Chemistry Exercises

Chapter 1: The Solid State

Class 12 Chemistry | 40 Questions

1short answerLOW

Why are solids rigid?

✅ Answer

Solids are rigid due to the strong intermolecular forces of attraction (or interatomic/interionic forces) between their constituent particles (atoms, ions, or molecules). These strong forces hold the particles in fixed positions within the crystal lattice. Although the particles can vibrate about their mean positions, they cannot move freely or translate from one position to another. This restricted movement and fixed, closely packed arrangement give solids their characteristic rigidity and definite shape, making them resistant to deformation.

NEET Relevance

This is a fundamental conceptual question, rarely asked directly in NEET. However, understanding this basic property is essential for comprehending the behavior of solids.

Key Concepts

Intermolecular forcesParticle arrangementRigidity
2short answerLOW

Why do solids have a definite volume?

✅ Answer

Solids have a definite volume because their constituent particles are closely packed and held together by strong intermolecular forces of attraction. These strong forces prevent the particles from moving away from each other significantly. The particles are confined to specific, fixed positions and can only vibrate about their mean positions. This fixed arrangement and the strong attractive forces ensure that the overall space occupied by the solid remains constant under given conditions of temperature and pressure, thus giving solids a definite volume.

NEET Relevance

Similar to question 1, this is a basic conceptual question. While not a direct NEET question, it forms the foundation for understanding the states of matter.

Key Concepts

Intermolecular forcesParticle arrangementDefinite volume
3short answer🎯 HIGH⭐ Important

Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, PVC, fibre glass, copper.

✅ Answer

To classify the given substances, we need to recall the defining characteristics of amorphous and crystalline solids.

Crystalline Solids: These solids have a long-range order in the arrangement of their constituent particles. They possess sharp melting points, are anisotropic (physical properties vary with direction), and have a definite heat of fusion.

Amorphous Solids: These solids have only short-range order, meaning their particles are arranged irregularly. They soften over a range of temperatures, are isotropic (physical properties are the same in all directions), and do not have a definite heat of fusion.

Based on these characteristics, the classification is as follows:

Crystalline Solids:

  1. Naphthalene: A molecular solid with a regular, ordered crystal lattice.
  2. Benzoic acid: A molecular solid with a well-defined crystal structure.
  3. Potassium nitrate: An ionic solid with a regular arrangement of K⁺ and NO₃⁻ ions.
  4. Copper: A metallic solid with a highly ordered crystal lattice.

Amorphous Solids:

  1. Polyurethane: A polymer, which typically forms amorphous structures.
  2. Teflon (Polytetrafluoroethylene): A polymer, generally amorphous.
  3. Cellophane: A regenerated cellulose film, which is amorphous.
  4. PVC (Polyvinyl chloride): A polymer, usually amorphous or semi-crystalline.
  5. Fibre glass: An amorphous solid, primarily composed of silica (SiO₂) in a disordered arrangement.

Summary:

  • Crystalline Solids: Naphthalene, Benzoic acid, Potassium nitrate, Copper
  • Amorphous Solids: Polyurethane, Teflon, Cellophane, PVC, Fibre glass

NEET Relevance

This type of classification question is very common in NEET. It tests the fundamental understanding of the differences between amorphous and crystalline solids, which is a core concept of the chapter. MCQs often ask to identify an amorphous/crystalline solid from a list or to state a property unique to one type.

Key Concepts

Crystalline solidsAmorphous solidsLong-range orderShort-range orderIsotropyAnisotropy

This question has appeared in previous NEET exams.

4short answer🎯 HIGH⭐ Important

Why is glass considered a supercooled liquid?

✅ Answer

Glass is considered a supercooled liquid (or pseudo-solid or amorphous solid) because it exhibits some properties characteristic of liquids, despite appearing solid. Here's why:

  1. Lack of Long-Range Order: Unlike true crystalline solids, glass does not have a long-range, regular, and repeating arrangement of its constituent particles (e.g., SiO₄ tetrahedra in silica glass). Instead, it possesses only short-range order, similar to liquids.
  2. Formation Process: When a liquid (like molten silica) is cooled rapidly, its particles do not get sufficient time to arrange themselves into a regular, long-range crystalline lattice. The liquid then solidifies into a glassy state, retaining the disordered structure of a liquid but with extremely high viscosity.
  3. No Sharp Melting Point: Crystalline solids have a sharp melting point, meaning they transition from solid to liquid at a specific temperature. Glass, being amorphous, softens gradually over a range of temperatures before becoming a true liquid, much like a very viscous liquid.
  4. Flow Property (Extremely Slow): Although imperceptible over short periods, glass can exhibit extremely slow flow over very long durations, especially at slightly elevated temperatures. This is why old glass panes in historical buildings are sometimes found to be slightly thicker at the bottom than at the top, indicating a very slow, gravitational flow over centuries. This flow is a characteristic property of liquids.
  5. Isotropic Nature: Glass is isotropic, meaning its physical properties (like refractive index, electrical conductivity) are the same in all directions, just like liquids. Crystalline solids, in contrast, are generally anisotropic.

Due to these reasons, glass is best described as a liquid that has been cooled to a temperature below its freezing point without crystallizing, resulting in an extremely viscous, non-flowing state that retains the disordered molecular arrangement of a liquid.

NEET Relevance

This is a very important conceptual question, frequently asked in board exams and often appears as MCQs in NEET. Questions might ask for the definition of a supercooled liquid, properties of glass, or reasons for its classification.

Key Concepts

Amorphous solidsSupercooled liquidLong-range orderShort-range orderViscosityIsotropyMelting point

This question has appeared in previous NEET exams.

5short answer🎯 HIGH⭐ Important

Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?

✅ Answer

Let's break down the answer based on the given observation:

Nature of the Solid
If the refractive index of a solid is observed to have the same value along all directions, it means the solid is isotropic in nature. Isotropic solids exhibit the same physical properties (such as refractive index, electrical conductivity, thermal expansion, etc.) regardless of the direction in which they are measured. This property is characteristic of amorphous solids.

Reason: Amorphous solids lack a long-range, ordered arrangement of their constituent particles. Their particles are arranged irregularly and randomly. As a result, the arrangement of particles is statistically uniform in all directions, leading to isotropic behavior. In contrast, crystalline solids have a highly ordered, anisotropic structure, meaning their physical properties vary with direction.

Cleavage Property
No, this solid would not show cleavage property.

Reason: Cleavage property refers to the tendency of a solid to break along definite, smooth planes when cut with a sharp-edged tool. This property is characteristic of crystalline solids. Crystalline solids have a highly ordered, regular arrangement of particles, which results in specific planes of weakness (cleavage planes) within their structure. When a crystalline solid is cut, it breaks along these planes, producing smooth, planar surfaces.

Amorphous solids, due to their irregular and disordered arrangement of particles, do not have such definite cleavage planes. When an amorphous solid is cut or broken, it fractures into pieces with irregular, uneven, or curved surfaces (this type of fracture is often called conchoidal fracture). Therefore, an amorphous solid (which is isotropic) would not exhibit cleavage property.

NEET Relevance

This is a highly important conceptual question for NEET. It directly tests the understanding of isotropic vs. anisotropic nature and the distinct properties (like refractive index and cleavage) of amorphous and crystalline solids. MCQs frequently involve identifying the type of solid based on these properties.

Key Concepts

IsotropyAnisotropyAmorphous solidsCrystalline solidsRefractive indexCleavage propertyConchoidal fracture

This question has appeared in previous NEET exams.

6long answer🎯 HIGH⭐ Important

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 containing a cation of higher valence, vacancies are introduced in the crystal lattice to maintain electrical neutrality. This is a type of impurity defect.

Consider an ionic solid like NaCl, which consists of Na⁺ and Cl⁻ ions. If a small amount of an impurity like SrCl₂ (containing Sr²⁺ ions) is added during the crystallization of NaCl, the Sr²⁺ ions (which have a higher valence than Na⁺ ions) occupy some of the lattice sites normally occupied by Na⁺ ions.

To maintain the overall electrical neutrality of the crystal, for every Sr²⁺ ion that replaces a Na⁺ ion, two Na⁺ ions must be removed from the lattice. Since only one Sr²⁺ ion occupies a Na⁺ site, the second Na⁺ ion site remains vacant. This creates a cation vacancy.

Example: In NaCl crystal, if one Sr²⁺ ion replaces one Na⁺ ion, the charge increases by +1 (Sr²⁺ vs Na⁺). To balance this extra positive charge and maintain electrical neutrality, another Na⁺ ion must leave its lattice site, creating a cation vacancy. Thus, for every Sr²⁺ ion introduced, one cation vacancy is created.

These vacancies enhance the electrical conductivity of the ionic solid as ions can move into these vacant sites.

Solution Steps

  1. Step 1: Introduction of Impurity

    Start with a pure ionic solid (e.g., NaCl) where cations (Na⁺) and anions (Cl⁻) are in their regular lattice positions. Introduce an impurity cation of higher valence (e.g., Sr²⁺ from SrCl₂) into the molten state or during crystallization.

  2. Step 2: Substitution of Cations

    The impurity cation (Sr²⁺) occupies some of the lattice sites normally held by the host cation (Na⁺). Since Sr²⁺ has a +2 charge and Na⁺ has a +1 charge, the substitution of one Sr²⁺ for one Na⁺ creates an excess positive charge of +1 in that region of the crystal.

  3. Step 3: Maintenance of Electrical Neutrality

    To maintain the overall electrical neutrality of the ionic crystal, the excess positive charge introduced by the higher valence cation must be compensated. This compensation occurs by the creation of cation vacancies.

  4. Step 4: Formation of Cation Vacancies

    For every Sr²⁺ ion that replaces a Na⁺ ion, an additional Na⁺ ion must leave its lattice site to balance the charge. Since the Sr²⁺ ion only occupies one Na⁺ site, the other Na⁺ site that was vacated remains empty, creating a cation vacancy. Thus, for every Sr²⁺ ion incorporated, one cation vacancy is created.

  5. Step 5: Consequences

    These cation vacancies allow for ionic conductivity, as other cations can move into these vacant sites. This type of defect is known as an impurity defect.

NEET Relevance

This concept is frequently tested in NEET, particularly in MCQs asking about the number of vacancies created upon doping with a higher valence ion, or the effect of such doping on conductivity. Understanding impurity defects is crucial.

Key Concepts

Impurity defectsIonic solidsCation vacanciesElectrical neutralityDoping

This question has appeared in previous NEET exams.

7short answer🎯 HIGH⭐ Important

Classify each of the following as being either a p-type or an n-type semiconductor:
(i) Ge doped with In

(ii) B doped with Si

✅ Answer

(i) Ge doped with In: p-type semiconductor
Germanium (Ge) belongs to Group 14 (valency 4). Indium (In) belongs to Group 13 (valency 3). When Ge is doped with In, an atom with one less valence electron (In) replaces a Ge atom. This creates an electron vacancy or a 'hole' in the crystal lattice. These holes act as positive charge carriers, making it a p-type semiconductor.

(ii) B doped with Si: n-type semiconductor
Boron (B) belongs to Group 13 (valency 3). Silicon (Si) belongs to Group 14 (valency 4). When B is doped with Si, an atom with one more valence electron (Si) replaces a B atom. This introduces an extra electron that is not involved in bonding and becomes delocalized, acting as a negative charge carrier. This makes it an n-type semiconductor.

NEET Relevance

This is a very common type of MCQ in NEET. Students must understand the concept of doping and how the relative valency of the dopant and host atom determines whether it's an n-type or p-type semiconductor.

Key Concepts

SemiconductorsDopingn-type semiconductorp-type semiconductorValence electronsGroup number

This question has appeared in previous NEET exams.

8long answerMEDIUM⭐ Important

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

Non-stoichiometric cuprous oxide, Cu₂O, with a copper to oxygen ratio slightly less than 2:1, indicates a metal deficiency defect. This means there are fewer copper ions than required for the ideal stoichiometric formula, leading to the presence of cation vacancies. This defect makes Cu₂O a p-type semiconductor.

Explanation:

  1. Ideal Stoichiometry: In ideal cuprous oxide (Cu₂O), copper exists as Cu⁺ ions, and the ratio of Cu⁺ to O²⁻ is exactly 2:1. 2. Metal Deficiency Defect: When the copper to oxygen ratio is less than 2:1, it implies that some Cu⁺ ions are missing from their lattice sites. This creates cation vacancies. 3. Maintenance of Electrical Neutrality: To maintain the overall electrical neutrality of the crystal, for every two missing Cu⁺ ions (total charge +2), one Cu²⁺ ion must be introduced into an adjacent lattice site. This is because if only Cu⁺ ions were missing, the crystal would become negatively charged. By replacing two Cu⁺ ions with one Cu²⁺ ion and one vacancy, the charge balance is maintained (2 × +1 = +2, and 1 × +2 = +2). 4. Formation of Holes: The presence of Cu²⁺ ions alongside Cu⁺ ions means that an electron can jump from a Cu⁺ ion to an adjacent Cu²⁺ ion. This movement of an electron effectively creates a 'hole' (an electron vacancy) at the site of the original Cu⁺ ion. 5. p-type Conduction: These 'holes' are mobile and can move through the crystal lattice as electrons jump from one Cu⁺ ion to another Cu²⁺ ion, effectively creating a positive charge movement. Since the conduction is due to the movement of these positive holes, non-stoichiometric Cu₂O acts as a p-type semiconductor.

Solution Steps

  1. Step 1: Understanding Non-stoichiometry

    The statement 'copper to oxygen ratio is slightly less than 2:1' for Cu₂O indicates a metal deficiency defect. This means there are fewer copper ions than the ideal stoichiometric formula requires.

  2. Step 2: Cation Vacancies

    Due to the deficiency of copper, some Cu⁺ ions are missing from their regular lattice sites, creating cation vacancies.

  3. Step 3: Charge Compensation

    To maintain the overall electrical neutrality of the crystal, the missing positive charge from the Cu⁺ vacancies must be compensated. This is achieved by some of the remaining Cu⁺ ions oxidizing to Cu²⁺ ions. For example, if two Cu⁺ ions are missing (total charge +2), one Cu²⁺ ion might occupy a site, leaving one vacancy and maintaining charge balance (2 × +1 = +2, and 1 × +2 = +2).

  4. Step 4: Formation of Holes

    The presence of both Cu⁺ and Cu²⁺ ions in the lattice allows for electron transfer. An electron from a Cu⁺ ion can jump to an adjacent Cu²⁺ ion. This movement of an electron effectively creates a 'hole' (an electron vacancy) at the site of the original Cu⁺ ion.

  5. Step 5: p-type Conduction

    These 'holes' are mobile and can move through the crystal lattice as electrons jump from one Cu⁺ ion to another Cu²⁺ ion. The movement of these positive holes constitutes electrical conduction. Since the charge carriers are positive holes, non-stoichiometric Cu₂O behaves as a p-type semiconductor.

