Answer

(b) At low temperature existence of a substance in solid state is due to

(a) slow molecular motion and

(b) strong cohesive forces

These two forces hold the constituent particles together thus causes existence of substance in solid state.

• High temperature: At high temperatures, the thermal energy of the particles increases, causing them to move more vigorously. This increased motion can overcome the cohesive forces holding the particles together, leading to a transition from solid to liquid or gas states.

• High thermal energy: High thermal energy implies that the particles have a lot of kinetic energy, which can disrupt the orderly arrangement of particles in a solid. This can cause the substance to melt or vaporize, making it less likely to exist in the solid state.

• Weak cohesive forces: Weak cohesive forces mean that the attractive forces between the particles are not strong enough to hold them together in a fixed, orderly structure. As a result, the substance is more likely to exist in a liquid or gaseous state rather than a solid state.

Answer

(b) Crystalline solid is anisotropic in nature as this solid shows different physical properties such as electrical resistance, refractive index in different directions.

Note Isotropic and anisotropic properties are related to amorphous solid and crystalline solid which can be clearly understood as

• (a) Definite and characteristic heat of fusion: Crystalline solids have a definite and characteristic heat of fusion because they have a well-ordered structure, which requires a specific amount of energy to break the bonds and change the state from solid to liquid.

• (c) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal: Crystalline solids are defined by their regular, periodic arrangement of constituent particles, which extends throughout the entire crystal, giving them a well-defined geometric shape.

• (d) A true solid: Crystalline solids are considered true solids because they have a definite shape and volume, and their particles are arranged in a highly ordered structure.

Answer

(b) Quartz glass $(SiO_2)$ is an amorphous solid due to its short range order of constituent particles.

Note Quartz is a crystalline solid while quartz glass is an amorphous solid.

• (a) Graphite (C): Graphite is a crystalline solid because it has a well-defined, long-range order of carbon atoms arranged in layers.

• (c) Chrome alum: Chrome alum is a crystalline solid as it forms a regular, repeating lattice structure in its solid state.

• (d) Silicon carbide (SiC): Silicon carbide is a crystalline solid due to its well-ordered, repeating arrangement of silicon and carbon atoms.

Answer

(d) The substances which have domain structure are oppositely oriented and cancel each other’s magnetic field are known as antiferromagnetic substances.

• Option (a): This arrangement shows all magnetic moments aligned in the same direction, which is characteristic of ferromagnetic substances, not antiferromagnetic substances.

• Option (b): This arrangement shows magnetic moments aligned in a parallel and antiparallel fashion but not in a way that they completely cancel each other out. This is more indicative of ferrimagnetic substances rather than antiferromagnetic substances.

• Option (c): This arrangement shows a random orientation of magnetic moments, which is characteristic of paramagnetic substances, not antiferromagnetic substances.

Answer

(a) Since, quartz glass is an amorphous solid having short range order of constituents. Hence, value of refractive index is same in all directions, can be measured and not be equal to zero always.

• (b) Different in different directions: This is incorrect because quartz glass is an amorphous solid, meaning it lacks a long-range ordered structure. In amorphous solids, the properties, including the refractive index, are isotropic, meaning they are the same in all directions.

• (c) Cannot be measured: This is incorrect because the refractive index of quartz glass can indeed be measured using standard optical techniques.

• (d) Always zero: This is incorrect because the refractive index of any material, including quartz glass, cannot be zero. The refractive index is a measure of how much light is bent, or refracted, when entering a material, and it always has a positive value greater than or equal to 1.

Answer

(d) Amorphous solids are isotropic in nature because it has no long range order and any physical property will be same on all direction. On the other hand, anisotropic nature is a characteristic feature of crystalline solid.

• (a) On heating, amorphous solids may become crystalline at certain temperatures because the added energy can allow the atoms or molecules to arrange into a more ordered, crystalline structure.
• (b) Amorphous solids may become crystalline over time as they slowly rearrange into a more stable, ordered state.
• (c) Amorphous solids can be moulded by heating because they do not have a fixed, rigid structure and can flow or deform when heated.

Answer

(b) Crystalline solid has regular arrangement of constituent particles observed over a long distance in the crystal lattice. Due to this regular arrangement crystalline solid have sharp melting point.

• (a) A regular arrangement of constituent particles observed over a short distance in the crystal lattice does not account for the sharp melting point because it implies only local order, which is characteristic of amorphous solids rather than crystalline solids.

• (c) The same arrangement of constituent particles in different directions does not necessarily lead to a sharp melting point. While it suggests isotropy, it does not address the long-range order required for a sharp melting point.

• (d) Different arrangement of constituent particles in different directions would result in anisotropy and lack of uniformity, which is not characteristic of crystalline solids and would not lead to a sharp melting point.

Answer

(a) lodine molecules are a class of non-polar molecular solid in which constituents molecules are held together by London or dispersion forces. These solids are soft and non-conductor of electricity.

• Dipole-dipole interactions: Iodine molecules (I2) are non-polar because they consist of two identical atoms sharing electrons equally. Dipole-dipole interactions occur between polar molecules, which have permanent dipole moments. Since iodine molecules are non-polar, dipole-dipole interactions are not applicable.

• Covalent bonds: Covalent bonds are strong bonds formed by the sharing of electrons between atoms within a molecule. In the case of iodine crystals, the iodine atoms within each I2 molecule are held together by covalent bonds, but the question pertains to the forces holding the molecules together in the crystal lattice. These intermolecular forces are not covalent bonds but rather weaker London dispersion forces.

• Coulombic forces: Coulombic forces, or electrostatic forces, occur between charged particles, such as ions. Iodine molecules are neutral and do not carry a charge, so coulombic forces are not responsible for holding the iodine molecules together in the crystal lattice.

Answer

(c) Diamond is a giant molecule in which constituent atoms are held together by covalent bond. Hence, this is a network solid.

• (a) $SO_2$ (solid): $SO_2$ in its solid form is composed of discrete molecular units held together by van der Waals forces, not by a continuous network of covalent bonds.

• (b) $I_2$: Solid iodine ($I_2$) consists of discrete diatomic molecules held together by van der Waals forces, not by a continuous network of covalent bonds.

• (d) $H_2O$ (ice): Ice is composed of water molecules held together by hydrogen bonds, not by a continuous network of covalent bonds.

Answer

(c) lodine is a non-polar molecular solid in which iodine molecules are held together by London force or dispersion force. This is soft and non-conductor of electricity.

Water is a hydrogen bonded molecular solid in which $H$ and $O$ are held together by polar covalent bond and each water molecule held together by hydrogen bonding. Due to non-ionic nature, they are not electrical conductor.

• Option 1 ($Mg(s)$): Magnesium is a metal, and metals are generally good conductors of electricity due to the presence of free-moving electrons within their structure.

