Chemistry Question Bank: Solid State (2 Marks)

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Section B – Questions of 2 Marks SA I
Q1. “Solid ice is lighter than water” – Explain.

  • According to X-ray studies, the structure of solid ice is almost identical to that of liquid water. Intermolecular hydrogen bonding present in ice forms a hexagonal three-dimensional crystal structure such that about one half of the total space remains unoccupied.
  • When ice melts, some of the hydrogen bonds in ice are broken and some water molecules occupy, some of the empty spaces available due to the breaking of the bonds. Thus, the hexagonal crystalline structure of the solid ice collapses and liquid water molecules attain closely spaced structure.
  • In solid ice vacant space is more than the water. Due to this, the density of liquid water is higher than that of solid ice. Lighter solids float on liquid having more density. Thus ice floats on water.

Q2. Write a note on ionic solid.

  • Ionic solids are crystalline solids in which the units occupying lattice points are positively and negatively charged ions. In such solids, the rep[eating units are positively and negatively charged ions.
  • Examples: Salts like NaCl, BaSO4, potassium bromide, copper nitrate, copper sulphate are ionic solids.

  • Each ion of ionic solid is surrounded by a number of oppositely charged ions. This number is called the coordination number. The coordination number for positive and negative ion may be the same or different. The general coordination numbers for ionic solids are 8, 6 and 4.
  • The charges on cations and anions balance each other hence the solid is electrically neutral.
  • Such solids are formed by the three-dimensional arrangements of cations and anions bound by strong coulombic (electrostatic) forces.
  • These solids are hard and brittle in nature. They have high melting and boiling points. They have high density due to close packing.
  • Since in solid state the ions are not free to move about, (due to strong electrostatic force) they are electrical insulators in the solid state. However, in the molten state or when dissolved in water, the ions become free to move about and they conduct electricity.

Q3. Write a note on a metallic solid.

  • Metallic solids are crystalline solids in which the units occupying lattice points are positive ions surrounded by a pool of electrons. (Concept of metallic bond)
  • Examples: The metals Na, Mg, Al are metallic solids.



  • In metallic solids, the units occupying lattice points are metal cations, surrounded by many mobile electrons.
  • They are good thermal and electrical conductor, malleable and ductile and are lustrous. This is also due to the presence of free electrons in them.
  • The atoms in a metal are held together by means of a special type of bond called the metallic bond.
  • Their ionization enthalpies are low. Their valence electrons are loosely held together and more vacant valency orbitals are available.

Q4. Write a note on covalent solid.

  • Covalent solids are crystalline solids in which unit lattice points are atoms.
  • Examples: Diamond, Silicon,silicon carbide (SiC), aluminium nitride (AlN), etc.

  • In covalent solids, the units occupying lattice points are atoms attached to each other by covalent bonding. They are also called giant molecules. Covalent solids are three-dimensional network solids.
  • Covalent bonds are strong and directional in nature, therefore atoms are held very strongly at their positions. Such solids are very hard and brittle. They have extremely high melting points and may even decompose before melting.
  • They are insulators and do not conduct electricity.

Q5. Explain why ionic solids are hard and brittle?

  • Ionic solids are formed by the three-dimensional arrangements of cations and anions bound by strong electrostatic forces.
  • Ionic crystals are hard due to strong electrostatic forces between them. They are brittle because the ionic bond is non-directional.

Q6. Distinguish between hexagonal close packing and cubic close packing in solids.



Hexagonal Close Packing Cubic Close Packing
In this three dimensional arrangement of the unit cell, spheres of the third layer are placed on triangular shaped tetrahedral voids of the second layer. In this three dimensional arrangement of the unit cell, spheres of the third layer are placed in the positions of tetrahedral voids having apices upward.
The spheres of the third layer lie exactly above the spheres of the first layer. In this arrangement, the spheres of the fourth layer lie exactly above the spheres of the first layer.
The arrangements of the first layer and third layer are identical. The arrangements of the first layer and fourth layer are identical.
The arrangement of hexagonal close packing is represented as ABAB type. The arrangement of cubic close packing is represented as ABCABC type.

