CBSE Class 12 Chemistry Solid State Assignment Set 04

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-I

Question 65. How will you distinguish between the following pairs of terms:
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell
Answer: (i) Tetrahedral and octahedral voids: A tetrahedral void is formed when four spheres are in contact with each other in a tetrahedral shape, making its coordination number 4. On the other hand, an octahedral void is created by six surrounding spheres, which gives it a coordination number of 6. Also, a tetrahedral void is smaller in size than an octahedral void.
(ii) Crystal lattice and unit cell: A crystal lattice is the complete three-dimensional pattern showing how particles are arranged in space. In contrast, a unit cell is the smallest portion of this lattice that, when repeated in different directions, builds up the whole crystal structure.
In simple words: A tetrahedral void is surrounded by 4 atoms, while an octahedral void is surrounded by 6 atoms. A crystal lattice is the complete grid of particles, while a unit cell is just one repeating block of that grid.
Exam Tip: Draw small diagrams or indicate the coordination numbers (4 for tetrahedral, 6 for octahedral) to clearly display the structural differences.

Question. How will you distinguish between the following pairs of terms :
(i) Tetrahedral and octahedral voids
(ii) Crystal lattice and unit cell (All India)
Answer:
(i) The differences between tetrahedral and octahedral voids are as follows:

(ii) A systematic order of the building particles of a crystal in three-dimensional space is named a crystal lattice.
The tiniest three-dimensional segment of a full crystal lattice, which generates the entire crystal lattice when repeated repeatedly in various directions, is called the unit cell.
In simple words: Tetrahedral voids are small gaps touched by 4 atoms, while octahedral voids are larger gaps touched by 6 atoms. A crystal lattice is the full 3D pattern of a crystal, and a unit cell is its smallest repeating building block.

Exam Tip: Highlighting the coordination numbers (4 for tetrahedral and 6 for octahedral) is crucial for scoring full marks when comparing these voids.

Question. (i) Write the type of magnetism observed when the magnetic moments are oppositively aligned and cancel out each other.
(ii) Which stoichiometric defect does not change the density of the crystal? (All India)
Answer:
(i) Diamagnetism is detected when the magnetic dipoles are aligned in opposite directions and completely neutralize each other.
(ii) Frenkel defect does not alter the density of the solid.
In simple words: Diamagnetism happens when magnetic forces point in opposite ways and cancel out. The Frenkel defect moves an ion to a different spot but keeps it inside the crystal, so the overall weight and density stay the same.

Exam Tip: Remember that Frenkel defect is a dislocation defect, meaning ions just change their positions, so density is conserved.

Question. (i) Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers.
(ii) Which stoichiometric defect decreases the density of the crystal? (All India)
Answer:
(i) Ferrimagnetism is seen under these conditions.
(ii) Schottky defect lowers the density of the solid.
In simple words: Ferrimagnetism happens when some magnetic parts point up and some point down but they do not balance out, leaving a small magnetic force. A Schottky defect leaves empty gaps where atoms should be, making the crystal less dense.

Exam Tip: Associate Schottky defects with missing pairs of cations and anions, which leads directly to a loss in density.

Question. Define the following terms: (Comptt. Delhi)
(i) n-type semiconductor
(ii) Ferrimagnetism
Answer:
(i) n-type semiconductor: This is formed when silicon or germanium is doped with an impurity from group 15.
(ii) Ferrimagnetism: This occurs when the magnetic domains are pointing in parallel and anti-parallel directions in unequal amounts.
In simple words: An n-type semiconductor is made by adding group 15 atoms to silicon or germanium to get extra moving electrons. Ferrimagnetism is when tiny magnets inside a material point in opposite directions but don't completely cancel each other out.

Exam Tip: Draw simple domain diagrams showing unequal numbers of upward and downward arrows to illustrate ferrimagnetism clearly.

