GSEB Class 12 Chemistry Solutions Chapter 1 The Solid State

 

Question 1. Why are solids rigid?
Answer: Solid substances exhibit rigidity because their constituent particles, such as atoms, ions, or molecules, are tightly bound by powerful interatomic, interionic, or intermolecular attractive forces.
In simple words: Solids are rigid because their particles are held very tightly in fixed positions by strong forces, preventing easy movement.

🎯 Exam Tip: Focus on the "strong attractive forces" and "fixed positions of constituent particles" as key points for explaining rigidity.

 

Question 2. Why do solids have a definite volume?
Answer: Solids possess a definite volume because their constituent particles are held firmly by strong attractive forces, ensuring that the distances between these particles remain constant at a given temperature.
In simple words: Solids have a fixed volume because strong forces keep their particles at constant distances from each other, even with temperature changes.

🎯 Exam Tip: Highlight the combination of strong attractive forces and constant interparticle distances as the reason for definite volume.

 

Question 3. Classify the following as amorphous or crystalline solids:
Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiberglass, copper?
Answer: Ionic solids: \((NH_4)_3PO_4\) and LiBr
Metallic solids: Brass and Rb
Molecular solids: \(P_4O_{10}\), \(P_4\)
Network (Covalent) solids: Graphite, SiC, Si
Amorphous solids: Plastics
In simple words: Solids are classified based on their internal structure and bonding: ionic (like salts), metallic (like metals), molecular (like sugar), covalent network (like diamond), or amorphous (like glass).

🎯 Exam Tip: Remember the key characteristics that differentiate crystalline (ionic, metallic, molecular, covalent network) from amorphous solids, especially regarding internal order.

 

Question 4. Why is glass considered a supercooled liquid?
Answer: Glass is classified as a supercooled liquid because, similar to conventional liquids, it exhibits a tendency to flow, albeit at an extremely slow rate.
In simple words: Glass is a supercooled liquid because it flows, although very slowly, just like liquids do.

🎯 Exam Tip: The ability of glass to flow, even if imperceptibly over long periods, is the defining characteristic that leads to its classification as a supercooled liquid.

 

Question 5. Refractive index of a solid is observed to have the same value along with all directions. Comment on the nature of this solid. Would it show cleavage property?
Answer: This solid is amorphous, as amorphous solids inherently display isotropic properties, meaning their physical characteristics, such as refractive index, are uniform in all directions. Consequently, it would not exhibit a cleavage property.
In simple words: The solid is amorphous because its properties are the same in all directions (isotropic). It won't cleave cleanly.

🎯 Exam Tip: Distinguish between isotropic (amorphous solids) and anisotropic (crystalline solids) properties, and how this relates to cleavage.

 

Question 6. Classify the following solids in different categories based on the nature of intermolecular forces operating in them: Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide?
Answer: 1. Ionic solids: Potassium sulphate, zinc sulphide
2. Covalent solids: Graphite, silicon carbide
3. Molecular solids: Benzene, urea, ammonia, water, argon
4. Metallic solids: Rubidium, tin.
In simple words: Solids are grouped by the type of forces holding them together: ionic (strong electrostatic), covalent (shared electrons forming a network), molecular (weak intermolecular forces), and metallic (delocalized electrons).

🎯 Exam Tip: Understand the primary bonding type (ionic, covalent, metallic) and intermolecular forces (van der Waals, hydrogen bonding) to correctly classify solids.

 

Question 7. Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
Answer: Given its properties – extreme hardness, electrical insulation in both solid and molten states, and a very high melting point – Solid A is characterized as a covalent or network solid.
In simple words: Solid A is a covalent or network solid because it's very hard, doesn't conduct electricity even when melted, and has a very high melting point.

🎯 Exam Tip: Correlate properties like hardness, electrical conductivity, and melting point to the different types of solids (ionic, metallic, molecular, covalent).

 

Question 8. Ionic solids conduct electricity in molten state but not in solid state Explain?
Answer: In their solid state, ionic solids are poor electrical conductors because their ions are fixed within the lattice and lack mobility. Conversely, when molten, these ions become dissociated and mobile, thereby enabling the conduction of electric current.
In simple words: Ionic solids don't conduct electricity when solid because ions are stuck, but they do when melted because the ions become free to move.

🎯 Exam Tip: The key is the mobility of charge carriers: ions are immobile in solid ionic compounds but become mobile in the molten state or in solution.

 

Question 9. What type of solids are electrical conductors, malleable and ductile?
Answer: Metallic solids
In simple words: Solids that conduct electricity, are malleable (can be hammered), and ductile (can be drawn into wires) are metallic solids.

🎯 Exam Tip: Remember that malleability, ductility, and electrical conductivity are characteristic properties of metals due to their electron sea model.

 

Question 10. Give the significance of a 'lattice point'?
Answer: A 'lattice point' signifies the position occupied by a single constituent particle within the solid's structure. This particle can be an individual atom, a molecule (which is a group of atoms), or an ion.
In simple words: A lattice point is simply a position in the crystal structure where an atom, molecule, or ion is located.

🎯 Exam Tip: Understand that a lattice point is a theoretical representation of the location of the fundamental building blocks of a crystal.

 

Question 11. Name the parameters that characterise a unit cell?
Answer: The parameters defining a unit cell are:
1. The lengths of its three edges, denoted as a, b, and c.
2. The angles between these edges, represented as \(\alpha\), \(\beta\), and \(\gamma\).
In simple words: A unit cell is defined by its three edge lengths (a, b, c) and the three angles (\(\alpha\), \(\beta\), \(\gamma\)) between them.

🎯 Exam Tip: These six parameters (three edge lengths and three interfacial angles) are crucial for describing the geometry of any crystal system.

 

Question 12. Distinguish between
1. Hexagonal and monoclinic unit cells
2. Face-centred and end-centered unit cells
Answer: 1. A hexagonal unit cell is characterized by edge lengths where a = b ≠ c, and axial angles \(\alpha = \beta = 90^\circ\), \(\gamma = 120^\circ\). In contrast, a monoclinic unit cell has edge lengths a ≠ b ≠ c, and axial angles \(\alpha = \gamma = 90^\circ\), \(\beta \neq 90^\circ\).
2. A face-centered unit cell features particles at all corners and also at the center of each face, resulting in 4 atoms per unit cell. An end-centered unit cell, on the other hand, has particles at all corners and at the centers of only two opposite faces, amounting to 2 atoms per unit cell.
In simple words: Hexagonal and monoclinic cells differ in their edge lengths and angles. Face-centered cells have particles on all faces and corners (4 atoms), while end-centered cells have particles on two opposite faces and corners (2 atoms).

🎯 Exam Tip: Clearly state the lattice parameters (edge lengths and angles) for distinguishing unit cells and the number of constituent particles per unit cell for different types of cubic arrangements.

 

Question 13. Explain how many portions of an atom located at
1. corner and
2. body centre of a cubic unit cell is part of its neighbouring unit cell?
Answer: 1. An atom positioned at a corner of a cubic unit cell contributes \( \frac{1}{8} \) of its volume to each of the eight adjacent unit cells.
2. An atom situated at the body center of a cubic unit cell is entirely contained within that single unit cell and is not shared with any neighboring unit cells.
In simple words: An atom at a corner is shared by 8 cells (1/8 each), while an atom in the body center belongs entirely to that one cell.

🎯 Exam Tip: Remember the contribution of atoms at different positions (corners, faces, edges, body center) to a single unit cell for calculations of the number of atoms per unit cell.

 

Question 14. What is the two-dimensional coordination number of a molecule in a square close-packed layer?
Answer: In a two-dimensional square close-packed layer, a molecule's coordination number is four. This means each molecule is in direct contact with four immediate neighbors.
In simple words: In a square close-packed layer, each molecule touches 4 others, so its coordination number is 4.

🎯 Exam Tip: Visualize the arrangement: in square packing, each central sphere touches one above, one below, and one on each side.

 

Question 15. A compound forms a hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer: In a hexagonal close-packed (hcp) structure, each atom gives rise to three voids: one octahedral and two tetrahedral.
The number of atoms in 0.5 mol is calculated as:
\( 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \)
Therefore, the total number of voids will be:
\( 3 \times 3.011 \times 10^{23} = 9.033 \times 10^{23} \)
The number of tetrahedral voids specifically will be:
\( 2 \times 3.011 \times 10^{23} = 6.022 \times 10^{23} \)
In simple words: In an hcp structure, there are three voids per atom (two tetrahedral, one octahedral). For 0.5 mol, calculate the total atoms, then multiply by 3 for total voids and by 2 for tetrahedral voids.

🎯 Exam Tip: Remember the ratio of voids to atoms in close-packed structures: for N atoms, there are N octahedral voids and 2N tetrahedral voids.

 

Question 16. A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy \( \frac{1}{3} \) rd of tetrahedral voids. What is the formula of the compound?
Answer: If N atoms form a ccp structure, for every 1 N atom (considering it for formula derivation), there are 2 tetrahedral voids.
The problem states that atoms of element M occupy \( \frac{2}{3} \) of these tetrahedral voids.
Therefore, the number of M atoms = \( \frac{2}{3} \times 2 = \frac{4}{3} \).
The ratio of M atoms to N atoms is \( \frac{4}{3} : 1 \).
To obtain whole numbers, multiply both sides by 3: \( 4:3 \).
Thus, the formula of the compound is \( M_4N_3 \).
In simple words: For every N atom in ccp, there are two tetrahedral voids. If M atoms occupy 2/3 of these voids, there are (2/3)*2 = 4/3 M atoms for every N atom. So, the formula is \(M_4N_3\).

🎯 Exam Tip: Remember that in ccp (or hcp) structures, for N particles, there are N octahedral voids and 2N tetrahedral voids. Always derive the simplest whole-number ratio for the empirical formula.

 

Question 17. Which of the following lattices has the highest packing efficiency
1. simple cubic
2. body- centered cubic and
3. hexagonal close-packed lattice?
Answer: Among the given lattice types, the hexagonal close-packed (hcp) lattice exhibits the highest packing efficiency at 74%. This is followed by body-centered cubic (bcc) at 68%, and then simple cubic (sc) with a packing efficiency of 52.4%.
In simple words: Hexagonal close-packed (hcp) has the highest packing efficiency (74%), then body-centered cubic (bcc) (68%), and simple cubic (sc) (52.4%) is the least efficient.

