• Diamond and graphite are — allotropes.

  • Diamond is — co-valent solid.

  • Number of atoms per unit cell:

    • Face-centred cubic (FCC) = 4 atoms

    • Body-centred cubic (BCC) = 2 atoms

  • Radius ratio for an ion to occupy tetrahedral site is between 0.225 to 0.414.

  • Isotropy refers to the property of a material where its physical properties are identical in all directions.

  • Anisotropy refers to the property of a material where its physical properties vary with direction.

  • Unit cell of a crystal is the smallest repeating unit of a crystal structure that, when repeated in space, forms the entire crystal.

  • Interstitials are atoms or ions that occupy spaces (interstices) between the regular lattice positions in a crystal.

  • Types of lattice imperfections found in crystals are:

    • Point defects (vacancies, interstitials, and substitutional defects)

    • Line defects (dislocations)

    • Planar defects (grain boundaries)

    1. Crystal lattice is a three-dimensional arrangement of points (atoms, ions, or molecules) in a repeating pattern, forming a structure that extends infinitely in space.

    2. Two types of close packing are known in crystals:

    • Hexagonal close packing (hcp)

    • Cubic close packing (ccp)

    1. Coordination number of a crystal is the number of nearest neighbor atoms or ions surrounding a central atom or ion in a crystal lattice.

    2. Interstitial voids are the empty spaces or gaps between the atoms or ions in a crystal lattice where smaller atoms or ions can be located.

    3. Types of lattice points in different cubic unit cells:

    • Simple cubic (SC) = 1

    • Body-centred cubic (BCC) = 2

    • Face-centred cubic (FCC) = 4

    1. Number of atoms in a body-centred cubic unit cell of a monoatomic substance is 2 atoms.

    2. Polycrystalline solid is a solid composed of many small crystals or crystallites, which are randomly oriented.

    3. Crystallites are small individual crystals within a polycrystalline material.

    4. Number of atoms in a body-centred cubic (BCC) cell is 2 atoms.

    5. Examples of ionic solids:

    • Sodium chloride (NaCl)

    • Magnesium oxide (MgO)

    1. Diamond, Graphite, and Quartz are — network solids.

    2. Metallic solids are solids in which atoms are held together by metallic bonds, with a sea of delocalized electrons, giving them properties like electrical conductivity and malleability.

    3. Molecular solids are solids where molecules are held together by intermolecular forces such as van der Waals forces or hydrogen bonding.

    4. Radius ratio is the ratio of the radius of the smaller ion to the radius of the larger ion in an ionic crystal.

    5. Carbon tetrachloride is immiscible in water because it is a non-polar molecule, while water is polar, leading to a lack of attraction between the two.

    6. The solubility of a solid not changed by increase of pressure.

    7. Effect of pressure on the solubility of a solid: The solubility of most solids is not affected significantly by changes in pressure.

    8. Solubility of calcium acetate generally increases with temperature.

    9. Examples of molecular solids:

    • Ice (solid water)

    • Dry ice (solid CO₂)

    1. Examples of hcp and bcc crystals:

    • hcp: Magnesium (Mg)

    • bcc: Iron (Fe)

    1. The commercial name of SiC is Carborundum.

    2. Coordination numbers of Cs+ & Cl– in CsCl lattice:

    • Cs+ has a coordination number of 8

    • Cl– has a coordination number of 8

  • Difference between crystalline and amorphous solids:

    • Crystalline solids have a well-defined, ordered structure and their atoms are arranged in a regular, repeating pattern. They have sharp melting points.

    • Amorphous solids lack long-range order, and their atoms are arranged randomly. They do not have a sharp melting point but soften over a range of temperatures.

  • Coordination number and crystal geometry for cation and anion radii:

    • Given radii:

      • Radius of cation = 95 pm

      • Radius of anion = 181 pm

    • The radius ratio rcationranion=95181=0.524\frac{r_{cation}}{r_{anion}} = \frac{95}{181} = 0.524.

      • For a radius ratio between 0.414 and 0.732, the coordination number is 6, and the crystal geometry is octahedral (like NaCl).

  • hcp and ccp giving the same identity:

    • hcp (hexagonal close packing) and ccp (cubic close packing) both have the same packing efficiency of about 74%, and both arrangements are based on a repeating pattern of atoms in space. In both types of packing, each atom is surrounded by 12 nearest neighbors, leading to the same coordination number and identical crystal structures. The only difference is the arrangement of layers: ccp has a face-centered cubic arrangement, while hcp has an ABAB stacking pattern.

  • Ideal radius of the anion in NaCl structure:

    • In a NaCl type of structure, the ideal radius ratio of the cation to anion is around 0.414 to 0.732 for a stable structure.

    • If the radius of the cation is r, then the ideal radius of the anion would be r × 1.414 (using the ideal ratio of 1:√2 for the geometry of NaCl).

  • Radius ratio and coordination number for Li+ and F- in LiF:

    • Given data:

      • rLi+=60 pmr_{\text{Li}^+} = 60 \, \text{pm}

      • rF−=136 pmr_{\text{F}^-} = 136 \, \text{pm}

    • Radius ratio:

      rLi+rF−=60136=0.441\frac{r_{\text{Li}^+}}{r_{\text{F}^-}} = \frac{60}{136} = 0.441

    • The radius ratio 0.441 falls within the range of 0.414 to 0.732, indicating a coordination number of 6, and the crystal structure would be octahedral (like NaCl).

  1. Given Data:

    • The unit cell is fcc (Face-Centered Cubic)

    • Edge length of the unit cell a=408.6 pm=408.6×10−12 ma = 408.6 \, \text{pm} = 408.6 \times 10^{-12} \, \text{m}

    • Density of silver ρ=10.5 g/cm3\rho = 10.5 \, \text{g/cm}^3

    • We need to calculate the atomic mass (M) of silver.

  2. FCC Unit Cell:

    • In an fcc unit cell, there are 4 atoms per unit cell.

    • Volume of the unit cell Vcell=a3V_{\text{cell}} = a^3.

  3. Density Formula:
    The formula to relate the density ρ\rho, the number of atoms in the unit cell, the atomic mass, and the unit cell volume is:

    ρ=n×MVcell×NA\rho = \frac{n \times M}{V_{\text{cell}} \times N_A}

    Where:

    • nn is the number of atoms in the unit cell (4 for fcc),

    • MM is the atomic mass of silver (what we need to find),

    • VcellV_{\text{cell}} is the volume of the unit cell,

    • NAN_A is Avogadro's number (6.022×1023 atoms/mol6.022 \times 10^{23} \, \text{atoms/mol}).

  4. Calculate the Volume of the Unit Cell:

    Vcell=a3=(408.6×10−12 m)3=6.799×10−29 m3V_{\text{cell}} = a^3 = (408.6 \times 10^{-12} \, \text{m})^3 = 6.799 \times 10^{-29} \, \text{m}^3

    To convert the volume to cm3\text{cm}^3 (since the density is given in g/cm3\text{g/cm}^3), we use the conversion factor 1 m3=106 cm31 \, \text{m}^3 = 10^6 \, \text{cm}^3:

    Vcell=6.799×10−29 m3=6.799×10−23 cm3V_{\text{cell}} = 6.799 \times 10^{-29} \, \text{m}^3 = 6.799 \times 10^{-23} \, \text{cm}^3

  5. Rearranging the Density Formula:
    Solving for MM (the atomic mass of silver):

    M=ρ×Vcell×NAnM = \frac{\rho \times V_{\text{cell}} \times N_A}{n}

  6. Substitute Values:

    M=10.5 g/cm3×6.799×10−23 cm3×6.022×1023 atoms/mol4M = \frac{10.5 \, \text{g/cm}^3 \times 6.799 \times 10^{-23} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{atoms/mol}}{4} M=10.5×6.799×6.0224 g/molM = \frac{10.5 \times 6.799 \times 6.022}{4} \, \text{g/mol} M=107.87 g/molM = 107.87 \, \text{g/mol}

Answer:

The atomic mass of silver is approximately 107.87 g/mol.

