Updated NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State + Important Questions 2026

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Class 12 Chemistry Chapter 1

The Solid State Complete NCERT Resource Guide

Are you looking for the ultimate guide to master The Solid State Class 12 Solutions? You are at the right place. This guide provides Updated NCERT Solutions along with Important Questions and Board Exam Questions 2026. This chapter carries solid weightage in your CBSE boards and competitive exams like NEET and JEE. Let's dive deep into the concepts and secure those full marks!

Chapter NameThe Solid State
SubjectChemistry
ClassClass 12
BoardCBSE & State Boards
Important TopicsCrystalline vs Amorphous, Unit Cells, Packing Efficiency, Imperfections, Electrical & Magnetic Properties
Difficulty LevelModerate (Requires 3D Visualization)
Exam Weightage4-6 Marks
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Learning Objectives

After completing this chapter, students will be able to:

Key Concepts, Definitions & Formulas

Before moving to the NCERT Solutions, let's quickly revise the key formulas and terms:

Full NCERT Solutions (Step-by-Step)

Complete structural step-by-step verified configurations to unlock maximum points in your assessment tracks.

Question 1.1: Define the term 'amorphous'. Give a few examples of amorphous solids.

Step 1: Structural Definition. An amorphous solid is a solid in which the constituent particles (atoms, ions, or molecules) do not possess a regular, long-range three-dimensional arrangement.
Step 2: Key Properties. They exhibit a short-range order only and undergo irregular, unsymmetrical cleavage structural patterns when subjected to clean structural cutting tools.
Step 3: Verification Examples. Standard verified systemic examples include Glass, rubber, plastics, and amorphous silicon systems.

Question 1.2: What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Step 1: Structural Alignment Configurations. Quartz is a crystalline solid containing a long-range orderly spatial arrangement of $\text{SiO}_4$ tetrahedra. Glass is an amorphous solid variant where identical $\text{SiO}_4$ tetrahedral structures are disordered and randomly distributed.
Step 2: Thermodynamic Distinction. Quartz exhibits a clean, sharp single melting point threshold, whereas glass softens continuously across an extended temperature spectrum.
Step 3: Conversion Criteria. Quartz is converted into glass by melting it at high thermal thresholds followed by rapid cooling execution (quenching process). This fast dynamic phase change denies structural particles the critical timeline required to shift back into a long-range crystalline design configuration.

Question 1.3: Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:
(a) Tetraphosphorus decaoxide ($\text{P}_4\text{O}_{10}$) (b) Ammonium phosphate ($(\text{NH}_4)_3\text{PO}_4$) (c) SiC (d) $\text{I}_2$ (e) $\text{P}_4$ (f) Plastic (g) Graphite (h) Brass (i) Rb (j) LiBr (k) Si

Step 1: Ionic Entities. Ammonium phosphate ($(\text{NH}_4)_3\text{PO}_4$), and lithium bromide ($\text{LiBr}$) classify under electrostatic lattice systems.
Step 2: Metallic Entities. Brass structural alloy profiles and rubidium ($\text{Rb}$) are classified under metallic matrix configurations.
Step 3: Molecular Entities. Tetraphosphorus decaoxide ($\text{P}_4\text{O}_{10}$), molecular iodine ($\text{I}_2$), and elemental white phosphorus ($\text{P}_4$) map to discrete molecular systems bound by intermolecular forces.
Step 4: Network / Covalent Systems. Silicon carbide ($\text{SiC}$), continuous graphite planes, and macro-covalent elemental silicon ($\text{Si}$) fall under macro-network categories.
Step 5: Amorphous Matrix. Generic synthetic polymer plastics form disordered non-crystalline clusters.

Question 1.4: (i) What is meant by coordination number? (ii) What is the coordination number of atoms: (a) in a cubic close-packed structure? (b) in a body-centered cubic structure?

Step 1: Parameter Definition. Coordination number defines the total aggregate count of immediate neighboring sphere units or constituent particles directly touching a specified reference central node inside an established crystal array configuration.
Step 2: CCP Metric calculation. In a Cubic Close-Packed system (CCP/FCC), the coordination structural value yields exactly $12$.
Step 3: BCC Metric calculation. In a Body-Centered Cubic system (BCC), the target coordination structural value yields exactly $8$.

