Its about chemistry class 12 chapter 1 solid state
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Chapter 1 – The Solid State Amorphous and Crystalline Solids Based on the nature of the order of arrangement of the constituent particles, solids are classified as amorphous and crystalline. Differences between amorphous and crystalline solids are listed in the given table. Amorphous solids Crystalline solids 1 Have irregular shape 1 Have definite characteristic geometrical shape 2 Have only short-range order in the arrangement of constituent particles 2 Have long-range order in the arrangement of constituent particles 3 Gradually soften over a range of temperature 3 Have sharp and characteristic melting point 4 When cut with a sharp- edged tool, they cut into two pieces with irregular shapes 4 When cut with a sharp-edged tool, they split into two pieces with plain and smooth newly generated surfaces. 5 Do not have definite heat of fusion 5 Have definite and characteristic heat of fusion 6 Isotropic in nature 6 Anisotropic in nature 7 Pseudo solids or super- cooled liquids 7 True solids Classification of Crystalline Solids Based on the nature of intermolecular forces, crystalline solids are classified into four categories − Molecular solids
Transcript
Chapter 1 – The Solid StateAmorphous and Crystalline Solids
Based on the nature of the order of arrangement of the constituent particles, solids are classified as amorphous and crystalline.
Differences between amorphous and crystalline solids are listed in the given table.
Amorphous solids Crystalline solids
1 Have irregular shape 1 Have definite characteristic geometrical shape
2 Have only short-range order in the arrangement of constituent particles
2 Have long-range order in the arrangement of constituent particles
3 Gradually soften over a range of temperature
3 Have sharp and characteristic melting point
4 When cut with a sharp-edged tool, they cut into two pieces with irregular shapes
4 When cut with a sharp-edged tool, they split into two pieces with plain and smooth newly generated surfaces.
5 Do not have definite heat of fusion 5 Have definite and characteristic heat of fusion
6 Isotropic in nature 6 Anisotropic in nature
7 Pseudo solids or super-cooled liquids 7 True solids
Classification of Crystalline Solids Based on the nature of intermolecular forces, crystalline solids are classified into four
categories −
Molecular solids
Ionic solids
Metallic solids
Covalent solids
Constituent particles are molecules
Ionic solids
Constituent particles are ions Hard but brittle Insulators of electricity in solid state, but conductors in molten state and in aqueous
solution High melting point Attractive forces are Coulombic or electrostatic Example − NaCl, MgO, ZnS
Metallic solids In metallic solids, positive ions are surrounded and are held together in a sea of
delocalised electrons. Hard but malleable and ductile Conductors of electricity in solid state as well as molten state Fairly high melting point Particles are held by metallic bonding Example − Fe, Cu, Mg
Covalent or network solids
Constituent particles are atoms Hard (except graphite, which is soft)
Insulators of electricity (except graphite, which is a conductor of electricity) Very high melting point and can decompose before melting Particles are held by covalent bonding Example − SiO2 (quartz), SiC, diamond, graphite
Add to your knowledge
The property by virtue of which two or more crystalline solids having similar chemical composition exist in the same crystalline form is called isomorphism. For example: Na3PO4.The property by virtue of which a particular substance exists in more than one crystalline form is called polymorphism. For example: existence of calcium carbonate in two crystalline forms called calcite and aragonite.
Q1)
Which of the following is a pseudo solid?
CaF2
NaCl
Glass
None of the above
Ans :
Amorphous solids have a tendency to flow, though very slowly. Therefore, they are called pseudo solids or super cooled liquids.
Q2)
Q2) Which of the following will show anisotropy?
Glass
Wood
Paper
Barium chloride
Ans :
Anisotropy is shown by crystalline solids.
Q3)
Harman was identifying the crystallizing pattern of magnesium when he noticed some empty space in the pattern. He determined the pattern to be hexagonal close packing and the percentage of empty space to be:
15.2%
21.3%
25.6%
28.7%
Ans :
Magnesium crystallizes in hexagonal close packing structure. In hcp, all the corners of one unit cell contain a sphere, also top and bottom face centres contain spheres and another three spheres are present inside the body.