🎯 Exam Tip: Memorize the packing efficiencies for common crystal structures: hcp and fcc (both 74%), bcc (68%), and simple cubic (52.4%).

 

Question 18. An element with a molar mass \( 2.7 \times 10^{-2} \) kg mol\(^{-1}\) forms a cubic unit cell with an edge length 405 pm. If its density is \( 2.7 \times 10^{23} \) kg\(^{-3}\), what is the nature of the cubic unit cell?
Answer: Given values:
Molar mass (M) = \( 2.7 \times 10^{-2} \) kg mol\(^{-1}\)
Edge length (a) = 405 pm = \( 405 \times 10^{-12} \) m
Density (d) = \( 2.7 \times 10^3 \) kg m\(^{-3}\)
Avogadro's number (\( N_A \)) = \( 6.022 \times 10^{23} \) mol\(^{-1}\)

The relationship between density, molar mass, edge length, and the number of atoms per unit cell (z) is given by:
\( d = \frac{zM}{a^3N_A} \)

Rearranging the formula to solve for z:
\( z = \frac{d \cdot a^3 \cdot N_A}{M} \)

Substituting the given values:
\( z = \frac{(2.7 \times 10^3 \text{ kg m}^{-3}) \times (405 \times 10^{-12} \text{ m})^3 \times (6.022 \times 10^{23} \text{ mol}^{-1})}{(2.7 \times 10^{-2} \text{ kg mol}^{-1})} \)

Performing the calculation yields:
\( z = 4 \)

Since the value of z is 4, the unit cell is a face-centered cubic (fcc) or cubic close-packed (ccp) structure.
In simple words: We use the density formula \( d = \frac{zM}{a^3N_A} \) to find 'z', the number of atoms per unit cell. After plugging in the given values (molar mass, edge length, density, Avogadro's number), we find \( z = 4 \), which indicates a face-centered cubic (fcc) or cubic close-packed (ccp) structure.

🎯 Exam Tip: Master the density formula \( d = \frac{zM}{a^3N_A} \) and be able to rearrange it to solve for any variable. Remember that \( z=1 \) for simple cubic, \( z=2 \) for bcc, and \( z=4 \) for fcc/ccp.

 

Question 19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Answer: When a solid is heated, a vacancy defect is typically created. This occurs because thermal energy causes some atoms or ions to depart permanently from their regular lattice positions. Consequently, the density of the substance decreases due to the loss of particles from the crystal structure.
In simple words: Heating a solid can create a vacancy defect, where some atoms or ions leave their spots. This defect causes the solid's density to decrease.

🎯 Exam Tip: Understand that heating increases atomic vibrations, making it more likely for particles to leave their lattice sites, leading to vacancy defects and a decrease in density.

 

Question 20. What type of stoichiometric defect is shown by:
1. ZnS
2. AgBr
Answer: 1. Zinc sulfide (ZnS) primarily exhibits a Frenkel defect.
2. Silver bromide (AgBr) is unique in that it can show both Schottky and Frenkel defects.
In simple words: ZnS typically shows a Frenkel defect, while AgBr can show both Frenkel and Schottky defects.

🎯 Exam Tip: Recall the conditions for Frenkel (large size difference between ions, low coordination number) and Schottky defects (similar ion sizes, high coordination number). AgBr is a common exception that exhibits both.

 

Question 21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it?
Answer: When an ionic solid is doped with an impurity cation of higher valence, vacancies are introduced. For instance, if \( Sr^{2+} \) is introduced into an NaCl crystal, each \( Sr^{2+} \) ion replaces two \( Na^{+} \) ions to maintain electrical neutrality. One \( Sr^{2+} \) occupies the site of one \( Na^{+} \) ion, while the site of the second \( Na^{+} \) ion remains vacant, creating a cation vacancy. The total number of cation vacancies created is equal to the number of \( Sr^{2+} \) ions introduced.
In simple words: When an impurity with a higher charge (like \( Sr^{2+} \)) is added to an ionic solid (like NaCl), each impurity ion replaces two host ions (like \( Na^{+} \)). One host ion site gets the impurity, and the other becomes an empty spot, creating a vacancy.

🎯 Exam Tip: Focus on the principle of electrical neutrality. A higher-valent impurity ion replacing a lower-valent host ion will create vacancies to balance the charge.

 

Question 22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example?
Answer: Ionic solids exhibiting anionic vacancies due to a metal excess defect often display color. This phenomenon can be explained using zinc oxide (ZnO) as an example. When ZnO is heated, it transforms from white to yellow. This color change occurs because heating causes ZnO to lose oxygen, leading to an excess of zinc within the crystal. The surplus \( Zn^{2+} \) ions migrate to interstitial positions, and the electrons released move to adjacent interstitial sites. These interstitial electrons are capable of absorbing visible light, which gives ZnO its characteristic yellow color. The process can be represented as:
\( ZnO \xrightarrow{\Delta} Zn^{2+} + \frac{1}{2} O_2 + 2e \)
In simple words: Ionic solids get color from anionic vacancies, a metal excess defect. For example, when ZnO is heated, it turns yellow because it loses oxygen, creating excess zinc ions and free electrons that absorb light.

🎯 Exam Tip: Relate metal excess defects (like F-centers from anionic vacancies) directly to the absorption of light and the resulting color in ionic crystals.

 

Question 23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Answer: To convert a Group 14 element into an n-type semiconductor, an impurity that is "electron-rich" must be used, meaning it should provide extra electrons. Therefore, this impurity should belong to Group 15 of the periodic table.
In simple words: To make an n-type semiconductor from a Group 14 element, you need to add an impurity from Group 15, which has extra electrons to act as charge carriers.

🎯 Exam Tip: For n-type semiconductors, dope with elements having more valence electrons than the host (e.g., Group 15 for Group 14). For p-type, dope with elements having fewer valence electrons (e.g., Group 13 for Group 14).

 

Question 24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic? Justify your answer?
Answer: Ferromagnetic substances are generally superior for creating permanent magnets. This is because when exposed to an external magnetic field, the magnetic moments within a ferromagnetic material align permanently in a single direction, producing a robust and lasting magnetic effect.
In simple words: Ferromagnetic substances make better permanent magnets because their magnetic moments align strongly and permanently in one direction when magnetized, giving a powerful and lasting magnetic field.

🎯 Exam Tip: Differentiate between ferromagnetism (strong, permanent alignment of domains) and ferrimagnetism (unequal, antiparallel alignment resulting in a net small moment). Ferromagnetic materials retain magnetism better.

GSEB Class 12 Chemistry The Solid State Text Book Questions And Answers

 

Question 1. Define the term 'amorphous'. Give a few examples of amorphous solids?
Answer: An amorphous solid is a non-crystalline substance in which the constituent particles (atoms, molecules, or ions) lack a definite, long-range ordered structure. Consequently, these solids do not possess a distinct geometric shape and are sometimes referred to as pseudo-solids. Examples include glass, rubber, and plastic.
In simple words: Amorphous solids are solids without a regular, ordered internal structure, meaning their particles are randomly arranged. Examples are glass and plastic.

🎯 Exam Tip: The key characteristic of amorphous solids is the lack of long-range order in their atomic or molecular arrangement, distinguishing them from crystalline solids.

 

Question 2. What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Answer: Glass is an amorphous solid characterized by only short-range order in its particle arrangement, whereas quartz is a crystalline solid with a distinct long-range order among its constituent particles. Quartz can be transformed into glass by melting it through heating and then rapidly cooling the molten substance, which disrupts its ordered long-range structure.
In simple words: Glass is amorphous with short-range order, while quartz is crystalline with long-range order. Quartz can become glass if it's melted and then cooled very quickly.

🎯 Exam Tip: The crucial difference lies in the extent of order: glass has short-range order (amorphous), while quartz has long-range order (crystalline). Rapid cooling prevents crystal formation.

 

Question 3. Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous?
1. Tetra phosphorus decoxide (\(P_4O_{10}\))
2. Ammonium phosphate (\((NH_4)_3PO_4\))
3. SiC
4. \(I_2\)
5. \(P_4\)
6. Plastic
7. Graphite
8. Brass
9. Rb
10. LiBr
11. Si
Answer: Ionic solids: Ammonium phosphate (\((NH_4)_3PO_4\)), Lithium bromide (LiBr)
Metallic solids: Brass, Rubidium (Rb)
Molecular solids: Tetra phosphorus decoxide (\(P_4O_{10}\)), Iodine (\(I_2\)), White phosphorus (\(P_4\))
Network (Covalent) solids: Graphite, Silicon carbide (SiC), Silicon (Si)
Amorphous solids: Plastics
In simple words: Solids are classified based on their bonding: ionic (salts), metallic (alloys, metals), molecular (non-metals, organic compounds), covalent network (strong covalent bonds), or amorphous (disordered structure like plastics).

🎯 Exam Tip: Understand the nature of bonding in each substance (e.g., metallic bonds in brass, ionic bonds in LiBr, covalent bonds in SiC, van der Waals forces in \(I_2\)) to correctly categorize them.

 

Question 4.
1. What is meant by the term 'coordination number'?
2. What is the coordination number of atoms
• In a cubic close-packed structure.
• In a body-centered cubic structure.
Answer: 1. The 'coordination number' refers to the total number of nearest neighboring particles that are in direct contact with a specific particle within a close-packed crystal structure.
2. The coordination number for atoms is:
• In a cubic close-packed (ccp) structure, it is 12.
• In a body-centered cubic (bcc) structure, it is 8.
In simple words: Coordination number is how many particles a central particle touches. In cubic close-packed, it's 12; in body-centered cubic, it's 8.

🎯 Exam Tip: Remember the coordination numbers for common crystal structures: 12 for hcp/ccp, 8 for bcc, and 6 for simple cubic.

 

Question 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?
Answer: To determine the atomic mass (M) of an unknown metal, given its density (d) and unit cell dimensions, we can utilize the following formula derived from crystal structure properties:
Let 'a' be the edge length of the cubic unit cell.
The volume of the unit cell is \( a^3 \).
The mass of the unit cell is calculated as \( z \times m \), where 'z' is the number of atoms per unit cell (1 for simple cubic, 2 for body-centered cubic, 4 for face-centered cubic) and 'm' is the mass of a single atom.
The mass of a single atom 'm' can be expressed as \( \frac{M}{N_A} \), where 'M' is the atomic mass and \( N_A \) is Avogadro's number (\( 6.022 \times 10^{23} \)).
Substituting 'm' into the unit cell mass equation, we get the mass of the unit cell as \( \frac{zM}{N_A} \).
Therefore, the density of the unit cell is given by:
\( d = \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} = \frac{zM}{a^3N_A} \)
From this formula, the atomic mass (M) can be calculated as:
\( M = \frac{d \cdot a^3 \cdot N_A}{z} \)
Ensure consistency in units (e.g., if 'a' is in cm and 'M' in g, then 'd' should be in g/cm\(^3\)).
In simple words: You can find an element's atomic mass by using its density and unit cell size. The formula \( d = \frac{zM}{a^3N_A} \) relates density (d), atomic mass (M), atoms per unit cell (z), edge length (a), and Avogadro's number (\( N_A \)). Rearrange it to solve for M.