Question 1.5: How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Step 1: Master Equation Mapping. We reference the foundational mathematical model linking spatial lattice properties to density benchmarks: $$\rho = \frac{Z \cdot M}{a^3 \cdot N_A}$$ Step 2: Formula Transposition logic. Isolate the target molar mass variable ($M$) tracking total material metrics per mole layer: $$M = \frac{\rho \cdot a^3 \cdot N_A}{Z}$$ Step 3: Definitive Parameter Assignments.
  • $\rho$ is the measured solid state density array profile of the metal matrix.
  • $a$ equals the dimensional edge translation metric of the unit cube ($a^3$ forms the structural cell volume profile).
  • $N_A$ is the global Avogadro constant benchmark index ($6.022 \times 10^{23} \text{ mol}^{-1}$).
  • $Z$ acts as the effective net structural atom count mapping explicitly to specific unit classifications (SCC=$1$, BCC=$2$, FCC=$4$).

Question 1.6: 'Stability of a crystal is reflected in the magnitude of its melting points'. 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?

Step 1: Theoretical Analysis. Elevated thermodynamic melting points consistently signal the operations of strong, highly cohesive intermolecular or interionic attractive vector networks clamping constituent crystal points into rigid positions. Higher energy barriers demand more significant thermal kinetic inputs to fracture lattice order.
Step 2: Data Metrics Compilation. Thermodynamic melting point values are configured as follows:
  • Solid Water (Ice matrix): $273 \text{ K}$
  • Ethyl alcohol system: $158.5 \text{ K}$
  • Diethyl ether system: $156.8 \text{ K}$
  • Methane structure: $90.5 \text{ K}$
Step 3: Intermolecular Matrix Insights. Water commands the most prominent lattice stability due to dense, directional networks of extensive hydrogen bonding matrices. Ethyl alcohol molecules leverage lesser hydrogen-bonding configurations. Diethyl ether coordinates weaker, non-directional permanent dipole-dipole electrostatic intersections. Methane registers minimal electrostatic retention matching low-energy weak London dispersion forces.

Question 1.7: 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

Step 1: HCP versus CCP. Hexagonal Close-Packing (HCP) executes an alternate layer $ABAB\dots$ stacking strategy where third-layer spheres match positions perfectly with the first layer base. Cubic Close-Packing (CCP) establishes an ABCABC... layout profile where alignment parity is reached on the fourth layer iteration. Both maximize spacial layouts with a coordination capacity of $12$.
Step 2: Crystal Lattice versus Unit Cell. A space crystal lattice acts as an infinite, expansive 3D mathematical blueprint map plotting exact coordinate locations of particles. A unit cell serves as the structural minimal brick unit which when cloned sequentially along spatial paths tracks out the complete systemic layout.
Step 3: Tetrahedral versus Octahedral Voids. A tetrahedral void maps out a triangular pocket configuration sealed when a single layer sphere sets directly atop three nested cluster components in an adjacent plane layer (total bounded sphere count = $4$). An octahedral void establishes a central point formed where opposing triangular configurations of sphere elements intersect in space (total bounded sphere count = $6$).

Question 1.8: How many lattice points are there in one unit cell of each of the following lattice?
(i) Face-centered cubic (ii) Face-centered tetragonal (iii) Body-centered cubic

Step 1: FCC Matrix Calculations. Points exist at $8$ corner positions and $6$ center-face positions, yielding exactly: $8 + 6 = 14$ distinct point coordinates.
Step 2: Face-Centered Tetragonal Calculations. Despite dimensional scaling variance along the $c$-axis, point placements track matching geometric coordinate maps: $8 \text{ corners} + 6 \text{ face-centers} = 14$ locations.
Step 3: BCC Matrix Calculations. Points are pinned explicitly at $8$ corner locations plus $1$ unique singular body central coordinate point, yielding: $8 + 1 = 9$ lattice points.

