Chapter 1: Solid State
Comprehensive Theory, Textbook Numericals, and Previous Year Questions
1. Introduction to Solid State
Matter exists in three states: solid, liquid, and gas. In the solid state, constituent particles (atoms, molecules, or ions) are closely packed due to strong intermolecular forces. Solids are characterized by a definite shape, definite volume, and high density.
Classification of Solids
Solids are broadly classified into two categories based on the arrangement of their constituent particles:
| Property | Crystalline Solids | Amorphous Solids |
|---|---|---|
| Arrangement of particles | Regular and long-range order. | Irregular and short-range order. |
| Shape | Definite characteristic geometrical shape. | Irregular shape. |
| Melting Point | Melt at a sharp and characteristic temperature. | Soften over a range of temperatures. |
| Cleavage Property | Cleave cleanly into two smooth pieces. | Cleave irregularly with rough surfaces. |
| Anisotropy | Anisotropic: Physical properties (like refractive index, electrical resistance) have different values in different directions. | Isotropic: Physical properties have the same value in all directions. |
| Examples | NaCl, Quartz, Diamond, Copper. | Glass, Rubber, Plastics. |
Isomorphism: Two or more substances having the same crystal structure are said to be isomorphous (e.g., $NaF$ and $MgO$).
Polymorphism: A single substance that exists in two or more forms or crystalline structures is said to be polymorphous. (e.g., Carbon exists as Diamond, Graphite, and Fullerene). In elements, polymorphism is called allotropy.
2. Classification of Crystalline Solids
Based on the nature of intermolecular forces, crystalline solids are classified into four types:
| Type of Solid | Constituent Particles | Bonding / Attractive Forces | Physical Nature & Conductivity | Examples |
|---|---|---|---|---|
| Ionic Solids | Cations and Anions | Strong Electrostatic (Coulombic) forces | Hard and brittle. Insulators in solid state but conductors in molten/aqueous state. High Melting Point (M.P.). | $NaCl, CaF_2, MgO$ |
| Molecular Solids | Molecules | Dispersion forces, Dipole-dipole interactions, Hydrogen bonding | Generally soft. Insulators. Low M.P. | Ice ($H_2O$), Solid $CO_2$ (Dry ice), $I_2$ |
| Covalent (Network) Solids | Atoms | Covalent bonds | Very hard (except graphite). Insulators (graphite is a conductor). Extremely high M.P. | Diamond, Quartz ($SiO_2$), SiC, Graphite |
| Metallic Solids | Positive metal ions in a sea of delocalized electrons | Metallic bonding | Hard but malleable and ductile. Excellent conductors of heat and electricity. | $Fe, Cu, Ag, Mg$ |
3. Crystal Lattices and Unit Cells
Crystal Lattice: A regular three-dimensional arrangement of points in space is called a crystal lattice.
Unit Cell: The smallest repeating fundamental portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
A unit cell is characterized by 6 parameters:
- Edge lengths: $a, b, c$
- Interfacial angles: $\alpha, \beta, \gamma$
Types of Cubic Unit Cells
The cubic crystal system ($a = b = c$, $\alpha = \beta = \gamma = 90^\circ$) has three common types of lattices:
- Simple Cubic (SC): Particles are present only at the corners of the cube.
- Body-Centered Cubic (BCC): Particles are at the corners and one particle is at the center of the body of the cube.
- Face-Centered Cubic (FCC): Particles are at the corners and one particle is at the center of each of the 6 faces.
Calculation of Number of Atoms per Unit Cell ($z$)
- SC: 8 corners $\times \frac{1}{8}$ per corner = 1 atom.
- BCC: (8 corners $\times \frac{1}{8}$) + (1 body center $\times 1$) = 1 + 1 = 2 atoms.
- FCC: (8 corners $\times \frac{1}{8}$) + (6 face centers $\times \frac{1}{2}$) = 1 + 3 = 4 atoms.
4. Packing in Solids and Voids
To understand how crystals are formed, constituent particles are considered as hard spheres. Packing occurs in steps: 1D to 2D to 3D.
- Close Packing in 3D: When 2D hexagonal close-packed layers are stacked, two types of 3D structures arise:
- Hexagonal Close Packing (HCP): ABAB... type arrangement. The coordination number is 12. Example: Mg, Zn.
- Cubic Close Packing (CCP or FCC): ABCABC... type arrangement. The coordination number is 12. Example: Cu, Ag.
Voids (Interstitial Sites)
The empty spaces left between the spheres during close packing are called voids.
- Tetrahedral Void: Surrounded by 4 spheres.
- Octahedral Void: Surrounded by 6 spheres.
Number of Octahedral Voids generated = $N$
Number of Tetrahedral Voids generated = $2N$
5. Packing Efficiency
Packing efficiency is the percentage of total space filled by the particles.
$$ \text{Packing Efficiency} = \frac{\text{Volume occupied by spheres in a unit cell}}{\text{Total volume of the unit cell}} \times 100 $$1. Simple Cubic Lattice (SC)
Let edge length be $a$ and radius of sphere be $r$. In SC, spheres touch along the edge.
