Maharashtra Board Class 12 Chemistry Chapter 1 Solid State Exercise Solutions

1. Choose The Most Correct Answer

Question i. Molecular solids are
(a) crystalline solids
(b) amorphous solids
(c) ionic solids
(d) metallic solids
Answer: (b) amorphous solids
In simple words: Molecular solids do not have a fixed, neat arrangement of particles. They are like a messy pile - not organized. That is why they are called amorphous (without shape).

📝 Teacher's Note: Show students a piece of glass and a piece of salt crystal. Glass is amorphous (no fixed shape inside). Salt crystal is crystalline (neat rows of particles). This makes the idea easy to see.

🎯 Exam Tip: Write "amorphous" clearly and explain it means no regular arrangement. Examiners look for this word.

Question ii. Which of the following is an n-type semiconductor?
(a) Pure Si
(b) Si-doped with As
(c) Si-doped with Ga
(d) Ge doped with In
Answer: (b) Si-doped with As
In simple words: As (arsenic) has 5 electrons in its outer shell. Si has 4. When As is added to Si, there is 1 extra electron which moves freely. This extra electron makes it n-type (negative type).

📝 Teacher's Note: Draw 5 dots around As and 4 dots around Si. Show students that the extra dot becomes a free electron. This visual helps them remember n-type means extra electrons.

🎯 Exam Tip: Remember As has 5 electrons, Si has 4. Extra electron = n-type. Write "donor impurity" for full marks.

Question iii. In Frenkel defect
(a) electrical neutrality of the substance is changed.
(b) density of the substance is changed.
(c) both cation and anion are missing
(d) overall electrical neutrality is preserved
Answer: (d) overall electrical neutrality is preserved
In simple words: In Frenkel defect, an ion just moves from its normal place to a gap. No ion is lost from the crystal. So the total positive and negative charges remain equal.

📝 Teacher's Note: Use coins to show this. Move one coin from a row to an empty space nearby. Total number of coins stays same. Same way, total charge stays same in Frenkel defect.

🎯 Exam Tip: Write "ion displacement, not removal" and "electrical neutrality preserved". These phrases get you marks.

Question iv. In crystal lattice formed by bcc unit cell the void volume is
(a) 68%
(b) 74%
(c) 32%
(d) 26%
Answer: (c) 32%
In simple words: In bcc (body-centred cubic), atoms touch along the body diagonal. This leaves 32% empty space between the atoms. Think of it like packing balls in a box - there are always gaps.

📝 Teacher's Note: Use ping pong balls in a box to show bcc packing. Students can see the empty spaces clearly. Tell them fcc has less empty space (26%), bcc has more (32%).

🎯 Exam Tip: Remember bcc void = 32%, fcc void = 26%. Write the percentage clearly with % symbol.

Question v. The coordination number of atoms in bcc crystal lattice is
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (d) 8
In simple words: Coordination number means how many atoms touch one atom. In bcc, the center atom touches 8 corner atoms. So coordination number is 8.

📝 Teacher's Note: Draw a cube with dots at corners and center. Show students that the center dot touches all 8 corner dots. This makes coordination number 8 very clear.

🎯 Exam Tip: Write "center atom touches 8 corner atoms" for explanation. Always write coordination number as a simple number, not a fraction.

Question vi. Which of the following is not correct ?
(a) Four spheres are involved in the formation of tetrahedral void.
(b) The centres of spheres in octahedral voids are at the apices of a regular tetrahedron.
(c) If the number of atoms is N the number of octahedral voids is 2N.
(d) If the number of atoms is N/2, the number of tetrahedral voids is N.
Answer: (c) If the number of atoms is N the number of octahedral voids is 2N.
In simple words: This statement is wrong. If there are N atoms, there are N octahedral voids (not 2N). The correct ratio is 1:1, not 1:2.

📝 Teacher's Note: Tell students the rule: For N atoms, octahedral voids = N, tetrahedral voids = 2N. Statement (c) mixes up these numbers.

🎯 Exam Tip: Remember the void formula: octahedral voids = N, tetrahedral voids = 2N for N atoms. Write this formula in your answer.

Question vii. A compound forms hcp structure. Number of octahedral and tetrahedral voids in 0.5 mole of substance is respectively
(a) \( 3.011 \times 10^{23} \), \( 6.022 \times 10^{23} \)
(b) \( 6.022 \times 10^{23} \), \( 3.011 \times 10^{23} \)
(c) \( 4.011 \times 10^{23} \), \( 2.011 \times 10^{23} \)
(d) \( 6.011 \times 10^{23} \), \( 12.022 \times 10^{23} \)
Answer: (a) \( 3.011 \times 10^{23} \), \( 6.022 \times 10^{23} \)
In simple words: 0.5 mole has \( 3.011 \times 10^{23} \) atoms. For these atoms: octahedral voids = same number of atoms = \( 3.011 \times 10^{23} \). Tetrahedral voids = 2 times atoms = \( 6.022 \times 10^{23} \).

📝 Teacher's Note: First find number of atoms in 0.5 mole. Then apply the void formula. Students often forget to start with finding the number of atoms first.

🎯 Exam Tip: Always write: "0.5 mole = \( 3.011 \times 10^{23} \) atoms" first. Then apply void rules. Show all calculation steps.

Question viii. Pb has fcc structure with edge length of unit cell 495 pm. Radius of Pb atom is
(a) 205 pm
(b) 185 pm
(c) 260 pm
(d) 175 pm
Answer: (d) 175 pm
In simple words: In fcc, atoms touch along the face diagonal. The formula is: face diagonal = 4r where r is radius. Face diagonal = \( a\sqrt{2} \) where a is edge length. So \( 4r = 495 \times \sqrt{2} \), which gives r = 175 pm.

📝 Teacher's Note: Draw the fcc face and show students how atoms touch along the diagonal. The formula \( 4r = a\sqrt{2} \) comes from this geometry.