NEET Relevance

This question tests the understanding of metal deficiency defects and their consequence on semiconductor type. It's a good conceptual question that can appear as an MCQ or a short answer type in NEET.

Key Concepts

Non-stoichiometric defectsMetal deficiency defectCation vacanciesElectrical neutralityp-type semiconductorHoles (electron vacancies)

This question has appeared in previous NEET exams.

9long answerMEDIUM⭐ Important

Analyze the given data and predict the type of magnetism:
(a) Cr₂O₃: Antiferromagnetic

(b) TiO₂: Diamagnetic

(c) MnO₂: Ferromagnetic

✅ Answer

The question provides examples of substances and their observed magnetic properties. We need to analyze this data by explaining the characteristics of each type of magnetism based on the given examples.

(a) Cr₂O₃: Antiferromagnetic

  • Explanation: Antiferromagnetic substances have domains similar to ferromagnetic substances, but their magnetic moments are oriented in opposite directions and cancel each other out. This results in a net magnetic moment of zero. When an external magnetic field is applied, they show very little or no magnetism. Cr₂O₃ is an example where the magnetic moments of the Cr³⁺ ions align in an antiparallel fashion, leading to cancellation.

(b) TiO₂: Diamagnetic

  • Explanation: Diamagnetic substances are those which are weakly repelled by an external magnetic field. They do not have any unpaired electrons. All their electrons are paired, and their magnetic moments cancel each other out. When placed in a magnetic field, they acquire an induced magnetic moment in the opposite direction to the applied field. TiO₂ (Titanium dioxide) is diamagnetic because Ti⁴⁺ has an electronic configuration of [Ar]3d⁰, meaning it has no unpaired electrons.

(c) MnO₂: Ferromagnetic

  • Explanation: Ferromagnetic substances are strongly attracted by an external magnetic field and can be permanently magnetized. In the solid state, the metal ions of ferromagnetic substances are grouped into small regions called domains. Within each domain, the magnetic moments of the ions are aligned in the same direction. In the absence of an external magnetic field, these domains are randomly oriented, and their magnetic moments cancel out. However, when placed in an external magnetic field, the domains align themselves in the direction of the applied field, and this alignment persists even after the removal of the field. MnO₂ is listed as ferromagnetic, implying strong, permanent magnetization due to aligned domains of Mn ions.

Solution Steps

  1. Step 1: Analyze Antiferromagnetism (Cr₂O₃)

    Define antiferromagnetism: Substances with domains where magnetic moments are aligned in opposite directions, cancelling each other out, resulting in zero net magnetic moment. Cr₂O₃ is an example where Cr³⁺ ions exhibit this behavior.

  2. Step 2: Analyze Diamagnetism (TiO₂)

    Define diamagnetism: Substances that are weakly repelled by a magnetic field. They have no unpaired electrons, meaning all electrons are paired, and their magnetic moments cancel. TiO₂ is diamagnetic because Ti⁴⁺ has no unpaired electrons (3d⁰ configuration).

  3. Step 3: Analyze Ferromagnetism (MnO₂)

    Define ferromagnetism: Substances strongly attracted by a magnetic field and can be permanently magnetized. They consist of domains where magnetic moments are aligned in the same direction. In an external field, these domains align with the field, and this alignment persists. MnO₂ is given as an example of a ferromagnetic substance.

NEET Relevance

Understanding different types of magnetic properties (diamagnetic, paramagnetic, ferromagnetic, antiferromagnetic, ferrimagnetic) and their characteristics is important for NEET. Questions often involve identifying the type of magnetism based on electron configuration or properties, or matching examples to types.

Key Concepts

Magnetic properties of solidsAntiferromagnetismDiamagnetismFerromagnetismMagnetic domainsUnpaired electrons

This question has appeared in previous NEET exams.

10short answerLOW

Why are solids rigid?

✅ Answer

Solids are rigid due to the strong intermolecular forces of attraction between their constituent particles (atoms, ions, or molecules). These strong forces hold the particles in fixed positions within the crystal lattice or amorphous structure. Although the particles can vibrate about their mean positions, they cannot move freely or translate past each other. This fixed arrangement and strong binding prevent solids from changing their shape easily, giving them their characteristic rigidity and definite shape and volume.

NEET Relevance

This is a very basic conceptual question from the introductory part of the chapter. While fundamental, it's rarely asked directly in NEET as an MCQ. However, understanding this concept is foundational for the entire chapter.

Key Concepts

Intermolecular forcesConstituent particlesFixed positionsCrystal latticeVibrational motion
11short answerMEDIUM

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:
(a) Tetraphosphorus decoxide (P₄O₁₀)

(b) Brass

(c) Rb

(d) LiBr

(e) P₄

(f) SiC

(g) Plastic

(h) I₂

(i) Graphite

(j) Tetraphosphorus (P₄)

(k) Si

(l) NH₃

(m) SiCl₄

(n) Amorphous SiO₂

(o) P₄O₆

✅ Answer

(a) Tetraphosphorus decoxide (P₄O₁₀): Molecular solid. It consists of discrete P₄O₁₀ molecules held together by weak van der Waals forces.
(b) Brass: Metallic solid. It is an alloy of copper and zinc, characterized by metallic bonding (sea of delocalized electrons).

(c) Rb (Rubidium): Metallic solid. It is an alkali metal with strong metallic bonds.

(d) LiBr (Lithium Bromide): Ionic solid. It is formed by electrostatic attraction between Li⁺ cations and Br⁻ anions.

(e) P₄ (White Phosphorus): Molecular solid. It consists of discrete P₄ molecules held by weak van der Waals forces.

(f) SiC (Silicon Carbide): Network (covalent) solid. Atoms are held by strong covalent bonds throughout the crystal lattice, forming a giant molecular structure.

(g) Plastic: Amorphous solid. It lacks a long-range ordered arrangement of its constituent molecules (polymers).

(h) I₂ (Iodine): Molecular solid. It consists of discrete I₂ molecules held by weak van der Waals forces.

(i) Graphite: Network (covalent) solid. Carbon atoms are covalently bonded in layers, with weak forces between layers. It exhibits characteristics of both covalent and metallic solids due to delocalized electrons within layers.

(j) Tetraphosphorus (P₄): Molecular solid. (Same as (e)). It consists of discrete P₄ molecules held by weak van der Waals forces.

(k) Si (Silicon): Network (covalent) solid. Similar to diamond, silicon atoms are covalently bonded in a giant three-dimensional network.

(l) NH₃ (Ammonia): Molecular solid. It consists of discrete NH₃ molecules held by hydrogen bonding and weak van der Waals forces.

(m) SiCl₄ (Silicon Tetrachloride): Molecular solid. It consists of discrete SiCl₄ molecules held by weak van der Waals forces.

(n) Amorphous SiO₂ (Silica Glass): Amorphous solid. It lacks a regular, long-range ordered structure, unlike crystalline quartz.

(o) P₄O₆ (Tetraphosphorus hexoxide): Molecular solid. It consists of discrete P₄O₆ molecules held by weak van der Waals forces.

NEET Relevance

Questions on classification of solids based on their bonding and properties often appear in NEET as MCQs, sometimes in 'match the following' format or 'identify the type of solid'.

Key Concepts

Types of solidsIonic solidsMetallic solidsMolecular solidsNetwork (covalent) solidsAmorphous solidsIntermolecular forces
12short answer🎯 HIGH⭐ Important

What is meant by the term ‘coordination number’? What is the coordination number of atoms:
(a) in a cubic close-packed structure?

(b) in a body-centred cubic structure?

✅ Answer

Coordination Number: The coordination number of an atom or ion in a crystal lattice is defined as the number of its nearest neighbours with which it is in direct contact. It represents the number of atoms or ions immediately surrounding a central atom or ion in a crystal structure.

(a) Coordination number of atoms in a cubic close-packed (ccp) structure:
In a ccp structure (which is equivalent to a face-centred cubic, FCC, lattice), each atom is surrounded by 12 nearest neighbours. For example, a corner atom is surrounded by 3 atoms in its own plane, 6 atoms in the plane above, and 3 atoms in the plane below (or vice versa for a face-centred atom).

Therefore, the coordination number is 12.

(b) Coordination number of atoms in a body-centred cubic (bcc) structure:
In a bcc structure, the atom located at the body-centre of the unit cell is in direct contact with the 8 atoms present at the corners of the cube. Similarly, each corner atom is in contact with 8 body-centred atoms in adjacent unit cells.

Therefore, the coordination number is 8.

NEET Relevance

Coordination number is a fundamental concept in solid state chemistry and is very frequently asked in NEET as direct MCQs or as part of questions involving crystal structures.

Key Concepts

Coordination numberCubic close-packed (ccp) structureFace-centred cubic (fcc) structureBody-centred cubic (bcc) structure

This question has appeared in previous NEET exams.

13short answer🎯 HIGH⭐ Important

How many octahedral voids are there in 1 mol of a compound forming hcp structure? What is the number of tetrahedral voids in 1 mol of a compound forming hcp structure?

✅ Answer

To determine the number of voids in 1 mol of a compound forming an hcp structure, we first need to find the number of atoms in 1 mol.

1. Number of atoms in 1 mol of a compound:
1 mol of any substance contains Avogadro's number (NA) of constituent particles. For atoms, NA = 6.022 × 10²³ atoms/mol.

So, the number of atoms (N) in 1 mol = 6.022 × 10²³ atoms.

2. Relationship between number of atoms and voids in close-packed structures (hcp or ccp):
If the number of close-packed spheres (atoms) is N:

* Number of octahedral voids = N

* Number of tetrahedral voids = 2N

3. Number of octahedral voids in 1 mol of a compound forming hcp structure:
Number of octahedral voids = N = 6.022 × 10²³

4. Number of tetrahedral voids in 1 mol of a compound forming hcp structure:
Number of tetrahedral voids = 2N = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

Summary:

  • Number of octahedral voids in 1 mol of a compound forming hcp structure = 6.022 × 10²³
  • Number of tetrahedral voids in 1 mol of a compound forming hcp structure = 1.2044 × 10²⁴

Solution Steps

  1. Step 1: Determine the number of atoms in 1 mol

    1 mol of any substance contains Avogadro's number (NA) of particles. So, for 1 mol of a compound, the number of atoms (N) is NA = 6.022 × 10²³.

  2. Step 2: Recall the relationship between atoms and voids in close-packed structures

    In close-packed structures (like hcp or ccp), if the number of constituent atoms is N, then:
    - Number of octahedral voids = N

    - Number of tetrahedral voids = 2N

  3. Step 3: Calculate the number of octahedral voids

    Using the relationship from Step 2, the number of octahedral voids = N = NA = 6.022 × 10²³.

  4. Step 4: Calculate the number of tetrahedral voids

    Using the relationship from Step 2, the number of tetrahedral voids = 2N = 2 × NA = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴.

NEET Relevance

This is a very common conceptual and numerical question in NEET. Understanding the relationship between the number of atoms and the number of voids in close-packed structures is crucial.

Key Concepts

HCP structureOctahedral voidsTetrahedral voidsAvogadro's numberStoichiometry of voids

This question has appeared in previous NEET exams.

14long answer🎯 HIGH⭐ Important

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

✅ Answer

What is a Semiconductor?
A semiconductor is a solid material that has electrical conductivity between that of a conductor (like copper) and an insulator (like glass). Its electrical conductivity can be significantly controlled by introducing impurities (a process called doping) or by changing its temperature. Common examples include silicon (Si) and germanium (Ge), which are Group 14 elements.

Types of Semiconductors and Conduction Mechanism
Semiconductors are broadly classified into two types:

1. Intrinsic Semiconductors
* Definition: These are pure semiconductors (e.g., pure silicon or germanium) that have not been doped with any impurities. Their electrical conductivity is solely due to the electrons excited from the valence band to the conduction band and the corresponding 'holes' created in the valence band.

* Conduction Mechanism: At absolute zero (0 K), intrinsic semiconductors behave as perfect insulators because all valence electrons are tightly held in covalent bonds. As the temperature increases, some electrons gain enough thermal energy to break free from their covalent bonds and move into the conduction band, becoming mobile charge carriers. This leaves behind a 'hole' (a vacant electron site) in the valence band. An electron from an adjacent bond can move into this hole, effectively causing the hole to move in the opposite direction. Thus, conduction occurs due to both electrons (in the conduction band) and holes (in the valence band). The conductivity of intrinsic semiconductors is generally low and increases with temperature.

2. Extrinsic Semiconductors
Extrinsic semiconductors are formed by adding a small, controlled amount of a suitable impurity (dopant) to a pure intrinsic semiconductor. This process, called doping, significantly enhances the conductivity by creating either excess electrons or holes. Extrinsic semiconductors are of two types:

  • a) n-type Semiconductors
  • Formation: An n-type semiconductor is formed when a pure semiconductor (like Si or Ge, which are Group 14 elements with four valence electrons) is doped with a pentavalent impurity (a Group 15 element) such as phosphorus (P), arsenic (As), or antimony (Sb). These impurity atoms have five valence electrons.
  • Conduction Mechanism: When a pentavalent impurity atom replaces a silicon atom in the crystal lattice, four of its five valence electrons form covalent bonds with the four surrounding silicon atoms. The fifth valence electron is extra and is very loosely bound to the impurity atom. This extra electron requires very little energy to move into the conduction band, even at room temperature, becoming a free electron. These free electrons are the majority charge carriers, while holes (formed due to thermal excitation) are the minority charge carriers. The impurity atoms are called 'donor' impurities because they donate extra electrons.
  • b) p-type Semiconductors
  • Formation: A p-type semiconductor is formed when a pure semiconductor (like Si or Ge) is doped with a trivalent impurity (a Group 13 element) such as boron (B), aluminium (Al), or gallium (Ga). These impurity atoms have three valence electrons.
  • Conduction Mechanism: When a trivalent impurity atom replaces a silicon atom in the crystal lattice, its three valence electrons form covalent bonds with three of the surrounding silicon atoms. The fourth bond with a silicon atom remains incomplete, creating an 'electron vacancy' or a 'hole' in the crystal lattice. An electron from an adjacent silicon atom can jump into this hole, thereby creating a new hole at its original position. This movement of holes under an electric field constitutes electrical conduction. The holes are the majority charge carriers, while electrons (formed due to thermal excitation) are the minority charge carriers. The impurity atoms are called 'acceptor' impurities because they accept electrons.

Solution Steps

  1. Step 1: Define Semiconductor

    Provide a concise definition of a semiconductor, highlighting its conductivity range and the methods to control it (doping, temperature).

  2. Step 2: Introduce Intrinsic Semiconductors

    Briefly describe intrinsic semiconductors as pure materials and explain their basic conduction mechanism involving thermally generated electron-hole pairs.