Q7. Calculate the number of atoms per unit cell of FCC and BCC crystal structure.

  • Face Centered Cubic Structure (fcc):

Bravais Lattice

  • From the structure, we can see that there are 8 particles at 8 corners of the unit cell. Each corner particle is shared by 8 other neighbouring unit cells. Hence each unit cell contains 1/8 th of the particle at its corner. The number of corners = 8. Hence the number of particles in unit cell at corners = 1/8 x 8 = 1
  • There are 6 particles at 6 faces of the unit cell. Each face particle is shared by 2 neighbouring unit cells. Hence each unit cell contains 1/2 of the particle at its face.
    The number of faces = 6. Hence the number of particles on face = 1/2 x 6 = 3

Hence number of particles in unit cell 1 + 3 = 4

  • Body Centered Cubic Structure (bcc):

  • From the structure, we can see that there are 8 particles at 8 corners of the unit cell. Each corner particle is shared by 8 neighbouring unit cells. Hence each unit cell contains 1/8 th of the particle at its corner. Number of corners = 8, hence number of particles in unit cell at corners = 1/8 x 8 = 1
  • At the same time, there is an atom at the centre of the cell,

Hence number of particles in unit cell 1 + 1 = 2



Q8. Metals are good conductors of heat and electricity. Why?

  • Metals are an orderly collection of positive ions (called kernels) surrounded by and held together by a sea of free electrons(delocalized). These electrons are mobile and are evenly spread out throughout the crystal.
  • Each metal atom contributes one or more electrons towards this sea of mobile electrons. These free and mobile electrons are responsible for the high electrical and thermal conductivity of metals. When an electric field is applied, these electrons flow through the network of positive ions. Similarly, when heat is supplied to one portion of a metal, the thermal energy is uniformly spread throughout by free electrons.

Q9. Write a note on radius ratio rule for ionic compounds.

  • The ratio of the radius of cations (r+) to the radius of the anion (r) of ionic compound is known as the radius ratio of the ionic solid.

  • The significance of the radius ratio:
  • It is useful in predicting the structure of ionic solid. In crystals, cations tend to get surrounded by the largest possible number of anions around it.
  • Greater the radius ratio, greater is the coordination number of cations and anions.
  • If cations are extremely small and anions are extremely large, then the radius ratio is very small. In such case packing of anions is very close to each other and due to repulsion between anions, the system becomes unstable. Hence the structure changes to some suitable stable arrangement.
  • The radius ratio at which anions just touch each other, as well as central cation, is called critical radius ratio.

Q10. Explain Ferromagnetism.

  • The substances which are attracted very strongly by a magnetic field are called ferromagnetic substances.
  • Examples: iron, cobalt, nickel, gadolinium, and CrO2
  • Besides strong attractions, these substances can be permanently magnetised.



  • In the solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an unmagnetized piece of a ferromagnetic substance, the domains are randomly oriented and their magnetic moments get canceled. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field and a strong magnetic effect is produced. The alignments of domains persist even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.

Q11. Explain Paramagnetism.

  • Paramagnetic substances are weakly attracted by a magnetic field.
  • Examples: O2, Cu2+, Fe3+, Cr3+, TiO, Ti2O3,VO, VO2, and CuO
  • They are magnetized in a magnetic field in the same direction.

  • They lose their magnetism in the absence of magnetic field. They are temporary magnets.
  • Paramagnetism is due to the presence of one or more unpaired electrons which are attracted by the magnetic field.

Q12. Explain Diamagnetism.

  • Diamagnetic substances are weakly repelled by a magnetic field. The magnetism exhibited by such substance is called diamagnetism.

  • Examples: H2O, TiO2, V2O5, NaCl, and C6H6
  • They are weakly magnetized in a magnetic field in opposite direction.
  • Diamagnetism is shown by those substances in which all the electrons are paired and there are no unpaired electrons. The pairing of electrons cancels their magnetic moments and they lose their magnetic character.

Q13. Write a note on impurity defect in solids.