Hence, total number of spheres per unit cell
(One corner sphere is showed by six unit cells) = 6
Suppose, the radius of each sphere = r
Hence, the volume of unit cell = Base area × height (c)
Base area of regular hexagon
Height,
Packing fraction
Hence, void = 1 – 0.744 = 0.256 or 25.6%.
Hence, option (C) is correct.
Q4)
While studying the crystal structure of chromium, Kanika recognized that the metal crystallizes with bcc lattice. Her observations of the side length gave a value of 287 pm. Determine the density of crystal in gm/c.c.
None of these
Ans :
For bcc,
Now,
Density
Hence, option (A) is correct.
Q5)
X-ray crystallographic data of an element A (atomic mass 100) showed that it has b.c.c structure. Yasuda wants to determine the number of atoms in 10 g of element A if its edge length is 400 pm.
None of these
Ans :
Hence, option (C) is correct.
Crystal Lattice
Regular three-dimensional arrangement of points in space
There are 14 possible three-dimensional lattices, known as Bravais lattices. Characteristics of a crystal lattice:
Each point in a lattice is called lattice point or lattice site. Each lattice point represents one constituent particle (atom, molecule or ion). Lattice points are joined by straight lines to bring out the geometry of the lattice.
Unit Cell
Smallest portion of a crystal lattice which, when repeated in different directions, generates the entire lattice
Characterised by −
(i) Its dimensions along the three edges a, b and c
(ii) Angles between the edges α, β and γ
The unit cells can be classified as follows:
Seven Crystal Systems
There are seven types of primitive unit cells, as given in the following table.
The given table lists seven primitive unit cells and their possible variations as centered unit cells.
7. Triclinic a ≠ b ≠ c α ≠ β ≠γ≠ Primitive K2Cr2O7, H3BO3
90°
Unit cells of 14 types Bravais lattices:
o Cubic lattices: All sides are of the same length, and the angles between the faces
are 90° each
Tetragonal lattices: One side is different in length from the other two, and the angles between the faces are 90° each
Orthorhombic lattices: Unequal sides; angles between the faces are 90° each
Monoclinic lattices: Unequal sides; two faces have angles not equal to 90°
Hexagonal lattice: One side is different in length from the other two, and the marked angles on two faces are 60°
Rhombohedral lattice: All sides are of equal length, and the marked angles on two faces are less than 90°
Triclinic lattice: Unequal sides; unequal angles, with none equal to 90°
Questions asked in previous years’ board examinations
Ques. Name the type of structure possessed by a unit cell of CsCl.
(1 mark)
−2004 CBSE Delhi
Sol: A unit cell of CsCl possesses body-centred cubic structure.
Q1)
Unit cell possible for tetragonal crystal system is
primitive and body centred
face centred
end centred
primitive and end centred
Ans :
For tetragonal crystal system, primitive and body centred structure is possible.
Q2)
Tetragonal crystal system has the unit cell dimensions as
a=b=c and α=β= γ=900
a=b≠c and α=β=γ=900
a≠b≠c and α=β=γ=900
a=b≠c and α=β=900, γ =1200
Ans :
Tetragonal crystal system has edge length as a=b≠c and axial angle as α=β=γ=900
Q3) The coordination number of a cation of body centred cubic lattice is
8
6
12
4
Ans :
Body centred cubic lattice has a coordination number of 8. The cation is surrounded by 8 anions. Alternatively, an anion is surrounded by 8 cations.
Q4)
Use the following information to answer the question.
Unit cell Axial distance
X Orthorhombic I a = b = c
Y Cubic II a = b ≠ c
Z Tetragonal III a ≠ b ≠ c
Which row correctly matches the unit cells with their respective axial distances?
X → II, Y → I, Z → III
X → III, Y → I, Z → II
X → I, Y → II, Z → III
X → II, Y → III, Z → I
Ans :
In orthorhombic crystal lattice, all the sides are unequal, i.e. a ≠ b ≠ c.