🎯 Exam Tip: Write the formula \( 4r = a\sqrt{2} \) for fcc. Substitute values and solve. Always include units (pm) in final answer.

2. Answer The Following In One Or Two Sentences

Question i. What are the types of particles in each of the four main classes of crystalline solids?
Answer: The smallest constituents or particles of various solids are atoms, ions or molecules.
In simple words: Crystalline solids are made of tiny building blocks. These can be atoms (single elements), ions (charged particles), or molecules (groups of atoms stuck together).

📝 Teacher's Note: Give examples: Salt has ions (Na+ and Cl-), Diamond has atoms (carbon), Ice has molecules (H2O). This makes the types very clear for students.

🎯 Exam Tip: Write all three types: atoms, ions, molecules. Give one example of each type for extra marks.

Question ii. Which of the three types of packing used by metals makes the most efficient use of space and which makes the least efficient use ?
Answer: fcc has the most efficient packing of particles while scc has the least efficient packing.
In simple words: fcc (face-centered cubic) packs atoms very tightly with less empty space. scc (simple cubic) has more gaps between atoms, so it wastes more space.

📝 Teacher's Note: Use oranges to show this. Stack them in a simple square pattern (scc) and then in fcc pattern. Students can see which one uses space better.

🎯 Exam Tip: Write "fcc - most efficient" and "scc - least efficient". Also mention that bcc is in between these two.

Question iii. The following pictures show population of bands for materials having different electrical properties. Classify them as insulator, semiconductor or a metal.
Answer: Picture A represents metal conductor, Picture B represents insulator, Picture C represents semiconductor.
In simple words: Picture A shows overlapping bands (metal - electricity flows easily). Picture B shows big gap between bands (insulator - no electricity). Picture C shows small gap (semiconductor - some electricity with heat).

[Diagram: This diagram shows three energy band diagrams side by side showing different band arrangements for metal, insulator, and semiconductor materials.]

📝 Teacher's Note: Draw the three band diagrams on board. Show students that big gap = insulator, small gap = semiconductor, no gap = metal. This visual helps a lot.

🎯 Exam Tip: Always label each diagram clearly as A, B, C. Write the key difference: "gap size determines electrical property".

Question iv. What is a unit cell?
Answer: Unit cell : It is the smallest repeating structural unit of a crystalline solid (or crystal lattice) which when repeated in different directions produces the crystalline solid (lattice). The crystal is considered to consist of an infinite number of unit cells. The unit cell possesses all the characteristics of the crystalline solid.
In simple words: A unit cell is like one building block. When you repeat this block many times in all directions, you get the whole crystal. It is the smallest part that shows all properties of the big crystal.

📝 Teacher's Note: Use Lego blocks or building tiles. Show how one block repeated makes a big structure. The block is the unit cell, the structure is the crystal.

🎯 Exam Tip: Write "smallest repeating structural unit" and "contains all characteristics of crystal". These are key phrases examiners want to see.

Question v. How does electrical conductivity of a semiconductor change with temperature ? Why?
Answer: Since the energy difference between valence band and conduction band in semiconductor is not large, the electrons from valence band can be promoted to conduction by heating. Hence electrical conductivity of a semiconductor increases with temperature.
In simple words: When you heat a semiconductor, electrons get more energy. They jump from the lower band to the upper band. More electrons in upper band means better electricity flow.

📝 Teacher's Note: Compare with a ladder. Heating gives electrons energy to climb from lower step to upper step. More electrons on upper step means better conductivity.

🎯 Exam Tip: Write "conductivity increases with temperature" and explain using "band gap" concept. Always mention that heating promotes electrons to conduction band.

Question vi. The picture represents bands of MOs for Si. Label valence band, conduction band and band gap.
Answer: In the given diagram, the lower filled band is the valence band, the upper empty band is the conduction band, and the gap between them is the band gap.
In simple words: The bottom filled box is valence band (where electrons normally stay). The top empty box is conduction band (where electrons go for electricity). The space between them is band gap.

[Diagram: This diagram shows the molecular orbital bands for silicon with a valence band (filled) at bottom, band gap in middle, and conduction band (empty) at top.]

📝 Teacher's Note: Draw three horizontal lines on board. Bottom line = valence (filled with electrons), middle = gap (empty space), top = conduction (empty, waiting for electrons).

🎯 Exam Tip: Label each part clearly in the diagram. Write "valence band - filled", "conduction band - empty", "band gap - energy difference".

Question vii. A solid is hard, brittle and electrically non-conductor. Its melt conducts electricity. What type of solid is it?
Answer: A solid crystalline electrolyte like NaCl is hard, brittle and electrically nonconductor. But its melt conducts electricity.
In simple words: This is an ionic solid like salt. When solid, the ions cannot move, so no electricity. When melted, ions can move freely and carry electricity.

📝 Teacher's Note: Show students table salt crystals. Explain that in solid form, the + and - ions are stuck in fixed positions. When heated to melt, they can move around.

🎯 Exam Tip: Write "ionic solid" or "crystalline electrolyte" as the answer. Always mention that ions are fixed in solid state but mobile when melted.

Question viii. Mention two properties that are common to both hcp and ccp lattices.
Answer: In hcp and ccp crystal lattices coordination number is 12 and packing efficiency is 74%.
In simple words: Both structures pack spheres very tightly. Each sphere touches 12 other spheres. About 74% of space is filled with spheres.

📝 Teacher's Note: Use oranges or ping pong balls to show close packing. Students can count how many balls touch one central ball - it will be 12.

🎯 Exam Tip: Always write both values: "coordination number = 12" and "packing efficiency = 74%". These are the key numbers examiners look for.

