Chapter 1: Solid State
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: A unit cell is the smallest repeating fundamental structural portion of a crystal lattice which, when repeated in different directions in three-dimensional space, generates the entire crystal lattice.
Answer: The formula for the density ($\rho$) is: $$\rho = \frac{z \cdot M}{a^3 \cdot N_A}$$ Where $z$ = 1 (for simple cubic), $M$ = Molar mass, $a$ = edge length, and $N_A$ = Avogadro's number.
Answer: Silver bromide ($AgBr$) is an example of a crystalline solid that exhibits both Schottky and Frenkel point defects.
Answer: The packing efficiency in a Face-Centered Cubic (fcc) or Cubic Close Packed (ccp) structure is 74%. (The remaining 26% is empty void space).
2 Mark Questions (Short Answer-I)
Answer:
| Property | Crystalline Solids | Amorphous Solids |
|---|---|---|
| 1. Arrangement | Regular and long-range order of particles. | Irregular and short-range order of particles. |
| 2. Melting Point | Have a sharp and characteristic melting point. | Soften over a range of temperatures. |
| 3. Anisotropy | They are anisotropic (physical properties vary with direction). | They are isotropic (physical properties are same in all directions). |
| 4. Cleavage | When cut with a sharp tool, they split into two smooth pieces. | When cut, they split into pieces with irregular/rough surfaces. |
Answer:
| Schottky Defect | Frenkel Defect |
|---|---|
| It arises when equal numbers of cations and anions are completely missing from their normal lattice sites. | It arises when an ion (usually a cation) leaves its normal site and occupies an interstitial site. |
| It decreases the overall density of the solid. | It does not change the density of the solid (no ions are lost). |
| Shown by ionic compounds having high coordination numbers and similar sizes of cations/anions (e.g., $NaCl, KCl$). | Shown by ionic compounds having low coordination numbers and a large difference in the sizes of cations/anions (e.g., $ZnS, AgCl$). |
Answer:
- An n-type semiconductor is formed when a pure Group 14 element (like Silicon or Germanium) is doped with a Group 15 element (like Phosphorus, Arsenic, or Antimony).
- Mechanism: Group 14 elements have 4 valence electrons, while Group 15 elements have 5. When doped, four out of the five valence electrons of the impurity atom form covalent bonds with surrounding Si/Ge atoms.
- The fifth electron remains unbonded, is loosely held, and is free to move, thereby increasing the electrical conductivity.
- Because the increase in conductivity is due to negatively charged electrons, it is called an n-type (negative-type) semiconductor.
3 Mark Questions (Short Answer-II)
Answer:
Consider a cubic unit cell of edge length $= a$.
1. Volume of the cubic unit cell ($V$) $= a \times a \times a = a^3$.
2. Let the mass of one single particle (atom) be $m$. If there are $z$ number of particles per unit cell, then:
Mass of the unit cell $= z \times m$
3. The mass of a single atom ($m$) is related to the molar mass ($M$) and Avogadro's number ($N_A$) by the formula:
$m = \frac{M}{N_A}$
4. Substituting the value of $m$ into the mass of the unit cell:
Mass of the unit cell $= z \times \frac{M}{N_A}$
5. The density of the unit cell ($\rho$) is defined as mass per unit volume:
$$\rho = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$$
6. Substituting the values:
$$\rho = \frac{z \cdot M}{a^3 \cdot N_A}$$
This is the required relationship between density ($\rho$) and edge length ($a$).
Answer:
In a BCC unit cell, spheres (atoms) touch each other along the body diagonal. Let the edge length of the cube be $a$ and the radius of each atom be $r$.
Step 1: Relation between $a$ and $r$
The length of the body diagonal of a cube with edge $a$ is $\sqrt{3}a$.
Since atoms touch along the body diagonal, its length is equal to $4r$ (radius of corner atom + diameter of center atom + radius of opposite corner atom).