• Option 2 ($TiO(s)$): Titanium monoxide (TiO) is a metallic compound and typically exhibits electrical conductivity due to the presence of delocalized electrons.

• Option 4 ($H_2O(s)$): Water in its solid form (ice) is a molecular solid held together by hydrogen bonds. It is not an electrical conductor because it lacks free ions or electrons to carry an electric current.

Answer

(a) lonic solids easily dissociated into its ions in molten state and show high electrical conductivity. So, statement (a) is incorrect while ionic solids are anisotropic and brittle linked with very strong force of interactions.

• (b) Brittle nature: Ionic solids are indeed brittle because when a force is applied, the ions of like charge are brought next to each other, causing repulsion and leading to the material breaking apart.
• (c) Very strong forces of interactions: Ionic solids have very strong electrostatic forces of attraction between the oppositely charged ions, which is a defining characteristic of these materials.
• (d) Anisotropic nature: Ionic solids are anisotropic, meaning their physical properties vary when measured along different directions due to the orderly arrangement of ions in a crystal lattice.

Answer

(b) Graphite is a good conductor of electricity due to presence of free valence electrons. In graphite, each carbon is $s p^{2}$ hybridised having one free electron which makes graphite a good conductor of electricity.

• (a) lone pair of electrons: Graphite does not have lone pairs of electrons that contribute to its electrical conductivity. Lone pairs are typically found in molecules where atoms have non-bonding pairs of electrons, which is not the case in the structure of graphite.

• (c) cations: Cations are positively charged ions, and their presence is not responsible for the electrical conductivity in graphite. Graphite’s conductivity is due to the delocalized electrons within its layers, not due to the movement of cations.

• (d) anions: Anions are negatively charged ions, and similar to cations, they do not play a role in the electrical conductivity of graphite. The conductivity is due to the free valence electrons that are delocalized across the layers of carbon atoms in graphite.

Answer

(c) Certain metal oxides like $VO_2, VO_2 VO_3$ and $TiO_3$ show metallic or insulating property depending upon temperature. As temperature varies metallic or insulating property varies. This is due to variation in energy gap between conduction band and valence band.

• (a) $TiO$: Titanium monoxide ($TiO$) is typically a metallic conductor and does not exhibit a transition to an insulating state with temperature changes. It does not have the temperature-dependent behavior of switching between conducting and insulating states.

• (b) $SiO_2$: Silicon dioxide ($SiO_2$) is an insulator and remains an insulator over a wide range of temperatures. It does not exhibit the temperature-dependent transition to a conducting state.

• (d) $MgO$: Magnesium oxide ($MgO$) is also an insulator and does not show a transition to a conducting state with changes in temperature. It remains an insulator regardless of temperature variations.

Answer

(d) $CrO_2, TiO$ and $ReO_3$ are some typical metal oxides which show electrical conductivity similar to metal. While $SiO_2, MgO$ and $SO_2$ are oxides of metal, semimetal and non-metal which do not show electrical properties.

• (a) $SiO_2$: Silicon dioxide is a non-metallic oxide and is an insulator, not a conductor. It does not exhibit electrical properties like metals.

• (b) $MgO$: Magnesium oxide is an ionic compound and an insulator. It does not conduct electricity like metals.

• (c) $SO_2(s)$: Sulfur dioxide is a molecular compound and a non-metallic oxide. It does not have metallic electrical properties and is not a conductor.

Answer

(c) Each point in a lattice is known as lattice point which are either atom or molecule or ion which are joined together by a straight line to bring out geometry of lattice in pure crystal constituents are arranged in fixed stoichiometric ratio.

Hence, existence of free electrons are not possible, it is possible on in case of imperfection in solid.

• (a) Molecule: In a pure crystal, lattice sites can be occupied by molecules. For example, in molecular crystals like ice or dry ice, the lattice points are occupied by water molecules or carbon dioxide molecules, respectively.

• (b) Ion: Lattice sites in ionic crystals are occupied by ions. For instance, in sodium chloride (NaCl), the lattice points are occupied by sodium ions (Na⁺) and chloride ions (Cl⁻).

• (d) Atom: In atomic crystals, lattice sites are occupied by atoms. For example, in a diamond crystal, the lattice points are occupied by carbon atoms.

Answer

(d) Graphite can’t be classified as ionic solid as graphite is not made up of ions. It is made up of carbon atoms covalently bonded to three carbon atoms so, it is a covalent solid.

Since, the formation of covalent bond occurs throughout the crystal therefore, it is a type of network solid. Due to presence of free electron graphite is also classified as conductor solid.

• (a) Graphite is not an incorrect option because it is a conducting solid due to the presence of free electrons that allow it to conduct electricity.
• (b) Graphite is not an incorrect option because it is a network solid, as the covalent bonds extend throughout the entire structure.
• (c) Graphite is not an incorrect option because it is a covalent solid, with carbon atoms covalently bonded to three other carbon atoms.

Answer

(a) When smaller ion (usually cation) is dislocated from its normal site in crystal and move to interstitial site is known as Frenkel defect as shown below

• Schottky defect: In a Schottky defect, both cations and anions are missing from their lattice sites, creating vacancies. This defect does not involve cations moving to interstitial sites.

• Vacancy defect: A vacancy defect occurs when an atom or ion is missing from its lattice site, creating a vacancy. It does not involve cations moving to interstitial sites.

• Metal deficiency defect: Metal deficiency defect typically occurs in compounds where there is a deficit of metal ions, often compensated by the presence of extra anions or by the oxidation of some metal ions to a higher oxidation state. This defect does not involve cations moving to interstitial sites.

Answer

(b) Schottky defect is observed in crystal when equal number of cations and anions are missing from the lattice. Thus, density of solid decreases.

When some cations move from their lattice site to interstitial site is known as Frenkel defect. When some impurity is present on crystal is known as impurity defect.

When lattice site is occupied by electron, this type of defect is known as metal excess defect. Hence, except (b) all statements are incorrect regarding Schottky defect.

• When some cations move from their lattice site to interstitial site is known as Frenkel defect.
• When some impurity is present on crystal is known as impurity defect.
• When lattice site is occupied by electron, this type of defect is known as metal excess defect.

Answer

(b) When group 13 elements are doped in group 14 element, it creates a hole in a molecule but the molecule as a whole remain neutral. Thus, $p$-type semiconductor is neutral in nature.

Positive hole (no electron)

• (a) Positive: This option is incorrect because although $p$-type semiconductors have holes (which are positive charge carriers), the overall charge of the semiconductor remains neutral. The holes are balanced by the negatively charged acceptor ions, resulting in no net charge.