Question ix. Sketch a tetrahedral void.
Answer:
[Diagram: Shows spheres arranged in close packing with a small triangular space (tetrahedral void) between four spheres. The void has a pyramid-like shape formed by four touching spheres.]
In simple words: A tetrahedral void is a small empty space between four spheres. It looks like a pyramid shape made when four balls touch each other.

📝 Teacher's Note: Take four tennis balls and show how they touch to form a pyramid. The empty space in the middle is the tetrahedral void.

🎯 Exam Tip: Draw four circles touching each other with a small triangular space in the center. Label it "tetrahedral void". Show it's surrounded by four spheres.

Question x. What are ferromagnetic substances?
Answer:

1. The substances which possess unpaired electrons and high paramagnetic character and when placed in a magnetic field are strongly attracted and show permanent magnetic moment even when the external magnetic field is removed are said to be ferromagnetic. They can be permanently magnetised.
2. In the solid state, the metal ions of ferromagnetic substance are grouped together into small regions called domains, where each domain acts as a tiny magnet.

For example: Fe, Co, Gd, Ni, CrO₂, etc.
In simple words: These are materials that become strong magnets. Even after removing a magnet, they stay magnetic. Iron and cobalt are common examples.

📝 Teacher's Note: Bring a piece of iron and a magnet to class. Show how iron gets magnetized and stays magnetic even after the magnet is removed.

🎯 Exam Tip: Write "permanent magnetic moment" and "domains act as tiny magnets". Give examples like Fe, Co, Ni. These are key marking points.

3. Answer The Following In Brief

Question i. What are valence band and conduction band?
Answer: There are two types of bands of molecular orbitals as follows:

• Valence band: The atomic orbitals with filled electrons from the inner shells form valence bands, where there are no free mobile electrons since they are involved in bonding.
• Conduction band: Atomic orbitals which are partially filled or empty on overlapping form closely placed molecular orbitals giving conduction bands where electrons are delocalised and can conduct heat and electricity.

In simple words: Valence band has electrons stuck in bonds. Conduction band has free electrons that can move and carry electricity. Think of valence as parked cars, conduction as moving traffic.

📝 Teacher's Note: Draw two boxes with gaps between them. Bottom box is valence (full of electrons), top box is conduction (few or no electrons). The gap decides if material conducts.

🎯 Exam Tip: Write "valence band - electrons involved in bonding" and "conduction band - free mobile electrons". Mention that conduction band conducts electricity.

Question ii. Distinguish between ionic solids and molecular solids.
Answer:

In simple words: Ionic solids are like salt - hard, high melting point, made of + and - ions. Molecular solids are like ice - soft, low melting point, made of molecules.

📝 Teacher's Note: Bring salt crystals and ice cubes. Let students feel the hardness difference. Heat both to show melting point difference. Salt needs much more heat.

🎯 Exam Tip: Make a proper table with all 6 points. Write examples clearly. Remember ionic = hard, high melting; molecular = soft, low melting.

Question iii. Calculate the number of atoms in fcc unit cell.
Answer:
Number of atoms in face-centred cubic (fcc) unit cell:
In this unit cell, there are 8 atoms at 8 corners and 6 atoms at 6 face centres.

Each corner contributes \( \frac{1}{8} \) atom to the unit cell, hence due to 8 corners:
Number of atoms = \( \frac{1}{8} \times 8 = 1 \)

Each face centre contributes half of the atom to the unit cell, hence due to 6 face centres:
Number of atoms = \( \frac{1}{2} \times 6 = 3 \)

\( \therefore \) Total number of atoms present in fcc unit cell = 1 + 3 = 4.
Hence the volume of the unit cell is equal to the volume of four atoms.
In simple words: Corner atoms are shared by 8 unit cells, so each gives \( \frac{1}{8} \). Face atoms are shared by 2 unit cells, so each gives \( \frac{1}{2} \). Total is 4 atoms per unit cell.

📝 Teacher's Note: Use a cube made of cardboard. Show 8 corners and 6 faces. Explain sharing concept with neighboring cubes. Each corner is shared by 8 cubes, each face by 2 cubes.

🎯 Exam Tip: Always write the formula: \( \frac{1}{8} \times 8 + \frac{1}{2} \times 6 = 4 \). Show the calculation step by step. Final answer is 4 atoms.

Question iv. How are the spheres arranged in first layer of simple cubic close-packed structures? How are the successive layers of spheres placed above this layer?
Answer:
(i) Stacking Of Square Close Packed Layers
In this arrangement, the two dimensional AAAA type square closed packed layers are placed one over the other in such a way that the crests of all spheres are in contact with successive layers in all directions. All spheres of different layers are perfectly aligned horizontally and vertically forming unit cells having primitive or simple cubic structure. Since all the layers are identical and if each layer is labelled as layer A, then whole three dimensional crystal lattice will be of AAAA… type. Each sphere is in contact with six surrounded spheres, hence the coordination number of each sphere is six.

(ii) Stacking Of Two Hexagonal Close Packed Layers
A close packed three dimensional structure can be generated by arranging hexagonal close packed layers in a particular manner. In this the spheres of second layer are placed in the depression of the first layer. In this if first layer is labelled as A then second layer is labelled as B since they are aligned differently. In this, all triangular voids of the first layers are not covered by the spheres of the second layer. The triangular voids which are covered by second layer spheres generate tetrahedral void which is surrounded by four spheres. The triangular voids in one layer have above them triangular voids of successive layers. The overlapping triangular voids from two layers together form an octahedral void which is surrounded by six spheres.
In simple words: In simple cubic, spheres are arranged in neat rows and columns like a brick wall. All layers look the same (AAAA). Each sphere touches 6 neighbors.

📝 Teacher's Note: Use marbles or ping pong balls. Show square arrangement (like floor tiles) and hexagonal arrangement (like oranges in a shop). Let students count neighbors.

🎯 Exam Tip: Write "square close packed" for simple cubic and mention "coordination number = 6". For hexagonal, write about A and B layer labeling and void formation.