  • This defect occurs when regular cation of a crystal is replaced by some different cation. The different cation may occupy regular lattice site or interstitial site.
  • If the impurity cation is substituted in place of regular cation then the defect is known as substitution impurity defect. If the impurity cation occupies interstitial space then the defect is called interstitial impurity defect.
  • Interstitial impurity in Stainless Steel

  • Substitutional impurity in Brass

Q14.  Explain the term Schottky defect.



  • When some of the lattice sites are vacant, the crystal is said to have a vacancy defect. The defect produced due to vacancies caused by an absence of anions and cations in the crystal lattice of ionic solid is called a Schottky defect. Thus in such defect, one positive ion and one negative ion are missing from their respective positions leaving behind a pair of holes.
  • This defect can also develop when a substance is heated.
  • Due to this defect, the observed density of crystal is found to be lower than expected density.
  • This defect is observed in ionic compounds with high coordination number, high radius and with cations and anions have almost equal size. like NaCl, KCl, CsCl, AgBr, KBr, etc.
  • Vacancy defects increase with the increase in temperature.

Q15. Explain the term Frenkel defect.

  • When cation and anion from ionic solid leave its regular site and moves to occupy a place between the lattice site (interstitial sites) is called an interstitial defect. This defect associated with the ionic compound is called Frenkel defect or dislocation defect. The ions occupying interstitial sites are called interstitials. The formation of this defect depends on the size of interstitials.
  • This defect is observed in case of an ionic compound having low coordination number and relatively smaller cations which can fit into interstitial space. This defect is common when the difference in ionic radii of two participating ions is large.
  • The presence of this defect does not alter the density of the solid.
  • The presence of ions in interstitial sites increases the dielectric constant of the crystal.
  • Example: In AgCl the defect is observed due to Ag+ ions. In ZnS the defect is observed due to Zn++ ions.

Q16. Classify the following species as paramagnetic, diamagnetic or ferromagnetic.

a) Sodium [Z = 11]



Electronic configuration of sodium atom is 1s2 2s2 2p6 3s1

The electronic configuration of the valence shell is

3s1 3px0 3py0 3pz0

There is one paired electron. Hence it is paramagnetic

b) Ca2+ cation [Z = 20]



Electronic configuration of Ca2+ cation is 1s2 2s2 2p6 3s2 3p6

The electronic configuration of the valence shell is

↓↑ ↓↑ ↓↑ ↓↑
3s2 3px2 3py2 3pz2

In Ca2+ cation all electrons are paired. Hence Ca2+ cation is diamagnetic.

c) Iron [Z = 26] 

Electronic configuration of iron atom is 1s2 2s2 2p6 3s2 3p6 3d4s2.

The electronic configuration of the last two shells is 3s2 3p6 3d4s2



  • Due to the presence of the unpaired electrons in the third orbit, it shows a peculiar magnetic property. There are certain regions in which all the iron atoms have magnetic moments in the same direction and form domain. Hence Iron shows ferromagnetism.

d) Cl anion [Z = 17]

Electronic configuration of Cl ion is 1s2 2s2 2p6 3s2 3p6

The electronic configuration of the valence shell is

↓↑ ↓↑ ↓↑ ↓↑
3s2 3px2 3py2 3pz2

All electrons are paired. Hence Cl ion is diamagnetic.

Q17. How many tetrahedral voids and octahedral voids are present per atom in a closed packed structure?

  • If the number of close-packed spheres (particles)is equal to ‘N’. Then, the number of octahedral voids formed = N And, the number of tetrahedral voids formed = 2N

Q18. Explain the term crystal lattice with suitable example.

  • A crystal is a homogenous portion of a solid substance made of a regular pattern of structural units bonded by plane surfaces making a definite angle with each other. The crystal lattice is a regular arrangement of constituent particles of a crystalline solid in three-dimensional space. It consists of a large number of unit cells.
  • The geometrical form consisting only of a regular array of points in space is called lattice or space lattice. It may be defined as an array of points showing how molecules, atoms or ions are arranged in different sites, in a three-dimensional space. Each point at the intersection of lines in the unit cell represents constituent particle viz. molecule or atom or ion. This point of intersection of lines in the unit cell is called lattice point or lattice site.
  • A crystal lattice can be obtained, handled and studied in a laboratory during experiments. It is a macroscopic property.