Question. Explain the following terms with suitable examples : (Comptt. All India)
(i) Frenkel defect (ii) F-centres
Answer:
(i) Frenkel defect: A type of defect where the smaller ion, usually a cation, moves from its normal position to a close interstitial space. Examples include ZnS and silver halides.
(ii) F-centres: An anionic gap that is filled by an unpaired electron is named an F-centre in alkali metal halides. Examples are KCl, NaCl, and LiCl.
In simple words: A Frenkel defect is when an atom slips out of its correct spot and squeezes into an empty space nearby. An F-centre is a missing negative ion spot that gets trapped with a free electron, which makes the crystal colorful.

Exam Tip: Clearly distinguish that Frenkel defect occurs in ionic compounds with a large difference in size between cations and anions.

Question. Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a face-centered cubic (f.c.c.) structure. (Atomic mass of Al = 27 g mol–1) (Comptt. All India)
Answer: We know that 1 mole of Aluminium equals 27 g, which contains \( 6.022 \times 10^{23} \) atoms.
Therefore, the count of atoms in 27 g of Al is:
\( \text{No. of atoms} = \frac{6.022 \times 10^{23}}{27} \)
Since a face-centered cubic (f.c.c.) unit cell consists of 4 atoms:
The number of f.c.c. unit cells in 8.1 g of Al is:
\[ \text{No. of unit cells} = \frac{6.022 \times 10^{23} \times 8.1}{27 \times 4} \]
\[ = 0.45165 \times 10^{23} = 4.5165 \times 10^{22} \]
In simple words: First, find the total number of aluminum atoms in 8.1 grams. Since each face-centered cubic unit cell is made of 4 atoms, divide the total atoms by 4 to get the total number of unit cells.

Exam Tip: Be precise with the value of atoms per unit cell based on the lattice type (Z = 4 for fcc, Z = 2 for bcc, Z = 1 for sc) to ensure correct calculations.

Short Answer Type Questions-II

Question. Iron has a body centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm–3. Use this information to calculate Avogadro’s number (At. mass of Fe = 56 g mol–1). (Delhi & All India)
Answer: From the given details:
Edge length \( a = 286.65\text{ pm} = 286.65 \times 10^{-10}\text{ cm} \)
Density \( d = 7.87\text{ g cm}^{-3} \)
Molar mass \( M = 56\text{ g mol}^{-1} \)
For a body-centred cubic (bcc) lattice, \( Z = 2 \)
Applying the relation:
\( d = \frac{Z \times M}{a^3 \times N_A} \)

\( \implies N_A = \frac{Z \times M}{a^3 \times d} \)
\[ N_A = \frac{2 \times 56}{(286.65 \times 10^{-10})^3 \times 7.87} \]
\[ N_A = \frac{112}{(2.87 \times 10^{-8})^3 \times 7.87} \]
\[ N_A = \frac{112}{23.63 \times 7.87 \times 10^{-24}} \]
\[ N_A = \frac{112}{185.97 \times 10^{-24}} \]
\[ N_A = \frac{112}{18.597} \times 10^{23} \]
\[ N_A = 6.022 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Use the density formula of a crystal. Put in the cell size, density, and mass of iron, then rearrange the formula to find Avogadro's number, which comes out to the standard constant.

Exam Tip: Always state the unit of Avogadro's number as \( \text{mol}^{-1} \) to secure full marks for the final calculated quantity.

Question. Silver crystallises with face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching the four corner atoms.) (All India)
Answer: The values given are:
Edge length \( a = 409\text{ pm} \)
For a face-centred cubic (fcc) unit cell, the connection between atomic radius \( r \) and edge length \( a \) is:
\[ r = \frac{a}{2\sqrt{2}} \]
Substituting the value of \( a \):
\[ r = \frac{409}{2 \times 1.414} = \frac{409}{2.828} \]
\[ r = 144.62\text{ pm} \]
Hence, the radius of the silver atom is 144.62 pm.
In simple words: For a face-centered cubic system, the radius of an atom is the side length divided by two times the square root of two. Just plug in 409 and solve.

Exam Tip: Writing down the general formula for the relation between radius and edge length before doing calculations is highly recommended by examiners.