Question v. Calculate the packing efficiency of metal crystal that has simple cubic structure.
Answer:
Step 1: Radius of sphere
In simple cubic lattice, the atoms (spheres) are present at eight corners and in contact along the edge in the unit cell. If 'a' is the edge length of the unit cell and 'r' is the radius of the atom, then:
\( a = 2r \) or \( r = \frac{a}{2} \)

Step 2: Number of atoms per unit cell
In simple cubic, number of atoms = \( \frac{1}{8} \times 8 = 1 \)

Step 3: Volume of atoms in unit cell
Volume of 1 atom = \( \frac{4}{3}\pi r^3 \)
Volume of atoms in unit cell = \( 1 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^3 \)

Step 4: Volume of unit cell
Volume of unit cell = \( a^3 = (2r)^3 = 8r^3 \)

Step 5: Packing efficiency
Packing efficiency = \( \frac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \times 100 \)
\( = \frac{\frac{4}{3}\pi r^3}{8r^3} \times 100 = \frac{4\pi}{3 \times 8} \times 100 = \frac{\pi}{6} \times 100 = 52.36\% \)
In simple words: We calculate how much space the spheres actually take up compared to the total box size. In simple cubic, only about 52% of space is filled with atoms.

📝 Teacher's Note: Use a box with one ball inside. Show that the ball doesn't fill the whole box - there's lots of empty space. This is why packing efficiency is only 52%.

🎯 Exam Tip: Write the formula clearly: \( \frac{\pi}{6} \times 100 = 52.36\% \). Show all steps: find r in terms of a, calculate volumes, then divide. Don't forget the percentage sign.

[Diagram: This diagram shows the scc (simple cubic close packed) structure with two spheres labeled A and B touching each other, with radius r and edge length a marked on a cubic unit cell.]

Step 2: Volume of sphere:
Volume of one particle = \( \frac{4\pi}{3} \times r^3 \)
= \( \frac{4\pi}{3} \times (a/2)^3 = \frac{\pi a^3}{6} \)

Step 3: Total volume of particles: Since the unit cell contains one particle. Volume occupied by one particle in unit cell = \( \frac{\pi a^3}{6} \)

Step 4: Packing efficiency:
Packing efficiency = \( \frac{\text{Volume occupied by particles in unit cell}}{\text{Volume of unit cell}} \times 100 \)
= \( \frac{\pi a^3/6}{a^3} \times 100 \)
= \( \frac{3.142 \times 100}{6} = 52.36\% \)
∴ Packing efficiency = 52.36%
Percentage of void space = 100 – 52.36 = 47.64%

📝 Teacher's Note: Show students how packing efficiency tells us how much space is actually filled by atoms. Like packing oranges in a box - some space is always empty between the oranges.

🎯 Exam Tip: Always write the formula first, then substitute values step by step. Remember to calculate void space as 100 minus packing efficiency.

Question vi. What are paramagnetic substances? Give examples.
Answer:

1. The magnetic properties of a substance arise due to the presence of electrons.
2. An electron while revolving around the nucleus, also spins around its own axis and generates a magnetic moment and magnetic properties.
3. If an atom or a molecule contains one or more unpaired electrons spinning in same direction, clockwise or anticlockwise, then the substance is associated with net magnetic moment and magnetic properties. They experience a net force of attraction when placed in the magnetic field. This phenomenon is called paramagnetism and the substance is said to be paramagnetic.

For example, \( O_2 \), \( Cu^{2+} \), \( Fe^{3+} \), \( Cr^{3+} \), NO, etc.
In simple words: Paramagnetic substances have unpaired electrons that act like tiny magnets. When you put them near a magnet, they get attracted to it. Think of iron filings getting pulled towards a magnet.

📝 Teacher's Note: Use a simple magnet and some iron filings to show attraction. Explain that unpaired electrons are like tiny spinning tops that create magnetic force.

🎯 Exam Tip: Always mention "unpaired electrons" and "attraction to magnetic field" in your answer. Give at least two examples with correct chemical symbols.

Question vii. What are the consequences of Schottky defect?
Answer:
Consequences of Schottky defect:

• Since the number of ions (cations and anions) decreases but volume remains unchanged, the density of a substance decreases.
• As the number of missing cations and anions is equal, the electrical neutrality of the compound remains same.
• This defect arises in ionic crystals like NaCl, AgBr, KCl, etc.

In simple words: Schottky defect means some positive and negative ions are missing from their places. The crystal becomes lighter but stays electrically balanced.

📝 Teacher's Note: Compare with removing equal numbers of boys and girls from a classroom. The class becomes smaller but the boy-girl ratio stays the same.

🎯 Exam Tip: Write three main points: density decreases, electrical neutrality maintained, and give examples. These are the key scoring points.

Question viii. Cesium chloride crystallizes in cubic unit cell with Cl⁻ ions at the corners and Cs⁺ ion in the centre of the cube. How many CsCl molecules are there in the unit cell?
Answer:
Number of Cs⁺ ion at body centre = 1
Number of Cl⁻ ions due to 8 corners = \( 8 \times \frac{1}{8} = 1 \)
Hence unit cell contains 1 CsCl molecule.
In simple words: Each corner ion is shared by 8 unit cells, so we count only 1/8 of each corner ion. The centre ion belongs completely to this unit cell.

📝 Teacher's Note: Use a dice to show corners. Each corner touches 8 cubes around it, so each cube gets only 1/8 share of that corner.

🎯 Exam Tip: Always show the calculation: 8 corners × 1/8 = 1. Write the final answer clearly as "1 CsCl molecule".

Question ix. Cu crystallizes in fcc unit cell with edge length of 495 pm. What is the radius of Cu atom?
Answer:
Given: a = 495 pm
Radius, r = ?
For fcc structure,
radius = \( r = \frac{a}{2\sqrt{2}} = \frac{495}{2 \times \sqrt{2}} = 175 \) pm.
Radius of Cu atom = 175 pm
In simple words: In fcc structure, atoms touch along the face diagonal. We use a special formula to find the radius from the edge length.

