Students can read the important questions given below for The Solid State Class 12 Chemistry. All The Solid State Class 12 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 12 Chemistry Important Questions provid ed for all chapters to get better marks in examinations. Chemistry Question Bank Class 12 is available on our website for free download in PDF.
Important Questions of The Solid State Class 12
Very Short Answer Questions
Question. Why are crystalline solids anisotropic?
Answer. Crystalline solids are anisotropic in nature because of different arrangements of particles in different directions. Some of their physical properties like electrical conductivity, refractive index, etc., shows different values in different directions in the same crystals.
Question. Some of the glass objects recovered from ancient monuments look milky instead of being transparent. Why?
Answer. Some of the glass objects from ancient monuments look milky instead of being transparent because it undergoes heating during the day and cooling at night i.e., annealing over a number of years. As a result, it acquires some crystalline character. they become a bit opaque.
Question. Why is glass considered supercooled liquid?
Answer. Glass is considered as super cooled liquid because glass is an amorphous solid and has tendency to flow very slowly like liquids.
Question. Give an example each of a molecular solid and an ionic solid.
Answer. Molecular solids : HCl, SO2, H2O
Ionic solids : KCl, CuSO4, ZnS
Question. Which one of the following is an example of molecular solid : CO2 or SiO2?
Answer. Carbon dioxide (CO2) is a molecular solid.
Quartz (SiO2) is a covalent solid.
Question. How will you distinguish between the following pair of terms :
Crystal lattice and unit cell?
Answer. Arrangement of lattice points in three dimensional space is called crystal lattice. The smallest repeating unit which represents arrangement of lattice points in a crystal lattice is called unit cell.
Question. Name the parameters that characterise a unit cell.
Answer. A unit cells is characterised by the two types of parameters :
(i) Dimensions along the three edges represented as a, b and c.
(ii) Angles between edges a, β and g.
Hence, a unit cell is represented by six parameters a, b, c, a, β and Υ.
Question. How many atoms constitute one unit cell of a face-centred cubic crystal?
Answer. In face centred cubic unit cell
(i) No. of atoms contributed from the corners of the unit cell
= 8 corner atoms × 1/8
atom per unit cell
Question. How many effective sodium ions are located at the edge centre of a unit cell in a sodium chloride crystal?
Answer.
Question. What is the coordination number of each type of ions in a rock-salt type crystal structure?
Answer. In a rock salt type crystal structure, coordination number is 6 : 6
⇒ Coordination number of M+ = 6
Coordination number of A– = 6
Question. Express the relationship between atomic radius
(r) and the edge length (a) in the bcc unit cell.
Answer.
Question. Express the relationship between atomic radius
(r) and the edge length (a) in the fcc unit cell.
Answer.
Question. Define the following term : Schottky defect
Answer. Schottky defect : e defect in which cations and anions are missing in the stoichiometric ratio of compound is called Schottky defect. Schottky defect is actually vacancy defect in ionic solids. In this defect, electrical neutrality is maintained. In this defect, density of solid decreases. Schottky defect is shown by ionic solids in which the cation and anion are of almost similar sizes. Examples : NaCl, KCl, CsCl, AgBr, etc.
Question. Which stoichiometric defect does not change the density of the crystal?
Answer. Frenkel defect does not change the density of the solid.
Question. Which point defect of its crystals decreases the density of a solid?
Answer. Schottky defect.
Question. What is meant by ‘doping’ in semiconductor?
Answer. Addition of an appropriate amount of suitable impurity in a crystalline solid is called doping. Doping is done to increase the conductivity of intrinsic semiconductors.
Question. How may the conductivity of an intrinsic semiconductor be increased?
Answer. The conductivity of intrinsic semiconductor like silicon is too low to be of practical use. This conductivity is increased by adding an appropriate amount of suitable impurity like Al or As which is electron deficient or electron rich. So, the electrical conductivity of silicon is increased.
Question. What are n-type semiconductors?
Answer. A semiconductor in which doped impurity has more valence electrons than the pure element is called n-type semiconductor. e.g., Ge or Si (Group-14) doped with P or As (Group-15).
Question. What is meant by an ‘intrinsic semiconductor’?
Answer. Pure substance which acts as semiconductor is called intrinsic semiconductor. e.g., Si and Ge at high temperature.