Tetrahedral

Q19. What is meant by the term coordination number of a crystal lattice?

  • The coordination number of the constituent particle of the crystal lattice is the number of particles surrounding a single particle in a crystal lattice.
  • More the coordination number more tightly the particles are packed in the crystal lattice. Coordination number is the measure of the hardness of the crystal.

Q20. Explain n-type semiconductors.

  • Let us suppose the germanium is doped with an element from the fifth group say phosphorous (pentavalent impurity). Phosphorus has five Valency electrons. Therefore, phosphorous can form four covalent bonds leaving one free electron unbonded. Due to pentavalent doping the number of free electrons increases. This impurity is called the donor impurity.

Semiconductors

  • Under the action of an electric field, free electron around phosphorous move through the crystal from negative end to positive end. Thus the conductivity of doped germanium increases.
  • The presence of an electron means the presence of a negative charge. Hence the doped material is called n-type semiconductor.

Q.21. Explain p-type semiconductors.

  • Let us suppose the germanium is doped with an element from the third group say boron (trivalent impurity). Boron has three Valency electrons. Therefore, boron can form only three covalent bonds with neighboring germanium atoms. One of the covalent bonds around each boron atom has an electron missing. The absence of an electron is called a hole. This impurity is called acceptor impurity.

  • Under the action of an electric field, an electron from a neighboring completely filled covalent bond jumps into this hole creating a hole in the bond from which electron has moved. The process is repeated continuously. Thus the hole appears to move through the crystal from positive end to negative end. Thus the conductivity of doped germanium increases.
  • The absence of an electron in the hole means the presence of a positive charge. Hence the doped material is called p-type semiconductor.

Q22. Draw the structures of SCC and FCC indicating lattice points.

  • SCC

  • FCC:

Q23. Distinguish between BCC unit cell and FCC unit cell.

BCC Unit Cell

FCC Unit Cell

The term BCC stands for the body-centered cubic arrangement of spheres (particles). The term FCC stands for the face-centred cubic arrangement of spheres. (particles).
This structure has spheres in the eight corners of a cube and one sphere in the centre of the cube. FCC has spheres in the eight corners of a cube and also in the centres of the cubic faces.
The packing efficiency of BCC is 0.68 (68%) The packing efficiency of FCC is 0.74 (74%)
Void space is 32 % Void space is 26 %
The coordination number of the BCC structure is 8. The coordination number of the FCC structure is 12.
There are two spheres (particles) in a unit cell. There are four spheres (particles) in a unit cell.
Examples: Lithium (Li), Sodium (Na), Potassium (K), Chromium (Cr) and Barium (Ba). Examples: Aluminum (Al), Copper (Cu), Gold (Au), Lead (Pb) and Nickel (Ni).

Q24. Distinguish between crystalline solids and amorphous solids.

Sr.No. Properties Crystalline Solids Amorphous Solids
1 Structural units Structural units are arranged orderly and repeating in three dimensions Structural units are not arranged orderly and repeating in three dimensions
2 Order Long order repeating Short order repeating
3 Melting points Very Sharp Not sharp (melts over temperature range)
4 Heat of fusion Definite and characteristics Neither definite nor characteristics
5 Cooling curve Not smooth and with breakpoints Smooth without a break
6 Cleavage using knife Regular and clean cut Irregular cut
7 Compressibility Generally incompressible Slightly compressible
8 Nature of bulk Anisotropic Isotropic
9 Nature True solids Supercooled liquids or pseudo solids

Q25. Complete the following:

Crystal System

Type Edge length Angle

i) Tetragonal

Body Centred a = b ≠ c

α = β = γ = 90°

ii) Monoclinic End Centred a ≠ b ≠ c

α =γ = 90° and  β ≠ 90°

Q26. Sodium metal crystallizes to form the BCC structure. The edge length of the unit cell is 4.29 x 10-8 cm. Calculate the radius of the sodium atom.

  • Solution:
  • Given: Edge length = a = 4.29  x 10-8 cm, Type of crystal structure = bcc
  • To Find: Atomic radius of sodium =?

