Question:

How will you distinguish between the following pairs of terms: (i) Hexagonal close-packing and cubic close-packing? (ii) Crystal lattice and unit cell? (iii) Tetrahedral void and octahedral void?

Answer:

(i) Hexagonal close-packing and cubic close-packing: Hexagonal close-packing is a type of arrangement of spheres in which spheres are packed in layers with each layer being slightly offset from the one below it, resulting in a hexagonal pattern. Cubic close-packing is a type of arrangement of spheres in which spheres are packed in layers with each layer being perfectly aligned with the one below it, resulting in a cubic pattern.

(ii) Crystal lattice and unit cell: A crystal lattice is a three-dimensional array of points in space where each point represents an atom, ion, or molecule. A unit cell is the smallest repeating unit of a crystal lattice, and it contains all the information necessary to describe the structure of the entire lattice.

(iii) Tetrahedral void and octahedral void: A tetrahedral void is a space between four spheres in a close-packed arrangement, and it has a tetrahedral shape. An octahedral void is a space between six spheres in a close-packed arrangement, and it has an octahedral shape.

Question:

Explain the following with suitable examples: (i) Ferromagnetism (ii) Paramagnetism (iii) Ferrimagnetism (iv) Antiferromagnetism

Answer:

(i) Ferromagnetism: Ferromagnetism is a type of magnetism that is exhibited by certain materials, such as iron, nickel, cobalt and some rare earth metals, in which the atomic magnetic moments (or the spins of the electrons) align in the same direction. This alignment results in a net magnetic field, which is responsible for the material’s strong attraction to other magnetic materials. Examples of ferromagnetic materials include iron, nickel, cobalt, and some rare earth metals.

(ii) Paramagnetism: Paramagnetism is a type of magnetism that is exhibited by certain materials, such as oxygen and aluminum, in which the atomic magnetic moments (or the spins of the electrons) are randomly oriented. This random orientation results in a weak attraction to other magnetic materials. Examples of paramagnetic materials include oxygen, aluminum, and some rare earth metals.

(iii) Ferrimagnetism: Ferrimagnetism is a type of magnetism that is exhibited by certain materials, such as barium ferrite, in which the atomic magnetic moments (or the spins of the electrons) are aligned in opposite directions. This alignment results in a net magnetic field, which is responsible for the material’s weak attraction to other magnetic materials. Examples of ferrimagnetic materials include barium ferrite and some rare earth metals.

(iv) Antiferromagnetism: Antiferromagnetism is a type of magnetism that is exhibited by certain materials, such as chromium dioxide, in which the atomic magnetic moments (or the spins of the electrons) are aligned in alternating directions. This alternating orientation results in a net magnetic field, which is responsible for the material’s weak attraction to other magnetic materials. Examples of antiferromagnetic materials include chromium dioxide and some rare earth metals.

Question:

Calculate the efficiency of packing in case of a metal crystal for the following crystal structure (with the assumptions that atoms are touching each other): (i) simple cubic (ii) body-centered cubic (iii) face-centered cubic

Answer:

(i) Simple cubic: Step 1: Calculate the total number of atoms in the crystal structure. This can be done by calculating the volume of the cube and then dividing it by the volume of one atom.

Step 2: Calculate the total number of atoms that can be placed in the cube. This can be done by calculating the total number of unit cells in the cube and then multiplying it by the number of atoms in one unit cell.

Step 3: Calculate the efficiency of packing by dividing the total number of atoms in the cube by the total number of atoms that can be placed in the cube.

(ii) Body-centered cubic: Step 1: Calculate the total number of atoms in the crystal structure. This can be done by calculating the volume of the cube and then dividing it by the volume of one atom.

Step 2: Calculate the total number of atoms that can be placed in the cube. This can be done by calculating the total number of unit cells in the cube and then multiplying it by the number of atoms in one unit cell, plus one additional atom at the center of the cube.

Step 3: Calculate the efficiency of packing by dividing the total number of atoms in the cube by the total number of atoms that can be placed in the cube.

(iii) Face-centered cubic: Step 1: Calculate the total number of atoms in the crystal structure. This can be done by calculating the volume of the cube and then dividing it by the volume of one atom.

Step 2: Calculate the total number of atoms that can be placed in the cube. This can be done by calculating the total number of unit cells in the cube and then multiplying it by the number of atoms in one unit cell, plus four additional atoms at the center of each face of the cube.

Step 3: Calculate the efficiency of packing by dividing the total number of atoms in the cube by the total number of atoms that can be placed in the cube.

Question:

A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body-centre. What is the formula of the compound? What are the coordination numbers of P and Q?