Question. Assign reason for the following : Phosphorus doped silicon is a semiconductor.
Answer. Phosphorus is pentavalent. When phosphorus is doped in silicon it increases the number of electrons which increases the conductivity of doped silicon and makes it semiconductor.
Question. Write the type of magnetism observed when the magnetic moments are aligned in parallel and anti-parallel directions in unequal numbers.
Answer. Ferrimagnetism
Question. What is meant by ‘anti ferromagnetism’?
Answer. Antiferromagnetism : Paramagnetic substances which show very poor magnetic effect are antiferromagnetic and this property is called anti ferromagnetism.
Examples : MnO, Fe2O3, Cr2O3.
Question. Account for the following : Fe3O4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K.
Answer. On heating at 850 K, the magnetic domains undergo realignment and become paramagnetic
Question. Explain the following with suitable example : Paramagnetism
Answer. Paramagnetism : Materials which are weakly attracted by magnetic elds are called paramagnetic materials and this property is called paramagnetism.
Paramagnetic substances contain unpaired electrons.
e.g., TiO, CuO, O2 and VO2, etc.
Short Answer Questions
Question. A compound forms hcp structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Answer. Number of atoms in 0.5 mol hcp lattice
= 0.5 × 6.022 × 1023 = 3.011 × 1023
∴ Number of octahedral voids = Number of atoms
= 3.011 × 1023
Number of tetrahedral voids = 2 × Number of atoms
= 2 × 3.011 × 1023 = 6.022 × 1023
Total number of voids = Number of octahedral voids
+ Number of tetrahedral voids
= 3.011 × 1023 + 6.022 × 1023
= 9.033 × 1023 voids
Question. An alloy of gold and cadmium crystallises with a cubic structure in which gold atoms occupy the corners and cadmium atoms fit into the face centres. Assign formula for this alloy.
Answer. When the particles are present not only corners but also at the centre of each face of the unit cell. It is called face-centred unit cell.
Number of atoms in a unit cell of the face-centredcubic (fcc) ⇒ 8 atoms on the corners (gold) and
6 atoms of cadmium on the face (one of each face). Contribution by atoms (gold) on the corners
Question. An element with density 2.8 g cm–3 forms a fcc unit cell with edge length 4 × 10–8 cm. Calculate the molar mass of the element. (Given : NA = 6.022 × 1023 mol–1)
Answer. Given, Density of solid, d = 2.8 g cm–3
For fcc unit cell, Z = 4
Edge length, a = 4 × 10–8 cm,
Molar mass, M = ?
NA = 6.022 × 1023 mol–1
Question. An element crystallises in structure having fcc unit cell of an edge 200 pm. Calculate the density if 200 g of this element contains 24 × 1023 atoms.
Answer. Edge length = 200 pm
Volume of the unit cell = (200 × 10–10 cm)3
= 8 × 10–24 cm3
In a fcc unit cell there are four atoms per unit cell.
Question. Explain how we can determine the atomic mass of an unknown metal if you know its density and the dimension of unit cell of its crystal.
Answer. We can determine the atomic mass of an unknown metal by using the formula of density of its unit cell. d (Density)
By knowing density d, dimension of unit cell we can calculate M, the atomic mass of metal as for a given unit cell, Z is Fixed and NA is a universal constant.
Question. (i) What type of non-stoichiometric point defect is responsible for the pink colour of LiCl? (ii) What type of stoichiometric defect is shown by NaCl?
Answer. (i) Metal excess defect is responsible for pink colour of LiCl. It is also known as anion vacancy defect.
(ii) NaCl crystal shows Schottky defect.
Question. If NaCl is doped with 10–3 mole percent SrCl2, what will be the concentration of cation vacancies? (NA = 6.02 × 1023 mol–1)
Answer. The number of cation vacancies created in the lattice of NaCl is equal to the number of divalent Sr2+ ions added.
Concentration of Sr2+ = 10–3 mol%
1 mole of Sr2+ = 6.023 × 1023Sr2+ ions
10–5 mol of Sr2+ = 6.023 × 1023 × 10–5
= 6.023 × 1018 Sr2+ ions.