  3. Step 3: Introduce Extrinsic Semiconductors

    Explain that extrinsic semiconductors are doped and are categorized into two types: n-type and p-type.

  4. Step 4: Describe n-type Semiconductors

    Explain how n-type semiconductors are formed by doping with pentavalent impurities (Group 15 elements). Detail the mechanism of conduction, emphasizing that extra electrons from the dopant become majority carriers.

  5. Step 5: Describe p-type Semiconductors

    Explain how p-type semiconductors are formed by doping with trivalent impurities (Group 13 elements). Detail the mechanism of conduction, emphasizing that the creation and movement of 'holes' become the majority carriers.

NEET Relevance

This topic is highly relevant for NEET. Questions frequently appear as MCQs on the definition of semiconductors, types of doping (n-type vs. p-type), identification of dopants, and the mechanism of conduction by electrons or holes.

Key Concepts

SemiconductorDopingIntrinsic semiconductorExtrinsic semiconductorn-type semiconductorp-type semiconductorElectron holesValence bandConduction bandDonor impurityAcceptor impurity

This question has appeared in previous NEET exams.

15short answerMEDIUM⭐ Important

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

In non-stoichiometric cuprous oxide (Cu₂O), the copper to oxygen ratio is slightly less than 2:1. This indicates a deficiency of copper ions in the crystal lattice, meaning some Cu⁺ ions are missing from their regular lattice sites. This type of defect is known as a metal deficiency defect.

To maintain the overall electrical neutrality of the crystal, for every two missing Cu⁺ ions, one Cu²⁺ ion is formed in the lattice. For example, if two Cu⁺ ions are missing, one Cu²⁺ ion can occupy one of the vacant sites, and the other site remains as a cation vacancy. Alternatively, if one Cu⁺ ion is missing, two adjacent Cu⁺ ions can convert to two Cu²⁺ ions to balance the charge.

The presence of Cu²⁺ ions (which have a higher positive charge than the original Cu⁺ ions) along with cation vacancies creates 'holes' in the valence band. An electron from an adjacent Cu⁺ ion can jump into a Cu²⁺ ion, effectively moving the positive charge (hole) to the original position of the Cu⁺ ion. This movement of positive holes under an applied electric field constitutes electrical conduction.

Since the electrical conduction in this non-stoichiometric Cu₂O is primarily due to the movement of these positive holes, it behaves as a p-type semiconductor.

NEET Relevance

Understanding the relationship between crystal defects (especially non-stoichiometric defects) and semiconductor properties (n-type/p-type) is relevant for NEET. Questions may involve identifying the type of semiconductor based on defect description.

Key Concepts

Non-stoichiometryMetal deficiency defectCation vacancyp-type semiconductorHole conductionElectrical neutrality

This question has appeared in previous NEET exams.

16numerical🎯 HIGH⭐ Important

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's derive the formula of ferric oxide based on the given information:

1. Assume the number of oxide ions (O²⁻):
Let the number of oxide ions (O²⁻) in the hexagonal close-packed (hcp) array be 'n'.

2. Determine the number of octahedral holes:
In a close-packed structure (like hcp), the number of octahedral holes is equal to the number of atoms (or ions) forming the close-packed array.

Therefore, the number of octahedral holes = n.

3. Calculate the number of ferric ions (Fe³⁺):
It is given that two out of every three octahedral holes are occupied by ferric ions (Fe³⁺).

Number of ferric ions (Fe³⁺) = (2/3) × (Number of octahedral holes)

Number of ferric ions (Fe³⁺) = (2/3) × n

4. Determine the ratio of ferric ions to oxide ions:
Ratio of Fe³⁺ ions : O²⁻ ions = (2/3)n : n

To simplify the ratio, divide both sides by 'n':

Fe³⁺ : O²⁻ = 2/3 : 1

Multiply both sides by 3 to get whole numbers:

Fe³⁺ : O²⁻ = 2 : 3

5. Write the chemical formula:
Based on the ratio of 2 ferric ions to 3 oxide ions, the formula of ferric oxide is Fe₂O₃.

Solution Steps

  1. Step 1: Assume Number of Anions

    Assume the number of oxide ions (O²⁻) in the hexagonal close-packed (hcp) array to be 'n'.

  2. Step 2: Determine Number of Octahedral Holes

    Recall that in a close-packed structure, the number of octahedral holes is equal to the number of particles forming the packing. So, the number of octahedral holes = n.

  3. Step 3: Calculate Number of Cations

    Use the given information that two out of every three octahedral holes are occupied by ferric ions (Fe³⁺). Calculate the number of Fe³⁺ ions as (2/3) × n.

  4. Step 4: Determine Simplest Ratio

    Find the ratio of the number of ferric ions to oxide ions (Fe³⁺ : O²⁻) and simplify it to the smallest whole numbers.

  5. Step 5: Write the Formula

    Based on the simplest whole-number ratio, write the chemical formula of the ferric oxide.

Final Answer: Verify units and significant figures.

NEET Relevance

Deriving the formula of a compound from its crystal structure and occupancy of voids is a very common and important type of question in NEET. It tests understanding of crystal packing and stoichiometry.

Key Concepts

Hexagonal close-packed (hcp) structureOctahedral holesStoichiometryFormula derivationIonic solids

This question has appeared in previous NEET exams.

17short answer🎯 HIGH⭐ Important

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀)

(ii) Graphite

(iii) Brass

(iv) Ammonium phosphate ((NH₄)₃PO₄)

(v) SiC

(vi) Rb

(vii) I₂

(viii) LiBr

(ix) P₄

(x) Plastic

✅ Answer

Here is the classification of each solid:

  • (i) Tetra phosphorus decoxide (P₄O₁₀): Molecular solid. It consists of discrete P₄O₁₀ molecules held together by weak van der Waals forces.
  • (ii) Graphite: Network (covalent) solid. It is a giant covalent network solid where carbon atoms are covalently bonded within layers, and these layers are held by weak van der Waals forces. It exhibits properties of both covalent and molecular solids due to its unique structure, but primarily classified as network covalent.
  • (iii) Brass: Metallic solid. Brass is an alloy of copper and zinc, characterized by metallic bonding (a 'sea' of delocalized electrons).
  • (iv) Ammonium phosphate ((NH₄)₃PO₄): Ionic solid. It is composed of ammonium cations (NH₄⁺) and phosphate anions (PO₄³⁻) held together by strong electrostatic forces.
  • (v) SiC (Silicon Carbide): Network (covalent) solid. It has a giant covalent network structure where silicon and carbon atoms are extensively bonded throughout the crystal.
  • (vi) Rb (Rubidium): Metallic solid. Rubidium is an alkali metal, and all metals are characterized by metallic bonding.
  • (vii) I₂ (Iodine): Molecular solid. It consists of discrete I₂ molecules held together by weak van der Waals forces.
  • (viii) LiBr (Lithium Bromide): Ionic solid. It is composed of lithium cations (Li⁺) and bromide anions (Br⁻) held together by strong electrostatic forces.
  • (ix) P₄ (White Phosphorus): Molecular solid. It consists of discrete P₄ molecules held together by weak van der Waals forces.
  • (x) Plastic: Amorphous solid. Plastics are polymers with a disordered arrangement of constituent molecules, lacking long-range crystalline order.

NEET Relevance

This type of classification question is highly relevant for NEET. It tests fundamental understanding of different types of solids based on their constituent particles and the forces holding them together, which is crucial for predicting properties.

Key Concepts

Types of solidsIonic solidMetallic solidMolecular solidNetwork (covalent) solidAmorphous solidChemical bondingIntermolecular forces

This question has appeared in previous NEET exams.

18short answerMEDIUM

Solid A is a very hard electrical insulator in solid as well as in molten state and has extremely high melting point. What type of solid is it?

✅ Answer

Let's analyze the given properties of Solid A:

  1. Very hard: This property suggests strong forces of attraction between the constituent particles.
  2. Electrical insulator in solid as well as in molten state: This indicates the absence of free mobile electrons or mobile ions in both states. If it were an ionic solid, it would conduct in the molten state. If it were a metallic solid, it would conduct in both solid and molten states.
  3. Extremely high melting point: This implies that a very large amount of energy is required to overcome the strong forces holding the particles together.

Considering these properties, the solid that fits this description is a covalent (or network) solid.

Covalent solids (like diamond, silicon carbide (SiC), or quartz (SiO₂)) are characterized by a continuous network of strong covalent bonds extending throughout the entire crystal. These strong bonds make them very hard, give them extremely high melting points, and because electrons are localized in covalent bonds (not free to move), they are excellent electrical insulators in both solid and molten states.

NEET Relevance

Questions that require correlating the physical properties of a solid (hardness, conductivity, melting point) with its type (ionic, metallic, molecular, covalent) are common in NEET. They test conceptual understanding of bonding and structure.

Key Concepts

Properties of solidsTypes of solidsCovalent solidNetwork solidElectrical conductivityMelting pointHardness
19numerical🎯 HIGH⭐ Important

If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

✅ Answer

When NaCl is doped with SrCl₂, each Sr²⁺ ion replaces two Na⁺ ions to maintain electrical neutrality. One Na⁺ site is occupied by Sr²⁺, and the other Na⁺ site becomes a cation vacancy. Therefore, the concentration of cation vacancies is equal to the mole percentage of SrCl₂ doped.

Given doping concentration = 10⁻³ mol % of SrCl₂.
This means 10⁻³ moles of SrCl₂ are present in 100 moles of NaCl.

Number of moles of SrCl₂ per mole of NaCl = (10⁻³/100) = 10⁻⁵ mol.

Since 1 Sr²⁺ ion introduces 1 cation vacancy, the concentration of cation vacancies is 10⁻⁵ mol per mole of NaCl.

To find the number of cation vacancies per mole of NaCl:
Number of vacancies = 10⁻⁵ mol × Avogadro's number

Number of vacancies = 10⁻⁵ mol × 6.022 × 10²³ mol⁻¹

Number of vacancies = 6.022 × 10¹⁸ vacancies.

Solution Steps

  1. Step 1: Understand the Doping Mechanism

    When NaCl (an ionic compound with Na⁺ and Cl⁻ ions) is doped with SrCl₂, the Sr²⁺ ions replace Na⁺ ions in the crystal lattice. To maintain electrical neutrality, for every Sr²⁺ ion (charge +2) that replaces an Na⁺ ion (charge +1), a second Na⁺ ion must be removed from the lattice, creating a cation vacancy. Thus, one Sr²⁺ ion introduces one cation vacancy.

  2. Step 2: Convert mol % to mole fraction

    Given doping concentration = 10⁻³ mol % of SrCl₂.
    This means 10⁻³ moles of SrCl₂ are present in 100 moles of NaCl.

    So, the number of moles of SrCl₂ per mole of NaCl = (10⁻³/100) = 10⁻⁵ mol.

  3. Step 3: Relate Doping to Vacancies

    As established in Step 1, each Sr²⁺ ion introduced creates one cation vacancy. Therefore, the concentration of cation vacancies is equal to the mole fraction of SrCl₂ doped.
    Concentration of cation vacancies = 10⁻⁵ mol per mole of NaCl.

  4. Step 4: Calculate the Number of Cation Vacancies

    To find the actual number of cation vacancies per mole of NaCl, multiply the mole fraction by Avogadro's number (NA = 6.022 × 10²³ mol⁻¹).
    Number of vacancies = (Concentration of vacancies in mol) × NA

    Number of vacancies = 10⁻⁵ mol × 6.022 × 10²³ mol⁻¹

    Number of vacancies = 6.022 × 10¹⁸ vacancies.

Final Answer: Verify units and significant figures.

NEET Relevance

This type of question frequently appears in NEET, testing the understanding of impurity defects, stoichiometry, and calculation of vacancies in ionic solids. It often involves relating doping concentration to the number of defects using Avogadro's number.

Key Concepts

DopingImpurity defectsCation vacanciesElectrical neutralityAvogadro's number

This question has appeared in previous NEET exams.

20long answer🎯 HIGH⭐ Important

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

✅ Answer

(i) Schottky Defect
Definition: A Schottky defect is a type of stoichiometric point defect in ionic crystals where an equal number of cations and anions are missing from their lattice sites, creating vacancies. This defect maintains the electrical neutrality of the crystal.

Characteristics:

  • Occurs in highly ionic compounds with high coordination numbers and where cations and anions are of nearly similar sizes (e.g., alkali halides).
  • Leads to a decrease in the density of the crystal because mass is lost while volume remains constant.
  • Increases the electrical conductivity of the crystal due to the movement of ions into the vacancies.

Example: NaCl, KCl, CsCl, AgBr.

(ii) Frenkel Defect
Definition: A Frenkel defect is a type of stoichiometric point defect in ionic crystals where an ion (usually the smaller cation) leaves its normal lattice site and occupies an interstitial site, creating a vacancy at its original position and an interstitial defect at the new position. The overall electrical neutrality of the crystal is maintained.

Characteristics:

  • Occurs in ionic compounds with a large difference in the size of cations and anions (cations are usually much smaller) and low coordination numbers.
  • Does not change the density of the crystal because no ions are missing from the crystal; they are just displaced within the lattice.
  • Increases the electrical conductivity due to the movement of ions and vacancies.

Example: AgCl, AgBr, AgI, ZnS.

(iii) Interstitials (Interstitial Defect)
Definition: An interstitial defect occurs when an atom or ion occupies an interstitial site (a void or space between the regular lattice sites) in the crystal structure. This can be either a foreign atom (impurity interstitial) or an atom of the host lattice itself (self-interstitial).

Characteristics:

  • Increases the density of the substance if it's a self-interstitial or an impurity atom.
  • Common in non-ionic solids (e.g., metals) where atoms can be squeezed into interstitial spaces.
  • In ionic solids, it is less common for host ions to occupy interstitial sites due to their larger size and charge repulsion, but it is a component of the Frenkel defect.

Example: In metals, a small atom like hydrogen occupying an interstitial site. In a Frenkel defect, the displaced cation becomes an interstitial.

(iv) F-centres
Definition: F-centres (from the German word 'Farbe' meaning colour) are anionic vacancies occupied by unpaired electrons. These defects are responsible for the colour of certain alkali halide crystals when they are heated in an atmosphere of the metal vapour.

Characteristics:

  • Formed when alkali metal halides are heated in an excess of the alkali metal vapour. For example, NaCl heated in sodium vapour turns yellow.
  • The excess metal atoms deposit on the surface, and their electrons diffuse into the crystal to occupy anionic vacancies.
  • These trapped electrons absorb light from the visible spectrum, exciting them to higher energy levels, and thus impart colour to the crystal.
  • They are a type of non-stoichiometric defect (metal excess defect).

Example:

  • NaCl crystal heated in Na vapour turns yellow.
  • KCl crystal heated in K vapour turns violet.
  • LiCl crystal heated in Li vapour turns pink.

