Solid State: CBSE Class 12 Chemistry NCERT Guide

Classification of Solids: Crystalline vs. Amorphous

Solids are broadly classified into two main types based on the arrangement of their constituent particles (atoms, molecules, or ions): Crystalline and Amorphous.

Crystalline Solids have a definite and regular arrangement of particles in a three-dimensional space. This long-range order means the pattern repeats itself periodically throughout the entire crystal. Think of it like a perfectly arranged brick wall. This regularity gives them sharp and characteristic melting points, as all bonds are of equal strength and break at the same temperature. They are anisotropic, meaning their physical properties like electrical resistance or refractive index have different values when measured along different directions. Examples include sodium chloride (NaCl), quartz, and diamond.

Amorphous Solids (from Greek 'amorphos' meaning no form) have a disordered, irregular arrangement of particles. The arrangement may have short-range order, but there is no long-range periodic pattern. Think of a pile of bricks. They do not have sharp melting points; instead, they soften gradually over a range of temperatures. Amorphous solids are isotropic, meaning their physical properties are the same in all directions. They are often called pseudo-solids or supercooled liquids because their structure resembles that of liquids. Examples include glass, rubber, and plastics.

Crystal Lattices and Unit Cells: The Building Blocks

Crystal Lattice

A regular three-dimensional arrangement of points in space representing the constituent particles (atoms, ions, or molecules) of a crystal.

Unit Cell

The smallest repeating portion of a crystal lattice which, when repeated in different directions, generates the entire lattice. It is characterized by its edge lengths (a, b, c) and the angles between them (α, β, γ).

Primitive Unit Cells

Unit cells where constituent particles are present only at the corner positions. A simple cubic (SC) cell is a primitive unit cell.

Centred Unit Cells

Unit cells where one or more constituent particles are present at positions other than corners. These include: 1. Body-Centred Cubic (BCC): Particles at all corners and one at the body centre. 2. Face-Centred Cubic (FCC): Particles at all corners and at the centre of each of the six faces. 3. End-Centred: Particles at all corners and at the centre of any two opposite faces.

Worked Example: Calculating the Density of a Unit Cell

• Problem 1: An element with a molar mass of 27 g/mol forms a cubic unit cell with an edge length of 405 pm. If its density is 2.7 g/cm³, what is the nature of the cubic unit cell (SC, BCC, or FCC)? Step 1: Understand the Formula The density (ρ) of a unit cell is given by the formula: ρ = (Z × M) / (a³ × Nₐ) Where: Z = Number of atoms per unit cell M = Molar mass (g/mol) a = Edge length of the cell (in cm) Nₐ = Avogadro's constant (6.022 × 10²³ mol⁻¹) Step 2: List the Given Values and Convert Units M = 27 g/mol ρ = 2.7 g/cm³ a = 405 pm. We need to convert this to cm. 1 pm = 10⁻¹⁰ cm. So, a = 405 × 10⁻¹⁰ cm = 4.05 × 10⁻⁸ cm. Nₐ = 6.022 × 10²³ mol⁻¹ Step 3: Rearrange the Formula to Solve for Z Z = (ρ × a³ × Nₐ) / M Step 4: Substitute the Values and Calculate Z Z = (2.7 g/cm³ × (4.05 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹) / 27 g/mol Z = (2.7 × (4.05)³ × 10⁻²⁴ × 6.022 × 10²³) / 27 Z = (2.7 × 66.43 × 10⁻¹ × 6.022) / 27 Z ≈ (179.36 × 0.6022) / 27 Z ≈ 108.01 / 27 Z ≈ 4 Final Answer: Since the number of atoms per unit cell (Z) is 4, the cubic unit cell is a Face-Centred Cubic (FCC) lattice. (Recall: SC has Z=1, BCC has Z=2, FCC has Z=4).
• Problem 2: Silver crystallises in an FCC lattice. If the edge length of the cell is 4.07 × 10⁻⁸ cm and the density is 10.5 g/cm³, calculate the atomic mass of silver. Step 1: Identify Given Information and the Target Variable Crystal structure = FCC, which means Z = 4 atoms per unit cell. Edge length (a) = 4.07 × 10⁻⁸ cm Density (ρ) = 10.5 g/cm³ Avogadro's constant (Nₐ) = 6.022 × 10²³ mol⁻¹ Target Variable = Atomic Mass (M) Step 2: Use the Density Formula and Rearrange for M ρ = (Z × M) / (a³ × Nₐ) M = (ρ × a³ × Nₐ) / Z Step 3: Substitute the values into the rearranged formula M = (10.5 g/cm³ × (4.07 × 10⁻⁸ cm)³ × 6.022 × 10²³ mol⁻¹) / 4 M = (10.5 × (4.07)³ × 10⁻²⁴ × 6.022 × 10²³) / 4 M = (10.5 × 67.42 × 10⁻¹ × 6.022) / 4 M = (707.91 × 0.6022) / 4 M = 426.3 / 4 M ≈ 106.57 g/mol Final Answer: The atomic mass of silver is approximately 106.6 g/mol.