Question. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F– ions and that Ca2+ ions are arranged in a fcc lattice. The F– ions fill all the tetrahedral holes in the face centred cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 × 10 –8 cm in length. The density of the solid is 3.18 g cm–3. Use this information to calculate Avogadro’s number (Molar mass of CaF2 = 78.08 g mol–1). (Delhi)
Answer: We have the following data:
Edge length of unit cell \( a = 5.46 \times 10^{-8}\text{ cm} \)
Density \( \rho = 3.18\text{ g cm}^{-3} \)
Number of formula units in a unit cell \( Z = 4 \) (since there are 4 \( \text{Ca}^{2+} \) and 8 \( \text{F}^- \) ions)
Molar mass of \( \text{CaF}_2 \), \( M = 78.08\text{ g mol}^{-1} \)
According to the crystal density relationship:
\( \rho = \frac{Z \times M}{a^3 \times N_A} \)

\( \implies N_A = \frac{Z \times M}{a^3 \times \rho} \)
Substituting the values:
\[ N_A = \frac{4 \times 78.08}{(5.46 \times 10^{-8})^3 \times 3.18} \]
\[ N_A = \frac{312.32}{1.628 \times 10^{-22} \times 3.18} \]
\[ N_A = 6.03 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Since there are 4 calcium fluoride formula units in the unit cell, use Z = 4. Put all the numbers into the density equation to find Avogadro's number.

Exam Tip: For calcium fluoride (fluorite structure), always set \( Z = 4 \) because the unit cell consists of 4 \( \text{CaF}_2 \) molecules.

Question. The density of copper metal is 8.95 g cm–3. If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure?
(Given : At. mass of Cu = 63.54 g mol–1 and NA = 6.02 × 1023 mol–1) (Delhi & All India)
Answer: Let us analyze each of the three potential crystal structures:

Case 1: If the copper unit cell is simple cubic:
For simple cubic, \( Z = 1 \)
\( a = 2r = 2 \times 127.8\text{ pm} = 255.6\text{ pm} = 2.556 \times 10^{-8}\text{ cm} \)
Applying the density formula:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} = \frac{1 \times 63.54}{(2.556 \times 10^{-8})^3 \times 6.02 \times 10^{23}} \]
\[ \rho \approx 6.34\text{ g cm}^{-3} \]
Since the calculated density is not equal to the real density (\( 8.95\text{ g cm}^{-3} \ )), copper does not have a simple cubic structure.

Case 2: If the copper unit cell is body-centred cubic:
For bcc, \( Z = 2 \)
\[ a = \frac{4r}{\sqrt{3}} = \frac{4 \times 127.8}{1.732} = 295.15\text{ pm} = 2.9515 \times 10^{-8}\text{ cm} \]
Applying the density formula:
\[ \rho = \frac{2 \times 63.54}{(2.9515 \times 10^{-8})^3 \times 6.02 \times 10^{23}} \]
\[ \rho \approx 8.21\text{ g cm}^{-3} \]
This value is also different from the actual density, so it is not body-centred cubic.

Case 3: If the copper unit cell is face-centred cubic:
For fcc, \( Z = 4 \)
\[ a = 2\sqrt{2}r = 2 \times 1.414 \times 127.8 = 361.4\text{ pm} = 3.614 \times 10^{-8}\text{ cm} \ ]
Applying the density formula:
\[ \rho = \frac{4 \times 63.54}{(3.614 \times 10^{-8})^3 \times 6.02 \times 10^{23}} \]
\[ \rho \approx 8.94\text{ g cm}^{-3} \]
Since this matches the actual density (\( 8.95\text{ g cm}^{-3} \ )), we conclude that copper crystallizes in a face-centred cubic structure.
In simple words: Test each of the three cell types (simple cubic, bcc, and fcc). Find the edge length for each type from the radius, then calculate the density. The type that gives a density close to 8.95 is fcc.

Exam Tip: Showing the step-by-step verification for at least two structures is crucial to get full marks for logical proof in descriptive exams.

