) Na₂O — cation:anion positions are reversed compared to fluorite.

Q17. The critical temperature (Tc) of YBa₂Cu₃O₇ superconductor is approximately:

(A) 23 K    (B) 39 K    (C) 77 K    (D) 93 K

Answer: (D) 93 K — this is above liquid nitrogen temperature (77 K), making it practically significant.

Q18. In BCC structure, the side of the unit cell 'a' in terms of atomic radius 'r' is:

(A) a = 2r    (B) a = 2.31r    (C) a = 2√2 r    (D) a = 4r/√3

Answer: (B) a = 2.31r — derived from body diagonal = 4r, so a = 4r/√3 = 2.31r. (Note: options B and D are the same value, correct answer is 4r/√3 ≈ 2.31r.)

Q19. Which silicate anion corresponds to a single-chain structure?

(A) [SiO₄]⁴⁻    (B) [Si₂O₇]⁶⁻    (C) [SiO₃]²⁻    (D) [Si₂O₅]²⁻

Answer: (C) [SiO₃]²⁻ — infinite chain with 2 corners shared per tetrahedron.

Q20. LiF and CsI are both less soluble than LiI. The best explanation is:

(A) Hard–hard and soft–soft combinations form more stable lattices
(B) Larger ions always form less soluble salts
(C) Smaller ions always form less soluble salts
(D) Solubility depends only on lattice energy

Answer: (A) — HSAB principle: hard–hard (LiF) and soft–soft (CsI) are both more stable than mixed combinations.

Q21. Boron doped into silicon creates:

(A) n-type semiconductor    (B) p-type semiconductor    (C) Intrinsic semiconductor    (D) Superconductor

Answer: (B) p-type — B has 3 valence electrons (less than Si's 4), creating holes.

Q22. In HCP, the atoms in the third layer are:

(A) In new positions (C layer)    (B) Directly above the second layer    (C) Directly above the first layer (A layer)    (D) In random positions

Answer: (C) — HCP = ABA pattern, so layer 3 = A = directly above layer 1.

Q23. The total number of atoms in the HCP unit cell is:

(A) 1    (B) 2    (C) 4    (D) 6

Answer: (B) 2 — (4 × 1/8 from top/bottom) + (4 × 1/8 again from other layer) + 1 inside = 2 total.

Q24. Zeolites function as molecular sieves because:

(A) They are highly acidic
(B) They have cavities and tunnels of specific sizes that selectively allow certain molecules
(C) They are magnetic
(D) They react chemically with unwanted molecules

Answer: (B)

Q25. The Meissner effect refers to:

(A) Emission of light by semiconductors
(B) Formation of Cooper pairs
(C) Expulsion of magnetic flux by a superconductor below Tc
(D) Increase in conductivity at low temperature in metals

Answer: (C)

Q26. NiAs structure is adopted by many MX compounds where M is a transition metal. The coordination of Ni in NiAs is:

(A) Tetrahedral    (B) Square planar    (C) Octahedral    (D) Trigonal prismatic

Answer: (C) — Ni sits in all octahedral holes of the HCP As lattice. As itself is in a trigonal prism of Ni.

Q27. The packing efficiency in a body-centered cubic structure is approximately:

(A) 52.4%    (B) 68.0%    (C) 74.0%    (D) 78.5%

Answer: (B) 68.0%

Q28. Which compound does NOT have the NaCl structure?

(A) MgO    (B) KCl    (C) CsCl    (D) FeO

Answer: (C) CsCl — it has its own distinct structure (simple cubic with interpenetrating lattices).

Q29. For a semiconductor, as temperature increases from 0 K to room temperature:

(A) Resistance increases
(B) Resistance decreases
(C) Resistance stays the same
(D) First decreases then increases

Answer: (B) — More electrons get thermally promoted to conduction band, increasing conductivity (decreasing resistance).

Q30. In the Born–Mayer equation, the value of ρ (repulsion constant) used for simple compounds is approximately:

(A) 10 pm    (B) 20 pm    (C) 30 pm    (D) 50 pm

Answer: (C) 30 pm

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