Relation: $a = 2r$
Volume of cubic unit cell: $V = a^3 = (2r)^3 = 8r^3$
Number of atoms ($z$) = 1. Volume of 1 atom = $\frac{4}{3}\pi r^3$
$$ \text{P.E.} = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 = \frac{\pi}{6} \times 100 \approx 52.36\% $$
Empty Space (Void): $100 - 52.36 = 47.64\%$
2. Body-Centered Cubic (BCC)
Spheres touch along the body diagonal. Body diagonal length = $\sqrt{3}a$.
Relation: $4r = \sqrt{3}a \implies a = \frac{4r}{\sqrt{3}}$
Volume of unit cell: $V = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}$
Number of atoms ($z$) = 2. Volume of 2 atoms = $2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$
$$ \text{P.E.} = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 = \frac{\sqrt{3}\pi}{8} \times 100 \approx 68\% $$
Empty Space (Void): $100 - 68 = 32\%$
3. Face-Centered Cubic (FCC / CCP)
Spheres touch along the face diagonal. Face diagonal length = $\sqrt{2}a$.
Relation: $4r = \sqrt{2}a \implies a = 2\sqrt{2}r$
Volume of unit cell: $V = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$
Number of atoms ($z$) = 4. Volume of 4 atoms = $4 \times \frac{4}{3}\pi r^3 = \frac{16}{3}\pi r^3$
$$ \text{P.E.} = \frac{\frac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 = \frac{\pi}{3\sqrt{2}} \times 100 \approx 74\% $$
Empty Space (Void): $100 - 74 = 26\%$
6. Density of Unit Cell
The density of a crystal is exactly equal to the density of its unit cell.
$$ \rho = \frac{z \cdot M}{a^3 \cdot N_A} $$
Where:
- $\rho$ = Density of the solid ($\text{g/cm}^3$ or $\text{kg/m}^3$)
- $z$ = Number of atoms per unit cell (1 for SC, 2 for BCC, 4 for FCC)
- $M$ = Molar mass of the substance ($\text{g/mol}$)
- $a$ = Edge length of the unit cell (If given in pm, convert to cm: $1 \text{ pm} = 10^{-10} \text{ cm}$)
- $N_A$ = Avogadro's number ($6.022 \times 10^{23} \text{ mol}^{-1}$)
7. Imperfections (Defects) in Solids
Any deviation from the perfectly ordered arrangement of constituent particles in a crystal is called a defect. Point defects are irregularities around a point (atom) in a crystal lattice.
A. Stoichiometric Defects
These defects do not disturb the stoichiometry (ratio of cations to anions) of the solid.
1. Schottky Defect
Arises when equal numbers of cations and anions are missing from their regular lattice sites.
- Consequence: Density of the solid decreases.
- Condition: Highly ionic compounds with high coordination number and similar sizes of cations and anions.
- Examples: $NaCl, KCl, AgBr$.
2. Frenkel Defect
Arises when an ion (usually cation, being smaller) leaves its normal site and occupies an interstitial site.
- Consequence: Density remains unchanged (no ions lost).
- Condition: Low coordination number, large difference in size of ions.
- Examples: $ZnS, AgCl, AgBr, AgI$.
*Note: Silver bromide ($AgBr$) shows both Schottky and Frenkel defects.
B. Impurity Defects
Occurs when foreign atoms are present at the lattice sites (Substitutional, e.g., Brass) or at interstitial sites (Interstitial, e.g., Stainless Steel).
C. Non-Stoichiometric Defects
The ratio of positive and negative ions differs from that indicated by the chemical formula.
- Metal Excess Defect (F-Centers): Anions are missing from their sites and the hole is occupied by an electron to maintain electrical neutrality. The electron trapped is called an F-center (Farbenzenter), which imparts color to the crystal (e.g., NaCl appears yellow, KCl appears violet).
8. Electrical Properties and Band Theory
Solids are classified into Conductors, Insulators, and Semiconductors based on electrical conductivity.
Band Theory: Atomic orbitals merge to form molecular orbitals which are so close in energy that they form a continuous band. The band containing valence electrons is the Valence Band (VB), and the next higher empty band is the Conduction Band (CB). The gap between them is the Energy Gap (Forbidden zone).
- Conductors: VB partially overlaps with CB. Electrons can easily flow. Zero energy gap.
- Insulators: Large energy gap between VB and CB. Electrons cannot jump.
- Semiconductors: Small energy gap. At absolute zero, they act as insulators, but at room temperature, some electrons jump to CB. (e.g., Si, Ge).
Extrinsic Semiconductors (Doping)
The conductivity of intrinsic (pure) semiconductors is increased by adding a suitable impurity. This process is called Doping.
- n-type Semiconductor: Formed when Group 14 element (Si or Ge) is doped with Group 15 element (P, As, Sb). The 5th valence electron is free to conduct electricity. "n" stands for negative (electron).
- p-type Semiconductor: Formed when Group 14 element is doped with Group 13 element (B, Al, Ga). A vacancy or "electron hole" is created. "p" stands for positive (hole).