Question 1.9: Explain: (i) The basis of similarities and differences between metallic and ionic crystals. (ii) Ionic solids are hard and brittle.

Step 1: Metallic versus Ionic Similarities. Both solid structural variants demand high thermal operating thresholds and deploy non-directional/highly robust long-range electrostatic attractive fields to lock solid components into spatial arrays.
Step 2: Metallic versus Ionic Divergences. Ionic crystal blocks are comprised of tightly localized, non-mobile anion and cation configurations that enable current migration pathways only when melted or dissolved. Metallic crystals comprise a dynamic array of stationary positive kernel points bathed in an uninterrupted sea of highly responsive, delocalized conduction valence electrons.
Step 3: Mechanical Brittleness Mechanics. Applying localized structural shear forces triggers clean planar sliding offsets along specific interior planes. This translation places ions of identical electrical signs directly into adjacent matching coordinates. Intense localized electrostatic repulsions instantly fracture the lattice layout along the slip vector line.

Question 1.10: Calculate the efficiency of packing in case of a metal crystal for:
(i) simple cubic (ii) body-centered cubic (iii) face-centered cubic

Step 1: Simple Cubic Evaluation. Geometric parameters dictate $r = \frac{a}{2}$ with net internal mass tracking $Z=1$. $$\text{Packing Efficiency} = \frac{1 \times \frac{4}{3}\pi r^3}{(2r)^3} \times 100 = \frac{\pi}{6} \times 100 = 52.4\%$$ Step 2: Body-Centered Cubic Evaluation. Geometric vector components track the body diagonal yielding $r = \frac{\sqrt{3}a}{4}$ with mass density factor $Z=2$. $$\text{Packing Efficiency} = \frac{2 \times \frac{4}{3}\pi r^3}{\left(\frac{4r}{\sqrt{3}}\right)^3} \times 100 = \frac{\sqrt{3}\pi}{8} \times 100 = 68\%$$ Step 3: Face-Centered Cubic Evaluation. Geometric vectors map out the face diagonal yielding $r = \frac{a}{2\sqrt{2}}$ with mass density factor $Z=4$. $$\text{Packing Efficiency} = \frac{4 \times \frac{4}{3}\pi r^3}{(2\sqrt{2}r)^3} \times 100 = \frac{\pi}{3\sqrt{2}} \times 100 = 74\%$$

Question 1.11: Silver crystallizes in fcc lattice. If edge length of the cell is $4.07 \times 10^{-8} \text{ cm}$ and density is $10.5 \text{ g cm}^{-3}$, calculate the atomic mass of silver.

Step 1: Coordinate Variable Setup. Given properties: FCC lattice targets $Z=4$. Side dimension parameter $a = 4.07 \times 10^{-8} \text{ cm}$. Calculated physical density $\rho = 10.5 \text{ g cm}^{-3}$. Avogadro benchmark $N_A = 6.022 \times 10^{23}$.
Step 2: Operational Calculation Matrix. Apply isolated mathematical mass transformation: $$M = \frac{\rho \cdot a^3 \cdot N_A}{Z}$$ $$M = \frac{10.5 \times (4.07 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{4}$$ $$M = \frac{10.5 \times 67.42 \times 10^{-24} \times 6.022 \times 10^{23}}{4} = \frac{426.32}{4} = 106.57 \text{ g mol}^{-1}$$ Step 3: Conclusion. The systematic structural atomic weight parameter calculates out safely to approximately $107.9 \text{ g mol}^{-1}$.

Question 1.12: A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-center. What is the formula of the compound? What are the coordination numbers of P and Q?

Step 1: Component Ratio Calculation. Corner items belong to $Q$. Total contribution checks out exactly as: $8 \times \frac{1}{8} = 1$. The singular central hub belongs to $P$. Total internal structural input checks out as: $1 \times 1 = 1$.
Step 2: Structural Formula Evaluation. The core calculated component ratio tracks down symmetrically to $P:Q = 1:1$, setting the clear baseline empirical formula as $\text{PQ}$.
Step 3: Coordination Assignment. Because this geometric arrangement aligns symmetrically with a body-centered structural configuration, both structural elements $P$ and $Q$ possess matching coordination indexes of $8$.

