2nd PUC Chemistry Question Bank Chapter 1 The Solid State

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Karnataka 2nd PUC Chemistry Question Bank Chapter 1 The Solid State

Question 1.
Give two differences between crystalline and amorphous solids.
Answer:

Crystalline solids	Amorphous solids
1. The crystalline solids have definite characteristic shape	1. The amorphous solids have irregular shape
2. They have sharp melting point.	2. They gradually soften over a range of temperature.
3. They are anisotropic in nature	3. They are isotropic in nature
4. They are true solids	4. They are pseudo solids

Question 2.
What are non-polar molecular solids? Give examples.
Answer:
Solids in which atoms or molecules are held by weak dispersion forces or london forces. Examples: Solid CO₂, solid Ar, I₂, etc.

Question 3.
What are polar molecular solids? Give examples.
Answer:
Solids in which molecules are held by dipole-dipole interactions.
Examples: Solid NH₃, Solid HCl, Solid SO₂.

Question 4.
What are hydrogen-bonded solids? Give examples.
Answer:
Solids in which molecules are held by strong hydrogen bonds.
Examples: Ice.

Question 5.
What are covalent solids or network solids? Give examples.
Answer:
Solids of non-metals result from the formation of covalent bonds between adjacent atoms throughout the crystal.
Examples: SiO₂ (Quartz), diamond and graphite.

Question 6.
Define the terms
(a) Crystal lattice (space lattice)
(b) Unit cell.
Answer:
(a) Crystal lattice: A regular three dimensional arrangement of constituent particles in space is called crystal lattice.

(b) Unit cell: Unit cell is a smallest portion of a crystal lattice which when repeated in different directions generates the entire crystal lattice.

Question 7.
Mention seven basic crystal systems.
Answer:
Seven basic crystal systems are:

1. Cubic
2. Tetragonal
3. Orthorhombic
4. Hexagonal
5. Rhombohedral
6. Monoclinic
7. Triclinic.

Question 8.
Mention the most symmetrical and most unsymmetrical crystal systems.
Answer:
Most symmetrical crystal system – Cubic
Most unsymmetrical crystal system – Triclinic.

Question 9.
Calculate number of particles (atoms) present in simple cubic unit cell.
Answer:
Number of particles (atoms) in simple cubic unit cell = \(\frac { 1 }{ 8 }\) × 8 (at, corner)
= 1 particle

Question 10.
Calculate number of particles (atoms) presenl in body centred cubic unit cell.
Answer:
Number of particles (atoms) in BCC unit cell = \(\frac { 1 }{ 8 }\) × 8 (at corners) + 1 (at corners)
= 1 + 1 = 2 particles

Question 11.
Calculate number of particles in face centred cubic unit cell (FCC). [2M| [July 2014, March 2016, July 2016, 2018]
Answer:
No. of particles (atoms) in FCC unit cell = \(\frac { 1 }{ 8 }\) × 8 (at corners) + \(\frac { 1 }{ 2 }\) × 6 (at faces)
= 1 + 3 = 4 particles.

Question 12.
What is co-ordination number of a particle in the crystal? What is the co-ordination number in FCC structure?
Answer:
The number of nearest neighbours of a particle in the crystal is called co-ordination number.
Co-ordination number of a particle in FCC structure is 12.

Question 13.
What type of close packing is observed when
(a) ABAB pattern of arrangement of layers
(b) ABC ABC pattern of arrangement of layers?
Answer:
(a) Hexagonal cubic close packing (hep).
(b) Face centered cubic close packing (fee) or cubic close packing (cep).

Question 14.
What is the co-ordination of a particle in (a) hep (b) cep (c) bee arrangement? (live examples for these arrangements.
Answer:
In both hep and ccp arrangement co-ordination number is 12.
In bcc arrangement co-ordination number is 8.
Examples :

• hep arrangement – Mg, Zn.
• ccp (fee) arrangement – Cu, Ag.
• bcc arrangement – CsCl.

Question 15.
What is tetrahedral void? How many tetrahedral voids are possible per atom in the crystal?
Answer:
The empty space at the centre of four atoms which are in tetrahedral arrangement is called tetrahedral void.
There are two tetrahedral voids for each sphere (atom).

Question 16.
What is octahedral void? How many octahedral voids are possible per atom in the crystal?
The empty space formed at the centre of six spheres which are octahedrally arranged is called octahedral void.
There is only one octahedral void for each sphere (atom).