Solution Steps

  1. Step 1: Schottky Defect

    Define Schottky defect as a stoichiometric point defect involving missing equal numbers of cations and anions from lattice sites. Explain its characteristics (density decrease, high coordination number, similar ion sizes) and provide examples like NaCl, KCl, CsCl, AgBr.

  2. Step 2: Frenkel Defect

    Define Frenkel defect as a stoichiometric point defect where an ion (usually cation) leaves its lattice site to occupy an interstitial site. Explain its characteristics (density unchanged, large size difference between ions, low coordination number) and provide examples like AgCl, AgBr, AgI, ZnS.

  3. Step 3: Interstitials

    Define interstitial defect as an atom or ion occupying an interstitial site. Differentiate between self-interstitials and impurity interstitials. Explain its characteristics (density increase) and provide examples, including its role in Frenkel defect.

  4. Step 4: F-centres

    Define F-centres as anionic vacancies occupied by unpaired electrons, responsible for crystal colour. Explain their formation (heating alkali halides in metal vapour), characteristics (absorption of light, imparting colour), and provide specific examples like NaCl turning yellow, KCl turning violet, LiCl turning pink.

NEET Relevance

This is a very important question for NEET. Definitions, characteristics, examples, and differences between these defects are frequently tested in MCQs. Understanding the impact on density and electrical conductivity is crucial.

Key Concepts

Point defectsStoichiometric defectsNon-stoichiometric defectsSchottky defectFrenkel defectInterstitial defectF-centresElectrical neutralityDensity changesColour centres

This question has appeared in previous NEET exams.

21short answerMEDIUM⭐ Important

Classify each of the following as being either a p-type or an n-type semiconductor: (i) Ge doped with In (ii) B doped with Si.

✅ Answer

(i) Ge doped with In:
* Germanium (Ge) belongs to Group 14 and has 4 valence electrons.

* Indium (In) belongs to Group 13 and has 3 valence electrons.

* When Ge is doped with In, an electron-deficient impurity (In) is introduced. Each In atom replaces a Ge atom, forming three covalent bonds with neighboring Ge atoms, but one bond remains incomplete (a 'hole' is created).

* These holes can accept electrons, making it a p-type semiconductor (p for positive holes).

(ii) B doped with Si:
* Boron (B) belongs to Group 13 and has 3 valence electrons.

* Silicon (Si) belongs to Group 14 and has 4 valence electrons.

* When B is doped with Si, an electron-rich impurity (Si) is introduced into the Group 13 element B. Each Si atom replaces a B atom. While B needs 3 electrons for bonding, Si provides 4. This creates an extra electron that is not involved in bonding.

* These extra electrons are delocalized and can move, making it an n-type semiconductor (n for negative electrons).

NEET Relevance

Questions on classifying semiconductors based on doping are common. It tests the understanding of valence electrons of host and dopant atoms and how they create either electron-rich (n-type) or electron-deficient (p-type) regions.

Key Concepts

SemiconductorsDopingn-type semiconductorp-type semiconductorValence electronsGroup 13, 14, 15 elements

This question has appeared in previous NEET exams.

22numerical🎯 HIGH⭐ Important

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

✅ Answer

For a face-centred cubic (FCC) unit cell, the atoms are in contact along the face diagonal. The relationship between the atomic radius (r) and the edge length (a) of the unit cell is given by:

Face diagonal = 4r
Also, by Pythagoras theorem, Face diagonal = √(a² + a²) = √(2a²) = a√2

Therefore, 4r = a√2
Or, a = 4r / √2

Simplifying, a = 2√2 r

Given atomic radius (r) = 0.144 nm

Substitute the value of r into the formula: a = 2 × √2 × 0.144 nm a = 2 × 1.414 × 0.144 nm a = 2.828 × 0.144 nm a = 0.407232 nm

Rounding to three significant figures (as per the given atomic radius): a ≈ 0.407 nm

Thus, the length of a side of the gold unit cell is approximately 0.407 nm.

Solution Steps

  1. Step 1: Identify the Crystal Structure and Given Information

    The problem states that gold crystallizes in a face-centred cubic (FCC) unit cell. The atomic radius (r) of gold is given as 0.144 nm. We need to find the length of a side (edge length, 'a') of the unit cell.

  2. Step 2: Recall the Relationship between 'a' and 'r' for FCC

    In an FCC unit cell, atoms are present at all corners and the centre of each face. The atoms along the face diagonal are in contact. The length of the face diagonal is equal to 4 times the atomic radius (4r).
    Using the Pythagorean theorem for a face diagonal: (Face diagonal)² = a² + a² = 2a².

    So, Face diagonal = a√2.

  3. Step 3: Equate the Expressions for Face Diagonal

    Equating the two expressions for the face diagonal:
    4r = a√2

    Rearrange to solve for 'a': a = 4r / √2

    To simplify, multiply numerator and denominator by √2: a = (4r × √2) / (√2 × √2) a = 4√2 r / 2 a = 2√2 r

  4. Step 4: Substitute the Given Value and Calculate

    Given r = 0.144 nm.
    Substitute this value into the derived formula: a = 2 × √2 × 0.144 nm a = 2 × 1.41421356 × 0.144 nm a = 2.82842712 × 0.144 nm a = 0.407293505 nm

  5. Step 5: Round to Appropriate Significant Figures

    The given atomic radius (0.144 nm) has three significant figures. Therefore, the answer should also be rounded to three significant figures. a ≈ 0.407 nm.

    Final Answer: The length of a side of the gold unit cell is 0.407 nm.

Final Answer: Verify units and significant figures.

NEET Relevance

Calculations involving the relationship between atomic radius and edge length for different cubic unit cells (SC, BCC, FCC) are very common in NEET. Students must memorize or be able to derive these relationships quickly.

Key Concepts

Face-centred cubic (FCC) unit cellAtomic radiusEdge lengthPythagorean theoremUnit cell parameters

This question has appeared in previous NEET exams.

23short answerLOW

In a ‘group 13-15’ compound semiconductor, the elements of group 13 and 15 are combined. For example, GaAs. What type of semiconductor is it? Explain.

✅ Answer

A 'group 13-15' compound semiconductor, such as Gallium Arsenide (GaAs), is an intrinsic semiconductor.

Explanation:

  • Elements from Group 13 (like Ga) have 3 valence electrons.
  • Elements from Group 15 (like As) have 5 valence electrons.
  • When these elements combine to form a compound semiconductor, each atom contributes its valence electrons to form covalent bonds. For example, in GaAs, each Ga atom forms four covalent bonds, one with each of its four nearest As atoms, and vice-versa.
  • The average number of valence electrons per atom in such a compound is (3 + 5) / 2 = 8 / 2 = 4. This average of 4 valence electrons per atom is similar to that of Group 14 elements (like Si or Ge), which are intrinsic semiconductors.
  • These compounds form a crystal lattice where each atom is tetrahedrally bonded to four atoms of the other type, resulting in a stable covalent network with no free electrons or holes at absolute zero. At higher temperatures, thermal energy can break some covalent bonds, creating electron-hole pairs, leading to intrinsic conductivity.

Therefore, 'group 13-15' compounds behave as intrinsic semiconductors, exhibiting properties similar to elemental semiconductors like silicon and germanium.

NEET Relevance

This question tests a basic understanding of compound semiconductors and their classification. While the concept is fundamental, direct questions on 'group 13-15' classification are less frequent in NEET compared to doping of elemental semiconductors.

Key Concepts

Compound semiconductorsIntrinsic semiconductorsValence electronsCovalent bondingGroup 13 elementsGroup 15 elements
24short answer🎯 HIGH⭐ Important

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous:
(a) Tetraphosphorus decoxide (P₄O₁₀)

(b) Brass

(c) Rb

(d) LiBr

(e) P₄

(f) SiC

(g) Plastic

(h) I₂

(i) P₄S₃

(j) Graphite

(k) H₂

(l) Si

✅ Answer

The classification of the given solids is as follows:

(a) Tetraphosphorus decoxide (P₄O₁₀): Molecular solid. It consists of discrete P₄O₁₀ molecules held together by weak van der Waals forces.

(b) Brass: Metallic solid. Brass is an alloy of copper and zinc, where metal atoms are held together by metallic bonds.

(c) Rb (Rubidium): Metallic solid. Rubidium is an alkali metal, and its atoms are held together by metallic bonds.

(d) LiBr (Lithium bromide): Ionic solid. It is formed by the electrostatic attraction between Li⁺ cations and Br⁻ anions.

(e) P₄ (White Phosphorus): Molecular solid. It consists of discrete P₄ molecules held together by weak van der Waals forces.

(f) SiC (Silicon Carbide): Network (covalent) solid. Silicon and carbon atoms are extensively linked by strong covalent bonds throughout the crystal lattice, forming a giant molecule.

(g) Plastic: Amorphous solid. Plastics are polymers with a disordered arrangement of constituent molecules, lacking a long-range ordered structure.

(h) I₂ (Iodine): Molecular solid. It consists of discrete I₂ molecules held together by weak van der Waals forces.

(i) P₄S₃ (Tetraphosphorus trisulphide): Molecular solid. It consists of discrete P₄S₃ molecules held together by weak van der Waals forces.

(j) Graphite: Network (covalent) solid. Carbon atoms are covalently bonded in layers, and these layers are held by weak van der Waals forces. However, within the layers, it's a covalent network.

(k) H₂ (Hydrogen): Molecular solid. At very low temperatures, hydrogen forms a solid consisting of discrete H₂ molecules held by weak van der Waals forces.

(l) Si (Silicon): Network (covalent) solid. Silicon atoms are extensively linked by strong covalent bonds throughout the crystal lattice, similar to diamond.

NEET Relevance

Classification of solids based on their bonding and constituent particles is a fundamental concept. Questions identifying the type of solid for a given substance or relating properties to solid type are common in NEET.

Key Concepts

Types of solidsIonic solidsMetallic solidsMolecular solidsNetwork (covalent) solidsAmorphous solidsIntermolecular forces

This question has appeared in previous NEET exams.

25long answer🎯 HIGH⭐ Important

Explain the following terms with suitable examples:
(a) Schottky defect

(b) Frenkel defect

(c) Interstitials

(d) F-centres

✅ Answer

Let's explain each term with suitable examples:

(a) Schottky Defect

Definition: A Schottky defect is a type of point defect in an ionic crystal that arises when an equal number of cations and anions are missing from their regular lattice sites. This maintains the electrical neutrality of the crystal.

Explanation: This defect is essentially a pair of vacancies (one cation vacancy and one anion vacancy). It occurs in highly ionic compounds where the cation and anion are of similar size, and the coordination number is high. The presence of Schottky defects decreases the density of the crystal.

Example: Alkali halides like NaCl, KCl, CsCl, and AgBr exhibit Schottky defects. In NaCl, if one Na⁺ ion and one Cl⁻ ion are missing from their lattice positions, it constitutes a Schottky defect.

(b) Frenkel Defect

Definition: A Frenkel defect is a type of point defect in an ionic crystal that arises when an ion (usually the smaller cation) leaves its regular lattice site and occupies an interstitial position, creating a vacancy at its original site.

Explanation: This defect is a combination of a vacancy defect and an interstitial defect. It occurs in ionic compounds where there is a large difference in the size of cations and anions, and the coordination number is low. The smaller ion (cation) can easily move into an interstitial site. Frenkel defects do not change the density of the crystal because no ions are missing from the crystal lattice.

Example: Silver halides like AgCl, AgBr, AgI, and ZnS exhibit Frenkel defects. In AgCl, an Ag⁺ ion might leave its lattice site and occupy an interstitial position, creating a vacancy at its original site.

(c) Interstitials (Interstitial Defect)

Definition: An interstitial defect occurs when an atom or ion occupies an interstitial site (a void space) in the crystal lattice that is normally vacant.

Explanation: This defect is common in non-ionic solids (like metals) where an extra atom is present in the interstitial space. In ionic solids, an interstitial defect usually refers to a cation occupying an interstitial site (as seen in Frenkel defect). The presence of interstitial atoms or ions increases the density of the crystal.

Example: In a metallic crystal, an extra atom of the same type (self-interstitial) or a different type (impurity interstitial) might occupy an interstitial site. For instance, carbon atoms occupying interstitial sites in iron to form steel.

(d) F-centres

Definition: F-centres (from the German word 'Farbe' meaning colour) are anionic vacancies in an ionic crystal that are occupied by unpaired electrons.

Explanation: These defects are responsible for the colour of certain alkali halide crystals. They are formed when alkali halide crystals are heated in an atmosphere of alkali metal vapour. For example, when NaCl crystals are heated in sodium vapour, Na atoms deposit on the surface. Cl⁻ ions diffuse to the surface to combine with Na atoms, forming NaCl. The electrons released from the Na atoms diffuse into the crystal and occupy the anionic vacancies, becoming trapped. These trapped electrons absorb light from the visible spectrum, causing the crystal to appear coloured (e.g., yellow for NaCl, violet for KCl, pink for LiCl).

Example:

  • NaCl crystals heated in Na vapour turn yellow due to F-centres.
  • KCl crystals heated in K vapour turn violet due to F-centres.
  • LiCl crystals heated in Li vapour turn pink due to F-centres.

Solution Steps

  1. Step 1: Schottky Defect

    Define Schottky defect as missing equal numbers of cations and anions, maintaining electrical neutrality. Explain its occurrence in compounds with similar ion sizes and high coordination numbers, and its effect on density. Provide examples like NaCl, KCl.

  2. Step 2: Frenkel Defect

    Define Frenkel defect as an ion leaving its lattice site to occupy an interstitial position. Explain its occurrence in compounds with large size differences between ions and low coordination numbers, and that it doesn't affect density. Provide examples like AgCl, ZnS.

  3. Step 3: Interstitials

    Define interstitial defect as an atom/ion occupying a normally vacant interstitial site. Explain its effect on density and its commonality in non-ionic solids or as part of Frenkel defects in ionic solids. Provide examples like carbon in iron.

  4. Step 4: F-centres

    Define F-centres as anionic vacancies occupied by unpaired electrons. Explain their formation (heating alkali halides in metal vapour) and their role in imparting colour to crystals. Provide examples like yellow NaCl, violet KCl.

NEET Relevance

Crystal defects are a very important topic for NEET. Questions on definitions, examples, effects on density, and conditions for their occurrence (e.g., ion size difference, coordination number) are frequently asked as MCQs.

Key Concepts

Point defectsStoichiometric defectsNon-stoichiometric defectsSchottky defectFrenkel defectInterstitial defectF-centresCrystal defects

This question has appeared in previous NEET exams.

26numerical🎯 HIGH⭐ Important

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(a) What is the length of the side of the unit cell?

(b) How many unit cells are there in 1.00 cm³ of aluminium?

✅ Answer

Given:
Metallic radius of Aluminium (r) = 125 pm

Aluminium crystallises in a cubic close-packed (ccp) structure, which is equivalent to a face-centred cubic (FCC) lattice.

