The Central Board of Secondary Education, CBSE, has scheduled the Chemistry Class 12th Examination on March 7, 2020. The students who are appearing for the CBSE Class 12th Chemistry Examination 2020 can go through the important questions and answers of the chapter Solid State. These important questions are based on the latest CBSE Class 12th Chemistry Syllabus.
Key Points to be mentioned while writing the answers to the below mentioned important questions:
Question 1- Determine the atomic mass of an unknown metal if you are given the mass density and the dimensions of the unit cell of its crystals.
Answer: Volume of unit cell = (a pm)3
= a3 x 10-30 cm3
Density of unit cell = Mass of unit cell/ Volume of unit cell
Mass of unit cell = Number of atoms in the unit cell x mass of each atom
= Z x m
Mass of each atom = M/ No
Molar mass can be calculated, M= d x a3 x No / Z
Question 2- Define the following terms:
- Unit Cell
- Coordination number
- Ferromagnetism and Ferrimagnetism
Answer: Unit Cell: It is the smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice.
The characteristics of a unit cell are as follows:
i) its dimensions along the three edges, a, b and c. The edges may or may not be mutually perpendicular.
ii) angles between the edges, α (between b and c) β (between a and c) and γ (between a and b).
Thus, a unit cell is characterised by six parameters, a, b, c, α, β and γ.
A unit cell is divided into two broad categories:
(i) Primitive Unit Cell: Constituent particles are present only at the corner positions of a unit cell.
(ii) Centred Unit Cell: One/ more constituent particles are present at positions other than corners in addition to those at corners.
The centred Unit cell is further divided into 3 types:
(i) Body-Centred Unit Cells: One constituent particle (atom, molecule or ion) at its body-centre besides the ones that are at its corners.
(ii) Face-Centred Unit Cells: One constituent particle present at the centre of each face, besides the ones that are at its corners.
(iii) End-Centred Unit cells: One constituent particle is present at the centre of any two opposite faces besides the ones present at its corners.
(For more details refer to NCERT textbook page number 7)
Coordination number: The number of nearest neighbours of a particle.
(i) In a one-dimensional close-packed arrangement, the coordination number is 2.
(ii) In the two-dimensional square close-packed arrangement, the coordination number is 4.
(iii) In the two-dimensional hexagonal close-packed arrangement, the coordination number is 6.
(For more details refer to NCERT textbook page number 12)
Ferromagnetism: The substances which are strongly attracted by magnetic fields and do not lose magnetism when the external magnetic field is removed. They can be permanently magnetised. Ex: Fe, Co, etc.
Ferrimagnetism: The substances which are weakly attracted by magnetic fields and lose magnetism upon heating becoming paramagnetic. Alignment of magnetic domains (a group of metal ions) in the parallel and anti-parallel direction with the direction of the magnetic field and in unequal in number. Ex. Fe3O4 (Magnetite), MgFe2O4 (Ferrite)
(For more details on Ferromagnetism and Ferrimagnetism, refer to NCERT textbook page number 28)
Semiconductor: Solids whose electrical conductivity lies between the typical metallic conductors and insulators. Possess conductivities in the intermediate range from 10–6 to 104 ohm–1m–1. They are of two types:
(i) n-type semiconductors
(ii) p-type semiconductors
Question 3- If NaCl is doped with 10-3 mole percent SrCl2, what will be the concentration of cation vacancies? (NA = 6.02 x 1023 mol-1)
Answer: 6.02 × 1018 mol-1
Question 4- Iron has a body-centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is 7.87 g cm-3. Use this information to calculate Avogadro’s number (At. mass of Fe = 56 g mol-1)
Answer: Avogadro’s number NA = 6.022 × 1023
Hint- Use the formula: d= Z x M/ a3 x NA
Question 5- The density of lead is 11.35 g cm-3 and the metal crystallizes with fee unit cell. Estimate the radius of the lead atom.
(At. Mass of lead = 207 g mol-1 and NA = 6.02 × 1023 mol-1)
Answer: Radius, r = 1.75 x 10-8 cm= 175 pm
Hint- Use the formulas: d= Z x M/ a3 x NA and for f.c.c unit cell, r = a/ 2 √2
Question 6- An element crystallizes in a b.c.c. lattice with a cell edge of 500pm. The density of the element is 7.5g cm-3. How many atoms are present in 300 g of the element?
Answer: 6.40 x 10 23 atoms
Hint- Use the formula: d= Z x M/ a3 x NA ; for b.c.c, z=2
Question 7- Silver crystallises in f.c.c. lattice. If the edge length of the cell is 4.07 × 10-8 cm and density is 10.5 g cm-3, calculate the atomic mass of silver.
Answer: M= 106.6 g mol -1
Hint- Use the formula: d= Z x M/ a3 x NA ; for f.c.c, z=4
Question 8- An element crystallises in bcc lattice with a cell edge of 400 pm. Calculate its density if 250 g of this element contains 2.5 × 1024 atoms.
Answer: d= 3.125 g cm-3
Hint- Use the formula: d= Z x M/ a3 x NA ; for b.c.c, z= 2
Question 9- An element has an atomic mass 93 g mol-1 and density 11.5 g cm-3. If the edge length of its unit cell is 300 pm, identify the type of unit cell.
Answer: Body centred cubic unit cell.
Hint- Use formula- Z= ρ x a3 x NA / M
Question 10- Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in an f.c.c. structure. (Atomic mass of Al = 27 g mol-1)
Answer: 4.5 x 1022 unit cells
Hint- Use formula- No. of atoms = η x NA
Question 11- Define:
- Frenkel defect
- Schottky defect
Answer: Frenkel defect: It is shown by ionic solids. The smaller ion is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. Frenkel defect is also called a dislocation defect. It does not change the density of the solid. Ex: ZnS, AgCl, etc.
F-Centre: The anionic sites occupied by unpaired electrons are called F-centres. They impart a yellow colour to the crystals of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Similarly, excess of lithium makes LiCl crystals pink and excess of potassium makes KCl crystals violet (or lilac).
Schottky defect: It is basically a vacancy defect in ionic solids and to maintain electrical neutrality, the number of missing cations and anions are equal. It decreases the density of the substance. Ex: In NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 1022 ions. Thus, there is one Schottky defect per 1016 ions. The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, NaCl, KCl, CsCl and AgBr.
The above-mentioned questions for CBSE Class 12th Chemistry Chapter 1- The Solid State are based on the previous year question papers, NCERT Textbook and sample papers. The students appearing for CBSE Class 12th Chemistry Examination 2020 are advised to go through the links below: