CBSE Class 12 Chemistry Solid State Assignment Set 05

Very Short Answer Type Questions

Question. Write a point of distinction between a metallic solid and an ionic solid other than metallic lustre.
Answer: Metallic solids conduct electric current in the solid form, but ionic substances do so only when melted or dissolved. Metals carry electricity via mobile electrons, whereas ionic compounds do so through free ions. Also, metallic solids are malleable and ductile, while ionic crystals are hard and break easily.
In simple words: Metals can conduct electricity when solid, but ionic solids must be melted or dissolved in water to conduct. Also, metals can be shaped, while ionic solids break easily.

Exam Tip: Highlight the state-dependent electrical conductivity of ionic solids as it is a major distinguishing point.

Question. How can the conductivity of an intrinsic semiconductor be increased?
Answer: The conductivity can be raised by introducing a proper amount of a suitable impurity, which is known as doping.
In simple words: You can make a semiconductor conduct better by adding a small amount of a specific impurity to it.

Exam Tip: Use the term "doping" and mention that it introduces extra charge carriers (electrons or holes).

Question. Which stoichiometric defect increases the density of a solid?
Answer: An interstitial defect raises the overall density of a solid.
In simple words: The density goes up during an interstitial defect because extra particles occupy the vacant spots in the crystal structure.

Exam Tip: Make sure to specify that it is the "interstitial defect" where mass increases while volume remains constant.

Question. What are n-type semiconductors?
Answer: n-type semiconductors: These are produced by doping silicon with elements from Group 15, such as phosphorus or arsenic.
In simple words: An n-type semiconductor is made by adding elements like phosphorus or arsenic to silicon, which adds extra negative charges (electrons).

Exam Tip: Clearly state that the majority charge carriers are electrons, which are introduced by pentavalent impurities.

Question. What type of stoichiometric defect is shown by AgBr and AgI ?
Answer: Silver bromide shows both Schottky and Frenkel defects, while silver iodide exhibits only the Frenkel defect.
In simple words: AgBr can have both Frenkel and Schottky defects, but AgI only shows the Frenkel defect.

Exam Tip: AgBr is a highly unique and important exception in solid state chemistry; always remember it shows both defects.

Question. What type of defect can arise when a solid is heated?
Answer: Vacancy-type defects can develop when a solid is heated.
In simple words: Heating a solid can cause some atoms to leave their places, creating empty spaces or vacancy defects.

Exam Tip: Explain that heating provides thermal energy, which allows atoms to leave their lattice positions, thereby lowering the density.

Question. Why does LiCl acquire pink colour when heated in Li vapours?
Answer: This happens because of a metal excess defect from anionic vacancies, where the vacant positions are filled by unpaired electrons, creating F-centres.
In simple words: When heated, lithium atoms leave behind electrons in empty spots of the crystal. These trapped electrons, called F-centres, absorb light and make the crystal look pink.

Exam Tip: The term "F-centres" (or Farbe centres) is crucial to include in your answer to get full marks.

Question. How many atoms constitute one unit cell of a face-centered cubic crystal?
Answer: There are 4 atoms that make up a single unit cell of an fcc crystal structure.
In simple words: An fcc unit cell has exactly four atoms inside it.

Exam Tip: Write the breakdown: \( 8 \text{ corners} \times \frac{1}{8} + 6 \text{ faces} \times \frac{1}{2} = 4 \text{ atoms} \) for clarity.

Question. What type of stoichiometric defect is shown by AgCl?
Answer: AgCl exhibits the Frenkel defect.
In simple words: AgCl shows the Frenkel defect because silver ions are small and can easily move into interstitial spaces.

Exam Tip: Frenkel defects occur in compounds with low coordination numbers and a large difference in size between cations and anions.

Question. What type of substances would make better Permanent Magnets: Ferromagnetic or Ferrimagnetic?
Answer: Ferromagnetic materials make more suitable permanent magnets. Examples include iron, cobalt, and nickel.
In simple words: Ferromagnetic materials make much better permanent magnets because they stay strongly magnetized even after the magnetic field is removed.

Exam Tip: Explain that their magnetic domains remain aligned in the same direction permanently.

Question. Calculate the number of atoms in a face centred cubic unit cell.
Answer: Within a face-centered cubic structure, the total lattice sites are: 8 at corners and 6 on faces.
Thus, the total atoms in one unit cell = \( 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4 \).
In simple words: In an fcc crystal, atoms are at the 8 corners and 6 face centers. They share space with neighboring cells, adding up to exactly 4 whole atoms.

Exam Tip: Always show the step-by-step contribution of corners and faces clearly in calculation questions.

Question. On heating a crystal of KCl in potassium vapour, the crystal starts exhibiting a violet colour. What is this due to?
Answer: The chloride ions migrate to the outer surface to react with potassium atoms, which ionize by releasing electrons. These free electrons then get trapped inside the anionic vacancies, acting as F-centres that give the crystal its violet hue.
In simple words: When heated, potassium atoms lose electrons that get trapped in the empty spaces left by chloride ions. These trapped electrons, called F-centres, make the crystal look violet.

Exam Tip: State that the excitation of these trapped electrons by absorbing energy from visible light causes the violet color.