(a) Length of the side of the unit cell (a)

For an FCC lattice, the atoms touch along the face diagonal. The relationship between the edge length (a) and the atomic radius (r) is:
Face diagonal = 4r

Also, by Pythagorean theorem, Face diagonal = √(a² + a²) = √(2a²) = a√2

Therefore, a√2 = 4r a = 4r / √2 a = 2√2 r

Substitute the given value of r: r = 125 pm a = 2√2 × 125 pm a = 2 × 1.414 × 125 pm a = 353.5 pm

To convert pm to cm:
1 pm = 10⁻¹² m = 10⁻¹⁰ cm a = 353.5 × 10⁻¹⁰ cm a = 3.535 × 10⁻⁸ cm

The length of the side of the unit cell is 353.5 pm or 3.535 × 10⁻⁸ cm.

    (b) Number of unit cells in 1.00 cm³ of aluminium

    First, calculate the volume of one unit cell (Vunitcell):
    Vunitcell = a³

    Vunitcell = (3.535 × 10⁻⁸ cm)³

    Vunitcell = (3.535)³ × (10⁻⁸)³ cm³

    Vunitcell = 44.13 × 10⁻²⁴ cm³

    Vunitcell = 4.413 × 10⁻²³ cm³

    Now, calculate the number of unit cells in 1.00 cm³ of aluminium:
    Number of unit cells = (Total volume) / (Volume of one unit cell)

    Number of unit cells = 1.00 cm³ / (4.413 × 10⁻²³ cm³)

    Number of unit cells = 0.2266 × 10²³

    Number of unit cells = 2.266 × 10²²

    There are approximately 2.27 × 10²² unit cells in 1.00 cm³ of aluminium.

      Solution Steps

      1. Step 1: Identify crystal structure and given radius

        Aluminium crystallises in ccp (FCC) structure. Metallic radius (r) = 125 pm.

      2. Step 2: Relate edge length (a) to radius (r) for FCC

        For FCC, atoms touch along the face diagonal. The face diagonal is 4r. Also, the face diagonal is a√2. Therefore, a√2 = 4r, which simplifies to a = 2√2 r.

      3. Step 3: Calculate edge length (a) in pm

        Substitute r = 125 pm into the formula: a = 2√2 × 125 pm = 2 × 1.414 × 125 pm = 353.5 pm.

      4. Step 4: Convert edge length (a) to cm

        Convert pm to cm: 1 pm = 10⁻¹⁰ cm. So, a = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm.

      5. Step 5: Calculate volume of one unit cell

        Volume of unit cell (Vunitcell) = a³. Vunitcell = (3.535 × 10⁻⁸ cm)³ = 4.413 × 10⁻²³ cm³.

      6. Step 6: Calculate number of unit cells in 1.00 cm³

        Number of unit cells = (Total volume) / (Volume of one unit cell) = 1.00 cm³ / (4.413 × 10⁻²³ cm³) = 2.266 × 10²².

      Final Answer: Verify units and significant figures.

      NEET Relevance

      This type of numerical problem is very common in NEET. It tests the understanding of crystal structures (FCC/BCC/SC), the relationship between atomic radius and edge length, and calculations involving unit cell volume and number of unit cells. Pay attention to units and significant figures.

      Key Concepts

      Cubic close-packed (ccp)Face-centred cubic (FCC)Relationship between edge length and atomic radiusVolume of unit cellUnit cell calculations

      This question has appeared in previous NEET exams.

      27numerical🎯 HIGH⭐ Important

      If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

      ✅ Answer

      Given:
      NaCl is doped with 10⁻³ mol % of SrCl₂.

      Understanding the Doping Mechanism:
      When SrCl₂ is added to NaCl, Sr²⁺ ions replace Na⁺ ions in the crystal lattice. To maintain electrical neutrality, for every Sr²⁺ ion (which has a +2 charge) that replaces two Na⁺ ions (each with a +1 charge), one Na⁺ ion site must become vacant. Effectively, one Sr²⁺ ion replaces one Na⁺ ion, and to balance the charge, one additional Na⁺ ion site must be vacant. Thus, for every Sr²⁺ ion introduced, one cation vacancy is created.

      Calculation:
      1. Concentration of SrCl₂:

      10⁻³ mol % means 10⁻³ moles of SrCl₂ per 100 moles of NaCl.

      So, in 100 moles of NaCl, there are 10⁻³ moles of SrCl₂.

      In 1 mole of NaCl, there are (10⁻³ / 100) moles of SrCl₂ = 10⁻⁵ moles of SrCl₂.

      2. Number of Sr²⁺ ions per mole of NaCl:
      Number of Sr²⁺ ions = Moles of SrCl₂ × Avogadro's number (NA)

      Number of Sr²⁺ ions = 10⁻⁵ mol × 6.022 × 10²³ ions/mol

      Number of Sr²⁺ ions = 6.022 × 10¹⁸ ions

      3. Concentration of Cation Vacancies:
      Since each Sr²⁺ ion introduces one cation vacancy, the number of cation vacancies will be equal to the number of Sr²⁺ ions.

      Concentration of cation vacancies = 6.022 × 10¹⁸ vacancies per mole of NaCl.

      Alternatively, if the question asks for concentration per unit volume or per unit number of Na⁺ ions, we can express it as:
      Concentration of cation vacancies = 10⁻⁵ mol of vacancies per mole of NaCl.

      If we consider 1 mole of NaCl to contain NA Na⁺ ions, then the concentration of cation vacancies is 10⁻⁵ vacancies per Na⁺ ion.
      Number of cation vacancies = 10⁻⁵ × 6.022 × 10²³ = 6.022 × 10¹⁸ vacancies.

      The concentration of cation vacancies is 6.022 × 10¹⁸ per mole of NaCl (or per 6.022 × 10²³ Na⁺ ions).

        Solution Steps

        1. Step 1: Understand the doping mechanism

          When SrCl₂ (Sr²⁺) dopes NaCl (Na⁺), each Sr²⁺ ion (charge +2) replaces two Na⁺ ions (each charge +1). To maintain electrical neutrality, one Sr²⁺ replaces one Na⁺, and an additional Na⁺ site becomes vacant. Thus, one Sr²⁺ ion creates one cation vacancy.

        2. Step 2: Calculate moles of SrCl₂ per mole of NaCl

          10⁻³ mol % of SrCl₂ means 10⁻³ moles of SrCl₂ in 100 moles of NaCl. Therefore, in 1 mole of NaCl, there are (10⁻³ / 100) = 10⁻⁵ moles of SrCl₂.

        3. Step 3: Relate moles of SrCl₂ to moles of cation vacancies

          Since 1 mole of SrCl₂ creates 1 mole of cation vacancies, 10⁻⁵ moles of SrCl₂ will create 10⁻⁵ moles of cation vacancies.

        4. Step 4: Convert moles of vacancies to number of vacancies

          Number of cation vacancies = Moles of vacancies × Avogadro's number (NA)
          Number of cation vacancies = 10⁻⁵ × 6.022 × 10²³ = 6.022 × 10¹⁸ vacancies.

        Final Answer: Verify units and significant figures.

        NEET Relevance

        This is a very important numerical problem for NEET. Questions on doping and calculation of defect concentration (especially cation vacancies) are frequently asked. Understanding the charge balance mechanism is crucial.

        Key Concepts

        DopingImpurity defectsCation vacanciesElectrical neutralityStoichiometryAvogadro's number

        This question has appeared in previous NEET exams.

        28long answer🎯 HIGH⭐ Important

        Explain the following with suitable examples:
        (a) Ferromagnetism

        (b) Paramagnetism

        (c) Ferrimagnetism

        (d) Antiferromagnetism

        (e) 13-15 compounds and 12-16 compounds.

        ✅ Answer

        Let's explain each term with suitable examples:

        (a) Ferromagnetism

        Definition: Ferromagnetism is a strong form of magnetism exhibited by certain materials, characterized by a spontaneous alignment of magnetic moments of atoms or ions in the same direction, even in the absence of an external magnetic field.

        Explanation: In ferromagnetic substances, the metal ions have unpaired electrons and are grouped together into small regions called domains. Within each domain, the magnetic moments are spontaneously aligned in the same direction. When the substance is placed in an external magnetic field, these domains align themselves in the direction of the field, resulting in a strong magnetic effect. This alignment persists even after the removal of the external field, making them permanent magnets. Ferromagnetic substances become paramagnetic above a certain temperature called the Curie temperature.

        Examples: Iron (Fe), Cobalt (Co), Nickel (Ni), Gadolinium (Gd), CrO₂.

        (b) Paramagnetism

        Definition: Paramagnetism is a weak form of magnetism exhibited by substances that have unpaired electrons. These substances are weakly attracted by an external magnetic field and lose their magnetism once the field is removed.

        Explanation: Paramagnetic substances have atoms, ions, or molecules with one or more unpaired electrons. Each unpaired electron has a magnetic moment. In the absence of an external magnetic field, these magnetic moments are randomly oriented due to thermal motion, so their net magnetic moment is zero. When placed in an external magnetic field, the magnetic moments align themselves in the direction of the field, causing a weak attraction. This alignment is temporary and disappears once the external field is removed.

        Examples: O₂, Cu²⁺, Fe³⁺, Cr³⁺, TiO, V₂O₃.

        (c) Ferrimagnetism

        Definition: Ferrimagnetism is a type of magnetism where the magnetic moments of the domains in a substance are aligned in parallel and anti-parallel directions in unequal numbers, resulting in a net magnetic moment.

        Explanation: In ferrimagnetic substances, the magnetic moments are aligned in opposite directions, similar to antiferromagnetism, but the magnitudes of the magnetic moments are unequal. This unequal alignment leads to a net spontaneous magnetic moment. These substances are weakly attracted by an external magnetic field and become paramagnetic on heating above their Curie temperature.

        Examples: Ferrites like MgFe₂O₄, ZnFe₂O₄, Fe₃O₄ (magnetite).

        (d) Antiferromagnetism

        Definition: Antiferromagnetism is a type of magnetism where the magnetic moments of the domains in a substance are aligned in parallel and anti-parallel directions in equal numbers, resulting in a net zero magnetic moment.

        Explanation: In antiferromagnetic substances, the magnetic moments of the constituent atoms or ions are aligned in an ordered fashion, but in opposite directions and with equal magnitudes. This perfect cancellation of magnetic moments leads to a net zero magnetic moment for the material. These substances show very little or no attraction to an external magnetic field.

        Examples: MnO, Mn₂O₃, FeO, NiO.

        (e) 13-15 compounds and 12-16 compounds

        These are types of semiconductor compounds formed by combining elements from specific groups of the periodic table. They are important in semiconductor technology as alternatives to elemental semiconductors like Si and Ge.

        13-15 Compounds

        Definition: These are compounds formed by combining elements from Group 13 (e.g., Al, Ga, In) and Group 15 (e.g., N, P, As, Sb) of the periodic table.

        Explanation: These compounds are isoelectronic with Group 14 elements (like Si and Ge), meaning they have an average of four valence electrons per atom. For example, in GaAs, Ga (Group 13) contributes 3 valence electrons and As (Group 15) contributes 5 valence electrons, averaging to (3+5)/2 = 4 electrons per atom. They form covalent bonds and exhibit semiconductor properties. Their band gap can be tuned by varying the composition, making them suitable for optoelectronic devices.

        Examples: GaAs (Gallium Arsenide), InSb (Indium Antimonide), AlP (Aluminium Phosphide), GaN (Gallium Nitride).

        12-16 Compounds

        Definition: These are compounds formed by combining elements from Group 12 (e.g., Zn, Cd, Hg) and Group 16 (e.g., S, Se, Te) of the periodic table.

        Explanation: Similar to 13-15 compounds, these are also isoelectronic with Group 14 elements, having an average of four valence electrons per atom. For example, in ZnS, Zn (Group 12) contributes 2 valence electrons and S (Group 16) contributes 6 valence electrons, averaging to (2+6)/2 = 4 electrons per atom. They also exhibit semiconductor properties and are used in various applications, including light-emitting diodes (LEDs), solar cells, and detectors.

        Examples: ZnS (Zinc Sulphide), CdS (Cadmium Sulphide), CdSe (Cadmium Selenide), HgTe (Mercury Telluride).

        Solution Steps

        1. Step 1: Ferromagnetism

          Define ferromagnetism as strong, permanent magnetism due to spontaneous parallel alignment of magnetic moments in domains. Explain domain alignment in external fields and persistence. Provide examples like Fe, Co, Ni.

        2. Step 2: Paramagnetism

          Define paramagnetism as weak, temporary magnetism due to unpaired electrons. Explain random orientation in absence of field and temporary alignment in external field. Provide examples like O₂, Cu²⁺.

        3. Step 3: Ferrimagnetism

          Define ferrimagnetism as unequal anti-parallel alignment of magnetic moments resulting in a net magnetic moment. Explain its weaker attraction compared to ferromagnetism. Provide examples like Fe₃O₄, ferrites.

        4. Step 4: Antiferromagnetism

          Define antiferromagnetism as equal anti-parallel alignment of magnetic moments resulting in a net zero magnetic moment. Explain the cancellation of moments. Provide examples like MnO, NiO.

        5. Step 5: 13-15 Compounds

          Define 13-15 compounds as those formed from Group 13 and Group 15 elements. Explain their isoelectronic nature with Group 14, covalent bonding, and semiconductor properties. Provide examples like GaAs, InSb.

        6. Step 6: 12-16 Compounds

          Define 12-16 compounds as those formed from Group 12 and Group 16 elements. Explain their isoelectronic nature with Group 14, semiconductor properties, and applications. Provide examples like ZnS, CdS.

        NEET Relevance

        Magnetic properties of solids and different types of semiconductors are frequently tested in NEET. Questions often involve definitions, examples, differences between types of magnetism, and the characteristics/applications of 13-15 and 12-16 compounds.

        Key Concepts

        Magnetic properties of solidsFerromagnetismParamagnetismFerrimagnetismAntiferromagnetismMagnetic domainsSemiconductors13-15 compounds12-16 compoundsIsoelectronic

        This question has appeared in previous NEET exams.

        29long answer🎯 HIGH⭐ Important

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

        ✅ Answer

        A semiconductor is a material that has electrical conductivity intermediate between that of a conductor (like metals) and an insulator (like glass). Their conductivity typically ranges from 10⁻⁶ to 10⁴ ohm⁻¹ cm⁻¹. The conductivity of semiconductors increases with increasing temperature, unlike metals where it decreases.

        Types of Semiconductors:
        Semiconductors can be broadly classified into two main types:

        1. Intrinsic Semiconductors:
        These are pure semiconductors (e.g., silicon, germanium) that have no impurities. Their electrical conductivity is due to the inherent properties of the material. At absolute zero, they behave as insulators. At higher temperatures, some electrons gain enough thermal energy to break free from covalent bonds and move into the conduction band, leaving behind 'holes' in the valence band. Both electrons and holes contribute to conduction, but their conductivity is generally low.

        2. Extrinsic Semiconductors:
        These are intrinsic semiconductors whose conductivity has been increased by adding a small amount of suitable impurity (doping). Doping creates either excess electrons or electron holes, significantly enhancing their conductivity. Extrinsic semiconductors are further divided into two types:

        (a) n-type Semiconductors:

        • Formation: When a pure semiconductor (like Si or Ge, Group 14 elements) is doped with an impurity from Group 15 (e.g., P, As, Sb), which has five valence electrons. * Conduction Mechanism: Four of the five valence electrons of the dopant atom form covalent bonds with the four surrounding silicon atoms. The fifth electron is extra and becomes delocalized. This extra electron requires very little energy to move into the conduction band. Thus, a large number of free electrons are available for conduction. * Charge Carriers: In n-type semiconductors, electrons are the majority charge carriers, and holes are the minority charge carriers. The 'n' stands for negative, referring to the charge of the majority carriers (electrons).

        (b) p-type Semiconductors:

        • Formation: When a pure semiconductor (like Si or Ge, Group 14 elements) is doped with an impurity from Group 13 (e.g., B, Al, Ga, In), which has three valence electrons. * Conduction Mechanism: The three valence electrons of the dopant atom form covalent bonds with three surrounding silicon atoms. The fourth bond with a silicon atom is incomplete, creating an 'electron hole' or a 'vacancy' in the valence band. An electron from an adjacent silicon atom can move into this hole, thereby creating a new hole at its original position. This movement of holes under an electric field constitutes electrical conduction. * Charge Carriers: In p-type semiconductors, holes are the majority charge carriers, and electrons are the minority charge carriers. The 'p' stands for positive, referring to the effective positive charge of the majority carriers (holes).

        Conduction Mechanism Summary:
        In both n-type and p-type semiconductors, the addition of impurities creates charge carriers (electrons or holes) that can move freely under the influence of an electric field, leading to increased electrical conductivity compared to intrinsic semiconductors.

        Solution Steps

        1. Step 1: Define Semiconductor

          Explain what a semiconductor is, its conductivity range, and how its conductivity changes with temperature.

        2. Step 2: Introduce Intrinsic Semiconductors

          Briefly describe intrinsic semiconductors and their limited conduction mechanism.

        3. Step 3: Introduce Extrinsic Semiconductors

          Explain the concept of doping and how it enhances conductivity, leading to extrinsic semiconductors.

        4. Step 4: Describe n-type Semiconductors

          Explain their formation (doping with Group 15 elements), the role of the extra electron, and identify majority/minority charge carriers. Mention the origin of 'n'.

        5. Step 5: Describe p-type Semiconductors

          Explain their formation (doping with Group 13 elements), the creation and movement of electron holes, and identify majority/minority charge carriers. Mention the origin of 'p'.

        6. Step 6: Summarize Conduction Mechanism

          Conclude by summarizing how doping leads to enhanced conductivity in both types due to the creation of mobile charge carriers.

        NEET Relevance

        This topic is highly relevant for NEET. Questions frequently appear on definitions of n-type/p-type, examples of dopants, majority/minority charge carriers, and the basic conduction mechanism. Energy band diagrams are also important for understanding.

        Key Concepts

        SemiconductorsIntrinsic semiconductorsExtrinsic semiconductorsDopingn-type semiconductorsp-type semiconductorsConduction mechanismElectronsHolesValence bandConduction band

        This question has appeared in previous NEET exams.

        30long answer🎯 HIGH⭐ Important

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

        ✅ Answer

        These terms describe various types of point defects, which are irregularities or deviations from the ideal arrangement around a point or an atom in a crystalline substance.

        (i) Schottky Defect:

        • Definition: A Schottky defect is a type of stoichiometric point defect in an ionic crystal. It arises when an equal number of cations and anions are missing from their regular lattice sites, maintaining the electrical neutrality of the crystal. * Conditions: This defect is typically found in highly ionic compounds with high coordination numbers and where the sizes of cations and anions are almost similar (e.g., NaCl, KCl, CsCl, AgBr). * Consequences: The presence of Schottky defects leads to a decrease in the density of the crystal because atoms/ions are missing. It also increases the electrical conductivity due to the movement of ions into the vacant sites. * Example: In NaCl, a Na⁺ ion and a Cl⁻ ion are missing from their respective lattice sites, creating two vacancies.

        (ii) Frenkel Defect:

        • Definition: A Frenkel defect is another type of stoichiometric point defect in an ionic crystal. It occurs when an ion (usually the smaller cation) leaves its regular lattice site and occupies an interstitial position, creating a vacancy at its original site and an interstitial defect at the new site. The overall electrical neutrality is maintained. * Conditions: This defect is found in ionic compounds with a large difference in the sizes of cations and anions (cations are usually much smaller) and low coordination numbers (e.g., AgCl, AgBr, AgI, ZnS). * Consequences: Frenkel defects do not change the density of the crystal because no ions are missing from the crystal; they are just displaced. It increases the electrical conductivity due to the movement of ions and vacancies. * Example: In AgCl, an Ag⁺ ion leaves its lattice site and occupies an interstitial position, creating a vacancy at its original site.

        (iii) Interstitials (Interstitial Defect):

        • Definition: An interstitial defect is a point defect where an atom or ion occupies an interstitial site (a void space) in the crystal lattice that is normally vacant. * Types: * Self-interstitial defect: When an atom of the host crystal occupies an interstitial site. This typically increases the density of the substance. * Impurity interstitial defect: When a foreign atom occupies an interstitial site. * Conditions: Common in non-ionic solids. In ionic solids, it's usually the smaller cation that occupies an interstitial site (as seen in Frenkel defects). * Consequences: Generally increases the density of the substance. * Example: In a metallic crystal, an extra atom of the same type or an impurity atom occupies a void space between the regularly arranged atoms.

        (iv) F-centres (Farbenzentren):

        • Definition: F-centres are non-stoichiometric point defects that arise in ionic crystals, particularly alkali halides, when an anion is missing from its lattice site and the resulting vacant site is occupied by an electron. These electrons are trapped in the anion vacancies. * Formation: They are typically formed when alkali halides are heated in an atmosphere of the alkali metal vapour. For example, when NaCl is heated in sodium vapour, Na atoms deposit on the surface. Cl⁻ ions diffuse to the surface to combine with Na atoms, forming NaCl. The electrons released by Na atoms (Na → Na⁺ + e⁻) diffuse into the crystal and occupy the anion vacancies. * Consequences: The trapped electrons absorb light in the visible region, causing the crystal to exhibit colour. The colour is characteristic of the metal and the halide. For example, NaCl with F-centres appears yellow, KCl appears violet, and LiCl appears pink. * Example: An anion vacancy in NaCl crystal occupied by an electron, giving the crystal a yellow colour.

        Solution Steps

        1. Step 1: Introduction to Point Defects

          Briefly introduce point defects as irregularities in crystal structure.

        2. Step 2: Explain Schottky Defect

          Define the defect, describe its formation (missing equal numbers of cations and anions), list conditions for its occurrence, explain its consequences (density decrease, conductivity increase), and provide an example (e.g., NaCl).

        3. Step 3: Explain Frenkel Defect

          Define the defect, describe its formation (ion displaced to interstitial site), list conditions for its occurrence, explain its consequences (density unchanged, conductivity increase), and provide an example (e.g., AgCl).

        4. Step 4: Explain Interstitials (Interstitial Defect)

          Define the defect (atom/ion in interstitial site), differentiate between self-interstitial and impurity interstitial, mention conditions, explain consequences (density increase), and provide an example (e.g., metallic crystal).

        5. Step 5: Explain F-centres

          Define F-centres (electron trapped in anion vacancy), describe their formation (heating in metal vapour), explain their consequences (colouration of crystals), and provide an example (e.g., yellow NaCl).

        NEET Relevance

        This is a very important topic for NEET. Questions on definitions, conditions for occurrence, consequences (density change, conductivity), and examples of each defect are very common. Distinguishing between Schottky and Frenkel defects is a frequent MCQ.

        Key Concepts

        Point defectsStoichiometric defectsNon-stoichiometric defectsSchottky defectFrenkel defectInterstitial defectF-centresCrystal latticeVacancyElectrical neutralityDensity

        This question has appeared in previous NEET exams.

        31numerical🎯 HIGH⭐ Important

        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

        Given:
        Metallic radius of Aluminium (r) = 125 pm

        Aluminium crystallises in a cubic close-packed (ccp) structure, which is equivalent to a Face-Centred Cubic (FCC) lattice.

        (i) Length of the side of the unit cell (a):
        For an FCC structure, the atoms touch along the face diagonal. The relationship between the atomic radius (r) and the edge length (a) of the unit cell is given by:

        4r = a√2

        Therefore, a = 4r / √2 = 2√2 r

        Substituting the given value of r: a = 2 × 1.414 × 125 pm a = 353.5 pm

        To convert pm to cm:
        1 pm = 10⁻¹⁰ cm a = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm

        The length of the side of the unit cell is 353.5 pm or 3.535 × 10⁻⁸ cm.

          (ii) Number of unit cells in 1.00 cm³ of aluminium:
          First, calculate the volume of one unit cell.

          Volume of unit cell (V) = a³

          V = (3.535 × 10⁻⁸ cm)³

          V = (3.535)³ × (10⁻⁸)³ cm³

          V = 44.13 × 10⁻²⁴ cm³

          Now, calculate the number of unit cells in 1.00 cm³ of aluminium:
          Number of unit cells = (Total volume) / (Volume of one unit cell)

          Number of unit cells = 1.00 cm³ / (44.13 × 10⁻²⁴ cm³)

          Number of unit cells = (1 / 44.13) × 10²⁴

          Number of unit cells = 0.02266 × 10²⁴

          Number of unit cells = 2.266 × 10²²

          There are approximately 2.27 × 10²² unit cells in 1.00 cm³ of aluminium.

            Solution Steps

            1. Step 1: Identify Crystal Structure and Given Data

              Recognize that cubic close-packed (ccp) is equivalent to Face-Centred Cubic (FCC). Note the given metallic radius (r = 125 pm).

            2. Step 2: Relate Edge Length (a) and Radius (r) for FCC

              Recall the relationship for FCC: the face diagonal is 4r, and the face diagonal is also a√2. So, 4r = a√2, which implies a = 4r/√2 = 2√2 r.

            3. Step 3: Calculate Edge Length (a)

              Substitute the value of r into the formula: a = 2√2 × 125 pm. Calculate the numerical value in pm. Convert pm to cm (1 pm = 10⁻¹⁰ cm) for consistency with the volume calculation later.

            4. Step 4: Calculate Volume of One Unit Cell

              Use the formula for the volume of a cube: V = a³. Substitute the calculated value of 'a' in cm and compute the volume in cm³.

            5. Step 5: Calculate Number of Unit Cells

              Divide the total given volume (1.00 cm³) by the volume of one unit cell to find the number of unit cells. Ensure units cancel out correctly.

            Final Answer: Verify units and significant figures.

            NEET Relevance

            Numerical problems involving the relationship between atomic radius and edge length for different crystal systems (FCC, BCC, Simple Cubic) and calculations of unit cell volume or number of unit cells are very common in NEET. This question tests fundamental concepts.

            Key Concepts

            Cubic close-packed (ccp) structureFace-Centred Cubic (FCC) latticeRelationship between atomic radius and edge length in FCCVolume of a unit cellUnit conversions (pm to cm)

            This question has appeared in previous NEET exams.

            32numerical🎯 HIGH⭐ Important

            If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

            ✅ Answer

            Given:
            NaCl is doped with 10⁻³ mol % of SrCl₂.

            Understanding the Doping Mechanism:
            When SrCl₂ is added to NaCl, Sr²⁺ ions replace Na⁺ ions in the crystal lattice. To maintain electrical neutrality, for every Sr²⁺ ion (which has a +2 charge) that replaces two Na⁺ ions (each with a +1 charge), one Na⁺ ion site must become vacant. However, the problem statement implies that one Sr²⁺ replaces one Na⁺, and to balance the charge, another Na⁺ site becomes vacant. More precisely, for every Sr²⁺ ion introduced, it replaces two Na⁺ ions, but only one Sr²⁺ ion occupies a lattice site. The extra positive charge of Sr²⁺ (+2) compared to Na⁺ (+1) is compensated by creating one cation vacancy for every Sr²⁺ ion incorporated into the lattice.

            So, 1 Sr²⁺ ion replaces 1 Na⁺ ion, and to balance the charge, 1 cation vacancy is created.

            Calculation of Cation Vacancy Concentration:
            1. Percentage of SrCl₂ doping: 10⁻³ mol %

            This means that in 100 moles of NaCl, there are 10⁻³ moles of SrCl₂.

            2. Moles of SrCl₂ per mole of NaCl:
            If 100 mol of NaCl contains 10⁻³ mol of SrCl₂,

            Then 1 mol of NaCl contains (10⁻³ / 100) mol of SrCl₂ = 10⁻⁵ mol of SrCl₂.

            3. Moles of Sr²⁺ ions:
            Since 1 mole of SrCl₂ provides 1 mole of Sr²⁺ ions, the concentration of Sr²⁺ ions is 10⁻⁵ mol per mole of NaCl.

            4. Concentration of Cation Vacancies:
            As established, for every one Sr²⁺ ion incorporated into the NaCl lattice, one cation vacancy (Na⁺ vacancy) is created to maintain electrical neutrality.

            Therefore, the concentration of cation vacancies = concentration of Sr²⁺ ions.

            Concentration of cation vacancies = 10⁻⁵ mol per mole of NaCl.

            5. Converting to number of vacancies per mole:
            To express this in terms of the number of vacancies per mole of NaCl, we multiply by Avogadro's number (NA = 6.022 × 10²³ mol⁻¹).

            Number of cation vacancies = 10⁻⁵ mol × 6.022 × 10²³ vacancies/mol

            Number of cation vacancies = 6.022 × 10¹⁸ vacancies.

            The concentration of cation vacancies is 10⁻⁵ mol per mole of NaCl, or 6.022 × 10¹⁸ cation vacancies per mole of NaCl.

              Solution Steps

              1. Step 1: Understand the Doping Mechanism

                Explain how doping NaCl with SrCl₂ creates cation vacancies. For every Sr²⁺ ion (charge +2) replacing a Na⁺ ion (charge +1), one additional Na⁺ vacancy must be created to maintain electrical neutrality. So, 1 Sr²⁺ ion introduces 1 cation vacancy.

              2. Step 2: Convert mol % to Moles of Dopant

                Interpret '10⁻³ mol % of SrCl₂'. This means 10⁻³ moles of SrCl₂ are present in 100 moles of NaCl. Calculate the moles of SrCl₂ per mole of NaCl.

              3. Step 3: Determine Moles of Sr²⁺ Ions

                Since 1 mole of SrCl₂ yields 1 mole of Sr²⁺ ions, the moles of Sr²⁺ ions are equal to the moles of SrCl₂ calculated in the previous step.

              4. Step 4: Relate Sr²⁺ Concentration to Cation Vacancy Concentration

                Based on the doping mechanism, state that the concentration of cation vacancies is equal to the concentration of Sr²⁺ ions.

              5. Step 5: Calculate Number of Vacancies (Optional but good for completeness)

                Multiply the molar concentration of vacancies by Avogadro's number to express the concentration in terms of the actual number of vacancies.

              Final Answer: Verify units and significant figures.

              NEET Relevance

              This type of numerical problem, involving doping and calculation of defect concentration, is very common in NEET. It tests the understanding of point defects, stoichiometry, and basic mole concepts. Students often get confused about the 1:1 or 1:2 relationship between dopant and vacancy.

              Key Concepts

              DopingIonic defectsCation vacanciesElectrical neutralityMole conceptAvogadro's number

              This question has appeared in previous NEET exams.

              33long answer🎯 HIGH⭐ Important

              Explain the following terms with suitable examples:
              (i) Schottky defect

              (ii) Frenkel defect

              (iii) Interstitials

              (iv) F-centres

              ✅ Answer

              (i) Schottky Defect
              Definition: A Schottky defect is a type of stoichiometric point defect in an ionic crystal where an equal number of cations and anions are missing from their respective lattice sites, creating vacancies. This defect maintains the electrical neutrality of the crystal.

              Effect: It decreases the density of the crystal.

              Conditions: It is typically found in highly ionic compounds with high coordination numbers and where the sizes of cations and anions are almost similar.

              Example: NaCl, KCl, CsCl, AgBr.

              (ii) Frenkel Defect
              Definition: A Frenkel defect is a type of stoichiometric point defect in an ionic crystal where an ion (usually the smaller cation) leaves its normal lattice site and occupies an interstitial site. This creates a vacancy at the original lattice site and an interstitial defect at the new location. The electrical neutrality of the crystal is maintained.

              Effect: It does not change the density of the crystal because the ions are merely displaced within the crystal.

              Conditions: It is typically found in ionic compounds with a large difference in the size of cations and anions and low coordination numbers.

              Example: AgCl, AgBr, AgI, ZnS.

              (iii) Interstitials (Interstitial Defect)
              Definition: An interstitial defect occurs when an atom or ion occupies an interstitial site (a void space) in the crystal lattice that is normally vacant. This defect increases the density of the substance.

              Types:

              * Interstitial atom: When an extra atom of the same type as the host atoms occupies an interstitial site.

              * Interstitial impurity: When an atom of a different type occupies an interstitial site.

              Example: In non-ionic solids, an atom might occupy an interstitial position. In ionic solids, a cation might occupy an interstitial position (as seen in Frenkel defect).

              (iv) F-centres (Farben centres)
              Definition: F-centres are anion vacancies occupied by unpaired electrons. These defects are responsible for the colour of alkali halide crystals. The term 'F' comes from the German word 'Farbe' meaning colour.

              Formation: They are formed when alkali halide crystals are heated in an atmosphere of the alkali metal vapour. For example, when NaCl crystals are heated in sodium vapour, sodium atoms deposit on the surface. Cl⁻ ions diffuse to the surface to combine with Na atoms, forming NaCl. The electrons released from the Na atoms diffuse into the crystal and occupy the anion vacancies, leading to the formation of F-centres.

              Effect: The electrons trapped in the anion vacancies absorb energy from the visible light spectrum, get excited, and then emit light of a characteristic colour upon de-excitation. This imparts colour to the crystal.

              Example:

              * NaCl with F-centres appears yellow.

              * KCl with F-centres appears violet.

              * LiCl with F-centres appears pink.

              Solution Steps

              1. Step 1: Schottky Defect

                Define Schottky defect as missing equal numbers of cations and anions, maintaining electrical neutrality. Explain its effect on density and conditions for its occurrence. Provide examples.

              2. Step 2: Frenkel Defect

                Define Frenkel defect as an ion leaving its lattice site to occupy an interstitial site, creating a vacancy and an interstitial defect. Explain its effect on density and conditions for its occurrence. Provide examples.

              3. Step 3: Interstitials

                Define interstitial defect as an atom/ion occupying a normally vacant interstitial site. Explain its effect on density and types (interstitial atom/impurity). Provide examples.

              4. Step 4: F-centres

                Define F-centres as anion vacancies occupied by unpaired electrons, responsible for colour. Explain their formation mechanism (e.g., heating NaCl in Na vapour) and the resulting colour. Provide examples.

              NEET Relevance

              Crystal defects (Schottky, Frenkel, F-centres) are very frequently asked topics in NEET. Questions can be definition-based, examples, effects on density, or conditions for their formation. Understanding the difference between Schottky and Frenkel defects is crucial.

              Key Concepts

              Point defectsStoichiometric defectsNon-stoichiometric defectsSchottky defectFrenkel defectInterstitial defectF-centresCrystal density

              This question has appeared in previous NEET exams.

              34numerical🎯 HIGH⭐ Important

              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
              Aluminium crystallises in a cubic close-packed (ccp) structure, which is equivalent to a face-centred cubic (FCC) unit cell.

              In an FCC structure, atoms touch along the face diagonal.

              The relationship between the edge length (a) and the atomic radius (r) for an FCC unit cell is:

              `4r = a√2`

              Given metallic radius (r) = 125 pm

              `a = 4r / √2 = 2√2 r`
              `a = 2 × 1.414 × 125 pm`

              `a = 353.5 pm`

              Therefore, the length of the side of the unit cell is 353.5 pm.

              (ii) Number of unit cells in 1.00 cm³ of aluminium
              First, convert the edge length 'a' from pm to cm:

              `1 pm = 10⁻¹⁰ cm`

              `a = 353.5 pm = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm`

              Volume of one unit cell (V) = `a³`
              `V = (3.535 × 10⁻⁸ cm)³`

              `V = (3.535)³ × (10⁻⁸)³ cm³`

              `V = 44.13 × 10⁻²⁴ cm³`

              `V = 4.413 × 10⁻²³ cm³`

              Total volume of aluminium given = 1.00 cm³

              Number of unit cells = `(Total volume) / (Volume of one unit cell)`
              Number of unit cells = `1.00 cm³ / (4.413 × 10⁻²³ cm³)`

              Number of unit cells = `0.2266 × 10²³`

              Number of unit cells = `2.266 × 10²²`

              Therefore, there are approximately 2.27 × 10²² unit cells in 1.00 cm³ of aluminium.

              Solution Steps

              1. Step 1: Identify crystal structure and relationship between 'a' and 'r'

                Aluminium crystallises in a cubic close-packed (ccp) structure, which is equivalent to a face-centred cubic (FCC) unit cell. For an FCC structure, the relationship between the edge length (a) and the atomic radius (r) is `4r = a√2`.

              2. Step 2: Calculate the length of the side of the unit cell (a)

                Rearrange the formula to solve for 'a': `a = 4r / √2 = 2√2 r`. Substitute the given metallic radius `r = 125 pm` into the equation: `a = 2 × 1.414 × 125 pm = 353.5 pm`.

              3. Step 3: Convert edge length to cm

                To calculate the volume in cm³, convert the edge length 'a' from picometers (pm) to centimeters (cm). `1 pm = 10⁻¹⁰ cm`. So, `a = 353.5 pm = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm`.

              4. Step 4: Calculate the volume of one unit cell

                The volume of a cubic unit cell (V) is `a³`. Substitute the value of 'a' in cm: `V = (3.535 × 10⁻⁸ cm)³ = 44.13 × 10⁻²⁴ cm³ = 4.413 × 10⁻²³ cm³`.

              5. Step 5: Calculate the number of unit cells

                The number of unit cells in a given volume is `(Total volume) / (Volume of one unit cell)`. Given total volume = 1.00 cm³. Number of unit cells = `1.00 cm³ / (4.413 × 10⁻²³ cm³) = 2.266 × 10²²`. Rounding to three significant figures, this is `2.27 × 10²²` unit cells.

              Final Answer: Verify units and significant figures.

              NEET Relevance

              This type of numerical problem is very common in NEET. It tests the understanding of crystal structures (FCC, BCC, simple cubic), the relationship between atomic radius and unit cell edge length, and basic volume calculations. Students must be proficient in unit conversions.

              Key Concepts

              Cubic close-packed (CCP) structureFace-centred cubic (FCC) unit cellRelationship between edge length and atomic radiusVolume of a unit cellUnit conversions

              This question has appeared in previous NEET exams.

              35numerical🎯 HIGH⭐ Important

              If NaCl is doped with 10⁻³ mol % of SrCl₂, what is the concentration of cation vacancies?

              ✅ Answer

              When NaCl is doped with SrCl₂, Sr²⁺ ions replace Na⁺ ions in the crystal lattice. To maintain electrical neutrality, for every Sr²⁺ ion introduced, two Na⁺ ions must be removed from the lattice. One Na⁺ site is occupied by Sr²⁺, and the other Na⁺ site remains vacant, creating a cation vacancy.

              Given doping concentration = 10⁻³ mol % of SrCl₂.
              This means 10⁻³ moles of SrCl₂ are added per 100 moles of NaCl.

              Concentration of SrCl₂ = `(10⁻³ / 100) mol` of SrCl₂ per mole of NaCl
              Concentration of SrCl₂ = `10⁻⁵ mol` of SrCl₂ per mole of NaCl

              Since each Sr²⁺ ion replaces two Na⁺ ions and creates one cation vacancy:
              Number of cation vacancies created per Sr²⁺ ion = 1

              Therefore, the concentration of cation vacancies is equal to the concentration of Sr²⁺ ions incorporated into the lattice.
              Concentration of cation vacancies = `10⁻⁵ mol` per mole of NaCl.

              To express this in terms of number of vacancies per mole, we multiply by Avogadro's number (NA):
              Number of cation vacancies = `10⁻⁵ mol × NA`

              Number of cation vacancies = `10⁻⁵ × 6.022 × 10²³ vacancies/mol`

              Number of cation vacancies = `6.022 × 10¹⁸ vacancies/mol`

              Thus, the concentration of cation vacancies is 6.022 × 10¹⁸ per mole of NaCl.

              Solution Steps

              1. Step 1: Understand the doping mechanism

                When NaCl is doped with SrCl₂, a divalent Sr²⁺ ion replaces a monovalent Na⁺ ion. To maintain electrical neutrality, for every Sr²⁺ ion occupying a Na⁺ site, one additional Na⁺ ion must leave its lattice site, creating a cation vacancy. Thus, 1 Sr²⁺ ion introduces 1 cation vacancy.

              2. Step 2: Convert mol % to molar fraction

                Given doping concentration = 10⁻³ mol % of SrCl₂. This means `10⁻³ moles of SrCl₂` are present in `100 moles of NaCl`. So, the molar fraction of SrCl₂ is `(10⁻³ / 100) = 10⁻⁵ mol` of SrCl₂ per mole of NaCl.

              3. Step 3: Relate SrCl₂ concentration to cation vacancy concentration

                Since each Sr²⁺ ion introduced creates one cation vacancy, the concentration of cation vacancies is equal to the concentration of SrCl₂ incorporated. Therefore, concentration of cation vacancies = `10⁻⁵ mol` per mole of NaCl.

              4. Step 4: Calculate the number of cation vacancies per mole

                To find the actual number of vacancies, multiply the molar concentration by Avogadro's number (NA = 6.022 × 10²³ mol⁻¹). Number of cation vacancies = `10⁻⁵ × 6.022 × 10²³ = 6.022 × 10¹⁸ vacancies/mol`.

              Final Answer: Verify units and significant figures.

              NEET Relevance

              This is a classic problem involving impurity defects in ionic solids. It frequently appears in NEET as MCQs, testing the understanding of how doping with ions of different valencies creates vacancies and how to calculate their concentration. It requires careful consideration of charge neutrality.

              Key Concepts

              DopingIonic defectsCation vacanciesElectrical neutralityAvogadro's number

              This question has appeared in previous NEET exams.

              36long answer🎯 HIGH⭐ Important

              Explain the following with suitable examples:
              (a) 12-16 and 13-15 group compounds

              (b) Amorphism

              (c) p-type and n-type semiconductors

              ✅ Answer

              (a) 12-16 and 13-15 Group Compounds
              These are compounds formed by combining elements from specific groups of the periodic table, primarily to create materials with specific semiconductor properties. They are often referred to as 'stoichiometric compounds' or 'intermetallic compounds' with specific ratios.

              13-15 Group Compounds:

              • Definition: These are compounds formed by combining elements from Group 13 (e.g., Al, Ga, In) and Group 15 (e.g., N, P, As, Sb) of the periodic table. They are typically covalent in nature and have an average of four valence electrons per atom, similar to Group 14 elements like Si and Ge, making them semiconductors.
              • Properties: They exhibit semiconductor properties and are often used in electronic devices. The bond between the Group 13 and Group 15 elements is predominantly covalent, but with some ionic character due to electronegativity differences.
              • Examples: GaAs (Gallium Arsenide), InSb (Indium Antimonide), AlP (Aluminium Phosphide). GaAs is widely used in high-speed electronics and optoelectronics.

              12-16 Group Compounds:

              • Definition: These are compounds formed by combining elements from Group 12 (e.g., Zn, Cd, Hg) and Group 16 (e.g., S, Se, Te) of the periodic table. Like 13-15 compounds, they also have an average of four valence electrons per atom and exhibit semiconductor properties.
              • Properties: They are also semiconductors, often with a higher degree of ionic character compared to 13-15 compounds due to larger electronegativity differences. They are used in various applications, including infrared detectors and solar cells.
              • Examples: ZnS (Zinc Sulfide), CdS (Cadmium Sulfide), CdSe (Cadmium Selenide), HgTe (Mercury Telluride).

              (b) Amorphism
              Definition: Amorphism refers to the state of a solid material that lacks a long-range, ordered arrangement of its constituent particles (atoms, ions, or molecules). Amorphous solids are also known as supercooled liquids or glassy solids.

              Characteristics:

              * Irregular arrangement: Particles are randomly arranged, similar to liquids, but are fixed in position.

              * Short-range order: They may possess some short-range order, meaning that particles are arranged in a regular fashion over a very short distance, but this order does not extend throughout the material.

              * Isotropic properties: Their physical properties (like refractive index, electrical conductivity, thermal expansion) are the same in all directions because there is no long-range order.

              * No sharp melting point: They gradually soften over a range of temperatures and flow, rather than melting sharply at a specific temperature.

              * Irregular cleavage: When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces.

              Examples: Glass, plastics, rubber, tar, amorphous silicon.