Question. Silver crystallises in face-centred cubic unit cells. Each side of the unit cell has a length of 409 pm. What is the radius of silver atom? (All India)
Answer: Given that the edge length \( a = 409\text{ pm} \).
Since silver has a face-centred cubic (fcc) lattice, the atomic radius \( r \) is calculated as:
\[ r = \frac{a}{2\sqrt{2}} \]
Substituting \( a = 409\text{ pm} \):
\[ r = \frac{409}{2 \times 1.414} = \frac{409}{2.828} \]
\[ r = 144.62\text{ pm} \]
Therefore, the atomic radius of silver is 144.62 pm.
In simple words: For a face-centered cubic structure, divide the edge length of the unit cell by two times the square root of two to find the radius of the atom.

Exam Tip: Always make sure to write down the formula before inserting numerical values to prevent loss of partial credit.

Question. Silver crystallizes in face-centered cubic unit cell. Each side of this unit cell has a length of 400 pm. Calculate the radius of the silver atom. (Assume the atoms just touch each other on the diagonal across the face of the unit cell. That is each face atom is touching the four corner atoms.) (Delhi)
Answer: Given:
Edge length of the unit cell \( a = 400\text{ pm} \)
For an fcc structure, we calculate the radius \( r \) using:
\[ r = \frac{a}{2\sqrt{2}} \]
Substituting the value:
\[ r = \frac{400}{2 \times 1.4142} = \frac{400}{2.828} \]
\[ r \approx 141.4\text{ pm} \]
Thus, the silver atom's radius is 141.4 pm.
In simple words: Use the formula for a face-centered cubic cell, where the atomic radius is the side length divided by two times the square root of two. Plug in 400 pm to find the radius.

Exam Tip: FCC geometry assumes the atoms touch along the face diagonal, which leads directly to the relationship \( 4r = \sqrt{2}a \).

Question. The density of lead is 11.35 g cm–3 and the metal crystallizes with fcc unit cell. Estimate the radius of lead atom.
(At. Mass of lead = 207 g mol–1 and NA = 6.02 × 1023 mol–1) (Delhi)
Answer: We are given the following information:
Density \( d = 11.35\text{ g cm}^{-3} \)
For an fcc structure, \( Z = 4 \)
Applying the equation for density:
\[ a^3 = \frac{Z \times M}{d \times N_A} \]
Substituting the values:
\[ a^3 = \frac{4 \times 207}{11.35 \times 6.02 \times 10^{23}} \]
\[ a^3 = \frac{828}{68.327 \times 10^{23}} = 12.118 \times 10^{-23}\text{ cm}^3 \]
\[ a^3 = 1.212 \times 10^{-22}\text{ cm}^3 \]
Taking the cube root of this value:
\[ a = (1.212 \times 10^{-22})^{1/3} = 4.95 \times 10^{-8}\text{ cm} \]
For an fcc structure, the relation for atomic radius is:
\[ r = \frac{a}{2\sqrt{2}} \]
\[ r = \frac{4.95 \times 10^{-8}}{2 \times 1.414} = \frac{4.95 \times 10^{-8}}{2.828} \]
\[ r = 1.75 \times 10^{-8}\text{ cm} = 175\text{ pm} \]
In simple words: First, use the density formula to find the volume of the unit cell, and take its cube root to get the edge length. Then, use the fcc radius formula to find the atomic radius.

Exam Tip: Convert the final radius from centimeters to picometers (\( 1\text{ pm} = 10^{-10}\text{ cm} \)) to present the answer in a standard atomic scale unit.

Question. Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom? (Delhi)
Answer: For a body-centred cubic (bcc) crystal structure, the atomic radius relation is:
\[ r = \frac{\sqrt{3}}{4}a \]
Given that the cell edge length \( a = 316.5\text{ pm} \):
\[ r = \frac{\sqrt{3} \times 316.5}{4} \]
\[ r = \frac{1.732 \times 316.5}{4} \]
\[ r = 137\text{ pm} \]
So, the atomic radius of tungsten is 137 pm.
In simple words: For a body-centered cubic cell, multiply the edge length by the square root of three, then divide by four to find the radius.