🎯 Exam Tip: This is a fundamental formula in solid-state chemistry. Understand each variable and unit conversions (pm to cm/m, kg to g) are critical for accurate calculations.

 

Question 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?
Answer: The assertion that 'the stability of a crystal is reflected in the magnitude of its melting point' is accurate. A higher melting point indicates greater crystal stability because a substantial amount of thermal energy is needed to disrupt the highly ordered crystalline structure and transition it into a disordered liquid state.
Here are approximate melting points for the given substances and observations regarding their intermolecular forces:
• Water (ice): 273 K ( \( 0^\circ C \) ) - Possesses strong hydrogen bonding.
• Ethyl alcohol: 158 K - Exhibits hydrogen bonding, though weaker compared to water.
• Diethyl ether: 157 K - Lacks hydrogen bonding; primarily held by weaker van der Waals forces of attraction.
• Methane: 90 K - Held together by very weak van der Waals forces.
**Observation on Intermolecular Forces:** The melting points decrease significantly from water to methane, which directly correlates with the decreasing strength of intermolecular forces. Water has the highest melting point due to strong hydrogen bonding. Ethyl alcohol also has hydrogen bonding, but less extensive than water, leading to a lower melting point. Diethyl ether, lacking hydrogen bonding, has a melting point similar to ethyl alcohol but relies on van der Waals forces. Methane, with only weak van der Waals forces, has the lowest melting point among them.
In simple words: A higher melting point means a more stable crystal, as more energy is needed to break its organized structure. Water (strong hydrogen bonds) has a much higher melting point than methane (weak van der Waals forces), showing how intermolecular forces affect melting points.

🎯 Exam Tip: Recognize the direct relationship between the strength of intermolecular forces (hydrogen bonding > dipole-dipole > London dispersion) and physical properties like melting point and boiling point.

 

Question 7. Distinguish between the following pairs of terms?
1. Hexagonal close-packing and cubic close-packing?
2. Crystal lattice and unit cell?
3. Tetrahedral void and octahedral void?
Answer: 1. **Hexagonal close-packing (hcp) and cubic close-packing (ccp):**
During crystal formation, constituent particles are arranged efficiently to maximize space occupancy, which influences the crystal's stability. In this close-packing model, solid particles are idealized as identical spheres.
*Close Packing in One Dimension:* Spheres can only be arranged in a single row, touching each other to form a one-dimensional close-packed structure.
*Close Packing in Two Dimensions:* From the first row, two methods can be employed to construct a crystal plane:
(i) Square Close Packing: Here, spheres are aligned both horizontally and vertically, forming a square pattern.
(ii) Hexagonal Close Packing: In this arrangement, spheres in the second row are placed within the depressions of the first row. This pattern continues, with spheres in the third row settling into the depressions of the second row, and so forth. This type of layering leads to hexagonal close packing.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो प्रकार की द्वि-आयामी (two-dimensional) संकुलन संरचनाओं को दर्शाता है। शीर्ष भाग वर्ग संकुलन (Square close packing) को दिखाता है जहाँ प्रत्येक गोला अपने सीधे ऊपर, नीचे, बाईं और दाईं ओर चार गोलों से घिरा होता है। निचला भाग षटकोणीय संकुलन (Hexagonal close packing) को दर्शाता है जहाँ प्रत्येक गोला छह पड़ोसी गोलों से घिरा होता है, जिसमें दूसरी पंक्ति के गोले पहली पंक्ति के अवसादों में स्थित होते हैं, जो अधिक सघन संकुलन है।
Hexagonal close packing achieves a more efficient packing density compared to square close packing. For single layers, spheres naturally arrange themselves in hexagonal close packing due to its superior efficiency. Even in this arrangement, several vacant spaces persist.
These vacant spaces within close-packed structures are termed 'voids'. There are two primary types:
A **tetrahedral void** is a vacant space enclosed by four touching spheres arranged in a tetrahedral geometry within a close-packed lattice. In any close-packed arrangement, the number of tetrahedral voids is precisely double the number of spheres.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो मुख्य प्रकार के छिद्रों (voids) को दिखाता है जो घनी संकुलित संरचनाओं में बनते हैं। बाईं ओर चतुष्फलकीय छिद्र (Tetrahedral void) दर्शाया गया है, जहाँ चार गोले एक-दूसरे को छूते हुए एक त्रिभुज बनाते हैं, और उनके केंद्र में एक छोटा सा खाली स्थान बनता है। दाईं ओर अष्टफलकीय छिद्र (Octahedral void) दर्शाया गया है, जो छह गोलों से घिरा होता है, जिसमें तीन गोले एक तल में ऊपर और तीन गोले दूसरे तल में नीचे होते हैं, जो उनके बीच एक बड़ा अष्टफलकीय खाली स्थान बनाते हैं।
An **octahedral void** is a void surrounded by six spheres in a close-packed arrangement. In a close packing, the number of octahedral voids is equal to the number of spheres. The total number of voids depends on the number of close-packed spheres (N) in the crystal. Specifically, for N particles in a crystal:
• The number of octahedral voids = N
• The number of tetrahedral voids = 2N

2. **Crystal lattice and unit cell:**
A **crystal lattice**, also known as a space lattice, is the three-dimensional, ordered arrangement of constituent particles (atoms, ions, or molecules) within a crystal. There are 14 distinct three-dimensional lattices, referred to as Bravais lattices. Key features of a crystal lattice include:
(i) Each specific position is called a lattice point or lattice site.
(ii) Each lattice point represents one constituent particle (which could be an atom, a molecule, or an ion).
(iii) Lattice points are connected by straight lines to illustrate the overall geometry of the lattice.
A **unit cell** is the smallest repeating segment of a crystal lattice that, when repeated in all three dimensions, generates the entire lattice. A unit cell is defined by six parameters: the lengths of its three edges (a, b, c) and the three angles (\(\alpha\), \(\beta\), \(\gamma\)) formed between these edges. Analysis of various crystal lattices reveals seven fundamental types of unit cells or crystal systems.
The seven basic unit cells or crystal systems are characterized by specific axial distances (a, b, c) and axial angles (\(\alpha\), \(\beta\), \(\gamma\)), as summarized in the table below:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र सात मूल क्रिस्टल प्रणालियों (क्यूबिक, टेट्रागोनल, हेक्सागोनल, रोम्बोहेड्रल, ऑर्थोरोम्बिक, मोनोक्लिनिक, और ट्राइक्लिनिक) के इकाई कोष्ठिकाओं (unit cells) की त्रि-आयामी ज्यामिति को दर्शाता है। प्रत्येक आरेख में, 'a', 'b', 'c' अक्षीय दूरियों को और 'α', 'β', 'γ' अक्षीय कोणों को दर्शाते हैं, जो प्रत्येक क्रिस्टल प्रणाली के विशिष्ट ज्यामितीय विन्यास को परिभाषित करते हैं। ये आरेख विभिन्न क्रिस्टल प्रणालियों के बीच के संरचनात्मक अंतर को समझने में मदद करते हैं।

System	Axial distance	Axial angles	Examples
1. Cubic	a = b = c	\(\alpha = \beta = \gamma = 90^\circ\)	ZnS, KCl, NaCl, Cu, Alums, diamond
2. Tetragonal	a = b \(\neq\) c	\(\alpha = \beta = \gamma = 90^\circ\)	Sn, \(SnO_2\), \(TiO_2\), \(CaSO_4\)
3. Orthorhombic or rhombic	a \(\neq\) b \(\neq\) c	\(\alpha = \beta = \gamma = 90^\circ\)	Rhombic Sulphur, \(KNO_3\), \(K_2SO_4\), \(BaSO_4\)
4. Monoclinic	a \(\neq\) b \(\neq\) c	\(\alpha = \gamma = 90^\circ\), \(\beta \neq 90^\circ\)	Monoclinic sulphur, \(Na_2SO_4 \cdot 10 H_2O\), \(FeSO_4\)
5. Trigonal or Rhombohedral	a = b = c	\(\alpha = \beta = \gamma \neq 90^\circ\)	\(NaNO_3\), Sb, As, Calcite (\(CaCO_3\)), HgS (Cinnabar)
6. Hexagonal	a = b \(\neq\) c	\(\alpha = \beta = 90^\circ\), \(\gamma = 120^\circ\)	HgS, Ice, ZnO, CdS, Graphite.
7. Triclinic	a \(\neq\) b \(\neq\) c	\(\alpha \neq \beta \neq \gamma \neq 90^\circ\)	\(CuSO_4 \cdot 5H_2O\), \(K_2Cr_2O_7\), \(H_3BO_3\)

3. **Tetrahedral void and octahedral void:**
(a) **Covering tetrahedral voids (Hexagonal Close-Packed Structure - hcp):** When a third layer of spheres is positioned directly over the tetrahedral voids of the second layer, the spheres of this third layer align precisely with those of the first layer. This results in an ABAB... stacking sequence, which defines the hexagonal close-packed (hcp) structure.
Elements like Magnesium (Mg), Zinc (Zn), and Cadmium (Cd) typically form hcp structures.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र हेक्सागोनल क्लोज-पैक्ड (hcp) संरचना को दर्शाता है। इसमें गोले की परतें इस तरह व्यवस्थित होती हैं कि तीसरी परत पहली परत के समान होती है, जिससे ABAB प्रकार की स्टैकिंग बनती है। यह व्यवस्था ठोसों में अधिकतम दक्षता वाले संकुलन को प्रदर्शित करती है, जहाँ प्रत्येक गोले को 12 पड़ोसी गोले छूते हैं।
(b) **Covering octahedral voids (Cubic Close-Packed Structure - ccp/fcc):** When the third layer of spheres is positioned to cover the octahedral voids of the second layer, a distinct arrangement emerges. In this configuration, the first, fourth, seventh, and subsequent layers become identical, forming an ABCABC... stacking sequence. This defines the cubic close-packed (ccp) or face-centered cubic (fcc) structure.
Elements such as Copper (Cu), Nickel (Ni), and Gold (Au) crystallize in ccp arrangements.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र क्यूबिक क्लोज-पैक्ड (ccp) या फेस-सेंटर्ड क्यूबिक (fcc) संरचना को दर्शाता है। इसमें गोले की परतें इस तरह व्यवस्थित होती हैं कि चौथी परत पहली परत के समान होती है, जिससे ABCABC प्रकार की स्टैकिंग बनती है। यह व्यवस्था भी उच्च संकुलन दक्षता दर्शाती है, जहाँ प्रत्येक गोले को 12 पड़ोसी गोले छूते हैं, लेकिन स्टैकिंग का क्रम hcp से भिन्न होता है।
In simple words: Hexagonal close-packing (hcp) uses an ABAB... layering by covering tetrahedral voids, while cubic close-packing (ccp/fcc) uses an ABCABC... layering by covering octahedral voids. Both maximize packing efficiency but differ in their stacking patterns.