Question. Which type of ionic substances show Schottky defect in solids?
Answer: Strongly ionic compounds that have high coordination numbers and minimal differences in the sizes of their cations and anions display Schottky defects.
In simple words: Crystals made of positive and negative ions of similar sizes and high coordination numbers usually show Schottky defects.

Exam Tip: Mention NaCl, KCl, or CsCl as classic examples to support your definition.

Question. How many atoms per unit cell (z) are present in bcc unit cell?
Answer: The quantity of atoms in a body-centered cubic unit cell is determined as follows:
The share of the 8 corner atoms = \( \frac{1}{8} \times 8 = 1 \).
The contribution from the single atom located at the center of the body = 1.
So, the total number of atoms in the unit cell = 1 + 1 = 2 atoms.
In simple words: A bcc unit cell has one atom shared from its corners and one whole atom in the middle, making 2 atoms in total.

Exam Tip: Show both corner and center contributions to avoid losing marks on steps.

Question. What type of stoichiometric defect is shown by NaCl?
Answer: NaCl displays the Schottky defect.
In simple words: NaCl shows the Schottky defect because the sodium and chloride ions are similar in size.

Exam Tip: Remember that Schottky defects involve an equal number of missing cations and anions to maintain electrical neutrality.

Short Answer Type Questions

Question. (a) What type of semiconductor is obtained when silicon is doped with boron?
(b) What type of magnetism is shown in the following alignment of magnetic moments?
↑ ↑ ↑ ↑ ↑ ↑
(c) What type of point defect is produced when AgCl is doped with CdCl2? (Delhi)
Answer:
(a) A p-type semiconductor is created when silicon is doped with boron.
(b) Ferromagnetism is displayed when all the magnetic dipoles are pointed in the same direction.
(c) An impurity defect in ionic solids is generated when \( \text{CdCl}_2 \) is introduced into \( \text{AgCl} \). This defect produces vacancies in the lattice, leading to enhanced electrical conductivity in the material.
In simple words: (a) Silicon with boron makes a p-type semiconductor because boron has fewer electrons, creating "holes". (b) Ferromagnetism is when all magnetic parts point the same way, making a very strong magnet. (c) Doping AgCl with CdCl2 leaves empty spaces because one Cd2+ replaces two Ag+ ions, boosting conductivity.

Exam Tip: Remember that doping silicon with a trivalent impurity (Group 13) yields a p-type semiconductor due to the creation of positive holes.

Question. An element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.0 g cm–3. Also calculate the radius of an atom of this element. (Comptt. Delhi)
Answer: We have the following parameter values:
For a body-centred cubic (bcc) system, \( Z = 2 \)
Edge length of unit cell \( a = 250\text{ pm} = 2.50 \times 10^{-8}\text{ cm} \)
Density of the solid \( \rho = 8.0\text{ g cm}^{-3} \)
Using the density relation to solve for molar mass \( M \):
\[ M = \frac{\rho \times a^3 \times N_A}{Z} \]
Substituting the values:
\[ M = \frac{8.0 \times (2.50 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{2} \]
\[ M \approx 37.6\text{ g mol}^{-1} \]
For a bcc crystal, the atomic radius is given by:
\[ r = \frac{\sqrt{3}}{4}a \]
\[ r = \frac{1.732 \times 250\text{ pm}}{4} = 108.25\text{ pm} \]
Therefore, the molar mass is \( 37.6\text{ g mol}^{-1} \) and the atomic radius is \( 108.25\text{ pm} \).
In simple words: First, use the density formula rearranged to solve for the molar mass of the element. Then, use the bcc radius formula to find the size of the atom from its edge length.

Exam Tip: For bcc, make sure to write \( 4r = \sqrt{3}a \) as the fundamental geometric relationship to earn complete step marks.

Question. Iron (II) oxide has a cubic structure and each unit cell has a size of 5 Å. If density of this oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell.
(Atomic mass of Fe = 56, O = 16, NA = 6.023 × 1023 and 1 Å = 10–8 cm) (Comptt. Delhi)
Answer: We have the following data:
Density \( \rho = 4\text{ g cm}^{-3} \)
Edge length \( a = 5\text{ Å} = 5 \times 10^{-8}\text{ cm} \)
Molar mass of FeO, \( M = 56 + 16 = 72\text{ g mol}^{-1} \)
Applying the crystal density relation:
\[ Z = \frac{\rho \times a^3 \times N_A}{M} \]
Substituting the values:
\[ Z = \frac{4 \times (5 \times 10^{-8})^3 \times 6.023 \times 10^{23}}{72} \]
\[ Z = \frac{301.15}{72} \approx 4.18 \approx 4 \]
Since the value of \( Z \approx 4 \), there are four formula units of FeO per unit cell. This means that \( 4\text{ Fe}^{2+} \) ions and \( 4\text{ O}^{2-} \) ions exist in every unit cell.
In simple words: Put the given values into the density formula to solve for Z. Since Z is approximately 4, each unit cell contains four iron ions and four oxide ions.

Exam Tip: Be sure to sum the individual masses of Fe (56) and O (16) to find the molecular mass (72) before using it in the formula.

Question. Niobium crystallizes in body-centred cubic structure. If its density is 8.55 g cm–3, calculate atomic radius of niobium, given its atomic mass 93u. (Comptt. Delhi)
Answer: We are given the following parameters:
Density \( \rho = 8.55\text{ g cm}^{-3} \)
For a body-centred cubic (bcc) system, \( Z = 2 \)
Molar mass \( M = 93\text{ g mol}^{-1} \)
Let us first calculate the edge length \( a \):
\[ a^3 = \frac{Z \times M}{\rho \times N_A} \]
Substituting the values:
\[ a^3 = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} \]
\[ a^3 = \frac{186}{5.149 \times 10^{24}} \approx 3.61 \times 10^{-23}\text{ cm}^3 = 36.1 \times 10^{-24}\text{ cm}^3 \]
Taking the cube root:
\[ a = (36.1 \times 10^{-24})^{1/3} \approx 3.304 \times 10^{-8}\text{ cm} = 330.4\text{ pm} \]
For a bcc system, the relation for atomic radius is:
\[ r = \frac{\sqrt{3}}{4}a \]
\[ r = \frac{1.732 \times 330.4\text{ pm}}{4} \approx 143.1\text{ pm} \]
Therefore, the atomic radius of niobium is 143.1 pm.
In simple words: Use the density formula to find the volume of the cube, take its cube root to get the edge length (330.4 pm), and then find the radius using the bcc formula.

Exam Tip: Always show the cube root working clearly, and note that \( a = 330.4\text{ pm} \) corresponds to \( 3.304 \times 10^{-8}\text{ cm} \) before finding the radius.

Question. The density of copper is 8.95 g cm–3. It has a face centred cubic structure. What is the radius of copper atom?
(Atomic mass Cu = 63.5 g mol–1, NA = 6.02 × 1023 mol–1) (Comptt. Delhi)
Answer: We use the crystal density equation:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
For a face-centred cubic (fcc) lattice, \( Z = 4 \).
Rearranging the equation to solve for the cell volume \( a^3 \):
\[ a^3 = \frac{Z \times M}{\rho \times N_A} \]
Substituting the values:
\[ a^3 = \frac{4 \times 63.5}{8.95 \times 6.02 \times 10^{23}} \]
\[ a^3 = \frac{254}{53.879 \times 10^{23}} \approx 4.714 \times 10^{-23}\text{ cm}^3 = 47.14 \times 10^{-24}\text{ cm}^3 \]
Taking the cube root to find the edge length \( a \):
\[ a = (47.14 \times 10^{-24})^{1/3} \approx 3.613 \times 10^{-8}\text{ cm} = 361.3\text{ pm} \]
For an fcc structure, the relationship between radius \( r \) and edge length \( a \) is:
\[ r = \frac{a}{2\sqrt{2}} \]
\[ r = \frac{361.3\text{ pm}}{2 \times 1.414} = \frac{361.3\text{ pm}}{2.828} \approx 127.8\text{ pm} \]
Hence, the radius of the copper atom is 127.8 pm.
In simple words: First, find the edge length by rearranging the density formula. Then, divide this edge length by two times the square root of two to find the radius of the copper atom.

Exam Tip: Always make sure to complete the radius calculation step (\( r = \frac{a}{2\sqrt{2}} \)) for fcc systems, even if the textbook solution stops at edge length.

Question. Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm–3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 55.84 g mol–1) (Comptt. Delhi)
Answer: From the given parameters:
Edge length \( a = 286.65\text{ pm} = 2.8665 \times 10^{-8}\text{ cm} \)
Density \( d = 7.874\text{ g cm}^{-3} \)
Molar mass of iron \( M = 55.84\text{ g mol}^{-1} \)
For a body-centred cubic (bcc) cell, \( Z = 2 \).
We apply the standard density relation:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
Rearranging the relation to find Avogadro's number \( N_A \):
\[ N_A = \frac{Z \times M}{a^3 \times d} \]
Substituting the values:
\[ N_A = \frac{2 \times 55.84}{(2.8665 \times 10^{-8})^3 \times 7.874} \]
\[ N_A = 6.022 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Put the given edge length, density, and molar mass of iron into the density equation to find Avogadro's number.

Exam Tip: Make sure to clearly state that \( Z = 2 \) for bcc lattice systems so that the examiner knows why that multiplier was chosen.

Question. Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron is 7.874 g cm–3. Use this information to calculate Avogadro’s number. (Gram atomic mass of Fe = 55.84 g mol^–1). (Comptt. All India)
Answer: From the given data:
Edge length of unit cell \( a = 286.65\text{ pm} = 2.8665 \times 10^{-8}\text{ cm} \)
Density of the crystal \( d = 7.874\text{ g cm}^{-3} \)
Gram atomic mass of iron \( M = 55.84\text{ g mol}^{-1} \)
For a body-centred cubic (bcc) system, the value of \( Z \) is 2.
Applying the crystal density formula:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
Rearranging the equation to find Avogadro's number \( N_A \):
\[ N_A = \frac{Z \times M}{a^3 \times d} \]
Substituting the values:
\[ N_A = \frac{2 \times 55.84}{(2.8665 \times 10^{-8})^3 \times 7.874} \]
\[ N_A \approx 6.022 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Use the density formula, input the given values of mass, density, and edge length, and solve to calculate Avogadro's number.

Exam Tip: Ensure that all numerical steps are shown clearly, as examiners assign specific credit to formula rearrangement and substitution.

Question. An element with molar mass 27 g mol–1 forms a cubic unit cell with edge length 4.05 × 10–8 cm. If its density is 2.7 g cm–3, what is the nature of the cubic unit cell?
Answer: We have the following parameter values:
Molar mass of the element \( M = 27\text{ g mol}^{-1} \)
Edge length of unit cell \( a = 4.05 \times 10^{-8}\text{ cm} \)
Density of the element \( d = 2.7\text{ g cm}^{-3} \)
We apply the density equation for a cubic lattice:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
Rearranging the equation to determine the number of atoms per unit cell, \( Z \):
\[ Z = \frac{d \times a^3 \times N_A}{M} \]
Substituting the values:
\[ Z = \frac{2.7 \times (4.05 \times 10^{-8})^3 \times 6.02 \times 10^{23}}{27} \]
\[ Z = \frac{2.7 \times 6.643 \times 10^{-23} \times 6.02 \times 10^{23}}{27} \]
\[ Z \approx 4 \ ]
Since the calculated number of atoms per unit cell \( Z = 4 \), the nature of the cubic unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).
In simple words: Rearrange the density formula to find the number of atoms per unit cell, Z. Since the calculated Z is 4, the crystal has a face-centered cubic structure.

Exam Tip: Remember to correlate the value of \( Z \) to the cell type clearly (Z = 1 for simple cubic, 2 for bcc, and 4 for fcc/ccp).

Question. An element with molar mass \( 27 \text{ g mol}^{-1} \) forms a cubic unit cell with edge length \( 4.05 \times 10^{-8} \text{ cm} \). If its density is \( 2.7 \text{ g cm}^{-3} \), what is the nature of the cubic unit cell?
Answer: Given data:
Molar mass, \( M = 27 \text{ g mol}^{-1} \)
Edge length, \( a = 4.05 \times 10^{-8} \text{ cm} \)
Density, \( d = 2.7 \text{ g cm}^{-3} \)
Avogadro's number, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)
Using the density formula:
\( d = \frac{Z \times M}{N_A \times a^3} \)

Rearranging the formula to find the number of atoms per unit cell (\( Z \)):
\( Z = \frac{d \times N_A \times a^3}{M} \)

Substituting the values:
\( Z = \frac{2.7 \text{ g cm}^{-3} \times 6.022 \times 10^{23} \text{ mol}^{-1} \times (4.05 \times 10^{-8} \text{ cm})^3}{27 \text{ g mol}^{-1}} \)
\( Z = \frac{6.022 \times 10^{23} \times (4.05)^3 \times 10^{-24}}{10} \)
\( Z = 6.022 \times 10^{-2} \times 66.43 = 4.0004 \approx 4 \)
Since the value of \( Z \) is 4, the crystal lattice is a face-centered cubic (fcc) unit cell.
In simple words: We calculate the number of atoms in the unit cell using the density formula. Since the result is 4, we know the crystal has a face-centered cubic structure.

Exam Tip: Remember to always check the value of Z: Z=1 for simple cubic, Z=2 for bcc, and Z=4 for fcc.

Question. Examine the given defective crystal:
\( X^+ \quad Y^- \quad X^+ \quad Y^- \quad X^+ \quad \)
\( Y^- \quad \text{O} \quad Y^- \quad X^+ \quad Y^- \quad \)
\( X^+ \quad Y^- \quad X^+ \quad \text{O} \quad X^+ \quad \)
\( Y^- \quad X^+ \quad Y^- \quad X^+ \quad Y^- \quad \)
Answer the following questions:
(i) Is the above defect stoichiometric or non stoichiometric?
(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.
(iii) How does this defect affect the density of the crystal?
Answer:
(i) This represents a stoichiometric defect.
(ii) It is known as a Schottky defect, and a common example is sodium chloride (\( \text{NaCl} \)).
(iii) The overall density of the crystal decreases due to this defect.
In simple words: This defect occurs when equal numbers of positive and negative ions are missing from their positions. This creates empty spaces, which lowers the overall density of the crystal.

Exam Tip: In Schottky defects, equal numbers of cations and anions are missing to maintain electrical neutrality, leading to decreased density.

Question. Define the following:
(i) Schottky defect
(ii) Frenkel defect
(iii) F-centre
Answer:
(i) Schottky defect: When equal numbers of positive cations and negative anions are absent from their regular sites in an \( A^+ B^- \) ionic solid, maintaining neutral electrical charge, it is referred to as a Schottky defect.(ii) Frenkel defect: When an ion departs from its normal lattice position and settles into an interstitial space while preserving overall electrical neutrality, this is designated as a Frenkel defect.(iii) F-centre: These are locations formed when free electrons get trapped inside anionic vacancies, which helps in giving color to the crystal structure (F stands for Fabre).
In simple words: Schottky defect is when matching pairs of positive and negative ions are missing. Frenkel defect is when an ion slips out of place into an empty space nearby. F-centre is when an electron gets trapped in an empty negative ion space, creating color.

Exam Tip: Be ready to compare defects—Schottky decreases density, Frenkel keeps density constant, and F-centres add colour to crystals.

