Revision Notes Chapter 1 The Solid State Class 12 Notes

Crystalline vs Amorphous Solids

Unit Cells and Crystal Lattices

A crystal lattice is a three-dimensional symmetrical arrangement of constituent particles (atoms, ions, or molecules) in space. The smallest repeating unit of this lattice which, when repeated in different directions, generates the entire space lattice is called a Unit Cell.

Unit cells are classified into Primitive (Simple Cubic) where particles are present only at the corners, and Centred Unit Cells (Body-Centred, Face-Centred, and End-Centred).

Dimensional Calculations of Unit Cells

When calculating the density of a crystal, we use the edge length ($a$) of the unit cell and its atomic mass ($M$). The basic equation of density ($d$) is given by:

$d = \frac{z \cdot M}{a^3 \cdot N_A}$

Where:

• $z$ is the number of atoms per unit cell.
• $M$ is the molar mass of the element/compound.
• $a$ is the edge length of the unit cell (typically in centimeters for density in $\text{g/cm}^3$; convert pm to cm by multiplying by $10^{-10}$).
• $N_A$ is Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$).

Must Remember Formulas and Relations

• Simple Cubic Unit Cell (SCC): Atoms per unit cell ($z$) = 1. Relation between edge length ($a$) and atomic radius ($r$) is $a = 2r$. Packing efficiency = $52.4\%$. Coordination number = 6.
• Body-Centred Cubic Unit Cell (BCC): Atoms per unit cell ($z$) = 2. Relation between edge length ($a$) and atomic radius ($r$) is $r = \frac{\sqrt{3}}{4} a$. Packing efficiency = $68\%$. Coordination number = 8.
• Face-Centred Cubic Unit Cell (FCC / CCP): Atoms per unit cell ($z$) = 4. Relation between edge length ($a$) and atomic radius ($r$) is $r = \frac{a}{2\sqrt{2}} = \frac{\sqrt{2}}{4} a$. Packing efficiency = $74\%$. Coordination number = 12.
• Number of Voids in CCP/FCC structure: If the number of close-packed particles is $N$, then the number of Octahedral voids = $N$, and the number of Tetrahedral voids = $2N$.
• Void Radius Ratios: Radius of tetrahedral void $r_{\text{tet}} = 0.225 R$ (where $R$ is lattice anion radius). Radius of octahedral void $r_{\text{oct}} = 0.414 R$.
• Schottky Defect: Occurs in highly ionic compounds with high coordination numbers where equal numbers of cations and anions are missing. This decreases the density of the crystal (e.g., NaCl, KCl, CsCl, AgBr).
• Frenkel Defect: Occurs when a smaller ion (usually cation) is dislocated to an interstitial site. It does not change the density of the crystal (e.g., ZnS, AgCl, AgBr, AgI). Note: AgBr shows both Schottky and Frenkel defects.

Deriving Chemical Formulas from Void Positions

1. Step 1: Identify the lattice forming element — Identify which atom forms the close-packed lattice (HCP, CCP, or FCC). Let the number of these atoms be $N$.
2. Step 2: Determine total available voids — Write down the potential number of voids. Total Octahedral Voids = $N$. Total Tetrahedral Voids = $2N$.
3. Step 3: Apply the fractional occupancy — Multiply the void counts by the fraction specified in the problem (e.g., 'cations occupy 2/3 of tetrahedral voids' translates to $\frac{2}{3} \times 2N = \frac{4}{3}N$).
4. Step 4: Establish the ratio and simplify — Write down the ratio of atoms (e.g., $A:B$). Multiply the whole ratio by the common denominator to convert fractional values into simple integers to get the empirical formula.

Key Terms and Definitions

Anisotropy

The property of crystalline solids whereby their physical properties (like refractive index, electrical conductivity) exhibit different values when measured along different directional axes.

Bravais Lattices

The 14 distinct space lattices possible in three dimensions, grouped under 7 crystal systems based on unit cell parameters.

Coordination Number

The number of immediate neighboring particles in direct contact with a particular particle in a crystal structure.

F-centres (Farbe centres)

Anionic sites occupied by unpaired electrons in non-stoichiometric metal-excess defects that impart colour to the crystals.