🎯 Exam Tip: Distinguish between hcp (ABABAB stacking, covering tetrahedral voids) and ccp/fcc (ABCABC stacking, covering octahedral voids) by their layering patterns and the types of voids they cover.

 

Question 8. How many lattice points are there in one unit cell of each of the following lattices?
1. Face-centered cubic
2. Face-centered tetragonal
3. Body-centered
Answer: The number of lattice points (or effective number of atoms) per unit cell for the specified lattices is:
1. Face-centered cubic (fcc): 4
2. Face-centered tetragonal: 4
3. Body-centered cubic (bcc): 2
In simple words: For a face-centered cubic or tetragonal cell, there are 4 effective lattice points. For a body-centered cell, there are 2.

🎯 Exam Tip: Remember how to calculate the effective number of atoms (z) per unit cell based on the contribution of atoms at corners (1/8), faces (1/2), edges (1/4), and body center (1).

Question 9. Explain?
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.

Answer:
(i) **Metallic and Ionic Crystals:**
* **Similarities:** Both metallic and ionic solids exhibit elevated melting points.
* **Differences:** * While ionic solids are typically rigid and fragile, metallic solids are also rigid but possess malleability and ductility. * Ionic solids are poor electrical conductors in their solid state but become good conductors when molten or dissolved. Conversely, metallic solids conduct electricity well in both solid and liquid states, and even as vapor. * Ionic solids are composed of cations and anions. Metallic solids, however, consist of kernels (positively charged ions) enveloped by a mobile 'sea' of delocalized electrons.
(ii) **Hardness and Brittleness of Ionic Solids:**
Ionic crystals are characterized by their hardness due to powerful electrostatic forces attracting oppositely charged ions. Their brittleness stems from the non-directional nature of ionic bonds, meaning a shift in alignment can lead to strong repulsions between like-charged ions, causing cleavage.
In simple words: Metallic and ionic solids both have high melting points, but metals are ductile conductors while ionic solids are brittle conductors only when molten. Ionic solids are hard because of strong forces, but brittle because these forces are non-directional.

🎯 Exam Tip: When comparing crystal types, focus on the nature of their bonding and constituent particles, as these determine physical properties like conductivity, malleability, and melting point.

Question 10. Calculate the efficiency of packing in the case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other)

Answer:
(i) **Simple Cubic Structure:**
In a simple cubic arrangement, atoms are in contact along the edges of the unit cell.
* The edge length of the unit cell, denoted 'a', is equivalent to \( 2r \), where 'r' is the atomic radius.
* A simple unit cell contains only one atom.
* Volume of the sphere(s) = \( \frac{4}{3} \pi r^3 \)
* Volume of the unit cell = \( a^3 = (2r)^3 = 8r^3 \)
* Packing efficiency = \( \frac{\text{Volume occupied by spheres}}{\text{Volume of the unit cell}} \times 100 \% \)
* Packing efficiency = \( \frac{\frac{4}{3} \pi r^3}{8r^3} \times 100 \% = \frac{\pi}{6} \times 100 \% \approx 52.4 \% \)
(ii) **Body-Centred Cubic (BCC) Structure:**
In a body-centred cubic structure, atoms touch along the body diagonal of the cube.
* The length of the body diagonal is \( 4r \).
* Thus, \( \sqrt{3}a = 4r \)
\( \implies r = \frac{\sqrt{3}}{4} a \)
* Each BCC unit cell contains 2 spheres.
* Total volume of 2 spheres = \( 2 \times (\frac{4}{3}) \pi r^3 = \frac{8}{3} \pi r^3 \)
* Volume of the cube = \( a^3 = (\frac{4r}{\sqrt{3}})^3 = \frac{64r^3}{3\sqrt{3}} \)
* Packing efficiency = \( \frac{\text{Volume occupied by spheres}}{\text{Volume of the unit cell}} \times 100 \% \)
* Packing efficiency = \( \frac{\frac{8}{3} \pi r^3}{\frac{64r^3}{3\sqrt{3}}} \times 100 \% = \frac{\pi \sqrt{3}}{8} \times 100 \% \approx 68 \% \)
(iii) **Face-Centred Cubic (FCC) Structure:**
In a face-centred cubic arrangement, the atoms make contact along the face diagonal.
* The length of the face diagonal is \( 4r \).
* So, \( 4r = \sqrt{2}a \)
\( \implies a = \frac{4r}{\sqrt{2}} = 2\sqrt{2}r \) (or \( r = \frac{a}{2\sqrt{2}} \))
* Each FCC unit cell contains 4 spheres.
* Total volume of four spheres = \( 4 \times \frac{4}{3} \pi r^3 = \frac{16}{3} \pi r^3 \)
* Volume of the cube = \( a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3 \)
* Packing efficiency = \( \frac{\text{Volume occupied by spheres}}{\text{Total volume of the unit cell}} \times 100 \% \)
* Packing efficiency = \( \frac{\frac{16}{3} \pi r^3}{16\sqrt{2}r^3} \times 100 \% = \frac{\pi}{3\sqrt{2}} \times 100 \% \approx 74 \% \)
* Both hexagonal close-packed (hcp) and cubic close-packed (ccp) structures exhibit equal packing efficiency of approximately 74%.
In simple words: Packing efficiency measures how much space atoms take up in a crystal. Simple cubic is least efficient (52.4%), BCC is better (68%), and FCC/HCP are most efficient (74%) because atoms are packed more closely.

🎯 Exam Tip: Remember the 'z' value (number of atoms per unit cell) for each lattice type: simple cubic (z=1), BCC (z=2), and FCC (z=4). These values are crucial for all packing efficiency and density calculations.

Question 11. Silver crystallises in fee lattice. If edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-³, calculate the atomic mass of silver?

Answer:Given: Edge length \( a = 4.07 \times 10^{-8} \) cm, density \( d = 10.5 \) g cm\(^{-3} \). For a face-centred cubic (FCC) lattice, the number of atoms per unit cell, \( z = 4 \). Avogadro's number \( N_A = 6.022 \times 10^{23} \) mol\(^{-1} \). The formula for density is: \[ d = \frac{zM}{a^3 N_A} \] To find atomic mass M, rearrange the formula: \[ M = \frac{d a^3 N_A}{z} \] Substitute the given values into the formula: \[ M = \frac{10.5 \text{ g cm}^{-3} \times (4.07 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{4} \] \[ M = \frac{10.5 \times (4.07)^3 \times 10^{-24} \times 6.022 \times 10^{23}}{4} \] \[ M \approx \frac{10.5 \times 67.419 \times 0.6022}{4} \text{ g mol}^{-1} \] \[ M \approx \frac{424.36}{4} \text{ g mol}^{-1} \] \[ M \approx 106.09 \text{ g mol}^{-1} \] The calculated atomic mass of silver is approximately 107 g mol\(^{-1} \) or 107 u.
In simple words: We used the crystal's density, unit cell size, and type (FCC) to calculate the atomic mass of silver, finding it to be about 107 grams per mole.

🎯 Exam Tip: Always ensure unit consistency (e.g., cm for length, g for mass) and use Avogadro's number correctly when performing calculations involving crystal density and atomic mass.

Question 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-centre. What is the formula of the compound? What are the numbers of P and Q?

Answer:
* In a cubic unit cell, atoms located at the corners contribute \( \frac{1}{8} \) to each cell. Since there are 8 corners in a cube, the total number of Q atoms per unit cell is \( 8 \times \frac{1}{8} = 1 \).
* An atom situated at the body-center of a cubic unit cell belongs entirely to that specific unit cell. Therefore, the number of P atoms per unit cell is 1.
* Based on these counts, the empirical formula of the compound is PQ.
* In this arrangement, the coordination number for both element P and element Q is 8, meaning each atom is surrounded by 8 nearest neighbors.
In simple words: With Q atoms at the cube corners and P at the center, each unit cell effectively has one P atom and one Q atom, making the compound formula PQ, and both P and Q have a coordination number of 8.

🎯 Exam Tip: Remember standard contributions of atoms at different lattice positions: corners (\( \frac{1}{8} \)), face centers (\( \frac{1}{2} \)), edge centers (\( \frac{1}{4} \)), and body center (1).

Question 13. Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm-³, calculate atomic radius of niobium using its atomic mass 93 u?

Answer:Given: For a BCC structure, the number of atoms per unit cell, \( z = 2 \). Density, \( d = 8.55 \) g cm\(^{-3} \). Molar mass, \( M = 93 \) u (which is \( 93 \) g mol\(^{-1} \)). Avogadro's number, \( N_A = 6.022 \times 10^{23} \) mol\(^{-1} \). The density formula is: \[ d = \frac{zM}{a^3 N_A} \] First, we calculate the edge length 'a' by solving for \( a^3 \): \[ a^3 = \frac{zM}{d N_A} \] Substitute the given values: \[ a^3 = \frac{2 \times 93 \text{ g mol}^{-1}}{8.55 \text{ g cm}^{-3} \times 6.022 \times 10^{23} \text{ mol}^{-1}} \] \[ a^3 = \frac{186}{51.4881 \times 10^{23}} \text{ cm}^3 \] \[ a^3 \approx 3.612 \times 10^{-24} \text{ cm}^3 \] Now, calculate 'a' by taking the cube root: \[ a = \sqrt[3]{3.612 \times 10^{-24} \text{ cm}^3} \] \[ a \approx 3.30 \times 10^{-8} \text{ cm} \] For a Body-Centred Cubic (BCC) structure, the relationship between the edge length 'a' and the atomic radius 'r' is: \[ \sqrt{3}a = 4r \] So, we can find 'r' as: \[ r = \frac{\sqrt{3}a}{4} \] Substitute the calculated value of 'a': \[ r = \frac{1.732 \times 3.30 \times 10^{-8} \text{ cm}}{4} \] \[ r = \frac{5.7156 \times 10^{-8} \text{ cm}}{4} \] \[ r \approx 1.429 \times 10^{-8} \text{ cm} \] The atomic radius of niobium is approximately \( 1.429 \times 10^{-8} \) cm.
In simple words: We used niobium's density and molar mass to first find its unit cell's edge length in a BCC structure. Then, using the specific BCC geometry relationship, we calculated its atomic radius.