Exam Tip: BCC calculations require using \( \sqrt{3} \approx 1.732 \), so be sure to memorize this value to speed up your calculations.

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. (At. mass of Fe = 55.845 u) (Delhi)
Answer: We use the standard density relationship:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
For a body-centred cubic (bcc) system, the number of atoms per unit cell is \( Z = 2 \).
Given details:
Edge length \( a = 286.65\text{ pm} = 2.8665 \times 10^{-8}\text{ cm} \)
Density \( d = 7.874\text{ g cm}^{-3} \)
Molar mass \( M = 55.845\text{ g mol}^{-1} \)
Rearranging the equation to solve for \( N_A \):
\[ N_A = \frac{2 \times 55.845}{(2.8665 \times 10^{-8})^3 \times 7.874} \]
\[ N_A = 6.02 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Rearrange the density formula to find Avogadro's number. Substitute the given mass, density, and cube of the edge length to get the standard value.

Exam Tip: Pay attention to the distinction between atomic mass in 'u' and molar mass in 'g/mol' when representing variables in the density formula.

Question. Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s number NA = 6.02 × 1023 mol–1) (All India)
Answer: For a face-centred cubic (fcc) lattice, the number of atoms per unit cell is \( Z = 4 \).
The standard formula for density is:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
For fcc geometry, the relation between the cell edge \( a \) and atomic radius \( r \) is:
\[ a = 2\sqrt{2}r \]
Given the radius of the copper atom \( r = 127.8\text{ pm} = 1.278 \times 10^{-8}\text{ cm} \):
\[ a = 2 \times 1.414 \times 1.278 \times 10^{-8}\text{ cm} \approx 3.614 \times 10^{-8}\text{ cm} \]
\[ a^3 = (3.614 \times 10^{-8})^3 \approx 4.723 \times 10^{-23}\text{ cm}^3 \]
Now, inserting these values into the density equation:
\[ d = \frac{4 \times 63.55}{4.723 \times 10^{-23} \times 6.02 \times 10^{23}} \]
\[ d = \frac{254.2}{28.43} \approx 8.94\text{ g cm}^{-3} \]
In simple words: First, find the edge length of the unit cell using the given radius for an fcc structure. Then, use that edge length along with Z = 4 to find the density.

Exam Tip: Make sure to convert atomic radius to centimeters before calculating the volume to get the density directly in \( \text{g cm}^{-3} \).

Question. Iron has a body centred cubic unit cell with the cell dimension of 286.65 pm. Density of iron is 7.87 g cm–3. Use this information to calculate Avogadro’s number. (Atomic mass of Fe = 56.0 u) (All India)
Answer: We use the density formula for a crystal system:
\[ d = \frac{Z \times M}{a^3 \times N_A} \]
For a body-centred cubic (bcc) system, the value of \( Z \) is 2.
Rearranging the equation to find Avogadro's number \( N_A \):
\[ N_A = \frac{Z \times M}{a^3 \times d} \]
Given values:
Edge length \( a = 286.65\text{ pm} = 2.8665 \times 10^{-8}\text{ cm} \)
Density \( d = 7.87\text{ g cm}^{-3} \)
Molar mass \( M = 56.0\text{ g mol}^{-1} \)
Substituting these values:
\[ N_A = \frac{2 \times 56.0}{(2.8665 \times 10^{-8})^3 \times 7.87} \]
\[ N_A \approx 6.043 \times 10^{23}\text{ mol}^{-1} \]
In simple words: Rearrange the density formula to find Avogadro's number. Put in the mass of iron, the cell size, and the density of the crystal to calculate the result.

Exam Tip: Make sure your exponent in the denominator is handled properly: \( (10^{-8})^3 = 10^{-24} \), which becomes \( 10^{24} \) in the numerator.

