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    Tyler
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    Q1.
    Which of the following statements regarding crystalline and amorphous solids is incorrect?
    A : Amorphous solids are of irregular shape.
    B : Crystalline solids are isotropic in nature.
    C : Amorphous solids soften over a range of temperature.
    D : Crystalline solids have characteristic geometrical shape.

    Solution
    Crystalline solids are not isotropic in nature. They are anisotropic in nature. That is, some of their physical properties such as electrical resistance or refractive index exhibit different values, when measured along different directions in the same crystals. The different arrangement of particles in different directions results in anisotropic nature of crystalline solids.

    The correct answer is B.

    Q.2
    Which of the following alternatives correctly classifies the materials as crystalline and amorphous solids?
    A : Crystalline Amorphous
    Sodium chloride Plastic

    B : Crystalline Amorphous
    Rubber Glass

    C : Crystalline Amorphous
    Polyvinyl chloride Benzoic acid

    D : Crystalline Amorphous
    Glass Benzoic acid

    Solution
    On the basis of the nature of order of arrangement of constituent particles, solids can be classified as crystalline and amorphous.
    Crystalline solids are solids consisting of large number of small crystals, each having a definite characteristic geometrical shape. The constituent particles are orderly arranged in a crystal of crystalline solid.
    Amorphous solids are solids consisting of particles of irregular shape. The arrangement of constituent particles in an amorphous solid has only short range order.
    Sodium chloride and benzoic acid are crystalline solids while plastic, rubber, glass, and polyvinyl chloride are amorphous solids.
    The correct answer is A.

    ________________________________________
    Q3.
    Use the following information to answer the next question.
    Ice is a molecular solid while SO2 is a molecular solid.
    The information in which alternative completes the given statement?
    A : i ii
    polar hydrogen bonded

    B : i ii
    polar non-polar

    C : i ii
    non-polar polar

    D : i ii
    hydrogen bonded polar

    Solution
    Molecular solids are divided into the following three categories.
    (i) Non-polar molecular solids
    (ii) Polar molecular solids
    (iii) Hydrogen bonded molecular solids
    Non-polar molecular solids comprise of either atoms (eg. Ar, He) or molecules formed by non-polar covalent bonds (eg. H2, Cl2, I2). In these solids, the atoms or molecules are held by weak dispersion forces, called London forces.
    Polar molecular solids comprise of molecules of substances such as HCl, SO2, which are formed by polar covalent bonds. In these solids, the molecules are held together by relatively stronger dipole-dipole interactions.
    Hydrogen bonded molecular solids comprise of molecules containing polar covalent bonds between H and F, O or N atoms. In these solids such as H2O (ice), the molecules are held together by strong hydrogen bonds.
    The correct answer is D.
    ________________________________________
    Q4.
    Which of the following solids is a covalent solid?
    A : CCl4
    B : SiO2
    C : ZnS
    D : CO2

    Ans
    Covalent solids are crystalline solids of non-metals resulting from the formation of covalent bonds between adjacent atoms throughout the crystal. Examples include SiO2 (quartz), SiC, C (diamond), etc.
    The correct answer is B.
    ________________________________________
    Q5.
    Use the following information to answer the next question.
    A compound is formed by two elements X and Y. The atoms of element X form hcp lattice and those of element Y occupy 1/3 of the tetrahedral voids.
    The formula of the given compound formed is
    A : XY
    B : XY2
    C : X2Y
    D : X3Y2

    solution
    It is known that the number of tetrahedral voids formed is equal to twice the number of atoms of element X.
    Given that,
    Only 1/3 of the tetrahedral voids are occupied by the atoms of element Y.
    Therefore, ratio of the number of atoms of X and Y = 1 : 2 × 1/3
    = 3 : 2
    Hence, the formula of the compound formed is X3Y2.
    The correct answer is D.
    ________________________________________

    Q6.
    The number of particles per unit cell of face-centred cubic (fcc) lattice is
    A : 1
    B : 2
    C : 4
    D : 5

    Solution
    In one unit cell of face-centred cubic lattice, there are 8 corner atoms and 6 face-centred atoms

    Particles per unit cell = 4
    The correct answer is C.
    ________________________________________
    Q7.
    Use the following information to answer the next question.
    i substances are strongly attracted while ii substances are weakly attracted by magnetic fields.
    The information in which alternative completes the given statement?
    A : i ii
    Ferromagnetic paramagnetic

    B : i ii
    Paramagnetic anti-ferromagnetic

    C : i ii
    Diamagnetic ferromagnetic

    D : i ii
    Paramagnetic ferromagnetic

    Solution
    Ferromagnetic substances are strongly attracted while paramagnetic substances are weakly attracted by a magnetic field. Fe, Co, Ni, Gd, CrO2 are some examples of ferromagnetic substances. They can be permanently magnetized.
    Paramagnetic substances are magnetized in a magnetic field in the same direction. In the absence of magnetic field, these substances lose their magnetism. O2, Cu2+, Fe3+, Cr3+ are some examples of paramagnetic substances.
    The correct answer is A.
    ________________________________________
    Q8.
    Use the following information to answer the next question.
    When a crystal of Si is doped with i , it leads to the formation of ii semi-conductor.
    The information in which alternative completes the given statement?
    A : i Ii
    B n-type