🎯 Exam Tip: Be careful with units and exponents during calculations. Remember the correct geometric relationship (\( \sqrt{3}a = 4r \)) for BCC structures to relate edge length to atomic radius.

Question 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?

Answer:The relationship between the radius of an octahedral void (r) and the radius of the atoms in close packing (R) can be derived by considering the geometry of an octahedral void. An octahedral void is formed at the center of six spheres. When these spheres touch each other, the small sphere (void) at the center also touches all six surrounding spheres. By analyzing the geometry, specifically by taking a cross-section through the center of the void and the centers of four surrounding spheres lying in a plane, we can establish this relationship. In such a plane, the large spheres touch along the diagonal, while the small sphere (void) touches the large spheres along the axes. From geometric considerations, the radius of an octahedral void (r) is related to the radius of the atoms in close packing (R) by the expression: \[ r = 0.414 R \]
In simple words: The radius of an octahedral void is 0.414 times the radius of the larger atoms that form the close-packed structure.

🎯 Exam Tip: Memorize the radius ratios for tetrahedral (\( r = 0.225 R \)) and octahedral (\( r = 0.414 R \)) voids, as these are frequently asked in objective and short-answer questions.

Question 15. Copper crystallises into a fcc lattice with edge length 3.61 × 10-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm-³.

Answer:Given: For a Face-Centred Cubic (FCC) lattice, the number of atoms per unit cell, \( z = 4 \). Molar mass of copper, \( M = 63.5 \) g mol\(^{-1} \). Edge length of the unit cell, \( a = 3.61 \times 10^{-8} \) cm. Avogadro's number, \( N_A = 6.022 \times 10^{23} \) mol\(^{-1} \). Using the density formula: \[ d = \frac{zM}{a^3 N_A} \] Substitute the given values into the formula: \[ d = \frac{4 \times 63.5 \text{ g mol}^{-1}}{(3.61 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \] First, calculate \( a^3 \): \( a^3 = (3.61)^3 \times (10^{-8})^3 \text{ cm}^3 = 47.045881 \times 10^{-24} \text{ cm}^3 \) Now, substitute \( a^3 \) into the density formula: \[ d = \frac{254 \text{ g}}{47.045881 \times 10^{-24} \text{ cm}^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \] \[ d = \frac{254}{47.045881 \times 0.6022} \text{ g cm}^{-3} \] \[ d = \frac{254}{28.339} \text{ g cm}^{-3} \] \[ d \approx 8.96 \text{ g cm}^{-3} \] The calculated density of copper is approximately 8.96 g cm\(^{-3} \), which is in excellent agreement with its measured value of 8.92 g cm\(^{-3} \).
In simple words: We calculated copper's density using its FCC structure, edge length, and molar mass, and the result closely matched the experimentally observed density, confirming the crystal structure.

🎯 Exam Tip: Double-check the number of atoms per unit cell ('z') for the given crystal structure and ensure all units are consistent before performing calculations.

Question 16. Analysis shows that nickel oxide has the formula Ni98O100. What fractions of nickel exist as Ni²+ and Ni³+ ions?

Answer:- For the given formula Ni\(_{98}\)O\(_{100}\), it implies that for every 100 oxygen atoms, there are 98 nickel atoms. - Since each oxygen atom (O) has a charge of -2, the total negative charge from 100 oxygen atoms is \( 100 \times (-2) = -200 \). - For the compound to be electrically neutral, the total positive charge from the nickel ions must be +200. - Let 'x' be the number of Ni\(^{2+}\) ions. - Then, the number of Ni\(^{3+}\) ions will be the total nickel atoms minus Ni\(^{2+}\) atoms, which is \( (98 - x) \). - The total positive charge from nickel ions can be expressed as: \( 2 \times (\text{number of Ni}^{2+}) + 3 \times (\text{number of Ni}^{3+}) \) - Equating the total positive charge to +200: \[ 2x + 3(98 - x) = 200 \] \[ 2x + 294 - 3x = 200 \] \[ 294 - x = 200 \] \[ x = 294 - 200 \] \[ x = 94 \] - Therefore, the number of Ni\(^{2+}\) ions is 94. - The number of Ni\(^{3+}\) ions is \( 98 - 94 = 4 \). - The fraction of Ni\(^{2+}\) ions = \( \frac{94}{98} \approx 0.959 \) - The fraction of Ni\(^{3+}\) ions = \( \frac{4}{98} \approx 0.0408 \) or approximately 4%.
In simple words: In Ni\(_{98}\)O\(_{100}\), we determined the relative amounts of Ni\(^{2+}\) and Ni\(^{3+}\) ions by ensuring the crystal remains electrically neutral, finding 94 Ni\(^{2+}\) and 4 Ni\(^{3+}\) ions.

🎯 Exam Tip: Problems involving non-stoichiometric compounds often require setting up an equation to balance the total positive and negative charges to determine the proportions of different oxidation states.

Question 17. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism?

Answer:- A semiconductor is a material with electrical conductivity that falls between that of a conductor (like metals) and an insulator. Their conductivity can be modified by temperature or by introducing impurities (doping). - There are two primary types of semiconductors: - (i) **n-type semiconductors:** These are formed by doping a group 14 element (e.g., Silicon or Germanium) with a group 15 element (e.g., Phosphorus or Arsenic). Group 15 elements have five valence electrons; four form covalent bonds with the host atoms, leaving one extra electron that becomes delocalized and contributes to electrical conductivity. In n-type semiconductors, electrons are the majority charge carriers. - (ii) **p-type semiconductors:** These are formed by doping a group 14 element (e.g., Silicon or Germanium) with a group 13 element (e.g., Boron, Aluminum, or Gallium). Group 13 elements have three valence electrons; when they replace a host atom, they form only three bonds, creating an electron deficiency or a 'hole' where the fourth electron should be. An electron from an adjacent bond can move to fill this hole, effectively making the hole appear to move in the opposite direction. In p-type semiconductors, positive holes are the majority charge carriers.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख n-प्रकार और p-प्रकार के अर्धचालकों में आवेश वाहकों की गति को दर्शाता है। n-प्रकार के अर्धचालक में, अतिरिक्त इलेक्ट्रॉन चालक होते हैं और धनात्मक प्लेट की ओर बढ़ते हैं। p-प्रकार के अर्धचालक में, इलेक्ट्रॉन-छिद्र (holes) धनात्मक आवेश वाहक होते हैं और ऋणात्मक प्लेट की ओर बढ़ते हैं।
In simple words: Semiconductors conduct electricity better than insulators but less than metals. n-type semiconductors have extra electrons from doping (group 15 impurities), while p-type semiconductors have "holes" (missing electrons) from doping (group 13 impurities).

🎯 Exam Tip: Remember that n-type semiconductors use electron-rich impurities (donors) to provide excess electrons, while p-type semiconductors use electron-deficient impurities (acceptors) to create holes.

Question 18. Non-stoichiometric cuprous oxide, Cu2O 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?

Answer:- When the copper to oxygen ratio in Cu\(_{2}\)O is slightly less than 2:1, it signifies the presence of a metal deficiency defect within the crystal lattice. - In this non-stoichiometric crystal, some Cu\(^{+}\) ions are missing from their regular sites, and to maintain electrical neutrality, an equivalent number of Cu\(^{2+}\) ions are formed in other positions. - When an electron moves from a Cu\(^{+}\) ion to a neighboring Cu\(^{2+}\) ion, a positive 'hole' is generated at the site previously occupied by the Cu\(^{+}\) ion. This hole is mobile and can effectively move through the crystal. - Consequently, the primary mechanism for electrical current flow in this material is through the movement of these positive holes, which is characteristic of a p-type semiconductor.
In simple words: Because Cu\(_{2}\)O has less copper than ideal (metal deficiency), some Cu\(^{+}\) ions become Cu\(^{2+}\) ions. This creates positive "holes" that can move and conduct electricity, making it a p-type semiconductor.

🎯 Exam Tip: Metal deficiency defects often lead to p-type semiconductor behavior as higher oxidation state ions create positive holes that act as charge carriers.

Question 19. Ferric oxide crystallises 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?

Answer:- Let the number of oxide ions (O\(^{2-}\)) in the hexagonal close-packed (hcp) array be 'n'. - In an hcp structure, the number of octahedral voids is equal to the number of spheres, so the number of octahedral voids is also 'n'. - Ferric ions (Fe\(^{3+}\)) are stated to occupy two out of every three octahedral holes. - Therefore, the number of Fe\(^{3+}\) ions = \( \frac{2}{3} \times n \). - The ratio of Fe\(^{3+}\) ions to O\(^{2-}\) ions can be written as: \[ \text{Fe}^{3+} : \text{O}^{2-} = \frac{2}{3}n : n \] - To simplify this ratio, divide both sides by 'n': \[ \frac{2}{3} : 1 \] - Multiply by 3 to get whole numbers: \[ 2 : 3 \] - Thus, for every 2 ferric ions, there are 3 oxide ions. - The formula of ferric oxide is Fe\(_{2}\)O\(_{3}\).
In simple words: In ferric oxide, for every 'n' oxide ions, there are 'n' octahedral voids. Since ferric ions occupy two-thirds of these voids, the ratio of ferric ions to oxide ions is 2:3, leading to the formula Fe\(_{2}\)O\(_{3}\).

🎯 Exam Tip: Remember the relationship between the number of spheres (N) in a close-packed array and the number of voids: N octahedral voids and 2N tetrahedral voids.