Question. Silver crystallises in fcc lattice. If edge length of the unit cell is \( 4.077 \times 10^{-8} \text{ cm} \), then calculate the radius of silver atom.
Answer: Given data:
Edge length of unit cell, \( a = 4.077 \times 10^{-8} \text{ cm} \)
For an fcc lattice, the relation between atomic radius (\( r \)) and edge length (\( a \)) is:
\( r = \frac{a}{2\sqrt{2}} \)

Substituting the given value:
\( r = \frac{4.077 \times 10^{-8} \text{ cm}}{2 \times 1.414} \)
\( r = \frac{4.077 \times 10^{-8} \text{ cm}}{2.828} \)
\( r = 1.441 \times 10^{-8} \text{ cm} \)
Hence, the atomic radius of silver is \( 1.441 \times 10^{-8} \text{ cm} \).
In simple words: For a face-centered cubic crystal, the radius of the atom is found by dividing the edge length by \( 2\sqrt{2} \). Doing this math gives us the atomic radius of silver.

Exam Tip: Be careful to use the correct relation between r and a depending on whether it is fcc (\( r = \frac{a}{2\sqrt{2}} \)) or bcc (\( r = \frac{\sqrt{3}a}{4} \)).

Question. An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains \( 2 \times 10^{24} \) atoms.
Answer: Given data:
Edge length, \( a = 250 \text{ pm} = 250 \times 10^{-10} \text{ cm} = 2.5 \times 10^{-8} \text{ cm} \)
Number of atoms per unit cell for fcc, \( z = 4 \)
Mass of the sample, \( m = 300 \text{ g} \)
Total number of atoms in sample = \( 2 \times 10^{24} \) atoms

First, we find the molar mass (\( M \)) of the element (mass of \( 6.022 \times 10^{23} \) atoms):
\( M = \frac{300 \text{ g} \times 6.022 \times 10^{23}}{2 \times 10^{24}} = 90.33 \text{ g mol}^{-1} \)

Now, we calculate the density (\( d \)) using the formula:
\( d = \frac{z \times M}{a^3 \times N_A} \)

Substituting the values:
\( d = \frac{4 \times 90.33 \text{ g mol}^{-1}}{(2.5 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \)
\( d = \frac{361.32}{1.5625 \times 10^{-23} \times 6.022 \times 10^{23}} \)
\( d = \frac{361.32}{9.409} = 38.4 \text{ g cm}^{-3} \)
Hence, the density of the element is \( 38.4 \text{ g cm}^{-3} \).
In simple words: First we find the molar mass of the element using the given number of atoms. Then we use the density formula for fcc to get the density of the crystal.

Exam Tip: Be sure to convert picometers to centimeters properly when calculating density in \( \text{g cm}^{-3} \).

Question. An element crystallizes in a b.c.c. lattice with cell edge of 500 pm. The density of the element is \( 7.5 \text{ g cm}^{-3} \). How many atoms are present in 300 g of the element?
Answer: Given data:
For a body-centered cubic (bcc) structure, number of atoms per unit cell, \( z = 2 \)
Edge length, \( a = 500 \text{ pm} = 5 \times 10^{-8} \text{ cm} \)
Density, \( d = 7.5 \text{ g cm}^{-3} \)
Mass of the sample = \( 300 \text{ g} \)

First, we calculate the molar mass (\( M \)) of the element using the density formula:
\( d = \frac{z \times M}{a^3 \times N_A} \)

Rearranging for \( M \):
\( M = \frac{d \times a^3 \times N_A}{z} \)

Substituting the values:
\( M = \frac{7.5 \text{ g cm}^{-3} \times (5 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{2} \)
\( M = \frac{7.5 \times 125 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \)
\( M = \frac{7.5 \times 125 \times 10^{-1} \times 6.022}{2} = 282.28 \text{ g mol}^{-1} \)

Now, we calculate the number of atoms in \( 300 \text{ g} \) of the element:
Since \( 282.28 \text{ g} \) of the element contains \( 6.022 \times 10^{23} \) atoms,
Then \( 300 \text{ g} \) of the element will contain:
\( \text{Number of atoms} = \frac{6.022 \times 10^{23}}{282.28} \times 300 \)
\( \text{Number of atoms} = 6.40 \times 10^{23} \text{ atoms} \)
In simple words: We first find the molar mass of the element using its density and cell edge. Then, we use the mass of the sample to find the total number of atoms in it.

Exam Tip: Be precise with powers of ten in calculations, especially when dealing with cell edge in pm cubed.

Question. If NaCl is doped with \( 10^{-3} \text{ mol \%} \) of \( \text{SrCl}_2 \), what is the concentration of cation vacancies?
Answer: Given concentration of \( \text{SrCl}_2 = 10^{-3} \text{ mol \%} = \frac{10^{-3}}{100} \text{ mol} = 10^{-5} \text{ mol} \).
Since doping with one mole of \( \text{Sr}^{2+} \) ions creates one mole of cation vacancies to maintain charge balance,
The number of vacancies per mole of doped salt is equal to the number of moles of \( \text{SrCl}_2 \).

Therefore, the concentration of cation vacancies is:
\( 10^{-5} \text{ mol} \times 6.022 \times 10^{23} \text{ vacancies mol}^{-1} = 6.022 \times 10^{18} \text{ vacancies} \)
Thus, the total concentration of cation vacancies is \( 6.022 \times 10^{18} \text{ vacancies} \).
In simple words: Adding \( \text{Sr}^{2+} \) ions into \( \text{NaCl} \) creates empty spots (vacancies) because each \( \text{Sr}^{2+} \) replaces two \( \text{Na}^+ \) ions but only takes up one spot. We calculate the number of these vacancies by multiplying the moles of the dopant by Avogadro's number.