    B : i ii
    Al p-type

    C : i ii
    P p-type

    D : i ii
    Ga n-type

    Solution
    The process of increasing conductivity of intrinsic conductors by adding an appropriate amount of suitable impurity is called doping. Doping can be done with an impurity, which is electron rich or electron deficient as compared to the intrinsic semi-conductor. Silicon and germanium are the intrinsic semi-conductors. The doping of crystals of Si or Ga with a group 13 element (electron deficient) such as B, Al, or Ga leads to the creation of p-type semi-conductor while with a group 15 element (electron rich) such as P or As leads to the creation of n-type semi-conductor.

    The correct answer is B.
    ________________________________________

    Q9.
    Which of the following defects can arise when a solid is heated?
    A : Frenkel defect
    B : Vacancy defect
    C : Schottky defect
    D : Interstitial defect
    Solution
    Vacancy defect in solids can arise when solids are heated. This is because on heating, some atoms or ions leave the lattice sites completely. As a result, the lattice sites are vacant. This defect leads to decrease in density of the substance.

    The correct answer is B.

    Question10
    Which type of unit cell does cinnabar (HgS) correspond to?
    Solution:
    Cinnabar (HgS) adopts a hexagonal unit cell.

    Question11
    Which type of molecular solid can CCl4 be classified into?
    Solution:
    CCl4 can be classified into a covalent molecular solid.

    Question 12
    What is the formula of a compound having a cubic structure formed by elements A and B in which the atoms of A are present at the corner and the atoms of B are present at the face centres of the cube?
    Solution:
    If the atoms of Aare present at the 8 corners of the cube, then the number of atoms present in the unit cell will be .
    If the atoms of B are present at the 6 face centres of the cube, then the number of atoms present in the unit cell will be 3 .
    Thus, the ratio of the atoms of A to the atoms of B is 1:3.
    Hence, the formula of the compound is AB3.

    Question 13
    What is meant by the term ‘coordination number’?
    Solution:
    The number of closest (or nearest) neighbours of any constituent particle in the crystal lattice is called the coordination number.

    Question 14
    In corundum, the oxide ions are arranged in hexagonal close-packing, while aluminium ions occupy two-thirds of the octahedral voids. What is the formula of corundum?
    Solution:
    Let the number of octahedral voids be N.
    Accordingly,
    Number of octahedral voids occupied by Al3+ ions = 2N/3
    In H.C.P., there is one octahedral void corresponding to each atom that constitutes the close-packing.
    Ratio of Al3+: 2:3
    Hence, the formula of corundum is Al2O3.

    Question 15
    Why does urea exhibit a definite heat of fusion, while glass exhibits an indefinite heat of fusion?
    Solution:
    Urea is a crystalline solid, whereas glass is an amorphous solid. Crystalline solids possess sharp melting points, while amorphous solids gradually soften over a temperature range. It is for this reason that urea has a definite heat of fusion, whereas glass does not.

    Question 16
    Calculate the number of atoms present in a Face-Centred Cubic Lattice (FCC).
    Solution:
    A Face-Centred Cubic Lattice contains 8 atoms at the corner and 6 atoms at the centre of faces.
    Contribution by atoms present at the corner = (1/8)*8 = 1
    Contribution by atoms present at the faces = (1/2)*6 = 3
    Thus, the total number of atoms in one unit cell can be calculated as:
    8 corner atoms *1/8 atom per unit cell + 6 face-centred atoms *1/2 atom per unit cell
    Number of atoms present in the unit cell = 1 + 3 = 4
    Hence, the number of atoms present in a Face-Centred Cubic Lattice (FCC) is 4.

    Question 17
    a. What is the total number of voids present in 1 mole of a compound forming hexagonal close-packed structure?
    b. Write the coordination number of each ion in CaF2.
    Solution:
    a. The number of atoms in the close-packing is 1 mole i.e., 1 × 6.022 × 1023 = 6.022 × 1023 atoms
    In closed-packing:
    Number of octahedral voids = Number of atoms = 6.022 × 1023
    Number of tetrahedral voids = 2 × Number of atoms in the close-packing
    = 2 × 6.022 × 1023
    = 12.044 × 1023
    Thus, total number of voids =6.022 × 1023 + 12.044 × 1023 = 18.066 × 1023
    Hence, the total number of voids in 1 mole of a compound forming hexagonal close-packed structure is 18.066 × 1023.
    b. In a molecule of CaF2, the coordination number of Ca2+is 8 and that of F−is 4.

    • This topic was modified 3 years, 2 months ago by  admin.
    #8622

    administrator
    Participant

    Good numerical on solid state chemistry, I would love to offer similar kind of help to the students in solid chemistry.

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