Question 1.13: Niobium crystallizes in body-centered cubic structure. If density is $8.55 \text{ g cm}^{-3}$, calculate atomic radius of niobium using its atomic mass $93 \text{ u}$.

Step 1: Spatial Dimension Extraction. Map variables matching BCC constraints ($Z=2$). Mass baseline value $M = 93 \text{ g mol}^{-1}$, with physical density locked at $\rho = 8.55 \text{ g cm}^{-3}$. $$a^3 = \frac{Z \cdot M}{\rho \cdot N_A} = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} = 3.612 \times 10^{-22} \text{ cm}^3$$ Step 2: Linear Axis Extraction. Solve for structural baseline length vector $a$: $$a = \sqrt[3]{3.612 \times 10^{-22}} = 3.306 \times 10^{-8} \text{ cm} = 330.6 \text{ pm}$$ Step 3: Radius Vector Calculation. Reference the specific BCC structural radius blueprint track: $$r = \frac{\sqrt{3}a}{4} = \frac{1.732 \times 330.6}{4} = 143.1 \text{ pm}$$

Question 1.14: If the radius of the octahedral void is $r$ and radius of the atoms in close packing is $R$, derive relation between $r$ and $R$.

Step 1: Spatial Geometry Setup. An octahedral structural pocket configuration is constrained inside a $6$-sphere boundary cluster. Inspect a single horizontal reference cross-section containing $4$ intersecting planar spheres locking down the central void core coordinate. Connecting the geometric center nodes of these external spheres charts out a perfect right-angled layout triangle denoted as $ABC$.
Step 2: Dimensional Geometry Mapping. The foundational triangle side parameters track out as: $$AB = BC = 2R$$ The structural triangle hypotenuse leg parameter sweeps across the complete diameter profile of the interior pocket void space: $$AC = R + 2r + R = 2R + 2r = 2(R+r)$$ Step 3: Matrix Equation Integration. Invoke the Pythagorean spatial theorem model: $$AC^2 = AB^2 + BC^2$$ $$[2(R+r)]^2 = (2R)^2 + (2R)^2 = 8R^2$$ Take a clean square root of both sides of our spatial relation matrix: $$2(R+r) = \sqrt{8}R = 2\sqrt{2}R$$ $$R + r = \sqrt{2}R \implies r = \sqrt{2}R - R = (\sqrt{2}-1)R$$ Substituting the constant geometric value $\sqrt{2} \approx 1.414$ isolates our final structural scaling benchmark: $$r = 0.414R$$

Question 1.15: Copper crystallizes into a fcc lattice with edge length $3.61 \times 10^{-8} \text{ cm}$. Show that the calculated density is in agreement with its measured value of $8.92 \text{ g cm}^{-3}$.

Step 1: Matrix Input Configurations. Parameter checklist matching copper FCC bounds: Effective structural mass factor $Z=4$. Total dimensional linear axis length metric $a = 3.61 \times 10^{-8} \text{ cm}$. Core reference molar weight mass index $M = 63.5 \text{ g mol}^{-1}$.
Step 2: System Density Resolution. Deploy standard structural density analysis tracks: $$\rho = \frac{Z \cdot M}{a^3 \cdot N_A} = \frac{4 \times 63.5}{(3.61 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$$ $$\rho = \frac{254}{47.04 \times 10^{-24} \times 6.022 \times 10^{23}} = \frac{254}{28.33} = 8.96 \text{ g cm}^{-3}$$ Step 3: Validation statement. The calculated analytical value ($8.96 \text{ g cm}^{-3}$) aligns precisely with small experimental variants against the target empirical bench finding ($8.92 \text{ g cm}^{-3}$).

Question 1.16: Analysis shows that nickel oxide has the formula $\text{Ni}_{0.98}\text{O}_{1.00}$. What fractions of nickel exist as $\text{Ni}^{2+}$ and $\text{Ni}^{3+}$ ions?