In cubic crystal lattice, all the sides are equal, i.e. a = b = c.
In tetragonal crystal lattice, two sides are equal, i.e. a = b ≠ c.
The correct answer is B.
Q5)
Use the following information to answer the question:
All the angles and edges of a crystal X are equal but the angles are not equal to 90º. Example of such type of crystal is cinnabar, HgS.
Which of the following structures correctly represent the crystal of X ?
Orthorhombic
Monoclinic
Tetragonal
Trigonal
Ans :
The answer is D.
Trigonal crystal is also known as rhombohedra. Its all the edges and angles equal. The angles of crystal are not equal to 90º.
Calculation of number of atoms in a unit cell
The number of atoms in a unit cell can be calculated, by using the following approximations.
An atom at the corner is shared by 8 unit cells. Hence, an atom at the corner contributes 1/8 to the unit cell.
An atom at the face is shared by 2 unit cells. Hence, an atom at the face contributes 1/2 to the unit cell.
An atom within the body of a unit cell is shared by no other unit cell. Hence, an atom at the body contributes singly, i.e., 1 to the unit cell.
Primitive Cubic Unit Cell
Open structure for a primitive cubic unit cell is shown in the given figure.
Actual portions belonging to one unit cell are shown in the given figure.
Total number of atoms in one unit cell
Body-Centred Cubic Unit Cell
Open structure for a body-centred cubic unit cell is shown in the given figure.
Actual portions belonging to one unit cell are shown in the given figure.
Total number of atoms in one unit cell
= 8 corners per corner atom + 1 body-centre atom
Face-Centred Cubic Unit Cell
Open structure for a face-centred cubic unit cell is shown in given figure.
Actual portions of atoms belonging to one unit cell are shown in the given figure.
Total number of atoms in one unit cell
= 8 corner atoms atom per unit cell + 6 face-centred atoms atom per unit cell
Questions asked in previous years’ board examinations
Ques. What is the total number of atoms per unit cell in a face-centred cubic (fcc) structure?
(1 mark)
−2008 CBSE Delhi
Sol: The total number of atoms per unit cell in a face-centred cubic (fcc) structure is 4 (8 corner
atoms atom per unit cell + 6 face-centred atoms atom per unit cell)
Ques. What is the number of atoms per unit cell in a body-centred cubic structure?
(1 mark)
−2007 CBSE Delhi
Sol: The number of atoms per unit cell in a body-centred cubic structure is 2 (8 corner atoms ×
atom per unit cell + 1 body-centre atom).
Q1)
Number of unit cells in 938 amu of sodium chloride is
4
8
16
24
Ans :
One unit cell contains 4 formula units
Mass of one unit cell = 58.5 × 4 amu
Number of unit cells is 938 amu
Q2)
In an FCC structure, a unit cell is shared equally by
4 unit cells
2 unit cells
6 unit cells
8 unit cells
Ans :
An FCC unit cell has 6 faces and is thus shared by 6 neighbouring unit cells.
Q3)
If the ionic radii of A+ and B− ions are 0.98×10-10m and 1.81×10-10m respectively, the coordination number of each ion is
6
8
4
8 and 4 respectively
Ans :
Radius ratio = radius of cation/radius of anion
Radius ratio = 0.98×10-10/1.81×10-10=0.54
For octahedral coordination (6), the radius ratio is 0.414 - 0.732
Q4)
The number of unit cells in 58.5 g of NaCl is nearly
6×1020
3×1025
1.5×1023
0.75×1024
Ans :
1 mole of NaCl = 58.5g = 6.023 × 1023 NaCl units
1 unit cell contains 4NaCl units. Hence, the number of unit cells present in 58.5g of NaCl = 6.023 × 1023/4 = 1.56 × 1023
Q5)
PQ is NaCl type solid. P occupies corners of cubic unit cell. If all the atoms present along face centered axis are removed, the resultant stoichiometry and % decrease in the total number of atoms per unit cell are
PQ2 and 25%
P2Q and 25%
P4Q3 and 12.5%
P3Q4 and 12.5%
Ans :
PQ has NaCl type structure so, there are 4 molecules per unit cell (4 atoms of P and Q each i.e. total 8 atom). P occupies 8 corners as well as centres of 6 faces. Removal of face centre atoms along the one axis means 2 atoms of P will be removed while Q remains same.