Exam Tip: Distinguishing Schottky and Frenkel Defects

A very common question in board exams asks you to differentiate between Schottky and Frenkel defects. Don't get confused! Here’s a clear breakdown:

Schottky Defect:

• What is it? It's a vacancy defect where an equal number of cations and anions are missing from their lattice sites to maintain electrical neutrality.
• Conditions: Occurs in ionic compounds with high coordination numbers and where the cation and anion are of similar size (e.g., NaCl, KCl, CsCl).
• Effect on Density: Since atoms are completely removed from the crystal lattice, the overall mass decreases while the volume remains the same. Therefore, the density of the crystal decreases.

Frenkel Defect:

• What is it? It's a dislocation defect where a smaller ion (usually the cation) is dislocated from its normal lattice site to an interstitial site.
• Conditions: Occurs in ionic compounds with low coordination numbers and a large difference in the size of the cation and anion (e.g., AgCl, AgBr, ZnS).
• Effect on Density: Since no atoms are removed from the crystal, the mass and volume remain the same. Therefore, the density of the crystal does not change.

Key Takeaway: Remember Schottky = Similar size, Skipped atoms, so density decreaSes. Frenkel = Far apart in size, ion moves to an interstitial site, so density is unaFfected.

Practice Questions with Solutions

• Q: A compound is formed by two elements X and Y. Atoms of the element Y (as anions) make ccp (cubic close-packed) lattice and those of the element X (as cations) occupy all the octahedral voids. What is the formula of the compound? A: Step 1: Identify the lattice structure and atom positions. The element Y forms a ccp lattice. A ccp lattice is equivalent to an FCC lattice. The number of atoms of Y in the unit cell is effectively 4 (8 corners × 1/8 + 6 faces × 1/2 = 1 + 3 = 4). Step 2: Determine the number of octahedral voids. The number of octahedral voids in a ccp/FCC lattice is equal to the number of atoms in the lattice. Therefore, the number of octahedral voids = 4. Step 3: Determine the number of atoms of X. The problem states that atoms of element X occupy all the octahedral voids. So, the number of atoms of X is 4. Step 4: Find the simplest ratio of X to Y. The ratio of atoms X : Y is 4 : 4, which simplifies to 1 : 1. Final answer: The formula of the compound is XY.
• Q: Niobium crystallises in a body-centred cubic (BCC) structure. If the density is 8.55 g/cm³, calculate the atomic radius of Niobium given its atomic mass is 93 u. A: Step 1: Find the edge length 'a' using the density formula. For BCC, Z = 2. ρ = (Z × M) / (a³ × Nₐ). Rearranging for a³: a³ = (Z × M) / (ρ × Nₐ). Step 2: Substitute the values. a³ = (2 × 93 g/mol) / (8.55 g/cm³ × 6.022 × 10²³ mol⁻¹). a³ = 186 / (51.488 × 10²³) cm³. a³ = 3.612 × 10⁻²³ cm³. To find 'a', take the cube root: a = (36.12 × 10⁻²⁴ cm³)¹/³ ≈ 3.30 × 10⁻⁸ cm. Step 3: Relate edge length 'a' to atomic radius 'r' for a BCC structure. For a BCC lattice, the atoms touch along the body diagonal. The relationship is √3a = 4r. Rearranging for r: r = (√3a) / 4. Step 4: Calculate the radius 'r'. r = (1.732 × 3.30 × 10⁻⁸ cm) / 4. r = 5.7156 × 10⁻⁸ cm / 4. r = 1.429 × 10⁻⁸ cm. Converting to picometers: r = 142.9 pm. Final answer: The atomic radius of Niobium is approximately 143 pm.