Answer:

Answer:

The formula of the compound is PQ2, as there are two atoms of Q at each corner of the cube.

The coordination number of P is 8, as it is located at the body-centre of the cube and is surrounded by 8 atoms of Q.

The coordination number of Q is 4, as each atom of Q is located at the corner of the cube and is surrounded by 4 atoms of P.

Question:

Copper crystallises into a fcc lattice with edge length 3.61×10^-8cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm³.

Answer:

1. Calculate the volume of the unit cell: V = (3.61 x 10^-8 cm)³ V = 4.16 x 10^-23 cm³

2. Calculate the molar mass of copper: M = 63.55 g/mol

3. Calculate the number of moles per unit cell: n = V/M n = 4.16 x 10^-23 cm³ / 63.55 g/mol n = 6.54 x 10^-25 mol

4. Calculate the mass of copper per unit cell: m = n x M m = 6.54 x 10^-25 mol x 63.55 g/mol m = 4.14 x 10^-22 g

5. Calculate the density of copper: d = m/V d = 4.14 x 10^-22 g / 4.16 x 10^-23 cm³ d = 8.92 g/cm³

6. Compare the calculated density with the measured value: The calculated density (8.92 g/cm³) is in agreement with the measured value (8.92 g/cm³).

Question:

Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous. (i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate (NH₄)3​PO₄ (iii) SiC (iv) I₂ (v) P₄ (vi) Plastic (vii) Graphite (viii) Brass (ix) Rb (x) LiBr (xi) Si

Answer:

(i) Tetra phosphorus decoxide (P₄O₁₀) - Network (Covalent)

(ii) Ammonium phosphate (NH₄)3​PO₄₄ - Ionic

(iii) SiC - Network (Covalent)

(iv) I₂ - Molecular

(v) P₄ - Network (Covalent)

(vi) Plastic - Amorphous

(vii) Graphite - Network (Covalent)

(viii) Brass - Metallic

(ix) Rb - Metallic

(x) LiBr - Ionic

(xi) Si - Network (Covalent)

Question:

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

Answer:

(i) Ge doped with In - n-type semiconductor (ii) Si doped with B - p-type semiconductor

Question:

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

Answer:

(i) Schottky Defect: A Schottky defect is a type of point defect in a crystal lattice that arises from an imbalance of cations and anions. It is named after Walter Schottky, who first described the defect in 1918. An example of a Schottky defect is when there are fewer cations than anions in a crystal lattice. This creates a region of positive charge within the lattice, which can attract electrons and cause an electrical imbalance.

(ii) Frenkel Defect: A Frenkel defect is a type of point defect in a crystal lattice that arises from an atom being displaced from its lattice site. It is named after Yakov Frenkel, who first described the defect in 1926. An example of a Frenkel defect is when an atom is displaced from its lattice site and is now located in a space between lattice sites. This creates a region of positive charge within the lattice, which can attract electrons and cause an electrical imbalance.

(iii) Interstitials: Interstitials are point defects in a crystal lattice that arise from an atom being located in an interstitial site, which is a space between lattice sites. An example of an interstitial is when an atom is located in a space between lattice sites. This can create a region of positive charge within the lattice, which can attract electrons and cause an electrical imbalance.

(iv) F-centres: F-centres are point defects in a crystal lattice that arise from an atom being replaced with an electron. It is named after the German physicist, Friedrich Hund, who first described the defect in 1927. An example of an F-centre is when an atom is replaced with an electron. This creates a region of negative charge within the lattice, which can repel electrons and cause an electrical imbalance.

Question:

If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between r and R.

Answer:

Step 1: The octahedral void is a space that is surrounded by six atoms in a close-packed structure.

Step 2: The distance between the atoms in the close-packed structure is equal to the sum of the radii of the atoms.

Step 3: Therefore, the radius of the octahedral void (r) is equal to the radius of the atoms in the close-packed structure (R) plus half the distance between the atoms (R/2).

Step 4: Thus, the relation between r and R can be expressed as: r = R + R/2 or r = 1.5R

Question:

Define the term ‘amorphous’. Give a few examples of amorphous solids.

Answer:

Definition: Amorphous solids are solids that lack a crystalline structure and have no long-range order.

Examples: Glass, plastics, rubber, some polymers, some gels, and some silicates.

Question:

How many lattice points are there in one unit cell of each of the following lattice? (i) Face-centred cubic (ii) Face-centred tetragonal (iii) Body-centred

Answer:

(i) Face-centred cubic: 8 lattice points

(ii) Face-centred tetragonal: 4 lattice points

(iii) Body-centred: 2 lattice points

Question:

Analysis shows that nickel oxide has the formula Ni0.98​O1.00​. What fractions of nickel exist as Ni₂⁺ and Ni₃⁺ ions?