• (c) Negative: This option is incorrect because $p$-type semiconductors are created by doping with group 13 elements, which introduce holes (positive charge carriers) into the material. There are no excess electrons to make the semiconductor negatively charged.

• (d) Depends on concentration of $p$ impurity: This option is incorrect because the overall charge of a $p$-type semiconductor does not depend on the concentration of the $p$ impurity. Regardless of the concentration, the semiconductor remains electrically neutral as the number of positive holes is balanced by the number of negatively charged acceptor ions.

Answer

(d) To get a $n$-type semiconductor from silicon, it should be doped with a substance with valence 5 .

e.g., Si doped with P lead to formation of $n$-type semiconductor as shown below

• (a) 2: Doping silicon with a substance that has a valency of 2 would not create an n-type semiconductor. Elements with a valency of 2, such as magnesium (Mg), do not have the extra electrons needed to contribute free electrons to the conduction band of silicon. Instead, they would create defects in the crystal structure, potentially leading to a different type of semiconductor behavior or even an insulator.

• (b) 1: Doping silicon with a substance that has a valency of 1, such as lithium (Li), would also not create an n-type semiconductor. Elements with a valency of 1 do not have enough electrons to donate to the conduction band. Instead, they would create holes in the silicon lattice, which could lead to the formation of a p-type semiconductor rather than an n-type.

• (c) 3: Doping silicon with a substance that has a valency of 3, such as boron (B), would result in the formation of a p-type semiconductor, not an n-type. Elements with a valency of 3 have fewer electrons than silicon and create “holes” in the crystal lattice, which act as positive charge carriers. This is the opposite of what is needed for an n-type semiconductor, which requires extra electrons as negative charge carriers.

Thinking Process

If number of atoms in fcc unit cell $=N$

Number of tetrahedral voids $=2 N$

Number of octahedral voids $=N$

Answer

(b) Fcc unit cell contains 8 tetrahedral voids at centre of each 8 smaller cube of an unit cell as shown below

Number of atoms in fcc unit cell $=4$

Number of octahedral voids $=4$

Number of tetrahedral voids $=8$

• Option (a) 6: This is incorrect because a face-centered cubic (fcc) unit cell does not have 6 tetrahedral voids. The correct number is 8, as each of the 8 smaller cubes within the unit cell contains one tetrahedral void.

• Option (c) 10: This is incorrect because an fcc unit cell does not have 10 tetrahedral voids. The correct number is 8, based on the structure of the unit cell and the distribution of voids.

• Option (d) 12: This is incorrect because an fcc unit cell does not have 12 tetrahedral voids. The correct number is 8, as determined by the arrangement of atoms and voids within the unit cell.

Answer

(a) $AgBr$ shows both Schottky as well as Frenkel defect. In $AgBr$, both $Ag^{+}$and $Br^{-}$ions are absent from the lattice causing Schottky defect. However, $Ag^{+}$ions are mobile so they have a tendency to move aside the lattice and trapped in interstitial site hence, cause Frenkel defect.

• Option (b) is incorrect because $AgBr$ does not typically exhibit metal excess defect or metal deficiency defect. Metal excess defect usually occurs in compounds with non-stoichiometric ratios due to an excess of metal ions, often caused by the presence of extra cations in interstitial sites or anion vacancies. Metal deficiency defect occurs when there is a deficit of metal ions, which is not characteristic of $AgBr$.

• Option (c) is incorrect because while $AgBr$ does show Schottky defect, it does not exhibit metal excess defect. Metal excess defect involves an excess of metal ions, which is not a typical behavior of $AgBr$.

• Option (d) is incorrect because $AgBr$ does not show metal deficiency defect. Metal deficiency defect involves a deficit of metal ions, which is not characteristic of $AgBr$. While $AgBr$ does show Frenkel defect, it does not exhibit metal deficiency defect.

Answer

(b) Packing efficiency is the percentage of total filled space by particles and it can be calculated as packing efficiency

$$ =\frac{\text { Volume occupied by four spheres in the unit cell }}{\text { Total volume of unit cell }} \times 100 $$

Since, packing efficiency for hcp or ccp is calculated to be $74 %$ which is maximum among all type of crystals.

• (a) hcp and bcc: The packing efficiency for bcc (body-centered cubic) is approximately 68%, which is lower than the 74% packing efficiency of hcp (hexagonal close-packed). Therefore, this pair does not have the most efficient packing.

• (c) bcc and ccp: While ccp (cubic close-packed) has a packing efficiency of 74%, the bcc structure has a lower packing efficiency of 68%. Thus, this pair does not represent the most efficient packing.

• (d) bcc and simple cubic cell: The simple cubic cell has a packing efficiency of only about 52%, which is significantly lower than both bcc (68%) and the most efficient structures (hcp and ccp at 74%). Therefore, this pair is not the most efficient in terms of packing.

Answer

(c) Packing efficiency for bcc arrangement is $68 %$ which represents total filled space in the unit cell. Hence, empty space in a body centred arrangement is $100-68=32 %$.

Note Here, empty space in bcc arrangement is asked therefore empty space in any crystal packing can be calculated as empty space in unit cell $=100-$ packing efficiency

• Option (a) 74: This is incorrect because 74% represents the packing efficiency of a face-centred cubic (fcc) arrangement, not a body-centred cubic (bcc) arrangement. The empty space in an fcc arrangement would be 100 - 74 = 26%.

• Option (b) 68: This is incorrect because 68% represents the packing efficiency of a body-centred cubic (bcc) arrangement, not the empty space. The empty space in a bcc arrangement is 100 - 68 = 32%.

• Option (d) 26: This is incorrect because 26% represents the empty space in a face-centred cubic (fcc) arrangement, not a body-centred cubic (bcc) arrangement. The packing efficiency of an fcc arrangement is 74%, leaving 26% as empty space.

Answer

(d) Hexagonal close packing can be arranging by two layers

$A$ and $B$ one over another which can be diagramatically represented as

Here, we can see easily that 1 st layer and 4 th layer are not exactly aligned.

Thus, statement (d) is not correct while other statements (a), (b) and (c) are true.

• (a) The coordination number is 12: This statement is true because, in hexagonal close packing (hcp), each sphere is in contact with 12 other spheres, giving it a coordination number of 12.

• (b) It has $74 %$ packing efficiency: This statement is true because the packing efficiency of hexagonal close packing is indeed 74%, meaning that 74% of the volume is occupied by the spheres, and the remaining 26% is empty space.

• (c) Tetrahedral voids of the second layer are covered by the spheres of the third layer: This statement is true because, in hcp, the tetrahedral voids created in the second layer are indeed covered by the spheres in the third layer, following the ABAB stacking pattern.

Answer

(a) $NaCl$ crystal have rock salt structure having fcc lattice in which $Cl^{-}$ions are present at fcc lattice points and face centre and $Na^{+}$occupies all the octahedral void of given unit cell.