NaCl type PQ structure PQ structure without atoms along axis
Hence, atoms of P per unit cell = (1/8) x 8 + (1/2) x 4 = 1+ 2 = 3
Hence, formula is P3Q4
% decrease in total no of moles per unit cell =
Coordination number − The number of nearest neighbours of an atom
Close-Packing in One dimension
Only one way of arrangement, i.e., the particles are arranged in a row, touching each other
Coordination number = 2
Close-Packing in Two Dimensions
Square close-packing in two dimensions
AAA type arrangement
The particles in the second row are exactly above those in the first row. Coordination number = 4
Hexagonal close-packing in two dimensions
ABAB type arrangement
The particles in the second row are fitted in the depressions of the first row. The particles in the third row are aligned with those in the first row.
More efficient packing than square close-packing Coordination number = 6
Close-Packing in Three Dimensions
Three-dimensional close-packing is obtained by stacking two-dimensional layers (square close-packed or hexagonal close-packed) one above the other.
By stacking two-dimensional square close-packed layers
The particles in the second layer are exactly above those in the first layer. AAA type pattern The lattice generated is simple cubic lattice, and its unit cell is primitive cubic unit cell.
Coordination number = 6
By stacking two-dimensional hexagonal close-packed layers
Placing the second layer over the first layer
The two layers are differently aligned. Tetrahedral void is formed when a particle in the second layer is above a void of the first
layer. Octahedral void is formed when a void of the second layer is above the void of the first
layer.
Here, T = Tetrahedral void, O = Octahedral void
Number of octahedral voids = Number of close-packed particles
Number of tetrahedral voids = 2 × Number of close-packed particles
Placing the third layer over the second layer: There are two ways −
Covering tetrahedral voids: ABAB … pattern. The particles in the third layer are exactly aligned with those in the first layer. It results in a hexagonal close-packed (hcp) structure. Example: Arrangement of atoms in metals like Mg and Zn
Covering octahedral voids: ABCABC … octahedral voids. The particles in the third layer are not aligned either with those in the first layer or with those in the second layer, but with those in the fourth layer aligned with those in the first layer. This arrangement is called ‘C’ type. It results in cubic close-packed (ccp) or face-centred cubic (fcc) structure. Example: Arrangement of atoms in metals like Cu and Ag
Coordination number in both hcp ad ccp structures is 12. Both hcp and ccp structures are highly efficient in packing (packing efficiency = 74%)
Questions asked in previous years’ board examinations
Ques. What is the coordination number of each type of ions in a rock-salt type crystal structure?
(1 mark)
−2008 CBSE Delhi
Sol: In rock salt or common salt, every Na+ ion is surrounded by 6 Cl− ions and each Cl− ion is surrounded by 6 Na+ ions. Thus, the coordination number of each type of ion in rock salt is 6.
Ques.What is the maximum possible coordination number of an atom in an hcp crystal structure of an element?
(1 mark)
−2005 CBSE Delhi
Sol: The maximum possible coordination number of an atom in an hcp crystal structure of an element is 12.
Q1)
Coordination number of Na in Na2O is
6
4
8
2
Ans :
In Na2O, Na+ is surrounded by four oxide ions and O2− is surrounded by eight sodium ions. So, the coordination number of Na+ is 4 and O2− is 8.
Coordination number of Na+ = 4
O2− = 8
Q2)
Coordination number of anion in the structure of fluorite is
12
8
6
4
Ans :
Fluorite type lattice is shown by compounds such as CaF2. Ca2+ ions are in ccp whereas F− ions occupy the tetrahedral voids. Hence, the coordination number of Ca2+ is 8 and that of F− is 4.