Exam Tip: Remember that adding a divalent cation like \( \text{Sr}^{2+} \) to a monovalent ionic solid like \( \text{NaCl} \) always creates one vacancy per added cation.

Question. Silver crystallises in f.c.c. lattice. If edge length of the cell is \( 4.077 \times 10^{-8} \text{ cm} \) and density is \( 10.5 \text{ g cm}^{-3} \), calculate the atomic mass of silver.
Answer: Given data:
Density, \( d = 10.5 \text{ g cm}^{-3} \)
Edge length, \( a = 4.077 \times 10^{-8} \text{ cm} \)
For an fcc structure, \( z = 4 \)
Avogadro's number, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)

Using the density formula:
\( d = \frac{z \times M}{a^3 \times N_A} \)

Rearranging the formula to find the atomic mass (\( M \)):
\( M = \frac{d \times a^3 \times N_A}{z} \)

Substituting the values:
\( M = \frac{10.5 \text{ g cm}^{-3} \times (4.077 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{4} \)
\( M = \frac{10.5 \times 67.77 \times 10^{-24} \times 6.022 \times 10^{23}}{4} \)
\( M = 106.6 \text{ g mol}^{-1} \)
Hence, the atomic mass of silver is \( 106.6 \text{ g mol}^{-1} \).
In simple words: We rearrange the standard crystal density formula to solve for the atomic mass of silver using the given density and unit cell dimensions.

Exam Tip: Be sure to carry out the cubing of the edge length carefully before performing the rest of the multiplication.

Question. (a) Based on the nature of intermolecular forces, classify the following solids: Silicon carbide, Argon
(b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example.
Answer:
(a) Silicon carbide is classified as a covalent network solid, whereas Argon belongs to non-polar molecular solids.
(b) Zinc oxide (\( \text{ZnO} \)) exhibits a metal excess defect due to extra zinc ions (\( \text{Zn}^{2+} \)) occupying interstitial positions. When heated, it loses oxygen and turns yellow due to the trapping of released electrons in the crystal lattice:
\[ \text{ZnO} \xrightarrow{\Delta} \text{Zn}^{2+} + \frac{1}{2}\text{O}_2 + 2\text{e}^- \]
(c) Group 12-16 compounds refer to semi-covalent materials formed by combining Group 12 and Group 16 elements, where the ionic character is determined by the electronegativity difference between the elements (e.g., \( \text{ZnS} \), \( \text{CdS} \)).
In simple words: Silicon carbide is a strong network solid like diamond, while argon is held by weak forces. Heating zinc oxide makes it lose oxygen, leaving extra zinc ions and electrons that turn it yellow. Group 12-16 compounds are pairs of elements from those groups with partly ionic, partly covalent bonds.

Exam Tip: Be sure to write the balanced chemical equation for the heating of \( \text{ZnO} \) to score full marks in part (b).

Question. (a) Based on the nature of intermolecular forces, classify the following solids: Benzene, Silver
(b) AgCl shows Frenkel defect while NaCl does not. Give reason.
(c) What type of semiconductor is formed when Ge is doped with Al?
Answer:
(a) Benzene is a non-polar molecular solid, while Silver is categorized as a metallic solid.
(b) The \( \text{Ag}^+ \) ion is small enough to fit into the empty interstitial spaces of the crystal, whereas the \( \text{Na}^+ \) ion is too large to do so. Hence, \( \text{AgCl} \) undergoes a Frenkel defect while \( \text{NaCl} \) does not.
(c) A p-type semiconductor is formed when a tetravalent element like Germanium (\( \text{Ge} \)) is doped with a trivalent impurity like Aluminum (\( \text{Al} \)).
In simple words: Benzene is a molecular solid, and silver is a metal. Silver ions can easily slip into interstitial spaces because they are small, but sodium ions cannot. Doping germanium with aluminum creates electron "holes," making a p-type semiconductor.

Exam Tip: Always compare the sizes of cations to explain why some compounds exhibit Frenkel defects instead of Schottky defects.

Question. (a) Based on the nature of intermolecular forces, classify the following solids: Sodium sulphate, Hydrogen
(b) What happens when CdCl2 is doped with AgCl?
(c) Why do ferrimagnetic substances show better magnetism than antiferromagnetic substances?
Answer:
(a) Sodium sulphate is classified as an ionic solid, while Hydrogen is a non-polar molecular solid.
(b) Doping \( \text{AgCl} \) with \( \text{CdCl}_2 \) introduces divalent \( \text{Cd}^{2+} \) ions. To preserve neutral charge, each \( \text{Cd}^{2+} \) ion displaces two monovalent \( \text{Ag}^+ \) ions. One site is occupied by the \( \text{Cd}^{2+} \) ion, while the other site remains empty, generating a cation vacancy (impurity defect).
(c) In ferrimagnetic substances, the magnetic dipole domains are aligned in opposite directions in unequal proportions, which yields a net magnetic moment. On the other hand, antiferromagnetic substances have domains oriented in opposite directions in equal numbers, totally canceling out any net magnetism.
In simple words: Sodium sulphate has ionic bonds, and hydrogen has simple non-polar molecular bonds. Adding cadmium ions to silver chloride knocks out twice as many silver ions, leaving empty spots in the lattice. Ferrimagnetic materials are magnetic because their domain spins do not completely cancel each other out, unlike antiferromagnetic ones which cancel out perfectly.