• Q: Classify the following solids into different categories based on the nature of intermolecular forces operating in them: Potassium sulfate, Tin, Benzene, Urea, Ammonia, Water (ice), Zinc sulfide, Graphite, Rubidium, Argon, Silicon carbide. A: Step 1: Recall the four types of crystalline solids: Ionic, Covalent (Network), Molecular, and Metallic. Step 2: Analyze each substance. - Potassium sulfate (K₂SO₄): Composed of K⁺ and SO₄²⁻ ions. It is an Ionic solid. - Tin (Sn): It is a metal. It is a Metallic solid. - Benzene (C₆H₆): Discrete non-polar molecules held by weak van der Waals forces. It is a Molecular solid (non-polar). - Urea (NH₂CONH₂): Polar molecules held by hydrogen bonds. It is a Molecular solid (hydrogen-bonded). - Ammonia (NH₃): Polar molecules held by hydrogen bonds. It is a Molecular solid (hydrogen-bonded). - Water (ice, H₂O): Polar molecules held by strong hydrogen bonds. It is a Molecular solid (hydrogen-bonded). - Zinc sulfide (ZnS): Composed of Zn²⁺ and S²⁻ ions, but with significant covalent character. It is primarily an Ionic solid (though some sources might mention its covalent nature). - Graphite: Carbon atoms linked in sheets by covalent bonds. It is a Covalent or Network solid. - Rubidium (Rb): It is an alkali metal. It is a Metallic solid. - Argon (Ar): Individual atoms held by very weak van der Waals forces. It is a Molecular solid (non-polar). - Silicon carbide (SiC): Silicon and Carbon atoms linked by strong covalent bonds. It is a Covalent or Network solid. Final answer: - Ionic: Potassium sulfate, Zinc sulfide. - Covalent (Network): Graphite, Silicon carbide. - Molecular: Benzene (non-polar), Urea (H-bonded), Ammonia (H-bonded), Water (H-bonded), Argon (non-polar). - Metallic: Tin, Rubidium.
• Q: What is the difference between ferromagnetism and ferrimagnetism? Give one example of each. A: Step 1: Define ferromagnetism. In ferromagnetic substances, the magnetic moments of the domains are all aligned in the same direction, even in the absence of an external magnetic field. This results in a very strong attraction to magnetic fields. These substances can be permanently magnetised. Example: Iron (Fe), Cobalt (Co), Nickel (Ni). Step 2: Define ferrimagnetism. In ferrimagnetic substances, the magnetic moments of the domains are aligned in parallel and anti-parallel directions in unequal numbers. This results in a net magnetic moment, and the substance is weakly attracted to a magnetic field compared to ferromagnetic substances. Example: Ferrites like MgFe₂O₄ or ZnFe₂O₄, and Magnetite (Fe₃O₄). Step 3: Summarize the key difference. The primary difference is the alignment of magnetic moments. In ferromagnetism, all moments are aligned in parallel, leading to a strong magnetic effect. In ferrimagnetism, moments are aligned anti-parallel in unequal numbers, leading to a weaker net magnetic effect. Final answer: Ferromagnetism involves the parallel alignment of all magnetic moments, causing strong magnetic attraction (e.g., Iron). Ferrimagnetism involves the anti-parallel alignment of magnetic moments in unequal numbers, causing a weaker net magnetic attraction (e.g., Magnetite, Fe₃O₄).

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