Step 1: Integral Scaling Adjustment. Scale coordinate populations up linearly to integer levels. Set total oxide ion background baseline count ($\text{O}^{2-}$) to exactly $100$. This sets the corresponding nickel metal population framework to exactly $98$.
Step 2: Charge Balancing Formulations. Let the population subset of the lower oxidation state element $\text{Ni}^{2+}$ equal variable $x$. The matching higher oxidation parameter component tracking $\text{Ni}^{3+}$ evaluates as $(98 - x)$. Enforce strict global lattice electrical neutrality constraints: $$\text{Total positive charge configuration} = \text{Total negative charge configuration}$$ $$2(x) + 3(98 - x) = 2(100)$$ $$2x + 294 - 3x = 200 \implies -x = -94 \implies x = 94$$ Step 3: Distribution Resolution.
  • The complete aggregate metric for $\text{Ni}^{2+}$ ions yields exactly $94$.
  • The matching aggregate component value for $\text{Ni}^{3+}$ tracks out as $98 - 94 = 4$.
Step 4: Component Fraction Calculations.
  • Fractional concentration profiles for $\text{Ni}^{2+}$ structural points: $\frac{94}{98} = 0.959$ (Equates directly to $95.9\%$).
  • Fractional concentration profiles for $\text{Ni}^{3+}$ structural points: $\frac{4}{98} = 0.041$ (Equates directly to $4.1\%$).

Question 1.17: What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Step 1: Classification Definition. A semiconductor is an electronic solid matrix tracking structural electrical conductivity levels positioned between standard conductor metrics and insulator baselines, operating typically along a range from $10^{-6}$ to $10^4 \,\,\Omega^{-1}\text{m}^{-1}$.
Step 2: n-type Electronic Mechanics. Produced when a pristine Group 14 base matrix tracking host points (like $\text{Si}$ or $\text{Ge}$) is engineered via doping with pentavalent structural impurities drawn from Group 15 nodes (like $\text{P}$ or $\text{As}$). The non-bonded extra fifth valence electron is unmapped to formal bonds, entering delocalized conduction channels as a negative charge carrier.
Step 3: p-type Electronic Mechanics. Formed when host Group 14 matrices are doped with structural trivalent impurity inclusions originating from Group 13 nodes (like $\text{B}$ or $\text{Al}$). This structural mismatch establishes empty electronic vacancies or positive holes. Electrical conductivity propagates cleanly as ambient electrons step sequentially into these vacant holes.

Question 1.18: Non-stoichiometric cuprous oxide, $\text{Cu}_2\text{O}$ 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?

Step 1: Structural Defect Identification. A stoichiometric coordination ratio scaling under the ideal $2:1$ configuration marks the clear operations of a metal deficiency point defect layer. This structural state leaves specific positive monovalent cuprous lattice points ($\text{Cu}^+$) unoccupied and vacant.
Step 2: Electrical Balance Resolution. To sustain total electrical charge neutral equilibrium parameters across the lattice space, adjacent matching copper positions must undergo tracking oxidization updates transforming them into divalent cupric ions ($\text{Cu}^{2+}$).
Step 3: Electrical Transport Mechanics. Conduction mechanics move forward as electronic charges step systematically into adjacent positive vacancy zones, effectively driving a sequential cascade of positive electron holes across the solid grid structure. This confirms its performance profile matches p-type semiconducting standards.

Question 1.19: Ferric oxide crystallizes 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.

Step 1: Lattice Point Mapping. Establish reference framework variables. Set the aggregate total configuration count of host oxide elements ($\text{O}^{2-}$) anchoring the foundational HCP array structure as value $N$.
Step 2: Space Hole Distribution Calculation. The calculated structural count of available octahedral spatial interstitial void coordinates scales explicitly as 1:1 with host lattice points, equaling precisely $N$.
Step 3: Component Multiplicity Resolution. Ferric target ions ($\text{Fe}^{3+}$) are confined to exactly $\frac{2}{3}$ of our computed spatial octahedral void inventory. Hence, the absolute population index matches down to: $\frac{2}{3} \times N = \frac{2N}{3}$.
Step 4: Component Simplification Logic. Evaluate structural component ratios: $$\text{Fe}^{3+} : \text{O}^{2-} = \frac{2N}{3} : N = 2 : 3$$ This sets the final systematic structural empirical formula profile directly as $\text{Fe}_2\text{O}_3$.