Question 20. Classify each of the following as being either a p-type or an n-type semiconductor:
(i) Silicon doped with Indium (In)
(ii) Silicon doped with Arsenic (As)

Answer:
(i) **Silicon doped with Indium (In):**
Indium (In) is a group 13 element, possessing three valence electrons. When it dopes silicon (Si), a group 14 element with four valence electrons, Indium forms three covalent bonds with neighboring silicon atoms but creates an electron deficiency or a 'hole' in the fourth position. This results in positive hole conduction, classifying it as a **p-type semiconductor**.
(ii) **Silicon doped with Arsenic (As):**
Arsenic (As) is a group 15 element, having five valence electrons. When it dopes silicon, four of Arsenic's valence electrons form covalent bonds with adjacent silicon atoms, leaving one extra electron. This excess electron becomes delocalized and readily available for conduction. This mechanism, where electrons are the majority charge carriers, classifies it as an **n-type semiconductor**.
In simple words: Doping silicon with Indium (Group 13) creates electron holes, making it a p-type semiconductor. Doping silicon with Arsenic (Group 15) provides extra electrons, making it an n-type semiconductor.

🎯 Exam Tip: To classify a doped semiconductor, identify the group number of the dopant relative to the host material (e.g., Group 13 dopant in Group 14 host leads to p-type; Group 15 dopant in Group 14 host leads to n-type).

Question 21. Gold (atomic radius = 0.144 nm) crystallises in a face-centered unit cell. What is the length of a side of the cell?

Answer:Given: Atomic radius \( r = 0.144 \) nm. For a face-centered cubic (FCC) unit cell, the relationship between the edge length 'a' and the atomic radius 'r' is: \[ a = 2\sqrt{2}r \] Substitute the given atomic radius into the formula: \[ a = 2 \times \sqrt{2} \times 0.144 \text{ nm} \] Using the approximate value of \( \sqrt{2} \approx 1.414 \): \[ a = 2 \times 1.414 \times 0.144 \text{ nm} \] \[ a = 2.828 \times 0.144 \text{ nm} \] \[ a \approx 0.407 \text{ nm} \] The length of a side (edge length) of the gold unit cell is approximately 0.407 nm.
In simple words: For gold, which has an FCC structure, we used its atomic radius and the specific geometric formula for FCC cells to calculate that the side length of its unit cell is 0.407 nm.

🎯 Exam Tip: Remember the geometric relationship for FCC structures: \( a = 2\sqrt{2}r \). This formula is key to solving problems involving edge length and atomic radius in FCC lattices.

Question 22. In terms of band theory, what is the difference
(i) between a conductor and an insulator
(ii) between a conductor and a semiconductor?

Answer:Based on band theory, the electrical conductivity of solids is determined by the availability of electrons in the conduction band and the energy gap between the valence band and the conduction band.
(i) **Difference between a conductor and an insulator:**
* **Conductor:** In conductors (like metals), the valence band and conduction band either overlap or the valence band is partially filled. This allows electrons to move freely into available energy states, leading to very high electrical conductivity (typically \( 10^6 \) to \( 10^8 \) ohm\(^{-1} \) cm\(^{-1} \)).
* **Insulator:** In insulators, there is a very large energy gap (forbidden zone) between a completely filled valence band and an empty conduction band. Electrons require a substantial amount of energy to jump across this gap, which is usually not available, resulting in extremely low electrical conductivity (typically \( 10^{-12} \) ohm\(^{-1} \) cm\(^{-1} \) or lower).
(ii) **Difference between a conductor and a semiconductor:**
* **Conductor:** As described above, conductors have overlapping or partially filled bands, allowing for easy electron flow and high conductivity. Their conductivity generally decreases with increasing temperature due to increased electron-phonon scattering.
* **Semiconductor:** Semiconductors have a small but finite energy gap between the valence band and the conduction band. At absolute zero, they behave like insulators. However, at room temperature or with a small amount of energy (like heat), some electrons can jump from the valence band to the conduction band, enabling moderate electrical conductivity (typically \( 10^{-2} \) to \( 10^{-9} \) ohm\(^{-1} \) cm\(^{-1} \)). Crucially, the conductivity of semiconductors increases with increasing temperature, as more electrons gain enough energy to cross the band gap.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख ऊर्जा बैंड सिद्धांत के आधार पर विभिन्न प्रकार के ठोस पदार्थों में चालन बैंड और संयोजकता बैंड को दर्शाता है। यह दर्शाता है कि कुचालकों में एक बड़ा वर्जित क्षेत्र होता है, अर्धचालकों में एक छोटा वर्जित क्षेत्र होता है, जबकि चालकों में आंशिक रूप से भरे हुए बैंड या अतिव्यापी बैंड होते हैं, जिससे इलेक्ट्रॉनों का मुक्त प्रवाह होता है।
In simple words: Conductors have accessible electrons for current due to overlapping or partially filled energy bands. Insulators have a huge energy gap preventing electron movement. Semiconductors have a small energy gap, so they conduct moderately, and their conductivity increases with temperature.

🎯 Exam Tip: When explaining band theory, focus on the size of the energy gap (forbidden zone) and the electron occupation of the valence and conduction bands for each material type.

Question 23. Explain the following terms with suitable examples:
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials
(iv) F-centres.

Answer:
(i) **Schottky defect:**
This defect occurs in ionic crystals when an equal number of cations and anions are missing from their regular lattice positions, creating vacant lattice sites. This leads to a decrease in the crystal's overall density. Schottky defects are typically found in highly ionic compounds with high coordination numbers and ions of similar sizes.
* **Example:** NaCl, KCl, CsCl.
(ii) **Frenkel defect:**
A Frenkel defect arises when an ion, usually the smaller cation, leaves its normal lattice site and occupies an interstitial position within the crystal. This defect does not change the overall density of the crystal because no ions are missing from the crystal entirely. It is observed in ionic compounds with a low coordination number where there is a significant size difference between cations and anions (cations being smaller).
* **Example:** AgCl, ZnS, AgBr.
(iii) **Interstitials:**
Interstitials are atoms or ions that occupy the normally vacant interstitial spaces within a crystal lattice. When extra particles are present in these positions, it leads to an increase in the density of the crystal. This defect is common in non-ionic solids.
* **Example:** Non-ionic metals where extra atoms are lodged in interstitial sites.
(iv) **F-centres (Farbenzentren):**
F-centres are anionic vacancies that have trapped electrons. These are formed when alkali halide crystals are heated in an atmosphere of the alkali metal vapor. The alkali metal atoms lose electrons, which then diffuse into the crystal and occupy the vacant anionic sites. These trapped electrons can absorb visible light, causing the crystal to exhibit a characteristic color.
* **Example:** Heating NaCl crystals in sodium vapor makes them appear yellow. Heating KCl in potassium vapor makes it violet.
In simple words: Schottky defect means missing pairs of ions, reducing density. Frenkel defect means an ion moves to an interstitial space, keeping density same. Interstitials are extra atoms in empty spaces, increasing density. F-centres are electron-trapped vacancies that give color to crystals.

🎯 Exam Tip: Distinguish Schottky and Frenkel defects based on their effect on crystal density and the characteristics of the ionic compounds (coordination number, ion size difference) in which they occur.

Question 24. Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1.00 cm³ of aluminium?

Answer:
(i) **Length of the side of the unit cell:**
For a cubic close-packed (ccp) structure, which is equivalent to a face-centered cubic (FCC) structure, the relationship between the edge length 'a' and the metallic radius 'r' is: \[ a = 2\sqrt{2}r \] Given metallic radius \( r = 125 \) pm. Substitute the value of 'r': \[ a = 2 \times \sqrt{2} \times 125 \text{ pm} \] \[ a = 2 \times 1.414 \times 125 \text{ pm} \] \[ a = 353.5 \text{ pm} \] The length of a side of the unit cell is approximately 354 pm.
(ii) **Number of unit cells in 1.00 cm³ of aluminium:**
First, convert the edge length 'a' from picometers (pm) to centimeters (cm): \( 1 \text{ pm} = 10^{-10} \text{ cm} \) So, \( a = 354 \text{ pm} = 354 \times 10^{-10} \text{ cm} = 3.54 \times 10^{-8} \text{ cm} \). Next, calculate the volume of one unit cell (\( a^3 \)): \( \text{Volume of unit cell} = a^3 = (3.54 \times 10^{-8} \text{ cm})^3 \) \( a^3 = 44.36 \times 10^{-24} \text{ cm}^3 = 4.436 \times 10^{-23} \text{ cm}^3 \). Now, determine the number of unit cells in 1.00 cm\(^{3}\) of aluminum: \( \text{Number of unit cells} = \frac{\text{Total volume}}{\text{Volume of one unit cell}} \) \( \text{Number of unit cells} = \frac{1.00 \text{ cm}^3}{4.436 \times 10^{-23} \text{ cm}^3/\text{unit cell}} \) \( \text{Number of unit cells} \approx 2.254 \times 10^{22} \) unit cells.
In simple words: For aluminum in a cubic close-packed structure, its unit cell side length is 354 pm. In 1 cubic centimeter of aluminum, there are approximately \( 2.25 \times 10^{22} \) such unit cells.

🎯 Exam Tip: Pay close attention to unit conversions (pm to cm) and correct exponent handling, especially when calculating volumes and counts of unit cells.

Question 25. If NaCl is doped with 10-3 mol % of SrCl2, what is the concentration of cation vacancies?

Answer:- When sodium chloride (NaCl) is doped with strontium chloride (SrCl\(_{2}\)), the Sr\(^{2+}\) impurity ions introduce vacancies into the NaCl lattice. - Each Sr\(^{2+}\) ion (with a +2 charge) replaces two Na\(^{+}\) ions (each with a +1 charge) to maintain electrical neutrality in the crystal. - One Sr\(^{2+}\) ion occupies a Na\(^{+}\) lattice site, and the second Na\(^{+}\) lattice site becomes vacant, thus creating one cation vacancy for every Sr\(^{2+}\) ion added. - Given that NaCl is doped with 10\(^{-3}\) mol % of SrCl\(_{2}\). - This implies that for every 100 moles of NaCl, there are 10\(^{-3}\) moles of SrCl\(_{2}\). - Since each mole of SrCl\(_{2}\) introduces one mole of cation vacancies, the concentration of cation vacancies is \( 10^{-3} \) mol per 100 moles of NaCl. - This can be expressed as: \( \frac{10^{-3} \text{ mol}}{100 \text{ mol}} = 10^{-5} \text{ mol of vacancies} \) per mole of NaCl. - To find the total number of vacancies per mole of NaCl, we multiply by Avogadro's number (\( N_A \)): - Number of vacancies = \( 10^{-5} \times N_A \) - Number of vacancies = \( 10^{-5} \times 6.022 \times 10^{23} \) - Number of vacancies = \( 6.022 \times 10^{18} \) vacancies.
In simple words: Doping NaCl with 10\(^{-3}\) mol % SrCl\(_{2}\) means for every Sr\(^{2+}\) ion introduced, one Na\(^{+}\) site becomes vacant to keep the charge balanced. This results in \( 6.022 \times 10^{18} \) cation vacancies per mole.