Exam Tip: When explaining magnetic properties, draw or mention the alignment of domain arrows to make your answer highly clear.

Question. An element crystallises in b.c.c. lattice with cell edge of 400 pm. Calculate its density if 500 g of this element contains \( 2.5 \times 10^{24} \) atoms.
Answer: Given data:
Edge length, \( a = 400 \text{ pm} = 400 \times 10^{-10} \text{ cm} = 4 \times 10^{-8} \text{ cm} \)
For a bcc lattice, number of atoms per unit cell, \( Z = 2 \)
Mass of sample = \( 500 \text{ g} \)
Number of atoms in the sample = \( 2.5 \times 10^{24} \)

First, we calculate the molar mass (\( M \)) of the element (mass of \( 6.022 \times 10^{23} \) atoms):
\( M = \frac{500 \text{ g} \times 6.022 \times 10^{23}}{2.5 \times 10^{24}} = 120.44 \text{ g mol}^{-1} \)

Now, we calculate density (\( d \)) using the formula:
\( d = \frac{Z \times M}{a^3 \times N_A} \)

Substituting the values:
\( d = \frac{2 \times 120.44 \text{ g mol}^{-1}}{(4 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \)
\( d = \frac{240.88}{64 \times 10^{-24} \times 6.022 \times 10^{23}} \)
\( d = \frac{240.88}{38.5408} = 6.25 \text{ g cm}^{-3} \)
Thus, the density of the element is \( 6.25 \text{ g cm}^{-3} \).
In simple words: First we find the molar mass of the element using the mass of the given number of atoms. Then, we substitute this molar mass along with the edge length into the bcc density formula to find the final density.

Exam Tip: Pay close attention to unit conversions—expressing edge length in cm ensures that the density is calculated in \( \text{g cm}^{-3} \).

Question. An element crystallises in fcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain \( 2.5 \times 10^{24} \) atoms.
Answer: Given data:
Edge length, \( a = 400 \text{ pm} = 4 \times 10^{-8} \text{ cm} \)
For an fcc structure, number of atoms per unit cell, \( Z = 4 \)
Mass of sample = \( 250 \text{ g} \)
Number of atoms in sample = \( 2.5 \times 10^{24} \)

First, we calculate the molar mass (\( M \)) of the element:
\( M = \frac{250 \text{ g} \times 6.022 \times 10^{23}}{2.5 \times 10^{24}} = 60.22 \text{ g mol}^{-1} \)

Now, we calculate the density (\( d \)):
\( d = \frac{Z \times M}{a^3 \times N_A} \)

Substituting the values:
\( d = \frac{4 \times 60.22 \text{ g mol}^{-1}}{(4 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \)
\( d = \frac{240.88}{64 \times 10^{-24} \times 6.022 \times 10^{23}} \)
\( d = \frac{240.88}{38.5408} = 6.25 \text{ g cm}^{-3} \)
Thus, the density of the element is \( 6.25 \text{ g cm}^{-3} \).
In simple words: We find the mass of one mole of atoms first. Then we calculate the density of the face-centered cubic structure using the density formula.

Exam Tip: Double-check the value of Z for fcc (which is 4) when working through numerical problems on crystal density.

Question. An element crystallises in bcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contains \( 2.5 \times 10^{24} \) atoms.
Answer: Given data:
Edge length, \( a = 400 \text{ pm} = 4 \times 10^{-8} \text{ cm} \)
For a bcc structure, number of atoms per unit cell, \( Z = 2 \)
Mass of sample = \( 250 \text{ g} \)
Number of atoms in sample = \( 2.5 \times 10^{24} \)

First, we calculate the molar mass (\( M \)) of the element:
\( M = \frac{250 \text{ g} \times 6.022 \times 10^{23}}{2.5 \times 10^{24}} = 60.22 \text{ g mol}^{-1} \)

Now, we calculate the density (\( d \)):
\( d = \frac{Z \times M}{a^3 \times N_A} \)

Substituting the values:
\( d = \frac{2 \times 60.22 \text{ g mol}^{-1}}{(4 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}} \)
\( d = \frac{120.44}{64 \times 10^{-24} \times 6.022 \times 10^{23}} \)
\( d = \frac{120.44}{38.5408} = 3.125 \text{ g cm}^{-3} \)
Thus, the density of the element is \( 3.125 \text{ g cm}^{-3} \).
In simple words: We find the molar mass first. Since the crystal is body-centered, we use Z = 2 in our density calculation to find the final density.

Exam Tip: Notice how the density of a bcc lattice is exactly half that of an fcc lattice when the edge length and molar mass are identical.