The atomic radius of sodium is 186 pm

Q27. In an ionic solid, the radii of cation and anion are 0.9858 Å respectively and 1.86 Å . Predict the type of geometry and coordination number of the solid.

Lattice Points

Radius ratio = 0.9858 Å/1.86 Å = 0.53

For radius ratio of 0.414 to 0.732, the coordination number is 6 and has octahedral geometry.

Q28. A unit cell of iron crystal has an edge length of 288 pm and density of 7.86 g cm-3. Determine the type of crystal lattice. [Atomic mass of Fe = 56]

  • Solution:
  • Given: Density = 7.86 g cm-3, edge length = 288 pm = 288 x 10-12 m = 288 x 10-10 cm = 2.88 x 10-8 cm, Avogadro’s number N = 6.022 x 1023 mol-1. Molecular mass of iron = M = 56 g mol-1,
  • To Find: the type of crystal lattice

Number of atoms per unit cell = 2,

Now body centred cubic structure has 2 atoms in the unit cell. Hence the crystal lattice type must be bcc

Q29. A compound is formed by two elements M & N. The element N forms CCP and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound?

In CCP crystal Number of N atoms (Z) in crystal = 4
Number of tetrahedral voids = 2 Z = 2 x 4 = 8
Number of M atoms = 1/3rd of 8 = 8/3
Thus the ratio of atoms N: M = 4: 8/3 = 12:8
The formula of the compound is N12Mor  N3M2

Q30. A compound forms hexagonal close-packed structure. What is the number of octahedral voids in 0.5 moles of it?

Number of moles of compound = 0.5 mol
1 mole of the compound has 6.022 ×10 23 particles
Number of particles in 0.5 mol compound = 0.5 × 6.022 × 1023 = 3.011 × 1023
The number of octahedral voids = number of atoms in closed packaging.
Number of octahedral voids in 0.5 mol = 3.011 × 1023

Q31. A compound forms hexagonal close-packed structure. What is the number of tetrahedral voids in 0.25 mole of it?

Number of moles of compound = 0.25 mol
1 mole of the compound has 6.022 ×10 23 particles
Number of particles in 0.5 mol compound = 0.25 × 6.022 × 1023 = 1.5055 × 1023
The number of tetrahedral voids = 2 x number of atoms in closed packaging.

The number of tetrahedral voids = 2 x 1.5055 × 1023
Number of tetrahedral voids in 0.25 mol = 3.011 × 1023

Q32. Copper crystallizes in FCC structure. The edge length of the unit cell is 3.8 Å. Calculate the radius of the copper atom.

  • Given: the edge length of the unit cell is 3.8 Å,
  • To Find: Radius of the copper atom =?
  • Solution:

For fcc structure,  r = a/8  =   3.8 Å/8  = 1.344 Å

The radius of the copper atom is 1.344 Å

Q33. Aluminium crystallizes in a CCP structure. Its metallic radius is 125 pm. What is the length of the side of the unit cell?

  • Given: Radius of the copper atom = 125 pm.
  • To Find: the edge length of the unit cell =?
  • Solution:

Closed cubic packed structure is similar to fcc

For fcc structure,  r = a/8

a = r8   = 125 pm x 8  = 353.5 pm

The length of the edge of aluminium crystal is 353.5 pm

Q34. The edge length of the unit cell is 353.5 pm. Calculate the number of unit cells present in 1 cm3 of Aluminium.

The volume of the unit cell of aluminium = (edge length)3 = (353.5 x 10-10 cm)3 =

The volume of the unit cell of aluminium = (3.535 x 10-8 cm)3 = 44.17 x 10-24 cm3

Number of unit cells = given volume/volume of each unit cell = 1 cm3/44.17 x 10-24 cm3

Number of unit cells in 1 cm3 = 2.264 x 1022.

Q35. A compound AmBn with A in the form of cation occupies the octahedral voids where B in the form of anion forms CCP type crystal lattice. What is the formula of the compound?

The closed cubic packed structure is similar to fcc

The number of atoms in unit cell for fcc = 4
therefore no of B atom present=4
no of A atom=tertrahedral void and octahedral void(8+4)=12
formula of the compound=A12B4

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