  1. Given Data:

    • The unit cell is fcc (Face-Centered Cubic)

    • Edge length of the unit cell a=408.6 pm=408.6×10−12 ma = 408.6 \, \text{pm} = 408.6 \times 10^{-12} \, \text{m}

    • Density of silver ρ=10.5 g/cm3\rho = 10.5 \, \text{g/cm}^3

    • We need to calculate the atomic mass (M) of silver.

  2. FCC Unit Cell:

    • In an fcc unit cell, there are 4 atoms per unit cell.

    • Volume of the unit cell Vcell=a3V_{\text{cell}} = a^3

  3. Density Formula:
    The formula to relate the density ρ\rho, the number of atoms in the unit cell, the atomic mass, and the unit cell volume is:

    ρ=n×MVcell×NA\rho = \frac{n \times M}{V_{\text{cell}} \times N_A}

    Where:

    • nn is the number of atoms in the unit cell (4 for fcc),

    • MM is the atomic mass of silver (what we need to find),

    • VcellV_{\text{cell}} is the volume of the unit cell,

    • NAN_A is Avogadro's number (6.022×1023 atoms/mol6.022 \times 10^{23} \, \text{atoms/mol}).

  4. Calculate the Volume of the Unit Cell:

    Vcell=a3=(408.6×10−12 m)3=6.799×10−29 m3V_{\text{cell}} = a^3 = (408.6 \times 10^{-12} \, \text{m})^3 = 6.799 \times 10^{-29} \, \text{m}^3

    To convert the volume to cm3\text{cm}^3 (since the density is given in g/cm3\text{g/cm}^3), we use the conversion factor 1 m3=106 cm31 \, \text{m}^3 = 10^6 \, \text{cm}^3

    Vcell=6.799×10−29 m3=6.799×10−23 cm3V_{\text{cell}} = 6.799 \times 10^{-29} \, \text{m}^3 = 6.799 \times 10^{-23} \, \text{cm}^3

  5. Rearranging the Density Formula:
    Solving for MM (the atomic mass of silver):

    M=ρ×Vcell×NAnM = \frac{\rho \times V_{\text{cell}} \times N_A}{n}

  6. Substitute Values:

    M=10.5 g/cm3×6.799×10−23 cm3×6.022×1023 atoms/mol4M = \frac{10.5 \, \text{g/cm}^3 \times 6.799 \times 10^{-23} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{atoms/mol}}{4} M=10.5×6.799×6.0224 g/molM = \frac{10.5 \times 6.799 \times 6.022}{4} \, \text{g/mol}
    M=107.87 g/molM = 107.87 \, \text{g/mol}

Answer:

The atomic mass of silver is approximately 107.87 g/mol.

1. Four important characteristics of solids:

  • Definite Shape and Volume: Solids have a fixed shape and volume due to the close packing of particles.

  • High Density: Solids generally have higher density than liquids and gases due to the closely packed arrangement of particles.

  • Incompressibility: Solids are rigid and resist changes in shape or volume when pressure is applied.

  • Strong Intermolecular Forces: The particles in solids are held together by strong forces, such as ionic, covalent, or metallic bonds, giving them their structural stability.

2. Two points to distinguish between Crystalline and Amorphous solids:

  • Order of Structure:

    • Crystalline solids have a well-defined, regular, and repeating arrangement of atoms or molecules in a three-dimensional lattice.

    • Amorphous solids have a random arrangement of particles, lacking long-range order.