Therefore, $\sqrt{3}a = 4r \implies a = \frac{4r}{\sqrt{3}}$
Step 2: Volume of the unit cell
Volume ($V$) $= a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}}$
Step 3: Volume occupied by atoms
A BCC unit cell contains a total of $z = 2$ atoms (8 corners $\times$ 1/8 + 1 center $\times$ 1).
Volume of 1 atom (sphere) $= \frac{4}{3}\pi r^3$
Volume of 2 atoms $= 2 \times \frac{4}{3}\pi r^3 = \frac{8}{3}\pi r^3$
Step 4: Packing Efficiency Calculation
$$\text{P.E.} = \frac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \times 100$$
$$\text{P.E.} = \frac{\frac{8}{3}\pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100$$
$$\text{P.E.} = \frac{8\pi \cdot 3\sqrt{3}}{3 \cdot 64} \times 100 = \frac{\sqrt{3}\pi}{8} \times 100$$
$$\text{P.E.} = \frac{1.732 \times 3.142}{8} \times 100 \approx 68\%$$
The packing efficiency of a BCC structure is 68%.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark or 3-mark numerical combined with a 1-mark or 2-mark theory question).
(b) What are paramagnetic substances? Give one example. [1 Mark] March 2019
Answer (a): Numerical on Density
Given:
Crystal structure = fcc $\implies z = 4$ atoms per unit cell.
Edge length ($a$) $= 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm}$.
Molar mass ($M$) $= 60 \text{ g/mol}$.
Avogadro's number ($N_A$) $= 6.022 \times 10^{23} \text{ mol}^{-1}$.
Formula: $$\rho = \frac{z \cdot M}{a^3 \cdot N_A}$$
Step 1: Calculate $a^3$
$a^3 = (4 \times 10^{-8} \text{ cm})^3 = 64 \times 10^{-24} \text{ cm}^3$
Step 2: Substitute values in the formula
$$\rho = \frac{4 \times 60}{64 \times 10^{-24} \times 6.022 \times 10^{23}}$$
$$\rho = \frac{240}{64 \times 6.022 \times 10^{-1}}$$
$$\rho = \frac{240}{385.408 \times 10^{-1}} = \frac{240}{38.54}$$
$$\rho = 6.227 \text{ g/cm}^3$$
The density of the metal crystal is $6.227 \text{ g/cm}^3$.
Answer (b): Theory
Paramagnetic substances are those substances which are weakly attracted by an external magnetic field. This property arises due to the presence of one or more unpaired electrons in their atoms, ions, or molecules.
Example: Oxygen gas ($O_2$), Copper(II) ion ($Cu^{2+}$), Iron(III) ion ($Fe^{3+}$).
(b) What is polymorphism? [1 Mark] Oct 2013, March 2021
Answer (a): Packing Efficiency Derivation
In a simple cubic lattice, the atoms (spheres) touch each other along the edges of the cube. Let the edge length be $a$ and the radius of each atom be $r$.
Step 1: Relation between $a$ and $r$
Since spheres touch along the edge, $a = 2r$.
Step 2: Volume of unit cell
$V = a^3 = (2r)^3 = 8r^3$
Step 3: Volume of atoms
A simple cubic unit cell has atoms only at the 8 corners. Total atoms $z = 8 \times \frac{1}{8} = 1$ atom.
Volume of 1 atom $= \frac{4}{3}\pi r^3$
Step 4: Calculation
$$\text{P.E.} = \frac{\text{Volume of 1 atom}}{\text{Volume of unit cell}} \times 100$$
$$\text{P.E.} = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 = \frac{4\pi}{3 \times 8} \times 100 = \frac{\pi}{6} \times 100$$
$$\text{P.E.} = \frac{3.142}{6} \times 100 = 52.36\%$$
The packing efficiency of a simple cubic lattice is 52.36%.
Answer (b): Theory
Polymorphism is the phenomenon where a single substance can exist in two or more different crystalline forms or structures under different conditions of temperature and pressure. (e.g., Carbon existing as Diamond and Graphite).
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