📝 Teacher's Note: Draw a square face of the cube and show how atoms touch along the diagonal. The formula comes from this diagonal relationship.

🎯 Exam Tip: Remember the fcc formula: r = a/(2√2). Always substitute values carefully and write the unit (pm) in your final answer.

Question x. Obtain the relationship between density of a substance and the edge length of unit cell.
Answer:

1. Consider a cubic unit cell of edge length 'a'. The volume of unit cell = \( a^3 \)
2. If there are 'n' particles per unit cell and the mass of particle is 'm', then Mass of unit cell = m × n.
3. If the density of the unit cell of the substance is ρ then Density of unit cell = \( \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} \)
   ρ = \( \frac{m \times n}{a^3} \)

In simple words: Density is mass divided by volume. For a unit cell, mass depends on how many atoms are there and how heavy each atom is. Volume is just the cube of edge length.

📝 Teacher's Note: Use a simple box analogy. If you know the size of the box and how many marbles are inside, you can find how heavy the box is per unit volume.

🎯 Exam Tip: Write the final formula clearly: ρ = mn/a³. This formula is used in many numerical problems, so remember it well.

Question 4. The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol.
Answer:
Given: Crystal structure of iridium = fcc
Molar mass of iridium = 192.2 gmol⁻¹
Density = ρ = 22.4 gcm⁻³
Radius of iridium = ?

Step 1: Find number of atoms per unit cell.
In fcc structure, there are 8 Ir atoms at 8 corners and 6 Ir atoms at 6 face centres.
∴ Total number of Ir atoms = \( \frac{1}{8} \times 8 + \frac{1}{2} \times 6 \) = 1 + 3 = 4

Step 2: Find mass of unit cell.
Mass of Ir atom = \( \frac{192.2}{6.022 \times 10^{23}} = 31.92 \times 10^{-23} \) g
∴ Mass of 4 Ir atoms = \( 4 \times 31.92 \times 10^{-23} = 1.277 \times 10^{-21} \) g
∴ Mass of unit cell = \( 1.277 \times 10^{-21} \) g

Step 3: Find edge length using density formula.
Density of unit cell = \( \frac{\text{Mass of unit cell}}{\text{Volume of unit cell}} \)
\( 22.4 = \frac{1.277 \times 10^{-21}}{a^3} \)
∴ \( a^3 = \frac{1.277 \times 10^{-21}}{22.4} = 57 \times 10^{-24} \) cm³
∴ a = \( (57 \times 10^{-24})^{1/3} = 3.848 \times 10^{-8} \) cm

Step 4: Find radius for fcc.
If r is the radius of iridium atom, then for fcc structure,
\( r = \frac{a}{2\sqrt{2}} = \frac{3.848 \times 10^{-8}}{2 \times 1.414} = 1.36 \times 10^{-8} \) cm = 136 pm

Radius of iridium atom = 136 pm
In simple words: We used the density formula to find how big the box (unit cell) is. Then we used the fcc formula to find how big each atom (sphere) is inside that box.

📝 Teacher's Note: Make students write the formula first, then substitute values. A common mistake is forgetting that fcc has 4 atoms per unit cell, not 1.

🎯 Exam Tip: Always write "Given:", list all values, then show each step clearly. Write the final answer with correct units (pm). You lose marks if units are missing.

Question 5. Aluminium crystallizes in cubic close packed structure with unit cell edge length of 353.6 pm. What is the radius of Al atom? How many unit cells are there in 1.00 cm³ of Al?
Answer:
Given: Structure of Al = Cubic close packed structure = ccp structure
Edge length of unit cell = a = 353.6 pm = \( 3.536 \times 10^{-8} \) cm
r = ?
Number of unit cells in 1.00 cm³ of Al = ?

Radius of Al atom = \( r = \frac{a}{2\sqrt{2}} = \frac{353.6}{2\sqrt{2}} = \frac{353.6}{2 \times 1.414} = 125 \) pm

Volume of one unit cell = \( a^3 = (3.536 \times 10^{-8})^3 = 4.421 \times 10^{-23} \) cm³

Number of unit cells = \( \frac{1.00}{4.421 \times 10^{-23}} = 2.26 \times 10^{22} \)

Radius of Al atom = 125 pm
Number of unit cells = \( 2.26 \times 10^{22} \)
In simple words: First we found how big each atom is using the fcc formula. Then we found how many tiny unit cells fit in 1 cubic cm by dividing total volume by volume of one unit cell.

📝 Teacher's Note: Remind students that ccp and fcc are the same structure. The huge number of unit cells shows how tiny atoms really are.

🎯 Exam Tip: Convert pm to cm when calculating volume. Write both answers clearly with correct units and scientific notation.

Question 6. In an ionic crystalline solid atoms of element Y form hcp lattice. The atoms of element X occupy one third of tetrahedral voids. What is the formula of the compound?
Answer:
In the given hcp lattice, Y atoms are present at 12 corners and 2 face centres.
∴ Number of Y atoms = \( \frac{1}{2} \times 12 + 2 \times \frac{1}{2} = 3 \)
There are 6 tetrahedral voids, the number of X atoms = \( \frac{1}{3} \times 6 = 2 \)
∴ Formula of the compound is \( X_2Y_3 \).
In simple words: We count how many Y atoms and X atoms are in one unit cell. Then we write the formula using the smallest whole number ratio.

📝 Teacher's Note: Explain that hcp has 6 tetrahedral voids per unit cell. Students often forget to count corner and face atoms correctly.

🎯 Exam Tip: Always count atoms per unit cell first, then find the ratio. Write the formula with correct subscripts.