Answer:

1. Calculate the total number of moles of Ni: 0.98 mol Ni x 1 mol Ni/1 mol Ni0.98O1.00 = 0.98 mol Ni

2. Calculate the fraction of Ni₂⁺ ions: 0.98 mol Ni x 1 mol Ni₂⁺/2 mol Ni = 0.49 mol Ni₂⁺

3. Calculate the fraction of Ni₃⁺ ions: 0.98 mol Ni x 1 mol Ni₃⁺/3 mol Ni = 0.33 mol Ni₃⁺

4. Calculate the percentage of Ni₂⁺ ions: 0.49 mol Ni₂⁺/0.98 mol Ni x 100% = 50% Ni₂⁺

5. Calculate the percentage of Ni₃⁺ ions: 0.33 mol Ni₃⁺/0.98 mol Ni x 100% = 33% Ni₃⁺

Question:

Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm. (i) What is the length of the side of the unit cell? (ii) How many unit cells are there in 1.00 cm³ of aluminium?

Answer:

(i) The length of the side of the unit cell can be calculated using the formula:

Side length = 4 x Metallic radius

Therefore, the length of the side of the unit cell is:

Side length = 4 x 125 pm = 500 pm

(ii) The number of unit cells in 1.00 cm³ of aluminium can be calculated using the formula:

Number of unit cells = Volume of aluminium / Volume of unit cell

The volume of the unit cell can be calculated using the formula:

Volume of unit cell = (Side length)3

Therefore, the volume of the unit cell is:

Volume of unit cell = (500 pm)³ = 1.25 x 10^-21 cm³

The number of unit cells in 1.00 cm³ of aluminium can be calculated as follows:

Number of unit cells = 1.00 cm³ / 1.25 x 10^-21 cm³ = 8.00 x 10²² unit cells

Question:

Silver crystallises in fcc lattice. If edge length of the cell is 4.07×10^-8cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.

Answer:

1. Calculate the volume of the unit cell: Volume = (edge length)3 = (4.07 x 10^-8 cm)³ = 6.9 x 10^-23 cm³

2. Calculate the mass of the unit cell: Mass = Volume x Density = 6.9 x 10^-23 cm³ x 10.5 g cm^-3 = 7.2 x 10^-22 g

3. Calculate the number of atoms in the unit cell: Number of atoms = Mass/Atomic mass = 7.2 x 10^-22 g / (Atomic mass of silver)

4. Solve for the atomic mass of silver: Atomic mass of silver = Mass/Number of atoms = 7.2 x 10^-22 g / 6.9 x 10^-23 cm³ = 10.5 g cm^-3

Question:

What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Answer:

1. A glass is different from a solid such as quartz because glass is an amorphous solid, meaning it does not have a regular atomic structure like quartz does.

2. Quartz can be converted into glass through a process called vitrification, which involves heating quartz to very high temperatures (above 1700 degrees Celsius) in an oxygen-free environment. This process causes the quartz to melt and form a viscous liquid, which then cools and solidifies into glass.

Question:

Stability of a crystal is reflected in the magnitude of its melting points. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?

Answer:

1. Stability of a crystal is reflected in the magnitude of its melting points because the higher the melting point, the more energy is required to break the intermolecular forces between the molecules, making the crystal more stable.

2. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book:

• Solid water: 0°C
• Ethyl alcohol: -114°C
• Diethyl ether: -116°C
• Methane: -182°C

3. From the melting points collected, we can say that the intermolecular forces between these molecules decrease in the order of solid water > ethyl alcohol > diethyl ether > methane. This is because the higher the melting point, the stronger the intermolecular forces are.

Question:

Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Answer:

1. Hexagonal close-packed array of oxide ions means that each oxide ion is surrounded by six other oxide ions in a hexagonal arrangement.

2. Two out of every three octahedral holes are occupied by ferric ions, which means that for every three oxide ions, two of them are replaced by ferric ions.

3. Therefore, the formula of ferric oxide can be written as Fe2O3.

Question:

Explain (i) The basis of similarities and differences between metallic and ionic crystals. (ii) Ionic solids are hard and brittle.

Answer:

(i) The basis of similarities and differences between metallic and ionic crystals: Similarities: • Both metallic and ionic crystals are composed of atoms held together in a lattice structure. • Both types of crystals have strong intermolecular forces.

Differences: • Metallic crystals are composed of a lattice of metal atoms, while ionic crystals are composed of positively and negatively charged ions held together by electrostatic forces. • Metallic crystals are malleable and ductile, while ionic crystals are brittle and non-malleable.