Hence, the concentration of cation vacancies is
6.023 × 1018
Question. A metal crystallises in a face centred cubic unit cell with a = 0.560 nm. Calculate the density of the metal if it contains 0.1% schottky defects.
Given : Atomic mass of metal = 40 g mol–1)
Answer. Due to Schottky defects, the vacant spaces will increase resulting in decrease in number of atoms per unit cell.
Question. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer. The solids which have conductivities between 10–6 to 104 ohm–1 m–1 are called semiconductors. e.g., Germanium and silicon. The two main types of semiconductors are as follows :
(i) n-type semiconductor : When a silicon crystal is doped with atoms of group-15 elements, such as P, As, Sb or Bi then only four of the five valence electrons of each impurity atom participate in forming covalent bonds and fifth electron is almost free to conduct electricity. Group-14 elements doped with a group-15 elements are called n-type semiconductors.
(ii) p-type semiconductor : When a silicon crystal is doped with atoms of group-13 elements, such as B, Al, Ga or In. Each impurity atoms form only three covalent bonds with the host atom. The place where the fourth electron is missing is caused a hole which move through the crystal like a positive charge and hence increases its conductivity. Group-14 elements doped with group-13 elements are called p-type semiconductors.
Question. What is a semiconductor? Describe the two main types of semiconductors.
Answer. Paramagnetism : Materials which are weakly attracted by magnetic fields are called paramagnetic materials and this property is called paramagnetism.
Paramagnetic substances contain unpaired electrons.
e.g., TiO, CuO, O2 and VO2, etc.
Question. Explain the following terms with suitable examples : Ferromagnetism and ferrimagnetism
Answer. Ferromagnetism : Substances which are very strongly attracted by the magnetic field are
called ferromagnetic and this property is called ferromagnetism.
Examples : Iron, cobalt, nickel, gadolinium and CrO2.
These substances can be permanently magnetised i.e., they retain magnetic property even in absence of applied magnetic field.
In ferromagnetic substance all the domains (tiny magnet formed by grouping of metal ions in small region) are oriented in the direction of the applied magnetic field and produce a strong magnetic effect. This ordered orientation of domain persists even when the magnetic field is removed.
Ferrimagnetism : Substances which are weakly attracted by the magnetic field as compared to ferromagnetic substances are called ferrimagnetic and this property is called ferrimagnetism.
Examples : Fe3O4 (Magnetite, Ferrites – MgFe2O4, ZnFe2O4.
Ferrimagnetism is observed when the magnetic domains are aligned in parallel and antiparallel directions in unequal numbers. Hence, net magnetic moment is never zero.
These substances lose ferrimagnetism and become paramagnetic on heating.Question. 113. Explain the following terms with suitable examples : Ferromagnetism and ferrimagnetism
Answer. Ferromagnetism : Substances which are very strongly attracted by the magnetic field are
called ferromagnetic and this property is called ferromagnetism.
Examples : Iron, cobalt, nickel, gadolinium and CrO2.
These substances can be permanently magnetised i.e., they retain magnetic property even in absence of applied magnetic field.
In ferromagnetic substance all the domains (tiny magnet formed by grouping of metal ions in small region) are oriented in the direction of the applied magnetic field and produce a strong magnetic effect. This ordered orientation of domain persists even when the magnetic field is removed.
Ferrimagnetism : Substances which are weakly attracted by the magnetic field as compared to ferromagnetic substances are called ferrimagnetic and this property is called ferrimagnetism.
Examples : Fe3O4 (Magnetite, Ferrites – MgFe2O4, ZnFe2O4.
Ferrimagnetism is observed when the magnetic domains are aligned in parallel and antiparallel directions in unequal numbers. Hence, net magnetic moment is never zero.
These substances lose ferrimagnetism and become paramagnetic on heating.
Long Answer Questions
Question. Tungsten crystallises in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm, what is the radius of tungsten atom?
Answer. Given : Structure = bcc, a = 316.5 pm, r = ?
Question. Silver crystallises in face cented cubic unit cell.
Each side of this unit cell has a length of 400 pm Calculate the radius of the silver atom. (Assume the atom just touch each other on the diagonal across the face of the unit cell, that is each face atom is touching the four corner atoms).
Answer. a = 400 pm
Question. Aluminium crystallises in a cubic close-packed structure. Radius of atom in the metal is 125 pm.