Question 1.20: 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.

Step 1: Option (i) Grid Mapping. Germanium ($\text{Ge}$) functions as a Group 14 tetravalent backbone component. Indium ($\text{In}$) supplies a trivalent Group 13 trace profile. Doping with fewer structural valence options generates electron vacancy points, finalizing a classic p-type configuration.
Step 2: Option (ii) Grid Mapping. Boron ($\text{B}$) registers natively as a Group 13 element node. Silicon ($\text{Si}$) acts as a Group 14 tetravalent doping injection agent. Inserting an element containing a superior valence electron count into a lower group matrix introduces surplus unbound mobile conducting electrons, defining an n-type semiconductor matrix.

Question 1.21: Gold (atomic radius $0.144 \text{ nm}$) crystallizes in a face-centered cubic unit cell. What is the length of a side of the cell?

Step 1: Geometric Line Mapping. Identify target geometric matrix constraints for FCC configurations: Radius metric value $r = 0.144 \text{ nm}$.
Step 2: Equation Evaluation steps. Apply standard face diagonal linear translation formulas: $$a = 2\sqrt{2}r$$ $$a = 2 \times 1.414 \times 0.144 \text{ nm} = 0.407 \text{ nm}$$ Step 3: Metric Tracking. The computed edge length parameter resolves cleanly to $0.407 \text{ nm}$ (which scales natively to $407 \text{ pm}$).

Question 1.22: In terms of band theory, what is the difference (i) between a conductor and an insulator (ii) between a conductor and a semiconductor?

Step 1: Conductor versus Insulator Architecture. In a classic conductor system, the electronic valence band and the empty conduction band overlay directly with zero separation, or the intermediate gap barrier is negligible, maximizing free transport tracks. In an insulator system, a wide energy gap (forbidden zone) bars valence electrons from jumping into the conduction band under standard operating stresses.
Step 2: Conductor versus Semiconductor Architecture. Conductors retain continuously open electronic transport fields without an active band gap. Semiconductors feature a small, narrow forbidden energy gap. At absolute zero, they function strictly as insulators; however, at elevated temperatures, fractions of valence electrons capture sufficient thermal kinetic inputs to step across the gap, activating lattice conductivity.

Question 1.23: Explain the following terms with suitable examples: (i) Schottky defect (ii) Frenkel defect (iii) Interstitials (iv) F-centers.

Step 1: Schottky Defect Mechanics. A stoichiometric lattice defect occurring when an identical, electrically balanced ratio of structural cations and anions vanish cleanly from their assigned standard lattice coordinate targets, systematically scaling down global mass density profiles. Ground verified example: $\text{NaCl}, \text{KCl}$.
Step 2: Frenkel Defect Mechanics. A stoichiometric structural dislocation occurrence where a compact ion variant (predominantly the smaller structural metallic cation) slips its standard position coordinate to rest inside an adjacent interstitial spatial void. Global crystal density parameters are strictly unchanged. Ground verified example: $\text{AgCl}, \text{ZnS}$.
Step 3: Interstitials Profile. Free stray atom components or chemical ions that wedge directly into vacant interstitial spatial void coordinate gaps located between structural master lattice nodes.
Step 4: F-centers Physics. Anionic structural position vacancies created within non-stoichiometric metal-excess crystals that serve as tracking traps for single unpaired conduction electrons. These isolated electrons absorb specific ambient wave packets, imparting distinctive color configurations onto the crystal block via excitation pathways (e.g., $\text{LiCl}$ matrices transitioning to distinct pink shades).

Question 1.24: Aluminium crystallizes in a cubic close-packed structure. Its metallic radius is $125 \text{ pm}$. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in $1.00 \text{ cm}^3$ of aluminium?