Where, coordination number of $Na^{+}=6$

coordination number of $Cl^{-}=6$

• (b) In the structure where $Ca^{2+}$ ions form an fcc lattice and $F^{-}$ ions occupy all the eight tetrahedral voids, the coordination number of $Ca^{2+}$ is 8 (since each $Ca^{2+}$ is surrounded by 8 $F^{-}$ ions in tetrahedral voids), while the coordination number of $F^{-}$ is 4 (each $F^{-}$ is surrounded by 4 $Ca^{2+}$ ions). Therefore, the coordination numbers are not the same.

• (c) In the structure where $O^{2-}$ ions form an fcc lattice and $Na^{+}$ ions occupy all the eight tetrahedral voids, the coordination number of $O^{2-}$ is 8 (since each $O^{2-}$ is surrounded by 8 $Na^{+}$ ions in tetrahedral voids), while the coordination number of $Na^{+}$ is 4 (each $Na^{+}$ is surrounded by 4 $O^{2-}$ ions). Therefore, the coordination numbers are not the same.

• (d) In the structure where $S^{2-}$ ions form an fcc lattice and $Zn^{2+}$ ions go into alternate tetrahedral voids, the coordination number of $S^{2-}$ is 4 (since each $S^{2-}$ is surrounded by 4 $Zn^{2+}$ ions in tetrahedral voids), while the coordination number of $Zn^{2+}$ is also 4 (each $Zn^{2+}$ is surrounded by 4 $S^{2-}$ ions). However, since only alternate tetrahedral voids are occupied, the overall structure does not maintain the same coordination number for all ions in the unit cell.

Answer

(c) Coordination number in a square closed packed structure in two dimensions is equal to 4 as shown below

• Option (a) 2: This option is incorrect because in a square close packed structure in two dimensions, each particle is surrounded by four other particles, not just two. Therefore, the coordination number cannot be 2.

• Option (b) 3: This option is incorrect because in a square close packed structure in two dimensions, each particle is surrounded by four other particles, not three. Therefore, the coordination number cannot be 3.

• Option (d) 6: This option is incorrect because a coordination number of 6 is characteristic of a hexagonal close packed structure in two dimensions, not a square close packed structure. In a square close packed structure, each particle is surrounded by four other particles, making the coordination number 4.

Answer

(d) When electron rich or electron deficient impurity is added to a perfect crystal it introduces electronic defect in them.

• (a) Dislocation defect: Dislocation defects are irregularities within the crystal structure that occur due to the misalignment of atoms or ions. These defects are typically introduced during the crystal growth process or due to mechanical deformation, not by doping.

• (b) Schottky defect: Schottky defects occur when equal numbers of cations and anions are missing from their lattice sites, maintaining electrical neutrality. These defects are intrinsic to the crystal and are not caused by the addition of impurities or doping.

• (c) Frenkel defect: Frenkel defects occur when an ion is displaced from its lattice site to an interstitial site, creating a vacancy and an interstitial defect. This type of defect is also intrinsic to the crystal and is not introduced by doping.

Answer

(b) Silicon doped with electron rich impurity such as phosphorus forms a $n$-type semiconductor. This is due to presence of mobile electron.

• (a) Silicon doped with electron rich impurity does not form a $p$-type semiconductor because $p$-type semiconductors are formed by doping silicon with electron-deficient impurities, which create holes as the majority charge carriers.
• (c) Silicon doped with electron rich impurity does not form an intrinsic semiconductor because intrinsic semiconductors are pure and undoped, having equal numbers of electrons and holes.
• (d) Silicon doped with electron rich impurity does not form an insulator because the doping introduces free electrons, which increase the material’s conductivity, making it a semiconductor rather than an insulator.

Answer

(b) Ferromagnetic substances can be magnetised by applying magnetic field to it and magnetic property persist within it even after removal of magnetic field.

Hence, choice (b) is the correct answer while other three choices are correct.

• (a) Paramagnetic substances are weakly attracted by magnetic field: This statement is true. Paramagnetic substances have unpaired electrons that align with an external magnetic field, causing a weak attraction.

• (c) The domains in antiferromagnetic substances are oppositely oriented with respect to each other: This statement is true. In antiferromagnetic substances, the magnetic moments of atoms or ions in the material are aligned in opposite directions, canceling each other out.

• (d) Pairing of electrons cancels their magnetic moment in the diamagnetic substances: This statement is true. In diamagnetic substances, all electrons are paired, and their magnetic moments cancel each other out, resulting in no net magnetic moment.

Answer

(d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions present at the lattice points. As we know the radii of octahedral or tetrahedral void is related to radii of atoms $(r)$ as

Radius of octahedral void $(R_0)=0.414 r$

Radius of tetrahedral void $(R_{t})=0.225 r$

Where, $\quad r=$ radius of bigger atom involved.

• (a) Bigger ions form the close packed structure: This statement is true. In ionic solids, the larger ions (usually anions) typically form the close-packed structure, while the smaller ions (usually cations) occupy the interstitial voids.

• (b) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size: This statement is true. The smaller ions (cations) occupy the voids in the close-packed structure formed by the larger ions (anions). The type of void (tetrahedral or octahedral) they occupy depends on their size relative to the voids.

• (c) Occupation of all the voids is not necessary: This statement is true. In ionic solids, not all the voids need to be occupied. The number of voids occupied depends on the stoichiometry and the specific structure of the ionic solid.

Answer

(a) When a ferromagnetic substance is placed in a magnetic field it becomes a permanent magnet because all the domains get oriented in the direction of magnetic field even after removal of applied magnetic field.

• (b) This option is incorrect because if all the domains were oriented in the direction opposite to the magnetic field, the substance would not become a permanent magnet. Instead, it would exhibit diamagnetic or antiferromagnetic behavior, which is not the case for ferromagnetic materials.

• (c) This option is incorrect because if the domains were oriented randomly, the substance would not exhibit a strong, unified magnetic field. The random orientation of domains would cancel out the magnetic effects, resulting in no permanent magnetization.

• (d) This option is incorrect because ferromagnetic substances are significantly affected by magnetic fields. The alignment of domains in response to an external magnetic field is a fundamental characteristic of ferromagnetic materials, leading to their strong magnetic properties.

Answer

(b) Packing efficiency in different types of unit cells can be tabulated as

Hence, correct order is fcc ($74 $%)> bcc ($68 $ %)>simple cubic ($52 $%).

• Option (a) fcc < bcc < simple cubic: This option is incorrect because it suggests that the packing efficiency of a face-centered cubic (fcc) unit cell is less than that of a body-centered cubic (bcc) unit cell, which in turn is less than that of a simple cubic unit cell. However, the actual packing efficiencies are 74% for fcc, 68% for bcc, and 52% for simple cubic. Therefore, fcc has the highest packing efficiency, followed by bcc, and then simple cubic.