(i) What is the length of the side of the unit cell?
(ii) How many unit cells are there in 1 cm3 of aluminium?
Answer. (i) For fcc (or ccp), a = 2√2r = 2 × 1.414 × 125 pm = 354 pm
(ii) a = 354 pm = 3.54 × 10–8 cm
Volume of one unit cell = a3 = (3.54 × 10–8 cm)3
= 4.436 × 10–23 cm3
Question. An element crystallises in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm–3. How many atoms are present in 300 g of the element?
Answer. Given, edge length a = 500 pm = 500 × 10–10 cm
Mass of substance = 300 g
Number of atoms = ?
Density, d = 7.5 g cm–3
Question. An element with molar mass 27 g mol–1 forms a cubic unit cell with edge length 4.05 × 10–8 cm.
If its density is 2.7 g cm–3, what is the nature of the cubic unit cell?
Answer. M = 27 g mol–1, a = 4.05 × 10–8 cm, d = 2.7 g cm–3
Question. An element X (molar mass = 60 g mol–1) has density of 6.23 g cm–3. Identify the type of cubic unit cell, if the edge length of the unit cell is 4 × 10–8 cm.
Answer. Given, d = 6.23 g cm–3, a = 4 × 10–8 cm,
M = 60 g mol–1
NA = 6.023 × 1023, Z = ?
Question. An element occurs in bcc structure. It has a cell edge length of 250 pm. Calculate the molar mass if its density is 8.55 g cm–3. Also calculate the radius of an atom of this element.
Answer. a = 250 pm
d = 8.55 gm–3; Z = 2
bcc unit cell body diagonal is equal to 4 times the
radius of atom.
∴ 4r = √3 a = √3 × 250
4 r = 1.732 × 250 ⇒ r = 108 pm
Question. Niobium crystallises in body-centred cubic structure. If its density is 8.55 g cm–3, calculate atomic radius of niobium, given its atomic mass 93 u.
Answer. Given density (d) = 8.55 g cm–3
Atomic radius (r) = ?
Atomic mass (M) = 93 u = 93 g mol–1
For bcc structure Z = 2
Question. Copper crystallises with face centred cubic unit cell. If the radius of copper atom is 127.8 pm, calculate the density of copper metal. (Atomic mass of Cu = 63.55 u and Avogadro’s number, NA = 6.02 × 1023 mol–1)
Answer. Given : Structure fcc, hence Z = 4, r = 127.8 pm,
M = 63.55 u, NA = 6.02 × 1023 mol–1, d = ?
Question. The density of lead is 11.35 g cm–3 and crystallise with fcc unit cell. Estimate the radius of lead atom.
(At. mass of lead = 207 g mol–1 and NA = 6.02 × 1023 mol–1
Answer.
Question. The well known mineral fluorite is chemically calcium fluoride. It is known that in one unit cell of this mineral there are 4Ca2+ ions and 8F– ions and that Ca2+ ions are arranged in a fcc lattice. The F– ions fll all the tetrahedral holes in the face centred cubic lattice of Ca2+ ions. The edge of the unit cell is 5.46 × 10–8 cm in length. The density of the solid is 3.18 g cm–3.
Use this information to calculate Avogadro’s number. (Molar mass of CaF2 = 78.08 g mol–1)
Answer. Given for fcc lattice Z = 4, d = 3.18 g cm–3
a = 5.46 × 10–8 cm, NA = ?
Question. Explain the following terms with suitable
examples :
(a) Crystal defects
(b) Metal deficiency defects
Answer. (a) Point defects are the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance, whereas the line defects are the irregularities or deviations from ideal arrangement in entire rows of lattice points. These irregularities are called crystal defects.
(b) This type of defects generally occurs when metal shows variable valency. The defect arises due to the missing of cation from its lattice site and the presence of the cation having higher charge in the adjacent lattice site, e.g., FexO.
Question. A metal crystallises in a face centred cubic unit cell with a = 0.560 nm. Calculate the density of the metal if it contains 0.1% schottky defects.
Given : Atomic mass of metal = 40 g mol–1)
Answer. Due to Schottky defects, the vacant spaces will increase resulting in decrease in number of atoms per unit cell.