Doping

The process of introducing an appropriate, small amount of an impurity into a pure semiconductor crystal to deliberately modify its electrical properties.

Ferromagnetism

A phenomenon where substances are strongly attracted by a magnetic field and can be permanently magnetized even in the absence of the magnetic field due to parallel alignment of domains (e.g., Fe, Co, Ni).

Solved Numerical and Conceptual Examples

• {"title":"Example 1: Formula determination","calculation":"Problem: A compound forms HCP structure. Atoms of element Y form the lattice and those of X occupy $2/3^{\\text{rd}}$ of the tetrahedral voids. What is the formula of the compound?\n\nSolution:\n1. Let the number of Y atoms forming the HCP lattice = $N$.\n2. Total number of tetrahedral voids = $2N$.\n3. Number of X atoms occupying tetrahedral voids = $\\frac{2}{3} \\times 2N = \\frac{4}{3}N$.\n4. Ratio of atoms $X : Y = \\frac{4}{3}N : N = 4 : 3$.\n5. Therefore, the formula of the compound is $X_4Y_3$."}
• {"title":"Example 2: Density Calculation","calculation":"Problem: An element has a BCC structure with a cell edge of $288\\text{ pm}$. The density of the element is $7.2\\text{ g/cm}^3$. How many atoms are present in $208\\text{ g}$ of the element?\n\nSolution:\n1. Volume of the unit cell, $V = a^3 = (288 \\times 10^{-10}\\text{ cm})^3 = 2.39 \\times 10^{-23}\\text{ cm}^3$.\n2. Volume of $208\\text{ g}$ of the element = $\\text{Mass} / \\text{Density} = 208\\text{ g} / 7.2\\text{ g/cm}^3 = 28.88\\text{ cm}^3$.\n3. Number of unit cells in this volume = $\\frac{\\text{Total Volume}}{\\text{Volume of one unit cell}} = \\frac{28.88}{2.39 \\times 10^{-23}} = 1.208 \\times 10^{24}\\text{ unit cells}$.\n4. Since BCC structure has $2$ atoms per unit cell ($z=2$):\nTotal number of atoms = $2 \\times 1.208 \\times 10^{24} = 2.416 \\times 10^{24}$ atoms."}

CBSE Board Exam Traps & Marking Cues

• Unit Conversion Trap: The edge length $a$ is almost always given in picometers (pm) or angstroms (Å), while density $d$ is in $\text{g/cm}^3$. Don't forget to convert $a$ into centimeters: $1\text{ pm} = 10^{-10}\text{ cm}$, $1\text{ Å} = 10^{-8}\text{ cm}$.
• AgBr Exception: Frequently asked in one-marker questions: Identify the compound that shows both Frenkel and Schottky defects. The answer is AgBr due to intermediate ionic character and size.
• F-centre color mechanism: In questions about why NaCl turns yellow or LiCl turns pink upon heating in metal vapors, explicitly state that 'unpaired electrons occupy anionic vacancies resulting in F-centres that absorb light in the visible region.' This gets you full marks on descriptive questions.

Practice Questions with Solutions

• Why does zinc oxide (ZnO) turn yellow on heating? On heating, ZnO loses oxygen gas: $\text{ZnO} \xrightarrow{\Delta} \text{Zn}^{2+} + \frac{1}{2}\text{O}_2 + 2e^-$. The excess $\text{Zn}^{2+}$ ions occupy interstitial sites and the electrons are trapped in neighboring interstitial vacancies (F-centres), which absorb visible light and impart a yellow colour.
• What is the coordination number of atoms in (a) CCP and (b) BCC structure? (a) In CCP (cubic close-packed / FCC) structure, the coordination number is 12. (b) In BCC (body-centred cubic) structure, the coordination number is 8.
• How do Schottky defects affect the density of ionic crystals? Schottky defects decrease the overall density of the crystal because an equal number of cations and anions are completely missing from their normal lattice sites, creating vacancies.
• What type of semiconductor is produced when Silicon is doped with Gallium? Silicon (group 14 element) doped with Gallium (group 13 element, trivalent impurity) produces electron-deficient holes, creating a p-type semiconductor.

Frequently Asked Questions