(ii) Ionic solids are hard and brittle: Ionic solids are held together by strong electrostatic forces, which makes them very hard and brittle. This is because the electrostatic forces between the ions are so strong that they cannot easily be broken apart or bent. As a result, ionic solids are very difficult to deform or break.

Question:

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

Answer:

Step 1: Determine the formula for SrCl₂. SrCl₂ = Strontium Chloride

Step 2: Calculate the concentration of cation vacancies. Concentration of cation vacancies = 10−3 mol percent of SrCl₂ x 2 (for each SrCl₂ molecule there are two cation vacancies) = 0.002 mol percent of cation vacancies

Question:

Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium using its atomic mass 93 U.

Answer:

Step 1: Calculate the atomic weight of niobium in grams.

Atomic weight = 93 U x 1.66 x 10^-24 g/U = 1.54 x 10^-22 g

Step 2: Calculate the volume of a single atom of niobium.

Volume = Density x (Atomic Weight/Avogadro’s Number) Volume = 8.55 g/cm³ x (1.54 x 10^-22 g/6.023 x 10^23 atoms) Volume = 2.54 x 10^-24 cm³

Step 3: Calculate the radius of a single atom of niobium.

Radius = (Volume/4π)^1/3 Radius = (2.54 x 10^-24 cm³/4π)^1/3 Radius = 0.867 x 10^-8 cm

Question:

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

Answer:

1. Calculate the volume of the unit cell: V = (4/3)πr³ = (4/3)π(0.144 nm)³ = 0.0019 nm³

2. Calculate the length of the side of the unit cell: a = (V)^(1/3) = (0.0019 nm³)^(1/3) = 0.288 nm

Question:

In terms of band theory, what is the difference (i) between a conductor and an insulator? (ii) between a conductor and a semiconductor?

Answer:

(i) The difference between a conductor and an insulator is that a conductor has a band gap that is small enough that electrons can move freely, allowing for the flow of electric current, while an insulator has a band gap that is too large for electrons to move, preventing the flow of electric current.

(ii) The difference between a conductor and a semiconductor is that a conductor has a band gap that is small enough that electrons can move freely, allowing for the flow of electric current, while a semiconductor has a band gap that is intermediate in size, allowing for the flow of electric current but at a lower rate than a conductor.

Question:

How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Answer:

1. Measure the density of the metal by using a scale or a graduated cylinder.

2. Measure the dimension of the unit cell of the metal using a ruler or caliper.

3. Calculate the volume of the unit cell by using the formula V=a^3, where a is the length of the side of the unit cell.

4. Calculate the atomic mass of the metal using the formula m=dV, where d is the density of the metal and V is the volume of the unit cell.

5. The atomic mass of the metal is now determined.

Question:

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

Answer:

(i) Coordination number is the number of atoms, ions, or molecules that are in contact with a central atom in a molecule or crystal lattice.

(ii) (a) In a cubic close-packed structure, the coordination number of atoms is 12.

(b) In a body-centred cubic structure, the coordination number of atoms is 8.

Question:

Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Answer:

1. Non-stoichiometric cuprous oxide, Cu₂O, is a p-type semiconductor because the ratio of copper to oxygen is slightly less than 2:1. This causes an imbalance of electrons and holes in the material, resulting in the presence of mobile electrons and immobile holes, which are the characteristics of a p-type semiconductor.

2. The imbalance of electrons and holes is caused by the presence of extra oxygen atoms in the material, which creates a deficiency of electrons. This deficiency of electrons causes the material to be a p-type semiconductor.

3. The extra oxygen atoms in the material also create a deficiency of holes, which results in an overall decrease in the conductivity of the material. This is why non-stoichiometric cuprous oxide is a p-type semiconductor.

Question:

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

Answer:

Answer:

A semiconductor is a material that has properties between those of an insulator and a conductor. It can be used to control and amplify electrical signals and is found in many everyday electronic devices.

The two main types of semiconductors are Intrinsic Semiconductors and Extrinsic Semiconductors.

Intrinsic Semiconductors are made from a pure element such as silicon or germanium. They have very few free electrons and are relatively poor conductors of electricity. They are able to conduct electricity when exposed to certain energy sources such as light or heat.

Extrinsic Semiconductors are made from a combination of elements such as silicon and boron. They have many more free electrons than intrinsic semiconductors and are better conductors of electricity. They are able to conduct electricity without the need for any energy source.

The main difference between the two types of semiconductors is their conduction mechanism. Intrinsic semiconductors require an external energy source to conduct electricity, while extrinsic semiconductors are able to conduct electricity without the need for an external energy source.