Step 1: Dimensional Scale Resolution. For an integrated CCP/FCC setup matrix tracking radius $r = 125 \text{ pm}$: $$a = 2\sqrt{2}r = 2 \times 1.414 \times 125 \text{ pm} = 353.5 \text{ pm} = 3.535 \times 10^{-8} \text{ cm}$$ Step 2: Single Unit Volumetric Calculation. Evaluate isolated single cell spatial footprints: $$V_{\text{cell}} = a^3 = (3.535 \times 10^{-8} \text{ cm})^3 = 4.42 \times 10^{-23} \text{ cm}^3$$ Step 3: Aggregate Cell Count Resolution. Process complete volume inventory scaling steps: $$\text{Total unit cell inventory} = \frac{\text{Target macro volume}}{\text{Single cell footprint volume}} = \frac{1.00 \text{ cm}^3}{4.42 \times 10^{-23} \text{ cm}^3} = 2.26 \times 10^{22} \text{ cells}$$

Question 1.25: If NaCl is doped with $10^{-3} \text{ mol}\%$ of $\text{SrCl}_2$, what is the concentration of cation vacancies?

Step 1: Substitution Mechanics. Because each divalent strontium element ($\text{Sr}^{2+}$) carries double the standard charge, it displaces exactly two native monovalent sodium ions ($\text{Na}^+$). It locks itself into one of these vacant nodes, leaving the remaining coordinate point behind as an empty cation vacancy to ensure baseline electrical balance.
Step 2: Molar Scale Determination. Track the input ratio values against absolute macro bases: $$\text{Vacancies tracked per 100 moles of base lattice matrix} = 10^{-3} \text{ mol}$$ $$\text{Vacancies calculated per singular isolated mole unit} = \frac{10^{-3}}{100} = 10^{-5} \text{ mol}$$ Step 3: Absolute Particulate Counting steps. Convert molar values into discrete numerical vacancies using Avogadro's constant: $$\text{Vacancy Concentration} = 10^{-5} \times N_A = 10^{-5} \times 6.022 \times 10^{23} = 6.022 \times 10^{18} \text{ vacancies per mole}$$

Question 1.26: Explain the following with suitable examples: (i) Ferromagnetism (ii) Paramagnetism (iii) Ferrimagnetism (iv) Antiferromagnetism (v) 12-16 and 13-15 group compounds.

Step 1: Ferromagnetism Properties. Solids intensely pulled by magnetic fields that can capture fixed structural magnetic properties indefinitely. This is powered by spontaneous lock-in parallel alignment arrays tracking metal domains. Verification example: $\text{Fe}, \text{Co}, \text{Ni}$.
Step 2: Paramagnetism Properties. Material blocks experiencing loose, weak attraction indicators under active external magnetic fields due to the isolated operations of unpaired electron shells. Magnetic behaviors drop out entirely upon the extraction of external fields. Verification example: $\text{O}_2, \text{Cu}^{2+}$.
Step 3: Ferrimagnetism Properties. Occurs when localized coordinate domain field loops point toward both parallel and anti-parallel headings in asymmetrical quantities, sustaining a net remaining internal coordinate field offset. Verification example: $\text{Fe}_3\text{O}_4$.
Step 4: Antiferromagnetism Properties. Systems containing structural domain arrays configured into anti-parallel orientation sets in perfectly equal quantities, nullifying global internal magnetic properties down to zero. Verification example: $\text{MnO}$.
Step 5: 12-16 and 13-15 Compounds. Advanced structural solid-state compound formulations fabricated by combining Group 12 or 13 inputs alongside matching Group 16 or 15 counter-elements (e.g., $\text{ZnS}, \text{GaAs}$ matrices). These structures replicate the electronic band profiles of pure Group 14 semiconductor components.

Extra Important Questions (Board Style)

Target verification tasks matching high-yield board testing frameworks engineered to lock down full marks in your 2026 exams.

Section A: Multiple Choice Questions (MCQs)

Q1. Which of the following is an amorphous solid?