Question. (a) Some of the glass objects recovered from ancient monuments look milky instead of being transparent. Why?
(b) Iron (ΙΙ) oxide has a cubic structure and each side of the unit cell is 5 Å. If density of the oxide is 4 g cm–3, calculate the number of Fe2+ and O2– ions present in each unit cell.
[Atomic mass : Fe = 56 u, O = 16 u; Avagadro's number = 6.023 × 1023 mol–1] (Comptt. All India)
Answer:
(a) A few of the glass items recovered from ancient structures exhibit a milky look due to the slow crystallization of glass over time.
(b) Let us find the volume and mass of a single unit cell:
Edge length of the cell \( a = 5\text{ Å} = 5 \times 10^{-8}\text{ cm} \)
Volume of the unit cell \( a^3 = (5 \times 10^{-8}\text{ cm})^3 = 1.25 \times 10^{-22}\text{ cm}^3 \)
Given density of FeO \( \rho = 4\text{ g cm}^{-3} \)
Total mass of one unit cell:
\[ \text{Mass of unit cell} = \text{Volume} \times \text{Density} = 1.25 \times 10^{-22}\text{ cm}^3 \times 4\text{ g cm}^{-3} = 5 \times 10^{-22}\text{ g} \]
Now, the mass of a single FeO unit (molar mass \( M = 56 + 16 = 72\text{ g mol}^{-1} \)) is:
\[ \text{Mass of one FeO unit} = \frac{72}{6.023 \times 10^{23}} = 1.195 \times 10^{-22}\text{ g} \ ]
Number of FeO formula units per unit cell:
\[ Z = \frac{5 \times 10^{-22}\text{ g}}{1.195 \times 10^{-22}\text{ g}} \approx 4.19 \approx 4 \]
Therefore, each unit cell contains \( 4\text{ Fe}^{2+} \) ions and \( 4\text{ O}^{2-} \) ions.
In simple words: (a) Ancient glass looks cloudy because it has slowly formed tiny crystal structures over hundreds of years. (b) Find the mass of one unit cell by multiplying its volume by its density, then divide it by the mass of a single FeO molecule to find that there are 4 of each ion inside.

Exam Tip: Since \( Z \approx 4 \), this confirms that iron(II) oxide has a rock-salt type (fcc) structure where both ions have a coordination number of 6.

Question. (a) What are intrinsic semi-conductors? Give an example.
(b) What is the distance between Na+ and Cl– ions in NaCl crystal if its density is 2.165 g cm–3? [Atomic Mass of Na = 23u, Cl = 35.5u; Avogadro's number = 6.023×1023] (Comptt. All India)
Answer:
(a) Intrinsic semiconductors: These substances act as electrical insulators under room temperature conditions but transform into semiconductors when the temperature is increased. Examples include silicon and germanium.
(b) Let us apply the crystal density formula:
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \]
For an NaCl crystal, the number of formula units per fcc unit cell is \( Z = 4 \).
Molar mass of NaCl is \( M = 23 + 35.5 = 58.5\text{ g mol}^{-1} \).
Rearranging the equation to find \( a^3 \):
\[ a^3 = \frac{Z \times M}{\rho \times N_A} \]
\[ a^3 = \frac{4 \times 58.5}{2.165 \times 6.023 \times 10^{23}} \]
Taking the cube root of the volume:
\[ a = 564 \times 10^{-10}\text{ cm} = 564\text{ pm} \]
The relationship between the edge length \( a \) and the distance between adjacent \( \text{Na}^+ \) and \( \text{Cl}^- \) ions is:
\[ a = 2 \times d_{\text{Na}^+ - \text{Cl}^-} \]

\( \implies d_{\text{Na}^+ - \text{Cl}^-} = \frac{a}{2} = \frac{564\text{ pm}}{2} = 282\text{ pm} \]
In simple words: (a) Intrinsic semiconductors are pure materials that don't conduct electricity at room temperature but start conducting when they get hot. (b) Find the edge length of the NaCl cube using the density formula, then divide it by two because the edge length contains two ion-to-ion distances.

Exam Tip: Be sure to use \( Z = 4 \) for NaCl, as it possesses an fcc structure. Always state the final distance in picometers for neatness.