Question 7. How are tetrahedral and octahedral voids formed?
Answer:
Tetrahedral void: The vacant space or void among four constituent particles having tetrahedral arrangement in the crystal lattice is called tetrahedral void. The arrangement of four spheres around the void is tetrahedral. A tetrahedral void is formed when a triangular void made by three coplanar spheres is in contact with fourth sphere above or below it.

Octahedral void: The vacant space or void at the centre of six spheres (or atoms) which are placed octahedrally is called octahedral void.
In simple words: Tetrahedral void is the empty space between 4 spheres arranged like a pyramid. Octahedral void is the empty space at the centre of 6 spheres arranged like two pyramids joined at their bases.

[Diagram: Two diagrams showing tetrahedral void formation between spheres and octahedral voids in cubic unit cells at body centre and edge centres.]

📝 Teacher's Note: Use balls or oranges to show students how voids form when spheres are packed together. The empty spaces between spheres are the voids.

🎯 Exam Tip: Define both types clearly. Mention that tetrahedral void has 4 spheres around it and octahedral void has 6 spheres around it.

Question 8. Third layer of spheres is added to second layer so as to form hcp or ccp structure. What is the difference between the addition of third layer to form these hexagonal close-packed structures?
Answer:

1. In the formation of hexagonal closed-packed (hcp) structure, the first one dimensional row shows depressions between neighbouring atoms.
2. When a second row is arranged so that spheres fit in these depressions then a staggered arrangement is obtained. If the first row is A then the second row is B.

In simple words: When we stack three layers of spheres, we can put the third layer in two different ways. This gives us two different crystal structures - hcp and ccp.

📝 Teacher's Note: Use coins or bottle caps to show students how layers can stack. The third layer position determines whether you get hcp or ccp structure.

🎯 Exam Tip: Explain that both start with AB layers, but the third layer goes in different positions. In hcp it goes as ABA, in ccp it goes as ABC.

Question 9. An element with molar mass 27 g/mol forms cubic unit cell with edge length of 405 pm. If density of the element is 2.7 g/cm³, what is the nature of cubic unit cell? (fcc or ccp)
Answer:
Given:
Molar mass = M = 27 g mol⁻¹
Nature of crystal = cubic unit cell
Edge length = a = 405 pm = \( 4.05 \times 10^{-8} \) cm
Density = ρ = 2.7 g cm⁻³
Nature of unit cell = ?

Step 1: Use the density formula to find number of atoms per unit cell.
\( \rho = \frac{n \times M}{a^3 \times N_A} \)

Step 2: Rearrange to find n.
\( n = \frac{\rho \times a^3 \times N_A}{M} \)

Step 3: Substitute the values.
\( n = \frac{2.7 \times (4.05 \times 10^{-8})^3 \times 6.022 \times 10^{23}}{27} \)

Step 4: Calculate.
= 3.997
≅ 4

Since n = 4 atoms per unit cell, the nature of unit cell is face-centred cubic (fcc).
The nature of cubic unit cell is fcc.
In simple words: We used the formula to find how many atoms are in one unit cell box. Since we got 4 atoms, it must be fcc structure. Simple cubic has 1 atom, bcc has 2 atoms, but fcc has 4 atoms.

📝 Teacher's Note: Make students remember: simple cubic = 1 atom, bcc = 2 atoms, fcc = 4 atoms per unit cell. This is the key to solving such problems. Show them the formula and practice substitution.

🎯 Exam Tip: Always write "Given", show the density formula, substitute values step by step, and clearly state the final answer. Write "fcc" not just "4". You get marks for showing work.

[Diagram: This shows hexagonal close packing structure with layers A, B, A stacked in a repeating pattern, plus a unit cell diagram showing the arrangement]

Question 10. An element has a bcc structure with unit cell edge length of 288 pm. How many unit cells and number of atoms are present in 200 g of the element? (\( 1.16 \times 10^{24} \), \( 2.32 \times 10^{24} \))

Question 11. Distinguish with the help of diagrams metal conductors, insulators and semiconductors from each other.
Answer:
Conductor:

1. A substance which conducts heat and electricity to a greater extent is called conductor.
2. In this, conduction bands and valence bands overlap or are very closely spaced.
3. There is no energy difference or very less energy difference between valence bands and conduction bands.
4. There are free electrons in the conduction bands.
5. The conductance decreases with the increase in temperature.
6. E.g., Metals, alloys.
7. The conducting properties can't be improved by adding third substance.

Insulator:

1. A substance which cannot conduct heat and electricity under any conditions is called insulator.
2. In this, conduction bands and valence bands are far apart.
3. The energy difference between conduction bands and valence bands is very large.
4. There are no free electrons in the conduction bands and electrons can't be excited from valence bands to conduction bands due to large energy difference.
5. No effect of temperature on conducting properties.
6. E.g., Wood, rubber, plastics.
7. No effect of addition of any substance.

Semiconductor:

1. A substance that has poor electrical conductance at low temperature but higher conductance at higher temperature is called semiconductor.
2. In this, conduction bands and valence bands are spaced closely.
3. The energy difference between conduction bands and valence bands is small.
4. The electrons can be easily excited from valence bands to conduction bands by heating.
5. Conductance increases with the increase in temperature.
6. E.g., Si, Ge
7. By doping, conducting properties improve. E.g. n-type, p-type semiconductors.

In simple words: Conductors let electricity flow easily (like water pipes). Insulators stop electricity completely (like plastic covers). Semiconductors are in between - they conduct electricity when heated or mixed with other elements.

📝 Teacher's Note: Use a torch, plastic ruler, and silicon chip to show the three types. Students can feel the difference. Also draw the band gap diagrams on board - big gap for insulator, small gap for semiconductor, no gap for conductor.

🎯 Exam Tip: Remember the key words: overlap (conductor), large gap (insulator), small gap (semiconductor). Draw band diagrams if asked. Temperature effect is important for semiconductors.