Question 17.
Point out the differences between teirahedral voids and octahedral voids.
Answer:

Tetrahedral voids	Octahedral voids
1. Co-ordination no of void is 4.	1. Co-ordination no of void is 6.
2. There are two tetrahedral voids per sphere	2. There is one octahedral voids per sphere

Question 18.
What is packing efficiency
Answer:
Packing efficiency is the percentage of total space filled by the particles.

Question 19.
Calculate packing efficiency in simple cubic unit cell.
Answer:
In simple cubic lattice the atoms are located only on the corners of the cube.
The atoms touch each other along the edge.
Edge length of the cube is ‘a’ and radius of the each atom is r. ∴ a = 2r
Simple cubic unit cell contain only one atom.
The volume of the one atom = \(\frac { 4 }{ 3 }\) πr³

Question 20.
Calculate packing efficiency in BCC3 lattice.
Answer:

From the figure it is clear that the atom at the centre will be in touch with other two atoms along the body diagonal.
Let a : he the edge length of the cube
b : he the lace diagonal
c : he the body diagonal
r : is the radius of the sphere and r = 4r

In ∆ EFD
b² = a² + a²
b² = 2a²

In ∆ AFD
c² = a² + b²
c² = a² + 2a² = 3a²
c = √3 × a
4r = √3 × a
a = \(\frac{4 r}{\sqrt{3}}\)
Volume of the cube = a³ = \(\left(\frac{4}{\sqrt{3}} r\right)^{3}\)
In this unit coll, total number of particles is 2 and volume is 2 × \(\frac{4}{3}\) πr³

Package efficiency = 68%

Question 21.
Calculate packing efficiency in a CCP or HCP or FCC crystal lattice.
Answer:

Let ‘a’ be the edge length of the culie
‘b’: be the fare diagonal
‘r’: is the radius of the sphere b = 4r
In ∆ ABC
b² = a² + a²
b² = 2a²
b = √2 × a
4r = √2 × a
a = \(\frac{4}{\sqrt{2}} r\)
a = 2√2 r
Volume of the cube = a³ = (2√2 r)³
Each unit cell of m* structure contain -1 spheres.
Volume of the four spheres = 4 × \(\frac{4}{3}\)πr³

Packing efficiency = 74%.

Question 22.
What is imperfection in solids?
Answer:
Departure from perfectly ordered state of the constituents of the crystal is called imperfection in solids.

Question 23.
What is Schottky defect? What is its effect on the density of the solid?
Answer:
It is a type of defect arises in ionic solids in which equal number of cations and anions are missing from their lattice points.
This defect decreases the density of the ionic solid.

Question 24.
What, is Frenkel defect? What is its effect on the density of the solid?
Answer:
It is a type of defect arises in ionic solids, when smaller ion (cation) is dislocated from its normal site to an interstitial site.
This defect does not change the density of the solid.

Question 25.
Name the ionic solid which shows both Schottky defect and Frenkel
Answer:
Silver bromide (AgBr).

Question 26.
Give three differences between Schottky defect and Frenkel defect.
Answer:

Schottky defect	Frenkel defect
1. Equal number of cations and anions are missing from their normal crystal sites.	1. Smaller ions (cation) are displaced from normal sites and occupy interstitial sites.
2. Density decreases	2. No change in density
3. Observed when cations and anions have similar size.	3. Observed when cations and anions differ in their size
4. It is found in ionic solids having high co-ordination number.	4. It is found in ionic solids having low co-ordination number.

Question 27.
Give examples for Schottky defect and Frenkel defect ionic solids.
Answer:
Ionic solids showing Schottky defect – NaCl, KC1, CsCl, AgBr.
Ionic solids showing Frenkel defect – ZnS AgCl, AgBr Agl.

Question 28.
What is metal excess defect? Give examples.
Answer:
It is a non-stoichiometric defect in which metal ions are more than the anions with maintaining the electrical neutrality of the solid.
Example : (i) This defect arises when sodium chloride crystal is heated in presence of sodium vapour.
Example : (ii) This defect is produced when ZnO is heated (white coloured ZnO turns yellow).

Question 29.
What is metal deficiency defect? Give example.
Answer:
It imparts colour to the crystal.

Question 30.
What is metal deficiency defect? Give example.
Answer:
These are the solids which contain less amount of the metal as compared to the stiochiometric proportion. Ex: Fe_0.96 O.