Q3)
A crystalline solid with formula XY2O4 has oxide ions in ccp lattice. Cations X are present in tetrahedral voids, while cations Y are present in octahedral voids. The percentage of octahedral voids occupied by Y is
12
34
50
78
Ans :
In ccp lattice of oxide ions, there are two tetrahedral voids and one octahedral void for each oxide ion.
For four oxide ions, there are 8 tetrahedral and 4 octahedral voids. Out of the 8 tetrahedral voids, 1 is occupied by X and out of the 4 octahedral voids, 2 are occupied by Y.
∴ Octahedral voids occupied by Y Therefore, the correct answer is C.
Q4)
A metal crystallizes with a face centred cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is
288pm
408pm
144pm
201pm
Ans :
For fcc, a=4R (2R=D)
a = 2D
D = ×408/2
D = 288pm
Q5)
In calcium fluoride structure, the coordination number of cation and anion is respectively
6,6
8,4
4,4
4,8
Ans :
In calcium fluoride structure CaF2, cation is surrounded by eight anions and each anion is coordinated to four cations.
Number of octahedral voids = Number of close-packed particles
Number of tetrahedral voids = 2 × Number of close-packed particles
In ionic solids, the bigger ions (usually anions) form the close-packed structure and the smaller ions (usually cations) occupy the voids.
If the latter ion is small enough, then it occupies the tetrahedral void, and if bigger, then it occupies the octahedral void.
Not all the voids are occupied. Only a fraction of the octahedral or tetrahedral voids are occupied.
The fraction of the octahedral or tetrahedral voids that are occupied depends on the chemical formula of the compound.
Example
A compound is formed by two elements X and Y. The atoms of element X form hcp
lattice and those of element Y occupy th of the tetrahedral voids. What is the formula of the compound formed?
Solution:
It is known that the number of tetrahedral voids formed is equal to twice the number of atoms of element X.
It is given that only of the tetrahedral voids are occupied by the atoms of element Y.
Therefore, ratio of the number of atoms of X and Y =
= 2: 1
Hence, the formula of the compound formed is X2Y.
Locating Tetrahedral Voids
A unit cell of ccp or fcc lattice is divided into eight small cubes. Then, each small cube has 4 atoms at alternate corners. When these are joined to each other, a regular tetrahedron is formed.
This implies that one tetrahedral void is present in each small cube. Therefore, a total of eight tetrahedral voids are present in one unit cell.
Since each unit cell of ccp structure has 4 atoms, the number of tetrahedral voids is twice the number of atoms.
Locating Octahedral Voids
When the six atoms of the face centres are joined, an octahedron is generated. This implies that the unit cell has one octahedral void at the body centre.
Besides the body centre, there is one octahedral void at the centre of each of the 12 edges.
But only of each of these voids belongs to the unit cell.
Now, the total number of octahedral voids in a cubic loose-packed structure
This means that in ccp structure, the number of octahedral voids is equal to the number of atoms in each unit cell.
Now, let us see the stacking of two layers and the formation of the two types of voids by means of this video.
Add to your knowledge
In NaCl, the Na+ ions occupy all the octahedral voids. In ZnS, Zn2+ are in alternate tetrahedral voids. In CaF2, F- ions occupy all the tetrahedral voids.
In Fe3O4, if Fe2+ ions are replaced by divalent cations such as Mg2+ and Zn2+, then the compounds obtained are called ferrites.
Questions asked in previous years’ board examinations
Ques. A cubic solid is made of two elements X and Y. Atoms Y are at the corners of the cube and X at the body centre. What is the formula of the compound?
(1 mark)
−2006 CBSE Delhi
Sol: The atom at the body centre makes a contribution of 1 to the unit cell, while the atom
at the corner makes a contribution of to the unit cell.
Thus, number of atoms Y per unit cell
= Number of atoms × Contribution per unit cell
= 8 (at the corners) × atoms per unit cell
= 1
Thus, number of atoms X per unit cell
= Number of atoms × contribution per unit cell
= 1 (at the body centre) × 1
= 1
Thus, the formula of the given compound is XY.
