The Solid State Class 12 Chemistry Exam Questions

Please refer to the below The Solid State important questions for Class 12 Chemistry. These questions and answers have been prepared as per the latest NCERT Book for Class 12 Chemistry. Students should go through chapter wise Class 12 Chemistry Important Questions designed as per the latest examination pattern issued by CBSE.

Very Short Answer Questions :

Question. Define primitive unit cells.
Answer : A unit cell in which constituent particles (lattice points) are present only at the corner positions is called primitive or simple unit cell.

Question. What type of point defect is produced when AgCl is doped with CdCl₂?
Answer : Addition of CdCl₂ to AgCl crystal causes impurity defect.
In this defect also one Cd²⁺ ion replaces two Ag⁺ ions from the crystal. One site is occupied by one Cd²⁺ ion and other site remains vacant.

Question. What type of stoichiometric defect is shown by NaCl? 
Answer : Schottky defect : The 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. 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. What type of stoichiometric defect is shown by AgCl?
Answer : Frenkel defect

Question. Account for the following :
Schottky defects lower the density of related solids. 
Answer : Schottky defects occur when cations and anions are missing from their lattice site. Mass of unit cell decreases which decreases the density of the solid.

Question. Define crystal lattice.
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. “Stability of a crystal is related to the magnitude of its melting point.” How?
Answer : The crystalline solids have sharp melting points.
Melting point of a solid is related to the force of attraction between its molecules. Higher the melting point of a crystalline substance, greater is the force between the constituent particles and hence, greater is the stability.

Question. How many atoms per unit cell (Z) are present in bcc unit cell?
Answer : In a body centred cubic unit cell.
(i) 8 corners × 1/8 per corner atom = 8 × 1/8 = 1 atom
(ii) 1 body centre atom = 1 × 1 = 1 atom
∴ Total number of atoms per unit cell = 1 + 1 = 2 atoms

Question. What type of stoichiometric defect is shown by KCl and why? 
Answer : Schottky defect : The 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. 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. How many effective sodium ions are located at the edge centre of a unit cell in a sodium chloride crystal?
Answer : Na units in edge =12 x 1/4 = 3

Question. Explain the following with suitable example :
Paramagnetism 
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. Which stoichiometric defect does not change the density of the crystal?
Answer : Frenkel defect does not change the density of the 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. “Crystalline solids are anisotropic in nature.” What does this statement mean?
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. What are diamagnetic susbtances? 
Answer : Substances which are weakly repelled by magnetic field are called diamagnetic and this property is called diamagnetism. Diamagnetism is shown by those substances in which all electrons are paired.
Examples : NaCl, H₂O, C₆H₆, Zn²⁺

Question. How many atoms are there in one unit cell of a body centred cubic crystal?
Answer : In a body centred cubic unit cell.
(i) 8 corners × 1/8 per corner atom = 8 × 1/8 = 1 atom
(ii) 1 body centre atom = 1 × 1 = 1 atom
∴ Total number of atoms per unit cell = 1 + 1 = 2 atom

Question. How do metallic and ionic substances differ in conducting electricity?
Answer : (i) Metallic solids are malleable and ductile whereas ionic solids are hard and brittle.
(ii) In solid state, ionic solids are electrical insulators because ions are not free to move about, e.g. NaCl, Na₂SO₄, etc. while metals are good electrical conductors in solid state because of the presence of free electrons e.g. copper, nickel etc. (any one)

Question. What change occurs when AgCl is doped with CdCl₂?
Answer : Addition of CdCl₂ to AgCl crystal causes impurity defect.
In this defect also one Cd²⁺ ion replaces two Ag⁺ ions from the crystal. One site is occupied by one Cd²⁺ ion and other site remains vacant.