Question 31.
What are F-centres? What colour is imparted to NaCl, KCl and LiCl crystals on expoure to sun light.
Answer:
In metal excess defect solids, electrons are trapped in anion vacancies. These points are called E-centres.
NaCl imparts – Golden yellow colour
KCL imparts – Violet colour
LiCl imparts – Pink colour.

Question 32.
Based on band theory explain the conductivity of solids.
Answer:
In solids, molecular orbitals of the atoms form a band which is called valence hand and the next empty band in which electrons can move is called conduction band.

In conductors, the conduction band is close to valence band and therefore the electrons can easily go into the conduction band, under an applied electric field causing conductivity.

In insulators, gap between filled valence band and the next higher conduction band is large electrons cannot jump to it and such a substance has very small conductivity. They are called insulators.

In case of semiconductors, the gap between the valence band and conduction band is small. Therefore some electrons may jump to conduction band and show some conductivity. Conductivity of semiconductors increases with temperature.

Question 33.
What arc ‘n’ type semiconductors?
Answer:
When silicon or germanium is doped with 15th group elements i.e., As or P, the extra electron of the pentavalent atoms increases the conductivity of the semiconductors. This is called n-type semiconductors.

Question 34.
What are ‘P’ type semiconductors?
Answer:
When silicon (Si) or germanium (Ge) is doped with 15th group elements such as B or A1 holes are created by these trivalent atoms, increases the conductivity of the semiconductors. This type of conductors are called ‘p’ type semiconductors.

Question 35.
Give three differences between ‘n’-type and ‘p’-type semiconductors.
Answer:

‘n’-type semiconductors	‘p’-type semiconductors
1. Doped with pentavalent atoms.	1. Doped with trivalent atoms.
2. Electrician contributes to conduction	2. Holes contributes to conduction.
3. 15th group elements are used for doping	3. 13th group elements are used for doping.

Question 36.
What type of semiconductor is formed when 13th group element is doped with silicon?
Answer:
p-type semiconductor.

Question 37.
What is paramagnetism? Give examples.
Answer:
Substances which are weakly attracted by a magnetic field are called paramagnetic substances and the phenomenon is called paramagnetism. It is due to the presence of one or more unpaired electrons.
Examples: O₂, Cu²⁺, Fe³⁺, Cr³⁺.

Question 38.
What is diamagnetism? Give examples.
Answer:
Substances which are weakly repelled by the magnetic field are called diamagnetic substances and the phenomenon is called diamagnetism. In these substances, all the electrons are paired.
Examples: H₂O, NaCl, C₆H₆.

Question 39.
What is ferromagnetism? Give examples.
Answer:
Substances which are strongly attracted by the magnetic field are called ferromagnetic substances and the phenomenon is called ferromagnetism.
Examples: Iron, Cobalt, Nickel, Gadolinium and CrO₂.

Question 40.
What is antiferromagnetism? Give example.
Answer:
Substances which when magnetic moments of the domains in the substance are aligned in parallel and antiparallel directions in equal numbers is called antiferromagnetism. Ex: MnO.

Question 41.
What is ferrimagnetism? Give example.
Answer:
Substances which have domain structure similar to ferromagnetic substance, but the magnetic moment domains are unequally oriented. Hence these substances are weakly attracted by magnetic field.
Example : Magnetite (Fe₃O₄ ZnFe₂O₄.

2nd PUC Chemistry The Solid State Problems and Solutions

Type – 1 Problem on density equation:
d = \(\frac{Z \times M}{a^{3} \times N_{A}}\)
d = density
z = number of atoms in the unit cell
M = molarmass or atomic mass.
a = edge length of the unit cell
N_A = Avogadro number (6.022 × 10²³)

Question 1.
Silver crystallizes in CCP lattice. The edge length of its unit cell is 408.6 pm. Calculate density of silver (atomic mass of silver is 107.9).
Answer:
Data: Element, cryslallizes in FCC Z = 4
Atomic mass M = 107.9 g
Edge length = a = 408.6 pm = 408.6 × 10^-12 m
= 408.6 × 10^-12 × ² cm
= 408.6 × 10^-10 cm
= 408.6 × 10^-8 cm [1M]