• Option (c) fcc < bcc > simple cubic: This option is incorrect because it implies that the packing efficiency of fcc is less than that of bcc, and bcc is greater than simple cubic. While it is true that bcc has a higher packing efficiency than simple cubic, fcc actually has a higher packing efficiency than bcc. The correct order is fcc > bcc > simple cubic.

• Option (d) bcc < fcc > simple cubic: This option is incorrect because it suggests that the packing efficiency of bcc is less than that of fcc, and fcc is greater than simple cubic, but it does not clearly establish the correct relationship between bcc and simple cubic. The correct order should explicitly state that fcc has the highest packing efficiency, followed by bcc, and then simple cubic. The correct order is fcc > bcc > simple cubic.

Answer

(a) Frenkel defect is also known as dislocation defect because in Frenkel defect atoms present in crystal lattice is dislocated to interstitial site.

• Schottky defect is incorrect because it involves the removal of equal numbers of cations and anions from the crystal lattice, creating vacancies, rather than the dislocation of atoms to interstitial sites.

• Non-stoichiometric defect is incorrect because it refers to a deviation from the ideal stoichiometric ratio of elements in a compound, which can occur due to various reasons such as the presence of extra atoms or vacancies, but it does not specifically involve the dislocation of atoms to interstitial sites.

• Simple interstitial defect is incorrect because it involves the placement of an extra atom in the interstitial space of the crystal lattice, rather than the dislocation of an atom from its original lattice site to an interstitial site.

Answer

(d) In the cubic close packing the unit cell has 8 tetrahedral voids within it and are located at each eight smaller cube of an unit cell.

• (a) 4 tetrahedral voids each of which is shared by four adjacent unit cells: This option is incorrect because in a cubic close packing (ccp) structure, there are actually 8 tetrahedral voids within a single unit cell, not 4. Additionally, these voids are not shared by four adjacent unit cells; they are entirely within the unit cell.

• (b) 4 tetrahedral voids within the unit cell: This option is incorrect because a cubic close packing (ccp) unit cell contains 8 tetrahedral voids, not 4. These voids are located at specific positions within the unit cell.

• (c) 8 tetrahedral voids each of which is shared by four adjacent unit cells: This option is incorrect because the 8 tetrahedral voids in a cubic close packing (ccp) unit cell are not shared by adjacent unit cells. They are entirely contained within the unit cell itself.

Answer

(a) Edge length for different types of unit cells can be tabulated as

• Option (b) is incorrect because it incorrectly assigns the edge length of $2 \sqrt{2} r$ to the bcc unit cell, which should actually be $\frac{4 r}{\sqrt{3}}$. Additionally, it assigns $\frac{4 r}{\sqrt{3}}$ to the fcc unit cell, which should be $2 \sqrt{2} r$.

• Option (c) is incorrect because it incorrectly assigns the edge length of $2 r$ to the fcc unit cell, which should actually be $2 \sqrt{2} r$. Additionally, it assigns $2 \sqrt{2} r$ to the bcc unit cell, which should be $\frac{4 r}{\sqrt{3}}$.

• Option (d) is incorrect because it incorrectly assigns the edge length of $2 r$ to the fcc unit cell, which should actually be $2 \sqrt{2} r$. Additionally, it assigns $2 \sqrt{2} r$ to the simple cubic unit cell, which should be $2 r$.

Thinking Process

Look at the option and choose the correct answer using the concept that metal have maximum value of $\kappa$ and insulator has minimum value.

Answer

(a) Conductivity of metal, insulator and semiconductors can be represented in the term of $\kappa$ (Kappa) which depends upon energy gap between valence band and conduction band.

Hence, correct order is

$\kappa_{\text {metals }} > > \kappa_{\text {insulators }} < \kappa_{\text {semiconductors }}$

• Option (b): $\kappa_{\text {metals }} < < \kappa_{\text {insulators }} < \kappa_{\text {semiconductors }}$

  • This option is incorrect because metals have the highest conductivity due to the presence of free electrons. Insulators have the lowest conductivity because they have a large energy gap between the valence band and the conduction band, making it difficult for electrons to move. Semiconductors have conductivity between that of metals and insulators, as they have a smaller energy gap that can be overcome under certain conditions (e.g., doping, temperature).

• Option (c): $\kappa_{\text {metals }}$., $\kappa_{\text {semiconductors }} > \kappa_{\text {insulators }}=$ zero

  • This option is incorrect because it implies that the conductivity of insulators is exactly zero, which is not true. While insulators have very low conductivity, it is not absolutely zero. Additionally, it does not clearly indicate the relative magnitudes of the conductivities of metals and semiconductors.

• Option (d): $\kappa_{\text {metals }} < \kappa_{\text {semiconductors }} > \kappa_{\text {insulators }} \neq$ zero

  • This option is incorrect because it suggests that the conductivity of metals is less than that of semiconductors, which is not true. Metals have the highest conductivity due to the abundance of free electrons, whereas semiconductors have moderate conductivity, and insulators have the lowest conductivity.

Answer

(c, $d)$

Tetrahedral voids are formed when the triangular void in the second layer lie exactly above the triangular voids in the first layer and the triangular shape of these voids oppositely overlap.

Octahedral voids are formed when triangular void of second layer is not exactly overlap with similar void in first layer.

• Option (a): A tetrahedral void is indeed formed when a sphere of the second layer is present above a triangular void in the first layer. This statement is true and hence not a reason for being incorrect.

• Option (b): It is true that not all triangular voids are covered by the spheres of the second layer. This statement is correct and thus not a reason for being incorrect.

Answer

(c, $d)$

In the case of antiferromagnetic substances, the magnetic moment becomes zero because the domains are oppositely oriented with respect to each other without the application of magnetic field which cancel out each other.

• (a) get oriented in the direction of the applied magnetic field: This option is incorrect because in antiferromagnetic substances, the domains do not align in the direction of the applied magnetic field. Instead, they are naturally aligned in opposite directions, canceling each other out.

• (b) get oriented opposite to the direction of the applied magnetic field: This option is incorrect because the zero magnetic moment in antiferromagnetic substances is not due to the domains aligning opposite to an applied magnetic field. Rather, it is due to the intrinsic property of the domains being oppositely oriented with respect to each other even without any external magnetic field.

Answer

(c, $d)$

Statements (c) and (d) can be correctly written as (c)Impurity defect changes the density of substance as impurity has different density than the ion present on perfect crystal. e.g., When $SrCl_2$ is added to the $NaCl$ crystal it causes impurity defect.

(d) Frenkel defect results neither decrease nor increase in density of substance.