[Diagram: Shows three energy band diagrams side by side - Metal (overlapping bands), Insulator (large energy gap between bands), and Semiconductor (small energy gap between bands)]

Question 12. What are n-type semiconductors? Why is the conductivity of doped n-type semiconductor higher than that of pure semiconductor? Explain with diagram.
Answer:
n-type semiconductor:
n-type semiconductor contains increased number of electrons in the conduction band. When Si semiconductor is doped with 15th group element phosphorus, P, the new atoms occupy some vacant sites in the lattice in place of Si atoms. P has five valence electrons, out of which four are involved in covalent bonding with neighboring Si atoms while one electron remains free and delocalised. These free electrons increase the electrical conductivity of the semiconductor. The semiconductors with extra non-bonding free electrons are called n-type semiconductors.

In simple words: We add phosphorus atoms to silicon. Phosphorus has 5 electrons, but only needs 4 to join with silicon. The extra electron becomes free and helps conduct electricity better.

📝 Teacher's Note: Show students that phosphorus has 5 valence electrons (group 15) while silicon has 4. The extra electron from phosphorus becomes mobile. Use the analogy of extra people in a room who can move around freely.

🎯 Exam Tip: Write "15th group element" and "extra free electrons". Draw the crystal structure showing P atom with one free electron. Mention that n stands for negative charge carriers.

[Diagram: Shows crystal lattice of silicon atoms with one phosphorus atom in place of silicon, with a mobile electron indicated]

Question 13. Explain with diagram. Frenkel defect. What are the conditions for its formation? What is its effect on density and electrical neutrality of the crystal?
Answer:
Frenkel defect: This defect arises when an ion of an ionic compound is missing from its regular site and occupies interstitial vacant position between lattice points. Cations have smaller size than anions, hence generally cations occupy the interstitial sites. This creates a vacancy defect at its original position and interstitial defect at new position. Frenkel defect is regarded as the combination of interstitial defect and vacancy defect.

Conditions for the formation of Frenkel defect:

1. This defect arises in ionic compounds with a large difference between the sizes of cation and anion.
2. The ionic compounds must have ions with low coordination number.

Consequences of Frenkel defect:

1. Since there is no loss of ions from the crystal lattice, the density of the solid remains unchanged.
2. The crystal remains electrically neutral.
3. This defect is observed in ZnS, AgCl, AgBr, AgI, CaF₂, etc.

In simple words: An ion jumps from its normal place to a gap between other ions. It is like a person leaving their chair and sitting on the floor between chairs. The total number of people stays same, so density stays same.

📝 Teacher's Note: Use chairs in classroom to demonstrate. One student leaves their chair (vacancy) and sits between chairs (interstitial). Total students same, but arrangement changed. This helps students visualize the defect easily.

🎯 Exam Tip: Write "combination of vacancy and interstitial defect". Mention "density unchanged" and "electrically neutral". Draw the diagram showing the ion in interstitial position and empty original site.

[Diagram: Shows ionic crystal structure with cations and anions, where one cation has moved from its regular lattice position to an interstitial site, leaving behind a vacancy]

Question 14. What is an impurity defect? What are its types? Explain the formation of vacancies through aliovalent impurity with example.
Answer: Impurity defect: This defect arises when foreign atoms, that is, atoms different from the host atoms are present in the crystal lattice.

There are two types of impurity defects namely

1. Substitutional defects and
2. Interstitial defects.

(1) Substitutional defects: These defects arises when foreign atoms occupy the lattice sites in place of host atoms, due to their displacements.

Examples: Solid solutions of metals (alloys). For example, Brass in which host atoms are of Cu which are replaced by impurity of Zn atoms. In this Zn atoms occupy regular sites while Cu atoms occupy substituted sites.

[Diagram: This diagram shows the crystal structure of brass with copper atoms and zinc atoms arranged in the lattice. Some positions show zinc atoms replacing copper atoms in the regular lattice sites.]

Vacancy through aliovalent impurity:
By addition of impurities of aliovalent ions:

When aliovalent ion like \( Sr^{2+} \) in small amount is added by adding \( SrCl_2 \) to \( NaCl \) during its crystallisation, each \( Sr^{2+} \) ion (oxidation state 2+) removes 2 \( Na^+ \) ions from their lattice points, to maintain electrical neutrality. Hence one of vacant lattice site is occupied by \( Sr^{2+} \) ion while other site remains vacant.

[Diagram: This diagram shows the NaCl crystal lattice with Sr²⁺ ions substituting Na⁺ ions, creating vacancies where Na⁺ ions were removed to maintain charge balance.]

(2) Interstitial impurity defect:
A defect in solid in which the impurity atoms occupy interstitial vacant spaces of lattice structure is called interstitial impurity defect.

For example, in steel, normal lattice sites are occupied by Fe atoms but interstitial spaces are occupied by carbon atoms.

[Diagram: This diagram shows the steel structure with iron atoms in regular lattice positions and smaller carbon atoms occupying the interstitial spaces between iron atoms.]

In simple words: Impurity defect means foreign atoms enter the crystal. They can either replace the original atoms (substitutional) or squeeze into empty spaces (interstitial). Think of it like guests in a house - they either take someone's chair or sit on the floor.

📝 Teacher's Note: Use coins of different sizes to show substitutional defects. Big coin replaces small coin. For interstitial, show how small objects fit between bigger ones without replacing them.

🎯 Exam Tip: Always write both types clearly - substitutional and interstitial. For aliovalent vacancy, mention charge balance is maintained. Give one example for each type.

Try This (Textbook Page No. 1)

Question 1. Will you call the arrangement of particles in this solid regular or irregular?
Answer: The arrangement of particles in this solid is regular.
In simple words: The particles are in a neat, repeated pattern - like tiles on a floor. So we call it regular.