Question. An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is \( 7.2 \text{ g/cm}^3 \).
Answer: Given data:
For a bcc unit cell, number of atoms per unit cell, \( Z = 2 \)
Edge length, \( a = 288 \text{ pm} = 288 \times 10^{-10} \text{ cm} = 2.88 \times 10^{-8} \text{ cm} \)
Density, \( d = 7.2 \text{ g/cm}^3 \)
Avogadro's number, \( N_A = 6.022 \times 10^{23} \text{ mol}^{-1} \)

Using the density formula:
\( d = \frac{Z \times M}{a^3 \times N_A} \)

Rearranging the formula to find molar mass (\( M \)):
\( M = \frac{d \times a^3 \times N_A}{Z} \)

Substituting the values:
\( M = \frac{7.2 \text{ g/cm}^3 \times (2.88 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{2} \)
\( M = \frac{7.2 \times 23.89 \times 10^{-24} \times 6.022 \times 10^{23}}{2} \)
\( M = \frac{7.2 \times 6.022 \times 2.39 \times 10^{-1}}{2} \)
\( M = \frac{103.626}{2} = 51.8 \text{ g mol}^{-1} \)
Hence, the molar mass of the element is \( 51.8 \text{ g mol}^{-1} \).
In simple words: We find the molar mass of the element by using the density formula for a body-centered cubic cell and rearranging it to solve for M.

Exam Tip: Be very careful when cubing decimal values like 2.88 to avoid rounding errors early in the steps.

Long Answer Type Questions (LA) (5 Marks)

Question. (a) An element has an atomic mass \( 93 \text{ g mol}^{-1} \) and density \( 11.5 \text{ g cm}^{-3} \). If the edge length of its unit cell is 300 pm, identify the type of unit cell.
(b) Write any two differences between amorphous solids and crystalline solids.
Answer:
(a) Given data:
Atomic mass, \( M = 93 \text{ g mol}^{-1} \)
Density, \( d = 11.5 \text{ g cm}^{-3} \)
Edge length, \( a = 300 \text{ pm} = 3 \times 10^{-8} \text{ cm} \)

Using the formula for finding number of atoms per unit cell (\( Z \)):
\( Z = \frac{d \times a^3 \times N_A}{M} \)

Substituting the values:
\( Z = \frac{11.5 \text{ g cm}^{-3} \times (3 \times 10^{-8} \text{ cm})^3 \times 6.022 \times 10^{23} \text{ mol}^{-1}}{93 \text{ g mol}^{-1}} \)
\( Z = \frac{11.5 \times 27 \times 10^{-24} \times 6.022 \times 10^{23}}{93} \)
\( Z = 2.01 \approx 2 \)
Since the value of \( Z \) is nearly equal to 2, the unit cell is a body-centered cubic (BCC) lattice.

(b) The main differences are:

In simple words: We calculate Z and find it is about 2, indicating a body-centered cubic unit cell. Crystalline solids are neat and highly ordered with directional properties, while amorphous solids are disordered and uniform in all directions.
Exam Tip: Be sure to write physical characteristics such as order type and isotropy clearly when asked to differentiate between crystalline and amorphous solids.

Question. (a) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a f.c.c. structure. (Atomic mass of Al = \( 27 \text{ g mol}^{-1} \))
(b) Give reasons:
(i) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect.
(ii) Silicon on doping with Phosphorus form n-type semiconductor.
(iii) Ferrimagnetic substances show better magnetism than antiferromagnetic substances.
Answer:
(a) Given data:
Mass of aluminum, \( m = 8.1 \text{ g} \)
Molar mass of aluminum, \( M = 27 \text{ g mol}^{-1} \)

First, we calculate the total number of aluminum atoms:
\( \text{Number of atoms} = \frac{m}{M} \times N_A \)
\( \text{Number of atoms} = \frac{8.1 \text{ g}}{27 \text{ g mol}^{-1}} \times 6.022 \times 10^{23} \text{ atoms mol}^{-1} \)
\( \text{Number of atoms} = 0.3 \times 6.022 \times 10^{23} = 1.8066 \times 10^{23} \text{ atoms} \)

An fcc unit cell contains 4 atoms. Therefore, the number of unit cells is:
\( \text{Number of unit cells} = \frac{1.8066 \times 10^{23}}{4} = 4.5 \times 10^{22} \text{ unit cells} \)

(b) Reasons:
(i) Schottky defects occur when the size differences between positive and negative ions are very small. In contrast, Frenkel defects appear when there is a significant size difference between ions. Since \( \text{Na}^+ \) and \( \text{Cl}^- \) are comparable in size, \( \text{NaCl} \) displays Schottky defects instead of Frenkel defects.
(ii) Phosphorus is a pentavalent element with 5 valence electrons. When added to silicon (which has 4 valence electrons), the extra electron remains unbound and is free to migrate, giving rise to an n-type semiconductor with high electrical conductivity.
(iii) In antiferromagnetic materials, domain magnetic dipoles are arranged in equal numbers in opposite directions, canceling out the overall magnetic moment. In ferrimagnetic substances, the domain dipoles align in opposite directions in unequal numbers, preserving a net magnetic field.
In simple words: First we calculate total atoms in the sample, then divide by 4 to get the unit cells. Sodium and chloride ions are similar in size, which causes Schottky defects. Phosphorus adds an extra free electron to silicon to create an n-type semiconductor. In ferrimagnetic substances, some spins point in opposite directions but don't cancel out fully, leaving a net magnetic force.

Exam Tip: Be sure to write out the mathematical steps in part (a) clearly, identifying the value of Z for fcc before dividing.