• (a) Vacancy defect results in a decrease in the density of the substance: This statement is actually true. A vacancy defect occurs when an atom or ion is missing from its lattice site, which reduces the overall mass of the crystal without changing its volume, thereby decreasing the density.

• (b) Interstitial defects result in an increase in the density of the substance: This statement is also true. An interstitial defect occurs when an extra atom or ion occupies a space in the crystal lattice that is normally not occupied, which increases the overall mass of the crystal without changing its volume, thereby increasing the density.

Answer

$(a, b, d)$

Option (a), (b) and (d) are true, option (c) can be correctly stated as the gap between valence band and conduction band can be determined. The gap between valence band and conduction band decide the conductivity of material.

• Option (c) is incorrect because the gap between the valence band and the conduction band can indeed be determined. This gap is a crucial factor in determining the electrical conductivity of a material.

Answer

$(a, b)$

In $p$-type semiconductor, the conductivity is due to existence of hole. When electric field is applied to $p$-type semiconductor hole starts moving towards negatively charged plate and electron towards positively charged plate.

• Option (c): This option is incorrect because in a p-type semiconductor, holes (which are the majority carriers) move towards the negatively charged plate, while electrons (which are the minority carriers) move towards the positively charged plate. Therefore, both electrons and holes do not appear to move towards the positively charged plate.

• Option (d): This option is incorrect because the movement of electrons and holes in a semiconductor is related. When an electric field is applied, electrons move towards the positively charged plate, and holes move towards the negatively charged plate. The movement of one type of charge carrier affects the overall conductivity and behavior of the semiconductor.

Answer

$(b, c)$

Silicon doped with an electron rich impurity is an $n$-type semiconductor. Conductivity of $n$-type semiconductor is due to presence of free electron delocalisation of electrons increases the conductivity of doped silicon due to increase in mobility of electron.

• (a) Silicon doped with electron rich impurity is a $p$-type semiconductor: This statement is incorrect because doping silicon with an electron-rich impurity (such as phosphorus) actually creates an $n$-type semiconductor, not a $p$-type. In $n$-type semiconductors, the majority charge carriers are electrons.

• (d) An electron vacancy increases the conductivity of $n$-type semiconductor: This statement is incorrect because in an $n$-type semiconductor, the conductivity is primarily due to the presence of free electrons, not electron vacancies (holes). Electron vacancies (holes) are the primary charge carriers in $p$-type semiconductors, not $n$-type.

Answer

$(a, d)$

When $KCl$ crystals are heated it leads the diposition of potassium ion on surface of $KCl$. The $Cl^{-}$ions diffuse to the surface of crystal and loss electron by potassium atom to form $K^{+}$ion released electron occupies anionic site which is known as F-centre and impart colour to the crystal.

• Option (b) is incorrect because the presence of a pair of electrons at anionic sites does not lead to the formation of F-centres, which are responsible for the violet or lilac color in $KCl$ crystals. F-centres are specifically related to unpaired electrons.

• Option (c) is incorrect because vacancies at anionic sites alone do not impart color to the crystals. The color is due to the presence of F-centres, which involve electrons occupying these vacancies.

Answer

( $b, c$ )

$NaCl$ has a rock salt type structure having fcc arrangement.

Total number of atoms per unit cell $=4$

$\therefore \quad$ Number of tetrahedral voids $=2 \times 4=8$

Number of octahedral voids $=4$

Hence, (b) and (c) are correct choices.

• Option (a) is incorrect because the number of tetrahedral voids in an NaCl crystal is not 4. The correct number is 8, as calculated by multiplying the total number of atoms per unit cell (4) by 2.

• Option (d) is incorrect because the number of tetrahedral voids is not four times the number of octahedral voids. The correct relationship is that the number of tetrahedral voids is twice the number of octahedral voids.

Answer

(a, c)

Amorphous solid has short range order which has a tendency to flow very slowly. Hence, it is also known as pseudo solids or super cooled liquids. Glass panes fixed to windows or doors of old buildings are invariably observed to be thicker at bottom than at the top. These are examples of amorphous solids.

• (b) true solids: True solids, also known as crystalline solids, have a long-range order and a definite geometric shape. They do not exhibit the tendency to flow like amorphous solids do. Therefore, amorphous solids cannot be classified as true solids.

• (d) super cooled solids: The term “super cooled solids” is not commonly used in the context of amorphous solids. While amorphous solids can be considered super cooled liquids due to their ability to flow very slowly, they are not referred to as super cooled solids.

Answer

(a, c)

n-type semiconductor When group 15 elements are doped into a perfect crystal it leads to formation of $n$-type semiconductor.

Here, in (a) as (group 15, period 3) is doped to perfect Si-crystal and in (c) as (group 15, period 2) is doped to perfect Si-crystal.

• Option (b) is incorrect because it involves doping with a group 13 element, which leads to the formation of a p-type semiconductor, not an n-type semiconductor.

• Option (d) is incorrect because it also involves doping with a group 13 element, which results in the formation of a p-type semiconductor, not an n-type semiconductor.

Answer

$(a, d)$

Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic. In ferrimagnetic substance domains are alligned in parallel and antiparallel direction in unequal numbers.

In ferromagnetic substances, all the domains get oriented in the direction of magnetic field and remain as such even after removing magnetic field.

Hence, (a) and (d) are correct choices.

• (b) Ferrimagnetic substances do lose their ferrimagnetism on heating and become paramagnetic. Therefore, the statement that they do not lose ferrimagnetism on heating is incorrect.

• (c) Antiferromagnetic substances have domain structures where the magnetic moments are aligned in opposite directions and cancel each other out. Therefore, the statement that their magnetic moments are not cancelled by each other is incorrect.

Answer

$(a, c)$

Quartz glass is an amorphous solid so it has not definite heat of fusion. This is due to short range order of molecule while quartz glass is also known as super cooled liquid and isotropic in nature.

• Quartz glass is not a crystalline solid because it lacks a long-range periodic atomic structure.
• Quartz glass does not have a definite heat of fusion because it is an amorphous solid, meaning it does not have a well-defined melting point.
• Quartz glass is also called a super cooled liquid because it behaves like a liquid that has been cooled below its normal freezing point without solidifying.
• Quartz glass is isotropic in nature, meaning its refractive index is the same in all directions.

Answer

$(a, b, c)$

$SiC$, AIN and diamond are examples of network solid as they have three dimensional structure while, $I_2$ is a molecular solid, because such solid particles are held together by dipole-dipole interactions.

• $SiC$, AIN, and diamond are examples of network solids as they have three-dimensional structures with strong covalent bonds throughout the lattice, making them different from molecular solids.
• $I_2$ is a molecular solid because its particles are held together by weaker dipole-dipole interactions rather than the strong covalent bonds found in network solids.