              (c) p-type and n-type Semiconductors
              Semiconductors are materials with electrical conductivity between that of conductors and insulators. Their conductivity can be significantly increased by adding small amounts of suitable impurities, a process called doping. This leads to extrinsic semiconductors, which are classified as p-type or n-type.

              n-type Semiconductors:

              • Definition: An n-type semiconductor is formed when a pure semiconductor (like silicon or germanium, Group 14) is doped with a pentavalent impurity (an element from Group 15, e.g., P, As, Sb).
              • Mechanism: The pentavalent impurity atom has five valence electrons. When it replaces a Group 14 atom, four of its valence electrons form covalent bonds with the surrounding four semiconductor atoms. The fifth valence electron is extra and becomes delocalised, contributing to electrical conductivity. These extra electrons are called 'donor electrons'.
              • Charge Carriers: The majority charge carriers are electrons (negative), and minority charge carriers are holes.
              • Example: Silicon doped with phosphorus or arsenic.

              p-type Semiconductors:

              • Definition: A p-type semiconductor is formed when a pure semiconductor (like silicon or germanium, Group 14) is doped with a trivalent impurity (an element from Group 13, e.g., B, Al, Ga).
              • Mechanism: The trivalent impurity atom has three valence electrons. When it replaces a Group 14 atom, it forms three covalent bonds with three surrounding semiconductor atoms. The fourth bond with the remaining semiconductor atom is incomplete, creating an electron deficiency or a 'hole'. These holes can accept electrons from neighbouring atoms, effectively moving through the crystal and contributing to electrical conductivity. These impurities are called 'acceptor impurities'.
              • Charge Carriers: The majority charge carriers are holes (positive), and minority charge carriers are electrons.
              • Example: Silicon doped with boron or aluminium.

              Solution Steps

              1. Step 1: Explain 12-16 and 13-15 Group Compounds

                Define 13-15 group compounds (elements from Group 13 and Group 15), their properties (semiconductors, covalent character), and provide examples (e.g., GaAs). Define 12-16 group compounds (elements from Group 12 and Group 16), their properties (semiconductors, more ionic character), and provide examples (e.g., ZnS).

              2. Step 2: Explain Amorphism

                Define amorphism as the lack of long-range order in solids. Describe its key characteristics: irregular arrangement, short-range order, isotropic properties, no sharp melting point, and irregular cleavage. Provide examples (e.g., glass, plastics).

              3. Step 3: Explain n-type Semiconductors

                Define n-type semiconductors as pure semiconductors doped with pentavalent impurities (Group 15). Explain the mechanism: extra electron from the impurity becomes delocalised. Identify majority charge carriers (electrons) and minority charge carriers (holes). Provide examples (e.g., Si doped with P).

              4. Step 4: Explain p-type Semiconductors

                Define p-type semiconductors as pure semiconductors doped with trivalent impurities (Group 13). Explain the mechanism: creation of an electron deficiency (hole) due to incomplete bonding. Identify majority charge carriers (holes) and minority charge carriers (electrons). Provide examples (e.g., Si doped with B).

              NEET Relevance

              All three parts of this question are highly relevant for NEET. Questions on 13-15 and 12-16 compounds often appear as examples or properties. Amorphous solids are frequently contrasted with crystalline solids. p-type and n-type semiconductors are a fundamental concept in solid-state physics and chemistry, with questions on their formation, charge carriers, and examples being very common.

              Key Concepts

              Group 12-16 compoundsGroup 13-15 compoundsSemiconductorsAmorphous solidsCrystalline solidsDopingExtrinsic semiconductorsn-type semiconductorp-type semiconductorMajority charge carriersMinority charge carriers

              This question has appeared in previous NEET exams.

              37short answerMEDIUM⭐ Important

              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, vacancy defects can arise. Heating provides thermal energy to the constituent particles (atoms, ions, or molecules) of the solid. This energy can cause some particles to leave their normal lattice sites, creating vacant sites or 'vacancies'.

              The physical property affected by this defect is density.

              Effect on Density: When vacancy defects are created, some particles leave the crystal lattice, but the overall volume of the solid remains largely unchanged (or expands slightly due to thermal expansion, but the primary effect on density from vacancies is due to mass loss). Since mass decreases while volume remains approximately constant, the density of the solid decreases.

              NEET Relevance

              Conceptual questions about the origin and consequences of defects are asked in NEET. Understanding how heating leads to vacancies and its effect on density is fundamental.

              Key Concepts

              Vacancy defectThermal energyDensity

              This question has appeared in previous NEET exams.

              38short answer🎯 HIGH⭐ Important

              What type of stoichiometric defect is shown by:
              (i) ZnS

              (ii) AgBr

              ✅ Answer

              (i) ZnS (Zinc Sulphide): ZnS shows Frenkel defect.
              * Reason: Zinc sulphide has a relatively large difference in the size of its ions (Zn²⁺ is much smaller than S²⁻). The Zn²⁺ ions are small enough to occupy interstitial sites, and they have a low coordination number (4). These conditions favor the formation of Frenkel defects, where a smaller ion (Zn²⁺) leaves its lattice site and occupies an interstitial site, creating a vacancy at its original position and an interstitial defect at the new position.

              (ii) AgBr (Silver Bromide): AgBr is unique as it shows both Frenkel defect and Schottky defect.
              * Frenkel Defect: AgBr exhibits Frenkel defect because the Ag⁺ ion is relatively small and can easily move into interstitial positions, similar to ZnS. It has a low coordination number (6).

              * Schottky Defect: AgBr also exhibits Schottky defect because the Ag⁺ and Br⁻ ions are of comparable size, which is a characteristic feature for compounds showing Schottky defects. The energy required to form a pair of cation and anion vacancies is relatively low.

              NEET Relevance

              Examples of compounds showing specific defects are very common multiple-choice questions in NEET. AgBr is a particularly important example due to showing both defects.

              Key Concepts

              Frenkel defectSchottky defectIonic radiiCoordination number

              This question has appeared in previous NEET exams.

              39long answer🎯 HIGH⭐ Important

              Explain the following terms with suitable examples:
              (i) Schottky defect

              (ii) Frenkel defect

              (iii) Interstitials

              (iv) F-centres

              ✅ Answer

              Let's explain each term with suitable examples:

              (i) Schottky Defect

              Definition: A Schottky defect is a type of stoichiometric point defect in ionic solids where an equal number of cations and anions are missing from their respective lattice sites, maintaining electrical neutrality of the crystal. It is essentially a pair of vacancies (one cation vacancy and one anion vacancy).

              Mechanism: This defect arises when ions leave their normal lattice positions, creating vacant sites. To maintain electrical neutrality, an equal number of positive and negative ions must be missing. This defect is common in highly ionic compounds with high coordination numbers and where the cations and anions are of nearly similar sizes.

              Consequences:

              • Density: The density of the crystal decreases because mass is lost from the crystal while the volume remains essentially the same.
              • Electrical Conductivity: It increases slightly due to the movement of ions into the vacant sites.
              • Stability: The presence of defects increases the entropy of the crystal, contributing to its stability.

              Examples: NaCl, KCl, CsCl, AgBr (AgBr shows both Schottky and Frenkel defects).

              (ii) Frenkel Defect

              Definition: A Frenkel defect is a type of stoichiometric point defect in ionic solids where an ion (usually the smaller cation) leaves its normal lattice site and occupies an interstitial site within the crystal lattice. It creates a vacancy at the original site and an interstitial defect at the new site.

              Mechanism: This defect occurs in ionic compounds where there is a large difference in the size of the ions (typically, the cation is much smaller than the anion) and the ions have a low coordination number. The smaller ion can easily move into an interstitial position without significantly distorting the lattice.

              Consequences:

              • Density: The density of the crystal remains unchanged because no ions leave or enter the crystal; they are merely displaced within the lattice.
              • Electrical Conductivity: It increases slightly due to the movement of ions and vacancies.
              • Dielectric Constant: The dielectric constant of the crystal increases due to the displacement of ions.

              Examples: ZnS, AgCl, AgBr, AgI (AgBr shows both Schottky and Frenkel defects).

              (iii) Interstitials (Interstitial Defect)

              Definition: An interstitial defect occurs when an atom or ion occupies an interstitial site (a void or space between the regular lattice sites) in the crystal structure. This defect can be found in both non-ionic (atomic/molecular) and ionic solids.

              Types:

              • Self-interstitial defect: When an atom of the host lattice occupies an interstitial position.
              • Impurity interstitial defect: When a foreign atom occupies an interstitial position.

              Mechanism: In non-ionic solids, interstitial atoms cause significant distortion to the lattice because they are typically larger than the interstitial voids. In ionic solids, interstitial defects are often associated with Frenkel defects, where a cation moves from its lattice site to an interstitial site.

              Consequences:

              • Density: In the case of self-interstitial defects, the density increases slightly if an extra atom is introduced into the lattice. If it's part of a Frenkel defect, the overall density doesn't change.
              • Lattice Distortion: Interstitial atoms cause strain and distortion in the crystal lattice.

              Examples:

              • Non-ionic solids: Extra carbon atoms occupying interstitial sites in iron to form steel.
              • Ionic solids: The interstitial Ag⁺ ion in AgCl (part of a Frenkel defect).

              (iv) F-centres (Farbenzentren)

              Definition: F-centres (from the German word 'Farbenzentren' meaning colour centre) are anionic vacancies in ionic crystals that are occupied by unpaired electrons. These defects are responsible for the colour of many alkali halide crystals.

              Mechanism: F-centres are typically formed when alkali halide crystals are heated in an atmosphere of alkali metal vapour. For example, when NaCl is heated in sodium vapour, Na atoms deposit on the surface. Cl⁻ ions diffuse to the surface to combine with Na atoms, forming NaCl. The electrons released by the Na atoms (Na → Na⁺ + e⁻) diffuse into the crystal and occupy the vacant anionic sites created by the migrating Cl⁻ ions. These electron-occupied anionic vacancies are F-centres.

              Consequences:

              • Colour: The unpaired electron in the F-centre absorbs light from the visible region of the spectrum, exciting the electron to higher energy levels. The complementary colour is transmitted, giving the crystal its characteristic colour (e.g., NaCl becomes yellow, KCl becomes violet, LiCl becomes pink).
              • Paramagnetism: Due to the presence of unpaired electrons, crystals with F-centres become paramagnetic.
              • Electrical Conductivity: F-centres can contribute to electrical conductivity at high temperatures.

              Examples:

              • NaCl crystal heated in Na vapour turns yellow.
              • KCl crystal heated in K vapour turns violet.
              • LiCl crystal heated in Li vapour turns pink.

              NEET Relevance

              This is a very important question covering fundamental concepts of crystal defects. Definitions, examples, and consequences of each defect are frequently tested in NEET, often as direct questions or statements in MCQs.

              Key Concepts

              Schottky defectFrenkel defectInterstitial defectF-centresStoichiometric defectsNon-stoichiometric defectsPoint defectsIonic solidsCrystal latticeDensityElectrical conductivityColour centres

              This question has appeared in previous NEET exams.

              40numerical🎯 HIGH⭐ Important

              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

              Given:
              Metallic radius of Aluminium (r) = 125 pm

              Structure: Cubic close-packed (ccp), which is equivalent to Face-Centred Cubic (FCC) structure.

              (i) What is the length of the side of the unit cell?

              For an FCC structure, the atoms are in contact along the face diagonal. The relationship between the edge length (a) and the atomic radius (r) is given by:
              Face diagonal = 4r

              Also, by Pythagoras theorem, Face diagonal = √(a² + a²) = √(2a²) = a√2

              Therefore, a√2 = 4r a = 4r / √2 a = 2√2 r

              Substitute the given value of r: a = 2√2 × 125 pm a = 2 × 1.414 × 125 pm a = 353.5 pm

              Converting pm to cm:
              1 pm = 10⁻¹² m = 10⁻¹⁰ cm a = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm

              Length of the side of the unit cell (a) = 353.5 pm or 3.535 × 10⁻⁸ cm

                (ii) How many unit cells are there in 1.00 cm³ of aluminium?

                First, calculate the volume of one unit cell:
                Volume of unit cell (V) = a³

                V = (3.535 × 10⁻⁸ cm)³

                V = (3.535)³ × (10⁻⁸)³ cm³

                V = 44.13 × 10⁻²⁴ cm³

                V = 4.413 × 10⁻²³ cm³

                Now, calculate the number of unit cells in 1.00 cm³ of aluminium:
                Number of unit cells = (Total volume of aluminium) / (Volume of one unit cell)

                Number of unit cells = 1.00 cm³ / (4.413 × 10⁻²³ cm³)

                Number of unit cells = 0.2266 × 10²³

                Number of unit cells = 2.266 × 10²²

                Number of unit cells in 1.00 cm³ of aluminium = 2.27 × 10²² (rounded to three significant figures)

                  Solution Steps

                  1. Step 1: Identify the crystal structure and given parameters

                    Aluminium crystallises in a cubic close-packed (ccp) structure, which is equivalent to a Face-Centred Cubic (FCC) structure. The metallic radius (r) is given as 125 pm.

                  2. Step 2: Relate edge length (a) to metallic radius (r) for FCC

                    In an FCC structure, atoms are in contact along the face diagonal. The length of the face diagonal is 4r. Using the Pythagorean theorem for a face of the cube, the face diagonal is also a√2. Therefore, a√2 = 4r.

                  3. Step 3: Calculate the length of the side of the unit cell (a)

                    From the relation a√2 = 4r, we get a = 4r/√2 = 2√2r.
                    Substitute r = 125 pm: a = 2 × 1.414 × 125 pm = 353.5 pm.

                    Convert pm to cm: 1 pm = 10⁻¹⁰ cm. So, a = 353.5 × 10⁻¹⁰ cm = 3.535 × 10⁻⁸ cm.

                  4. Step 4: Calculate the volume of one unit cell

                    The volume of a cubic unit cell (V) is a³.
                    V = (3.535 × 10⁻⁸ cm)³ = 44.13 × 10⁻²⁴ cm³ = 4.413 × 10⁻²³ cm³.

                  5. Step 5: Calculate the number of unit cells in 1.00 cm³

                    Number of unit cells = (Total volume) / (Volume of one unit cell)
                    Number of unit cells = 1.00 cm³ / (4.413 × 10⁻²³ cm³) = 2.266 × 10²².

                    Rounding to three significant figures, the number of unit cells is 2.27 × 10²².

                  Final Answer: Verify units and significant figures.

                  NEET Relevance

                  Numerical problems involving the relationship between atomic radius and edge length for different crystal systems (especially FCC, BCC, Simple Cubic), calculation of unit cell volume, and then determining the number of unit cells or density are very common in NEET. This question tests multiple such concepts.

                  Key Concepts

                  Cubic close-packed (ccp)Face-Centred Cubic (FCC)Metallic radiusUnit cell edge lengthVolume of unit cellNumber of unit cells

                  This question has appeared in previous NEET exams.