  • Melting Point:

    • Crystalline solids have a sharp, definite melting point.

    • Amorphous solids do not have a sharp melting point and instead soften over a range of temperatures.

3. Chief characteristics of ionic crystals:

  • Strong Electrostatic Forces: Ionic crystals are held together by strong electrostatic forces of attraction between oppositely charged ions (cation and anion).

  • Hard and Brittle: Due to the strong ionic bonds, ionic crystals are hard but brittle and break along cleavage planes when stress is applied.

  • High Melting and Boiling Points: The strong forces between ions result in high melting and boiling points.

  • Electrical Conductivity: Ionic crystals conduct electricity only when they are molten or dissolved in water, as ions are free to move in these states.

4. General characteristics of covalent crystals:

  • Strong Covalent Bonds: The atoms in covalent crystals are held together by strong covalent bonds.

  • Hard and Rigid: These crystals are usually hard and rigid due to the presence of strong covalent bonds throughout the structure.

  • High Melting Points: Covalent crystals generally have high melting points because it requires a large amount of energy to break the covalent bonds.

  • Non-conductors: Covalent crystals do not conduct electricity as there are no free electrons or ions to carry charge.

  • Examples: Diamond, quartz (SiO₂), and graphite.

5. Molecular crystals and their types:

  • Molecular Crystals: These crystals consist of molecules held together by weaker intermolecular forces such as Van der Waals forces, dipole-dipole interactions, or hydrogen bonding, rather than by strong covalent or ionic bonds.

  • Types of Molecular Crystals:

    1. Non-polar molecular crystals: Held together by London dispersion forces (e.g., solid noble gases like Argon, and non-polar compounds like iodine).

    2. Polar molecular crystals: Held together by dipole-dipole interactions (e.g., HCl).

    3. Hydrogen-bonded molecular crystals: Held together by hydrogen bonds (e.g., ice, H₂O).

6. Metallic crystals and their properties:

  • Metallic Crystals: Metallic crystals are made up of metal atoms that are held together by metallic bonds, where electrons are delocalized and move freely within the structure.

  • Properties:

    • Electrical Conductivity: Metals conduct electricity due to the presence of free electrons that move throughout the lattice.

    • Malleability and Ductility: Metals can be hammered into sheets (malleable) or drawn into wires (ductile) because the layers of atoms can slide over one another without breaking the metallic bond.

    • Luster: Metallic crystals have a shiny appearance due to the reflection of light from the delocalized electrons.

    • High Melting and Boiling Points: Metals generally have high melting and boiling points due to strong metallic bonds.

    • Examples: Iron, copper, aluminum.

7. Space lattice and unit cell:

  • Space Lattice: A space lattice is a three-dimensional arrangement of points (atoms, ions, or molecules) in space, which repeats periodically. Each point in the lattice represents the position of one particle, and the lattice forms the overall structure of the crystal.

  • Unit Cell: The unit cell is the smallest repeating unit or building block of a crystal. It is the portion of the lattice that, when repeated in space, forms the entire crystal. The unit cell contains the complete symmetry of the crystal and helps define its overall geometry and properties.

(vi) State Kohlrausch law.
(vii) What is electrode potential ?
(viii) What is the unit of equivalent conductance ?
(ix) Define electrochemical equivalent.
(x) What is the unit of molar conductance ?
(xi) What is cell constant ?
(xii) What is electromotive force ?
(xiii) In the outside circuit of a cell electron flows from which electrode ?
(xiv) Which gas is liberated at anode due to electrolysis of molten sodium hydride ?
(xv) Which gas is liberated at cathode due to the electrolysis of aqueous solution of
sodium chloride ?
(xvi) What is the unit of e.m.f. ?
(xvii) What is an electrochemical cell ?
(xviii) Define degree of dissociation of an electrolyte ?
(xix) What is the unit of specific conductance ?
(xx) What is the basis on which anode or cathode is identified in a cell ?

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