📝 Teacher's Note: Ask students to look at a brick wall or a tiled floor. That regular pattern is what "crystalline" means in solids.

🎯 Exam Tip: Write "regular arrangement" for crystalline and "irregular arrangement" for amorphous. Use these exact words.

Question 2. Is the arrangement of constituent particles in directions \( \overrightarrow{AB} \), \( \overrightarrow{CD} \) and \( \overrightarrow{EF} \) same or different?
Answer: \( \overrightarrow{AB} \) represents arrangement of identical particles of one type.
\( \overrightarrow{CD} \) represents arrangement of identical particles of another type.
\( \overrightarrow{EF} \) represents regular arrangement of two different particles in alternate positions.
In simple words: Each direction shows a different pattern. AB has all same type particles. CD has all same but different from AB. EF has both types mixed in order.

📝 Teacher's Note: Use colored beads or buttons in rows to show different arrangements. Students can see how direction matters in crystals.

🎯 Exam Tip: Write clearly about each direction separately. Mention "identical particles" and "alternate positions" - these are key exam words.

Use Your Brain Power (Textbook Page No. 2)

Question 1. Identify the arrangements A and B as crystalline or amorphous.
Answer: Arrangement in image A indicates the substance is crystalline.
Arrangement in image B indicates the substance is amorphous.
In simple words: Picture A shows neat rows like books on a shelf. Picture B looks messy like books thrown in a box. Neat = crystalline. Messy = amorphous.

📝 Teacher's Note: Show students a neat stack of books (crystalline) and a messy pile (amorphous). This visual helps them understand the difference easily.

🎯 Exam Tip: Look for regular patterns (crystalline) or random arrangements (amorphous). Write both words clearly in your answer.

Try This (Textbook Page No. 3)

Question 1. Graphite is a covalent solid yet soft and good conductor of electricity. Explain.
Answer:

1. Each carbon atom in graphite is \( sp^2 \) hybridised and covalently bonded to other three \( sp^2 \) hybridised carbon atoms forming σ bonds and the fourth electron in \( 2p_z \) orbital of each carbon atom is used in the formation of a π bond. This results in the formation of hexagonal rings in two dimensions.
2. In graphite, the layers consisting of hexagonal carbon network are held together by weak van der Waal's forces imparting softness.
3. The electrons in π bonds in the ring are delocalised and free to move in the delocalised molecular orbitals giving good electrical conductance.

[Diagram: This diagram shows the layer structure of graphite with hexagonal rings of carbon atoms. The layers are separated by weak van der Waals forces, and mobile electrons move freely between layers.]

In simple words: Graphite has layers like pages in a book. The pages (layers) slide easily making it soft. Free electrons move between layers like cars on a highway, making it conduct electricity.

📝 Teacher's Note: Use a book to show how pages slide over each other. This explains why graphite is soft. Draw simple hexagons to show the ring structure.

🎯 Exam Tip: Write three points clearly - layer structure, weak forces between layers (soft), and mobile electrons (conductor). Use key words like "delocalised electrons."

Use Your Brain Power (Textbook Page No. 13)

Question 1. Which of the three lattices scc, bcc and fcc has the most efficient packing of particles? Which one has the least efficient packing?
Answer: fcc has the most efficient packing of particles while scc has the least efficient packing.
In simple words: fcc packs spheres most tightly like oranges in a shop. scc has the most empty space like balls loosely arranged in a box.

📝 Teacher's Note: Show students how oranges are stacked in fruit shops (fcc-like). Compare with loose arrangement in a bag (scc-like). Visual helps understanding.

🎯 Exam Tip: Remember: fcc = most efficient (74%). scc = least efficient (52%). Writing percentages gets extra marks.

Can You Think (Textbook Page No. 20)

Question 1. When ZnO is heated it turns yellow and returns back to original white colour on cooling. What could be the reason?
Answer: When colourless ZnO is strongly heated, the metal atoms are deposited on crystal surface and anions \( O^{2-} \) migrate to the surface producing vacancies at anion lattice points.

These anions combine with Zn atoms forming ZnO and release electrons.
\( Zn + O^{2-} \rightarrow ZnO + 2e^- \)

These released electrons diffuse into the crystal and occupy vacant sites of anions and produce F-centres. Due to these colour centres, ZnO turns yellow.
In simple words: Heating makes some oxygen leave their places, creating empty spots. Electrons fill these spots and create color centers that make ZnO look yellow. Cooling brings oxygen back, so it becomes white again.

📝 Teacher's Note: Compare to a parking lot - when cars (oxygen) leave, empty spaces form. Electrons park in these spaces causing color change.

🎯 Exam Tip: Write key words: "vacancies," "F-centres," and "color centres." Explain the electron equation clearly. Mention heating creates defects, cooling removes them.

Can You Tell (Textbook Page No. 23)

Let a small quantity of phosphorus be doped into pure silicon.

Question 1. Will the resulting material contain the same number of total number of electrons as the original pure silicon?
Answer: Total number of electrons in doped silicon will be more than in original silicon.
In simple words: Phosphorus has more electrons than silicon. When we add phosphorus atoms, we add extra electrons too.

📝 Teacher's Note: Phosphorus has 5 outer electrons, silicon has 4. Adding phosphorus means adding extra electrons to the material.

🎯 Exam Tip: Write "more electrons" clearly. Explain that phosphorus is electron-rich compared to silicon.

Question 2. Will the material be electrically neutral or charged?
Answer: Material will be electrically neutral.
In simple words: Even though we add extra electrons, we also add the same number of protons (in phosphorus nuclei). So positive and negative charges still balance out.

📝 Teacher's Note: Explain that adding atoms means adding both electrons and protons. The total charge stays zero because equal positive and negative charges are added.

🎯 Exam Tip: Write "electrically neutral" clearly. Mention that both electrons and protons are added in equal numbers.

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