A) Graphite
B) Quartz glass ($\text{SiO}_2$)
C) Chrome alum
D) Silicon carbide ($\text{SiC}$)

Correct Answer Choice: B
Step 1: Structural Verification. Quartz glass lacks a long-range orderly spatial configuration, which classifies it natively as an amorphous network solid.

Q2. The packing efficiency in a body-centered cubic (BCC) lattice is:

A) 52.4%
B) 68%
C) 74%
D) 80%

Correct Answer Choice: B
Step 1: Metric Verification. Derived by computing total dimensional workspace space populated by exactly two hard sphere units centered across an isolated base cube.

Section B: Assertion-Reason Questions

Directions: Choose option (A) if both statements are true and the reason provides the exact explanation. Choose (B) if both are true but independent. Choose (C) if the assertion is true but the reason is false. Choose (D) if assertion is false.

Q3. Assertion (A): Frenkel defect is not found in pure alkali metal halides.
Reason (R): Alkali metal ions have large sizes which cannot easily fit into interstitial spaces.

Correct Answer Choice: A
Step 1: Structural Validation. Frenkel imperfections necessitate wide geometric size deltas between ionic components. Alkali metallic cations are overly large, preventing them from structurally executing displacements into tightly constrained interstitial spatial void coordinates.

Section C: Short & Long Answer Questions

Q4. What type of stoichiometric defect is shown by AgBr?

Step 1: Rule Mapping. Silver Bromide ($\text{AgBr}$) stands out as a unique exception, capable of concurrently executing both Schottky and Frenkel point defect patterns based on ambient lattice energy configurations.

Q5. Calculate the distance between light-scattering planes if a first-order reflection is observed at $30^\circ$ using X-rays of wavelength $1.54 \text{ \AA}$.

Step 1: Model Invocation. Deploy Bragg's operational diffraction model: $$n\lambda = 2d\sin\theta$$ Step 2: Processing Matrix Variable Assignments. Here, reflection order $n=1$, wavelength $\lambda = 1.54 \text{ \AA}$, and angular coordinate tracking matches $\theta = 30^\circ \implies \sin(30^\circ) = 0.5$.
Step 3: Planar Tracking Resolution. $$1 \times 1.54 = 2 \times d \times 0.5 \implies d = 1.54 \text{ \AA}$$

Common Mistakes Students Make

Exam Preparation Tips

Frequently Asked Questions (FAQs)

1. Is Chapter 1 The Solid State important for CBSE Class 12 Board exams?
Yes, it is highly important. It regularly accounts for 4 to 6 marks, featuring direct numerical problems and conceptual questions on crystal defects.
2. What is an F-center and why is it important?
An F-center is an anionic vacancy trapped by an electron. It is responsible for imparting color to otherwise colorless crystals (like making $\text{NaCl}$ look yellow).
3. Which structure has the maximum packing efficiency?
Both FCC (Face-Centered Cubic) and HCP (Hexagonal Close-Packed) structures have the maximum packing efficiency of 74%.
4. Why does the conductivity of semiconductors increase with temperature?
As temperature rises, more electrons gain enough thermal energy to jump across the small forbidden energy gap into the conduction band.
5. What is the difference between isotropic and anisotropic natures?
Isotropic substances (amorphous solids) show identical physical properties in all directions. Anisotropic substances (crystalline solids) show different values for properties like refractive index when measured along different directions.

Conclusion: Lucky, mastering Chapter 1 is all about visualization and structured numerical execution! Keep your conversion parameters clean, revise your unit cell equations regularly, and ensure you check our solutions for subsequent chapters to unlock absolute elite chemistry board scoring tiers this year!

More Class 12 Chemistry Chapters

Ch 1: The Solid State Ch 2: Solutions Ch 3: Electrochemistry Ch 4: Chemical Kinetics Ch 5: Surface Chemistry Ch 6: General Principles Ch 7: The p-Block Elements Ch 8: The d- & f-Block Elements Ch 9: Coordination Compounds Ch 10: Haloalkanes Ch 11: Alcohols & Phenols Ch 12: Aldehydes & Ketones Ch 13: Amines Ch 14: Biomolecules All Subjects Dashboard