🎯 Exam Tip: Remember the 1:1 relationship between the moles of higher-valent impurity ions added (e.g., Sr\(^{2+}\) in NaCl) and the moles of cation vacancies created.

Question 26. Explain the following with suitable examples:
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Antiferromagnetism

Answer:
(i) **Ferromagnetism:**
Ferromagnetic substances exhibit a very strong attraction to external magnetic fields and can retain their magnetism even after the external field is removed, thus becoming permanent magnets. This behavior is due to the spontaneous alignment of magnetic moments of atoms or ions in the same direction, forming magnetic domains.
* **Examples:** Iron (Fe), Cobalt (Co), Nickel (Ni), CrO\(_{2}\) (Chromium dioxide).
(ii) **Paramagnetism:**
Paramagnetic substances are weakly attracted by an external magnetic field and contain unpaired electrons. They lose their magnetic properties once the external magnetic field is removed. The strength of paramagnetism is directly proportional to the number of unpaired electrons present.
* **Examples:** Oxygen (O\(_{2}\)), Copper(II) oxide (CuO), Fe\(^{3+}\) ions, Aluminum (Al), Manganese (Mn), Nitric oxide (NO).
(iii) **Ferrimagnetism:**
Ferrimagnetic substances are expected to have large magnetic moments due to unpaired electrons, but they actually exhibit a small net magnetic moment. This phenomenon occurs because their magnetic moments are aligned in opposite directions, but in unequal magnitudes, resulting in a net, though reduced, magnetic effect.
* **Examples:** Magnetite (Fe\(_{3}\)O\(_{4}\)), Magnesium ferrite (MgFe\(_{2}\)O\(_{4}\)).
(iv) **Antiferromagnetism:**
Antiferromagnetic substances possess unpaired electrons but exhibit a zero net magnetic moment. This is because their magnetic moments are aligned in opposite directions in a perfectly compensatory manner, effectively canceling each other out.
* **Examples:** Manganese oxide (MnO), Manganese dioxide (MnO\(_{2}\)).
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विभिन्न चुंबकीय पदार्थों में चुंबकीय क्षणों के संरेखण को दर्शाता है। फेरोमैग्नेटिक पदार्थों में सभी क्षण एक ही दिशा में संरेखित होते हैं, फेरिमैग्नेटिक पदार्थों में क्षण विपरीत दिशाओं में असमान रूप से संरेखित होते हैं, और एंटीफेरोमैग्नेटिक पदार्थों में क्षण विपरीत दिशाओं में समान रूप से संरेखित होते हैं, जिससे शुद्ध चुंबकीय क्षण शून्य हो जाता है।
In simple words: Ferromagnetism is strong, permanent magnetism (moments aligned). Paramagnetism is weak, temporary attraction (unpaired electrons). Ferrimagnetism is weak magnetism from unequally opposed moments. Antiferromagnetism has zero net magnetism due to equally opposed moments.

🎯 Exam Tip: Focus on the alignment and magnitude of magnetic moments (all parallel, unequal antiparallel, or equal antiparallel) to differentiate between ferromagnetic, ferrimagnetic, and antiferromagnetic substances.

GSEB Class 12 Chemistry The Solid State Additional Important Questions And Answers

Question 1.
(i) Find out the odd one from each group and justify your answer.
(a) NaCl, Iodine, Plastic, Ice
(b) Glass, Sugar, Rubber, Asphalt
(ii) From group (a) identify crystal (s).
(a) having a sharp melting point
(b) which is/are isotropic
(iii) What do you understand by isotropic properties?
(iv) List out any four points of differences between amorphous and crystalline solids?

Answer:
(i) **Identifying the odd one out:**
(a) **Plastic:** This is the odd one because it is an amorphous solid, whereas NaCl, Iodine, and Ice are all crystalline solids.
(b) **Sugar:** This is the odd one because it is a crystalline solid, while Glass, Rubber, and Asphalt are all amorphous solids.
(ii) **Identifying crystals from group (a) and their properties:**
(a) **Having a sharp melting point:** NaCl, Iodine, and Ice are crystalline solids, and all crystalline solids exhibit sharp melting points.
(b) **Which is/are isotropic:** Plastic is an amorphous solid and therefore exhibits isotropic properties (properties are the same in all directions).
(iii) **Understanding isotropic properties:**
Isotropic properties refer to characteristics of a material (such as electrical conductivity, thermal conductivity, refractive index, and mechanical strength) that have the same values regardless of the direction of measurement within the material. Amorphous solids typically display isotropic behavior.
(iv) **Differences between amorphous and crystalline solids:**

Crystalline solids	Amorphous solids
1. Possess long-range order in their atomic arrangement.	1. Possess only short-range order in their atomic arrangement.
2. Exhibit sharp and definite melting points.	2. Do not possess a sharp melting point; they soften gradually over a range of temperatures.
3. They are anisotropic, meaning their physical properties (e.g., electrical conductivity, thermal conductivity, refractive index) differ when measured along different directions.	3. They are isotropic, meaning their physical properties are the same in all directions.
4. When cut with a sharp-edged tool, they cleave into two pieces with smooth, definite planes.	4. When cut, they undergo irregular cleavage, resulting in pieces with uneven or irregular surfaces.
5. Are generally more rigid and incompressible.	5. Are generally less rigid and more compressible.

In simple words: The odd one out in the first group is Plastic (amorphous), and in the second group is Sugar (crystalline). Crystalline solids like NaCl, Iodine, and Ice have sharp melting points and are anisotropic, while amorphous solids like Plastic are isotropic. Isotropic means properties are the same in all directions. Crystalline solids have long-range order and definite melting points, unlike amorphous solids which have short-range order and melt over a range.

🎯 Exam Tip: A good way to remember isotropic vs. anisotropic is that "iso" means same (same properties in all directions for amorphous solids), while "aniso" means not same (different properties in different directions for crystalline solids).

Question 2. In a classroom seminar, Ramu explains the voids in crystal packing by suitably arranging lemon on the table.
(i) What do you understand by the term 'void'?
(ii) Using diagrams explain different types of voids?
(iii) If a small sphere is placed in the 'void' in each case, what would be the coordination number of the sphere? Explain.
(iv) What is the number of tetrahedral voids and octahedral voids in a crystal having 'n' particles?

Answer:
(i) **Definition of 'void':**
In a close-packed arrangement of a crystal, the vacant spaces or unoccupied regions between the constituent particles (atoms, ions, or molecules) are referred to as 'voids' or interstitial sites. These voids represent the empty spaces left after the spheres are packed as closely as possible.
(ii) **Types of voids (with diagram explanation):**
* **Tetrahedral void:** A tetrahedral void is a vacant space surrounded by four touching spheres. Three spheres lie in one plane, and the fourth sphere is placed above or below the center of the triangular space formed by the first three. If a small sphere is placed in a tetrahedral void, it is in contact with these four surrounding spheres.
* **Octahedral void:** An octahedral void is a vacant space enclosed by six touching spheres. It is formed by superimposing two equilateral triangular voids, one inverted relative to the other. If a small sphere is placed in an octahedral void, it is in contact with these six surrounding spheres.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो मुख्य प्रकार के अंतराकाशीय रिक्ति (voids) को दर्शाता है जो क्रिस्टल जालक में पाए जाते हैं। टेट्राहेड्रल रिक्ति को चार निकटवर्ती गोले से घिरा हुआ दिखाया गया है, जबकि ऑक्टाहेड्रल रिक्ति को छह निकटवर्ती गोले से घिरा हुआ दर्शाया गया है।
(iii) **Coordination number of a sphere placed in a void:**
* If a small sphere is placed in a **tetrahedral void**, its coordination number (the number of nearest touching neighbors) will be **4**, as it touches the four spheres forming the void.
* If a small sphere is placed in an **octahedral void**, its coordination number will be **6**, as it touches the six spheres forming the void.
(iv) **Number of tetrahedral and octahedral voids in a crystal with 'n' particles:**
In a crystal containing 'n' constituent particles (spheres) in a close-packed arrangement (like hcp or ccp):
* The number of tetrahedral voids is \( 2n \).
* The number of octahedral voids is \( n \).
In simple words: Voids are empty spaces between packed spheres. Tetrahedral voids are surrounded by four spheres, and octahedral voids by six. A small sphere in a tetrahedral void has a coordination number of 4, and in an octahedral void, it's 6. For 'n' particles, there are 'n' octahedral and '2n' tetrahedral voids.

🎯 Exam Tip: Clearly differentiate between tetrahedral and octahedral voids by the number of surrounding spheres and their relative positions. Remember the 2:1 ratio of tetrahedral to octahedral voids for 'n' close-packed particles.

 

Question 3.
1. The dimensions of a crystal system determined is a = b = c, α = β = γ = 90°. Identify the crystal system and give the diagrammatic representation of it.
2. Explain the term 'space lattice' and 'unit cell'.
Answer:
1. This system represents a cubic crystal.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक क्यूबिक क्रिस्टल सिस्टम को दर्शाता है। इसमें तीन अक्ष (a, b, c) हैं जो लंबाई में समान हैं और एक-दूसरे के लंबवत हैं, जिसका अर्थ है कि उनके बीच के कोण (α, β, γ) सभी 90° हैं। यह सबसे सरल और सबसे सममित क्रिस्टल प्रणाली है।
2. A crystal lattice, also known as a space lattice, refers to the organized three-dimensional arrangement of constituent particles (atoms, ions, or molecules) within a crystalline solid. A unit cell is defined as the smallest repeating segment of this space lattice, which, when repeated in all directions, generates the entire crystal structure.
In simple words: 1. A cubic system has equal sides and all 90-degree angles. 2. A space lattice is the overall arrangement of particles in a crystal, and a unit cell is the smallest piece that repeats to form that pattern.