9. Magnetic Properties
Depending on their response to magnetic fields, solid substances are classified into:
| Property | Diamagnetic | Paramagnetic | Ferromagnetic |
|---|---|---|---|
| Behavior in Magnetic Field | Weakly repelled. | Weakly attracted. | Strongly attracted. |
| Electrons | All electrons are paired. | Presence of one or more unpaired electrons. | Unpaired electrons; magnetic domains align in the direction of the field. |
| Magnetism retention | Loses magnetism when field is removed. | Loses magnetism when field is removed. | Retains magnetism permanently even after field is removed. |
| Examples | $H_2O, NaCl, C_6H_6, TiO_2$ | $O_2, Cu^{2+}, Fe^{3+}, Cr^{3+}$ | $Fe, Co, Ni, CrO_2, Gd$ |
10. Solved Textbook Problems (Numericals)
Problem 1: Density Calculation
When gold crystallizes, it forms face-centered cubic cells. The unit cell edge length is 408 pm. Calculate the density of gold. (Molar mass of gold = $197 \text{ g/mol}$)
Solution:
For FCC lattice, $z = 4$
$M = 197 \text{ g/mol}$
$a = 408 \text{ pm} = 408 \times 10^{-10} \text{ cm} = 4.08 \times 10^{-8} \text{ cm}$
$N_A = 6.022 \times 10^{23} \text{ atoms/mol}$
Using formula: $$ \rho = \frac{z \times M}{a^3 \times N_A} $$
$$ \rho = \frac{4 \times 197}{(4.08 \times 10^{-8})^3 \times 6.022 \times 10^{23}} $$
$$ \rho = \frac{788}{67.91 \times 10^{-24} \times 6.022 \times 10^{23}} $$
$$ \rho = \frac{788}{40.89} = 19.27 \text{ g/cm}^3 $$
Answer: Density of gold is $19.27 \text{ g/cm}^3$.
Problem 2: Voids and Formula
A compound is formed by two elements A and B. The atoms of element B forms ccp lattice and those of element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound?
Solution:
Let the number of atoms of element B forming ccp lattice = $N$
We know, number of tetrahedral voids generated = $2N$
Atoms of element A occupy 2/3 of the tetrahedral voids.
Therefore, number of atoms of A = $\frac{2}{3} \times 2N = \frac{4N}{3}$
Ratio of A : B = $\frac{4N}{3} : N = 4 : 3$
Answer: The formula of the compound is $A_4B_3$.
Problem 3: Calculating Radius from Density
Niobium crystallizes in BCC structure. If density is $8.55 \text{ g cm}^{-3}$, calculate atomic radius of niobium using its atomic mass 93 u.
Solution:
For BCC lattice, $z = 2$
$\rho = 8.55 \text{ g/cm}^3$, $M = 93 \text{ g/mol}$
Using formula: $\rho = \frac{z \cdot M}{a^3 \cdot N_A} \implies a^3 = \frac{z \cdot M}{\rho \cdot N_A}$
$$ a^3 = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} = \frac{186}{51.48 \times 10^{23}} = 3.61 \times 10^{-23} \text{ cm}^3 $$
$$ a = (36.1 \times 10^{-24})^{1/3} \approx 3.3 \times 10^{-8} \text{ cm} = 330 \text{ pm} $$
For BCC, relation between radius ($r$) and edge length ($a$) is $r = \frac{\sqrt{3}}{4}a$
$$ r = \frac{1.732 \times 330}{4} = 142.8 \text{ pm} $$
Answer: Atomic radius of Niobium is $142.8 \text{ pm}$.
11. Previous Year Board Questions (HSC)
Important questions specifically compiled from Maharashtra Board examinations.
1 Mark Questions (VSA)
- Write the formula for the density of a unit cell. (March 2014)
- Name the defect in which an equal number of cations and anions are missing from the crystal lattice. (Oct 2015)
- What are ferromagnetic substances? (March 2016)
- Define: Unit cell. (Oct 2013, March 2017)
- What is the packing efficiency in a CCP structure? (Oct 2017)
- What is a point defect? (March 2022)
2 Mark Questions (SA-I)
- Distinguish between crystalline and amorphous solids. (March 2013, Oct 2014, March 2018)
- Distinguish between Schottky and Frenkel defect. (March 2015, Oct 2019)
- Explain n-type semiconductor with an example. (March 2017)
- A face-centered cubic crystal has an edge length of $400\text{ pm}$. Calculate its density. (Given molar mass = $60\text{ g/mol}$) (March 2019)
- Classify the following solids into different types: Plastic, Diamond, NaCl, Brass. (March 2021)
3 Mark Questions (SA-II)
- Derive the relationship between the density of a substance and the edge length of a unit cell. (March 2020, March 2022)
- Explain ferromagnetic and paramagnetic substances with one example each. (Oct 2013, March 2017)
- Calculate the packing efficiency in a body-centered cubic (BCC) lattice. (Oct 2014, March 2018, March 2023)
- Explain Metal deficiency defect and Metal excess defect with suitable examples. (Oct 2021)
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