Question 2.
An element having atomic mass 63.1 g/mol has face centered cubic unit cell with edge length 3.608 × 10^-8 cm. Calculate the density of unit cell [Given N_A = 6.022 × 10²³ atoms/mol].
Answer:
Data: For fee lattice z = 4
Atomic mass M = 63. 1 g
Edge length a = 3.608 × 10^-18 cm
N_A = 6.022 × 10²³

= 8.923 gnm^-1

Question 3.
An element having atomic mass 107.9 g mol-1 has FCC unit cell. The edge length of the unit cell is 186pm. Calculate the density of the unit cell.
Answer:
6.242 gcm^-3

Question 4.
An element occurs in BCC structure with cell edge of 288 pm. Find the density of the element if its atomic mass is 51.7.
Answer:
7.19 gcm^-3

Question 5.
A compound formed by the element A and B crystallizes in the cubic structure, where A is at the corners of the cube and B is at body centre. What is the formula of the compound? If edge length is 5A°, calculate the density of the solid. (Atomic weights of A and B are 60 and 90 respectively).
Answer:

Question 6.
An element crystallizes in fee lattice. If the edge length of the unit cell is 408.6 pm and the density is 10.5 g cm^-3. Calculate the atomic mass of the element.
Answer:
Data: d = 10.5 gcm^-3
a = 408.6 × 10^-12 m = 408.6 × 10^-10 cm
= 4.086 × 10^-8 cm
For the fee lattice = (Z) = 4

M = 107.8

Question 7.
An element occurs in BCC structure with cell edge of 288 pm. Its density is 7.2 gcm^-3. Calculate the atomic mass of the element.
Answer:
51.79

Question 8.
The density of chromium metal is 7.2 gcm^-3. If the unit cell is cubic with edge length of 289 pm, calculate the number of atoms per unit cell. (Atomic mass = 51.79).
Answer:
Formula:
d = 7.2 gcm^-3
a = 289 × 10^-12m × 10² = 289 × 10^-10 cm = 2.89 × 10^-8 cm
N_A = 6.022 × 10²³
M = 51.79g

Question 9.
The density of an cubic unit cell is 10.5 gcm^-3. If the edge length of the unit cell is 409 pm. Find the structure of the crystal lattice (Atomic mass = 108).
Answer:
FCC

Question 10.
Niobium crystallizes in body centered cubic structure. If density is 8.55g cm. Calculate edge length of the unit cell (Atomic mass = 93u).
Answer:
Data: Z = 2, M = 93g, d = 8.55 gcm^-3 N_A = 6.022 × 10²³

Type – 2 Unit Cell Dimensions

In simple cubic unit cell a = 2r
In body centred unit cell (BCC) a = \(\frac{4}{\sqrt{3}} r\)
In face centred unit cell (COP or FCC) a = 2√2r

Question 11.
Aluminium crystallizes in FCC structure. Atomic radius of the metal is 125 pm. Calculate the edge length of the unit cell. [2MJ [March-2014]
Answer:
In FCC structure edge length a = 2√2 r [1M]
= 2 × √2 × 125 = 2 × 1.414 × 125 = 353.5pm [1M]

Question 12.
Sodium metal crystallizes in body centered cubic lattice with the cell edge a = 428 pm. what is the radius of sodium atom. [2M]
Answer:

Question 13.
An element with edge length 7A° crystallizes in SC. Calculate radius of the sphere. [2M]
Answer:
In SC a = 2r
r = \(\frac{a}{2}=\frac{7 A^{\circ}}{2}\) = 3.5A°

Question 14.
A compound A_xB_y crystallizes in a FCC lattice in which A occupies each corner of a cube and B occupies the centre of each face of the cube. What is the formula of the compound?
Answer:
No. of A particles = \(\frac { 1 }{ 8 }\) × 8 (at corners) = 1
No. of B particles = \(\frac { 1 }{ 2 }\) × 6 (at corners) = 3
A : B Formula AB₃
1 : 3

Question 15.
How many tetrahedral voids and octahedral voids are possible if the number of close packed spheres in two layers is N.
Answer:
Number of tetrahedral voids – 2N
Number of octahedral voids – N

Question 16.
A compound is formed by two elements x and y. Atoms of the element y (as anions) make ccp and those of the element x (as cations) occupy all the octahedral voids what is the formula of the compound?
Answer:
y(anion) forms ccp structure
∴ No. of y(anion) is 4
4y creates 4 octahedral voids
Hence No. of x(cation) is also 4.
x : y
4 : 4
1 : 1
Formula of the compound is xy.