🎯 Exam Tip: Understanding the seven crystal systems and their characteristic axial lengths and angles is crucial for crystallography problems. Defining fundamental terms like "space lattice" and "unit cell" accurately demonstrates foundational knowledge.

 

Question 4.
(a) The unit cell dimensions of a particular crystal system is given below.
a = b = c α = β = γ ≠ 90°
Identify the crystal system and suggest a suitable example.
(b) A certain metal M crystallises in fee lattice. It forms a compound with elements A and B. A occupies all the octahedral voids and B occupies half of the tetrahedral voids. Find out the empirical formula of the compound.
Answer:
(a) The crystal system described, with equal axial lengths (a = b = c) but angles not equal to 90 degrees (α = β = γ ≠ 90°), is the rhombohedral system. Examples include calcium carbonate (CaCO3) and cinnabar (HgS).
(b) For a metal M crystallizing in a face-centered cubic (fcc) lattice, the number of M atoms per unit cell is 4. In an fcc structure, the number of octahedral voids equals the number of atoms (n), and the number of tetrahedral voids is twice the number of atoms (2n).
Since A occupies all octahedral voids, the number of A atoms is 4.
Since B occupies half of the tetrahedral voids (half of 2n), the number of B atoms is \( \frac{1}{2} \times (2 \times 4) = 4 \).
Therefore, the ratio of M:A:B is 4:4:4, simplifying to 1:1:1. The empirical formula of the compound is MAB.
In simple words: (a) When all sides are equal but angles are not 90 degrees, it's a rhombohedral crystal. (b) In an fcc structure, if M has 4 atoms, A occupies 4 octahedral holes, and B occupies half of the 8 tetrahedral holes (which is 4). So, the formula is MAB.

🎯 Exam Tip: Memorizing the seven crystal systems with their axial parameters and examples is essential. For solid-state chemistry, understanding the relationship between the number of atoms, octahedral voids, and tetrahedral voids in different packing arrangements (like fcc) is key to determining empirical formulas.

 

Question 5.
(a) NaCl does not exhibit Frenkel defect. Do you agree? Justify.
(b) Assume that a fresh piece of KCl crystal is heated in an atmosphere of potassium vapours. Is there any colour change? Substantiate your answer.
Answer:
(a) Yes, this statement is accurate. Sodium chloride (NaCl) typically does not display the Frenkel defect because the ionic radii of Na+ and Cl- are very similar. For a Frenkel defect to occur, there must be a significant size difference allowing a smaller ion (usually the cation) to move from its lattice position into an interstitial site without distorting the lattice excessively. Since the Na+ ion is not significantly smaller than the Cl- ion, it cannot easily fit into an interstitial site, thus making the Frenkel defect uncommon in NaCl.
(b) Yes, a distinct color change will be observed. When potassium chloride (KCl) crystals are heated in an environment of potassium vapors, the crystal will acquire a violet coloration. This phenomenon occurs because potassium atoms from the vapor deposit on the crystal's surface. Chloride ions (Cl-) from the interior of the crystal then migrate to the surface to combine with these potassium atoms, forming KCl. In this process, the potassium atoms lose electrons, which then diffuse into the crystal lattice. These liberated electrons occupy the vacant sites created by the migrating Cl- ions. These electron-occupied anion vacancies are known as F-centers. These F-centers absorb specific wavelengths of visible light, leading to the crystal appearing violet.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक आयनिक क्रिस्टल में एक एनायन रिक्ति (anion vacancy) की उपस्थिति को दर्शाता है, जिसे F-सेंटर भी कहा जाता है। इसमें क्रिस्टल जाली में एक क्लोराइड आयन (Cl-) अपने सामान्य स्थान से गायब है, और उसकी जगह एक इलेक्ट्रॉन (e-) ने ले ली है। यह क्रिस्टल में रंग पैदा करने वाले दोष को दर्शाता है।
In simple words: (a) NaCl doesn't get Frenkel defects because its Na+ and Cl- ions are roughly the same size, so Na+ can't easily move into a tiny gap. (b) Yes, the KCl crystal turns violet. When heated in potassium vapor, potassium atoms stick to the surface. Chloride ions leave their spots to join potassium, and the electrons they leave behind get trapped in those vacant spots, creating "F-centers" that absorb light and give the violet color.

🎯 Exam Tip: Differentiating between Schottky and Frenkel defects based on ionic radii and coordination numbers is important. Understanding the formation of F-centers and their role in imparting color to ionic crystals, along with the associated mechanism, is a frequently tested concept.

 

Question 6.
Name the parameters that characterize a unit cell.
Answer:
A unit cell is uniquely defined by a set of six characteristic parameters, which include:
1. Edge lengths: These are the lengths of the unit cell along its three axes, typically denoted as a, b, and c. These edges can be either mutually perpendicular or not, depending on the crystal system.
2. Interfacial angles: These are the angles formed between the crystallographic axes. Alpha (α) is the angle between edges b and c; Beta (β) is the angle between edges a and c; and Gamma (γ) is the angle between edges a and b.
In simple words: A unit cell is defined by its three side lengths (a, b, c) and the three angles (alpha, beta, gamma) between those sides.

🎯 Exam Tip: Knowing the six parameters (three edge lengths a, b, c; and three angles α, β, γ) is fundamental for understanding crystal systems and unit cell geometry. Be prepared to recall these for different crystal types.

 

Question 7.
(a) The following diagram shows the alignment of magnetic moments for some magnetic properties.
Which figure corresponds to antiferromagnetism? Justify your answer.
(b) Study the list of substances given below. Classify them on the basis of magnetic properties and name each class.
Al, TiO2, Co, MnO2, Benzene, O2, NaCl, NH3, Fe3+, Ni
Answer:
(a) Figure (ii) represents antiferromagnetism. This is because in antiferromagnetic materials, the magnetic moments of the constituent atoms or ions are aligned in an antiparallel fashion with equal magnitudes. This perfect cancellation results in a net magnetic moment of zero for the material, even though individual atoms possess magnetic moments.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख विभिन्न प्रकार के चुंबकीय संरेखणों को दर्शाता है। चित्र (i) फेरोमैग्नेटिज़्म को इंगित करता है जहाँ सभी चुंबकीय क्षण एक ही दिशा में संरेखित होते हैं। चित्र (ii) एंटीफेरोमैग्नेटिज़्म को दर्शाता है जहाँ चुंबकीय क्षण समान परिमाण के साथ विपरीत दिशाओं में संरेखित होते हैं, जिसके परिणामस्वरूप शुद्ध चुंबकीय क्षण शून्य होता है। चित्र (iii) फेरिमैग्नेटिज़्म को दर्शाता है जहाँ चुंबकीय क्षण विपरीत दिशाओं में असमान परिमाण के साथ संरेखित होते हैं, जिसके परिणामस्वरूप शुद्ध चुंबकीय क्षण गैर-शून्य होता है।
(b) The substances can be categorized based on their magnetic properties as follows:
(i) Diamagnetic substances – titanium dioxide (TiO2), benzene, sodium chloride (NaCl), and ammonia (NH3).
(ii) Paramagnetic substances – aluminum (Al), oxygen (O2), and ferric ions (Fe3+).
(iii) Ferromagnetic substances – cobalt (Co) and nickel (Ni).
(iv) Antiferromagnetic substances – manganese dioxide (MnO2).
In simple words: (a) Figure (ii) shows antiferromagnetism because magnetic spins line up perfectly opposite to each other, so they cancel out and there's no overall magnetism. (b) Materials are grouped by how they react to magnets: diamagnetic (repelled), paramagnetic (weakly attracted, have unpaired electrons), ferromagnetic (strongly attracted, can become permanent magnets), and antiferromagnetic (magnetic moments cancel out).

🎯 Exam Tip: Be able to identify and differentiate between ferromagnetic, paramagnetic, and antiferromagnetic substances based on their magnetic moment alignments. Provide relevant examples for each class of magnetic materials.

 

Question 8.
1. A small number of K+ ions in KCl are substituted by Pr³+ ions. Draw the possible structure of the solid with a defect in the crystal lattice.
2. Semiconductors can be obtained by doping silicon with As or with Ga. What is the basic difference between these?
Answer:
1. When a small quantity of Pr3+ ions substitutes K+ ions in a KCl crystal lattice, it introduces vacancies to maintain electrical neutrality. Since each Pr3+ ion has a charge of +3 and replaces a K+ ion with a charge of +1, for every Pr3+ ion introduced, two K+ ions must be removed from the lattice. One K+ site will be occupied by Pr3+, and another K+ site will remain vacant to balance the charge. This creates a cation vacancy for every two K+ ions replaced by one Pr3+ ion.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह आरेख एक पोटेशियम क्लोराइड (KCl) क्रिस्टल जाली में Pr³⁺ आयन के प्रतिस्थापन से उत्पन्न दोष को दर्शाता है। एक Pr³⁺ आयन (जो +3 आवेश वहन करता है) दो K⁺ आयनों को प्रतिस्थापित करता है (प्रत्येक +1 आवेश वहन करता है)। एक K⁺ स्थान पर Pr³⁺ कब्जा करता है, जबकि दूसरा K⁺ स्थान एक रिक्ति के रूप में खाली रह जाता है, जिससे क्रिस्टल की विद्युत तटस्थता बनी रहती है।
2. The fundamental difference lies in the type of charge carriers produced. Doping silicon (a group 14 element) with arsenic (As, a group 15 element) results in an n-type semiconductor. Arsenic contributes an extra electron, which becomes a mobile charge carrier, leading to electron-rich conduction. In contrast, doping silicon with gallium (Ga, a group 13 element) produces a p-type semiconductor. Gallium has one fewer valence electron than silicon, creating an "electron hole" or positive vacancy. Conduction in p-type semiconductors occurs due to the movement of these positive holes.
In simple words: 1. If Pr3+ (charge +3) replaces K+ (charge +1) in KCl, one Pr3+ takes a K+ spot, and another K+ spot must be left empty as a "vacancy" to keep the overall charge balanced. 2. Doping silicon with arsenic adds extra electrons, making it an n-type semiconductor (negative charge carriers). Doping silicon with gallium creates "holes" (missing electrons), making it a p-type semiconductor (positive charge carriers).

🎯 Exam Tip: Understanding the concept of doping in ionic solids to create defects (like cation vacancies due to impurity ions of higher valence) is crucial. For semiconductors, clearly explain the role of donor (n-type) and acceptor (p-type) impurities and the resulting charge carriers.