Question. What is the total number of atoms per unit cell in a face centred cubic (fcc) structure?
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
= 8 × 1/8 = 1 atom
(ii) No. of atoms contributed from the faces
= 6 face centred atoms × 1/2 atoms per unit cell
= 6 x  1/2 = 3 atoms
∴ Total no. of atoms per unit cell = 4 atoms

Question. Write a distinguishing feature of metallic solids.
Answer : Malleable, ductile, lustrous and conductor of heat and electricity.

Short Answer Questions :

Question. In a face centred lattice of X and Y, X atoms are present at the corners while Y atoms are at face centres.
(a) What is the formula of the compound?
(b) What would be the formula of the compound if
(i) one of the X atom is missing from a corner in each unit cell.
(ii) two atoms of X are missing from the corners.
(iii) one atom of X from a corner is replaced by Z atom (also monovalent)?
Answer : 

Question. A copper crystal has a face-centred cubic lattice structure. Atomic radius of the copper atom is 128 pm. Calculate the density of copper.
Answer :

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 , b and c.  
(ii) Angles between edges α, β and γ.
Hence, a unit cell is represented by six parameters a, b, c, α, β and γ.

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
                                        = 1/8 x 8 = 1
Contribution by atom (cadmium) on the faces
                                        = 1/2 x 6 = 3
Number of atoms present in the unit cell = 1 + 3 = 4 Hence, formula for the given alloy = AuCd₃

Question. Copper crystallises in face-centred cubic lattice and has density of 8.930 g cm^–3 at 293 K.
Calculate the radius of copper atom.
[Atomic mass of Cu = 63.55 u, N_A = 6.02 × 10²³ mol^–1]
Answer :

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. Explain the following terms with suitable example of each.
(a) Ferromagnetism
(b) Antiferromagnetism 
Answer : (a) 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.

Schematic alignment of magnetic moments in
(a) ferromagnetic (b) ferrimagnetic.
(b) Antiferromagnetism : Paramagnetic substances which show very poor magnetic effect are antiferromagnetic and this property is called antiferromagnetism.
Examples : MnO, Fe₂O₃, Cr₂O₃.

Question. Copper crystallises into an fcc lattice with edge length 3.61 × 10^–8 cm. Calculate the density of copper. [Given : Cu = 63.5 g mol^–1, N_A = 6.022 × 10²³ mol^–1] 
Answer : 

Question. Examine the given defective crystal.
A+  B –  A+  B –  A+
B –  O    B –  A+  B –
A+  B –  A+   O    A+
B –  A+  B –  A+  B –
Answer the following questions :
(i) What type of stoichiometric defect is shown by the crystal?
(ii) How is the density of the crystal affected by this defect?
(iii) What type of ionic substances show such defect? 
Answer : (i) Schottky defect
(ii) Density of the crystal decreases.
(iii) This defect is shown by ionic substances in which the cation and anion are of almost similar sizes.

Question. The radius of Na⁺ ion is 95 pm and that of Cl^– ion is 181 pm. Predict whether the coordination number of Na⁺ ion is 6 or 4.
Answer : Radius of Na⁺ = 95 pm
Radius of Cl^– = 181 pm
Radius ratio, r₊/r_– = r (Na⁺)/r(Cl^–) 95/181 = 0.524
The radius ratio lies between 0.414 – 0.732.
Hence, Na+ ions prefer to occupy octahedral holes having coordination number 6.

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 fill 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 : 

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 : N_A = 6.022 × 10²³ mol^–1)
Answer : 

Question. Aluminium crystallises in a fcc structure. Atomic radius of the metal is 125 pm. What is the length of the side of the unit cell of the metal?
Answer : For fcc (or ccp), a = 2√2r
= 2 × 1.414 × 125 pm = 354 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 : 

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 : 

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.

Schematic alignment of magnetic moments in
(a) ferromagnetic
(b) ferrimagnetic.

Question. The unit cell of an element of atomic mass 108 u and density 10.5 g cm^–3 is a cube with edge length, 409 pm. Find the type of unit cell of the crystal.
[Given : Avogadro’s constant = 6.023 × 10²³ mol^–1]
Answer :