Answer

$(a, d)$

In hcp and fcc arrangement octahedral voids are formed. In fcc the octahedral voids are observed at edge and centre of cube while in bcc and simple cubic, no any octahedral voids are observed.

• In bcc (body-centered cubic) arrangement, there are no octahedral voids because the atoms are arranged in such a way that the voids formed are not of octahedral geometry. Instead, bcc has tetrahedral voids.

• In simple cubic arrangement, the atoms are arranged in a simple cubic lattice, which does not create octahedral voids. The voids in a simple cubic structure are not large enough to be considered octahedral.

Answer

$(a, b)$

Frenkel defect arises when the smaller ion (usually cation) is dislocated from its original site to interstitial site, this is also known as dislocation defect. Since, stoichiometry of substance persist so, it is categorised as stoichiometric defect.

• Impurity defect: This type of defect occurs when foreign atoms are present in the crystal lattice, either substituting for the host atoms or occupying interstitial sites. Frenkel defect does not involve foreign atoms, hence it is not an impurity defect.

• Non-stoichiometric defect: This defect occurs when the ratio of the number of atoms of different elements in the compound deviates from the ideal stoichiometric ratio. Frenkel defect does not alter the stoichiometry of the compound, so it is not a non-stoichiometric defect.

Answer

$(b, d)$

Vacancy and Schottky defects which lead to decrease the density both are the types of a stoichiometric defect. In case of Frenkel defect and interstitial defect, there is no change in density of substance.

• Interstitial defect: In an interstitial defect, extra atoms are inserted into the crystal lattice at positions that are normally unoccupied. This does not remove any atoms from the lattice, so the overall mass of the crystal increases while the volume remains the same, leading to no change in density.

• Frenkel defect: In a Frenkel defect, an atom or ion is displaced from its normal lattice site to an interstitial site within the crystal. This defect does not change the overall number of atoms in the crystal, so the mass and volume remain constant, resulting in no change in density.

Answer

Liquids and gases have the tendency to flow, i.e., their molecules can move freely from one place to another. Therefore, they are known as fluids. e.g., glass panes fixed to windows or doors of old buildings are sometimes found to be thicker at bottom. This is due to ability of glass to flow.

Answer

The distance between the constituent particles is very less in solids. On bringing them still closer repulsion will start between electron clouds of these particles. Hence, they cannot be brought further close together and are incompressible.

Answer

Crystals have long range repeated pattern of arrangement of constituent particles but in the process of crystallisation some deviations from the ideal arrangement (i.e., defects) may be introduced, therefore, crystals are usually not perfect.

Answer

Yellow colour in $NaCl$ is due to metal excess defect due to which unpaired electrons occupy anionic sites, known as F-centres. These electrons absorb energy from the visible region for the excitation which makes crystal appear yellow.

Answer

In the crystals of $FeO$, some of the $Fe^{2+}$ cations are replaced by $Fe^{3+}$ ions. Three $Fe^{2+}$ ions are replaced by two $Fe^{3+}$ ions to make up the loss of positive charge. Eventually there would be less amount of metal as compared to stoichiometric composition.

Answer

On heating $ZnO$ loses oxygen as follows

$$ ZnO \xrightarrow{\text { Heat }} Zn^{2+}+\frac{1}{2} O_2+2 e^{-} $$

$Zn^{2+}$ ions and electrons move to interstitial sites and F-centres are formed which impart yellow colour to $ZnO(s)$.

Answer

The gap between conduction band and valence band is small in semiconductors. Therefore, electrons from the valence band can jump to the conduction band on increasing temperature. Thus, they become more conducting as the temperature increases.

Answer

On doping germanium with galium some of the positions of lattice of germanium are occupied be galium. Galium atom has only three valence electrons. Therefore, fourth valency of nearby germanium atom is not satisfied and the place remains vacant.

This place is deficient of electrons and is therefore called electron hole or electron vacancy. Electron from neighbouring atom comes and fills the hole, thereby creating a hole in its original position.

Under the influence of electric field electrons move towards positively charged plates through these and conduct electricity. The holes appear to move towards negatively charged plates.

Answer

Let the number of $N$ atoms in $ccp$ is $x$

$\therefore \quad$ Number of tetrahedral voids $=2 x$

$$ \begin{aligned} & \text { Number of } M \text { atoms }=\frac{1}{3} \times 2 x \\ & \frac{\text { Number of } N \text { atoms }}{\text { Number of } M \text { atoms }}=\frac{3 x}{2 x}=\frac{3}{2} \end{aligned} $$

So, the formula of the compound is $M_2 N_3$.

Answer

On heating, amorphous substances change to crystalline form at some temperature some objects from ancient civilisation are found to be milky in appearance. This is due to crystallisation.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow(2)$

A. When some of lattice sites are vacant in any non-ionic solid, the crystal is said to have vacancy defect and due to decrease in number of particles present in crystal lattice the density of crystal decreases.

B. Simple interstitial defect are shown by non-ionic solids in which constituent particles is displaced from its normal site to an interstitial site. Hence, density of solid increases.

C. Frenkel defect is shown by ionic solids in which smaller ions get dislocated from its normal site to its interstitial site which lead to decrease its density.

D. Schottky defect is shown by ionic solids in which equal number of cation and anion get missed from ionic solids and thus, density of solid decreases.

Answer

A. $\rightarrow(2,3)$

B. $\rightarrow(3,4)$

C. $\rightarrow(3,5)$

D. $\rightarrow(1,4)$

A. For primitive unit cell, $a=b=c$

Total number of atoms per unit cell $=1 / 8 \times 8=1$

Here, $1 / 8$ is due to contribution of each atom present at corner.

B. For body centred cubic unit cell, $a=b=c$

This lattice contain atoms at corner as well as body centre. Contribution due to atoms at corner $=1 / 8 \times 8=1$ contribution due to atoms at body centre $=8$

C. For face centred unit cell, $a=b=c$

Total constituent ions per unit cell present at corners $=\frac{1}{8} \times 8=1$

Total constituent ions per unit cell present at face centre $=\frac{1}{2} \times 6=3$

D. For end centered orthorhombic unit cell, $a \neq b \neq c$

Total contribution of atoms present at corner $=\frac{1}{8} \times 8=1$

Total contribution of atoms present at end centre $=\frac{1}{2} \times 2=1$

Hence, other than corner it contain total one atom per unit cell.

Answer

A. $\rightarrow(3) \quad$

B. $\rightarrow(1) \quad$

C. $\rightarrow(2)$

A. Impurity defect arises due to replacement of one common ion present in any crystal by another uncommon ion.

B. Metal excess defect is due to missing of cation from ideal ionic solid which lead to create a F-centre generally occupied by unpaired electrons. e.g., $NaCl$ with anionic site.

C. Metal deficiency defect $\ln FeO, Fe^{3+}$ exists along with $Fe^{2+}$ which lead to decrease in metal ion(s) so this is a type of metal deficiency defect.

Answer

A. $\rightarrow(4)$

B. $\rightarrow$ (3)

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. Mg in solid state show electronic conductivity due to presence of free electrons hence, they are known as electronic conductors.

B. $MgCl_2$ in molten state show electrolytic conductivity due to presence of electrolytes in molten state.

C. Silicon doped with phosphorus contain one extra electron due to which it shows conductivity under the influence of electric field and known as p-type semiconductor.

D. Germanium doped with boron contain one hole due to which it shows conductivity under the influence of electric field and known as $n$-type semiconductor.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow(2)$

A. Square close packing in two dimensions each sphere have coordination number 4 , as shown below

B. Hexagonal close packing in two dimensions each sphere have coordination number 6 as shown below and creates a triangular void

C. Hexagonal close packing in 3 dimensions is a repeated pattern of sphere in alternate layers also known as $A B A B$ pattern

D. Cubic close packing in a 3 dimensions is a repeating pattern of sphere in every fourth layer

Answer

(a) In simple cubic unit cell each atom is present at corners having contribution 1/8. Hence, total number of atoms present per unit cell in scc is $\frac{1}{8} \times 8=1$.

Answer

(b) Assertion and Reason both are correct but Reason is not correct explanation of Assertion.

Correct explanation is that graphite have layered structure with free electrons due to which it is a good conductor of electricity. On the other hand, diamond have tetrahedral arrangement with no unpaired electron. Therefore, diamond is hard and brittle but insulator.

Answer

(c) Assertion is correct statement but Reason is incorrect statement.

Assertion is true as in ccp atom present at face centre and corner of each unit cell which creates octahedral void at each body centre and all twelve edges of a unit cell as shown below

Correct reason is that beside the body centre there is one octahedral void at centre of each of 12 edges which is surrounded by six atoms.

Out of six atoms four belongs to same unit cell (2 at corner and 2 at face centre) and 2 atoms belongs to adjacent unit cell.

Answer

(b) Assertion and Reason both are correct statements but reason is not the correct explanation of Assertion.

Correct reason is that, packing efficiency is maximum for fcc structure because it consists of total four atoms per unit cell. Packing efficiency is maximum in fcc structure which is equal to $74 %$.

Answer

(c) Assertion is correct statement but reason is incorrect statement.

Semiconductors are solids with conductivities in the intermediate range varie from $10^{-6}-10^{4} \Omega^{-1} m^{-1}$. Intermediate conductivity is due to small energy gap between valence band and conduction band.

(Also, refer to $Q .60$ )

Answer

Cubic close packed structure contains one atom at each of eight corners of a unit cell and one atom at each of six faces which can be represented below

As we know any atom surrounded by six atoms (hard sphere) creates an octahedral void. In case of fcc body centre is surrounded by six identical atoms present at face centre hence, there is a octahedral void at body centre of each unit cell.

Beside the body centre there is one octahedral void at centre of each of 12 edge as shown below

Since, each void is shared by 4 unit cell. Therefore, contribution of octahedral void to each edge of a unit cell is $\frac{1}{4}$.

Number of octahedral void at centre of 12 edge $=\frac{1}{4} \times 12=3$

Number of octahedral void at body centre $=1$

Therefore, total number of octahedral void at each $ccp$ lattice $=3+1=4$

Answer

Cubic close packed structure contains one atom at each corner of an unit cell and at face centre of each unit cell. Each unit cell consists of 8 small cubes.

Each small cube contains 4 atoms at its alternate corner when these atoms are joined to each other lead to creation of a tetrahedral void as shown below

Since, there are total 8 smaller cubes present at one unit cell and each smaller cube has one tetrahedral void hence, total number of tetrahedral void present in each unit cell is equal to eight.

As we know ccp structure has 4 atoms per unit cell. Thus, total number of tetrahedral void in one ccp unit cell is equal to 8 .

Answer

Conductivity of a semiconductor is too low for practical use. The conductivity of a semiconductor can be increased by adding a suitable amount of impurity to perfect crystal. This process is known as doping. It can be done by adding either of two types of impurity to the crystal.

(A) By adding electron rich impurities i.e., group 15 elements to the silicon and germanium of group 14 elements. Out of 4 valence electrons of group 14 elements and 5 valence electrons of group 15 elements, four electrons of each element led to formation of four covalent bonds while the one extra electron of group 15 elements become delocalised.

Thus, increases conductivity of semiconductor. This type of semiconductor is known as $n$-type semiconductor.

(B) By adding electron deficient impurity i.e., group 14 to the perfect crystal of group 14 elements when group 13 element is doped to group 14 element it lead to create a hole in the ideal crystal which is known as electron hole or electron vacancy.

An electron from the neighbouring atom come and fill the electron hole in doing so an electron from the neighbour leaves an electron hole to its original position. Thus, it increases conductivity of semiconductor. This type of semiconductor is known as p-type semiconductor.

Thinking Process

Consider the number of $Fe^{2+}$ and $Fe^{3+}$ ions as $x$ and $y$ then write their sum equal to 0.93 . Write another equation in terms of $x$ and $y$ by taking the sum of their total charge equal to 2 [charge on oxygen]. Now using the substitution method. Calculate the value of $x$ and $y$ then calculate fraction of $Fe^{2+}$ ion present in this sample.

Answer

Let the formula of the sample be $(Fe^{2+}) _{x}(Fe^{3+}) _{y} O$

On looking at the given formula of the compound

$$ x+y=0.93 \quad \quad …(i) $$

Total positive charge on ferrous and ferric ions should balance the two units of negative charge on oxygen

Therefore,

$$ \begin{matrix} \quad 2 x+3 y=2 \quad \quad …(ii) \\ \Rightarrow \quad x+\frac{3}{2} y=1 \quad \quad …(iii) \end{matrix} $$

On subtracting equation (i) from equation (iii) we have

$$\frac{3}{2} y-y =1-0.93 $$

$$ \Rightarrow \quad \frac{1}{2} y =0.07 $$

$$ \Rightarrow \quad y =0.14 $$

$$ \text { On putting the value of } y \text { in equation (i), we get } $$

$$ \Rightarrow \quad x+0.14 =0.93 $$

$$ \Rightarrow \quad x =0.93-0.14 $$

$$ \Rightarrow \quad x =0.79 $$

$$ \text { Fraction of } Fe^{2+} \text { ions present in the sample } =\frac{0.79}{0.93}=0.849 $$

Metal deficiency defect is present in the sample because iron is less in amount than that required for stoichiometric composition.