Q1)
In corundum, the oxide ions are arranged in ccp array and the aluminium ions occupy of the octahedral voids. The formula of corundum is
AlO3
Al3O2
Al2O3
AlO
Ans :
In ccp, there is one octahedral void corresponding to each atom constituting the close packing.
In corundum, only of the octahedral voids are occupied. It means, corresponding to each
oxide ion, there are aluminium ions.
So, the whole number ratio of oxide and aluminium ions in corundum is 3:2.
∴ Formula of corundum = Al2O3
Therefore, the correct answer is C.
Q2)
The number of octahedral voids per atom present in a cubic close packed structure is
1
3
2
4
Ans :
Number of octahedral voids in ccp is equal to the effective number of atoms in ccp. The effective number of atoms is 4 and so there are 4 octahedral voids so 1 octahedral void per atom.
Q3)
In a solid AB having NaCl structure, atoms occupy the corners of the cube unit cell. If all the face-centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid will be
AB2
A4B3
A3B4
A2B
Ans :
In one NaCl-type unit cell of solid, B− ions are present at the corners and face centres while A+ ions are present at the edge centres and body centre.
Contribution of a face-centred B− ions = 1/2
In one NaCl-type unit cell of solid, B− ions are present at the corners and face the centres and A+ ions are present at the edge centres and body centres. Thus, a unit cell will contain 4A+ and 4B-
ions.In this unit, two face-centered B− ions lie along one axis, and they are removed.
Contribution of two face-centred B− ions = 2 × 1/2 = 1Therefore, in the resultant unit cell, the number of A+ ions present per unit cell is 4 and the number of B− ions present per unit cell is 4−1, i.e., 3.
The resultant stoichiometry is A4B3.Hence, the correct option is B.
Q4)
In a solid structure, W atoms are located at corners of the cubic lattice, O atoms at the centre of edges and Na atom at centre of cube. The formula for the compound is
NaWO2
NaWO3
Na2WO3
NaWO4
Ans :
W atoms per unit cell = 8 × 1/8 = 1
O atoms per unit cell = 12 × ¼ = 3
Na atoms per unit cell = 1
Hence, the formula is NaWO3
Q5)
A compound has cubic close (ccp) arrangement of X. Its unit cell structure shown below. The empirical formula of the compound is
MX
Ans :
Arrangement of atoms in a cubic crystal system:
PC (primitive cubic)
FCC (face centered cubic)
BCC (body centered cubic)
Number of
Number of
Packing Efficiency
Percentage of total space filled by particles
Calculations of Packing Efficiency in Different Types of Structures
Simple cubic lattice
In a simple cubic lattice, the particles are located only at the corners of the cube and touch each other along the edge.
Let the edge length of the cube be ‘a’ and the radius of each particle be r.
Then, we can write:
a = 2r
Now, volume of the cubic unit cell = a3
= (2r)3
= 8r3
The number of particles present per simple cubic unit cell is 1.
Therefore, volume of the occupied unit cell
Hence, packing efficiency
Body-centred cubic structures
It can be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.
From ΔFED, we have
From ΔAFD, we have
Let the radius of the atom be r.
Length of the body diagonal, c = 4r
or,
Volume of the cube,
A body-centred cubic lattice contains 2 atoms.
hcp and ccp Structures
Let the edge length of the unit cell be ‘a’ and the length of the face diagonal AC be b.
From ΔABC, we have
Let r be the radius of the atom.
Now, from the figure, it can be observed that:
Now, volume of the cube,
We know that the number of atoms per unit cell is 4.
Thus, ccp and hcp structures have maximum packing efficiency.
Calculations Involving Unit Cell Dimensions
In a cubic crystal, let
a = Edge length of the unit cell
d = Density of the solid substance
M = Molar mass of the substance
Then, volume of the unit cell = a3
Again, let
z = Number of atoms present in one unit cell
m = Mass of each atom
Now, mass of the unit cell = Number of atoms in the unit cell × Mass of each atom
= z × m
But, mass of an atom, m
Therefore, density of the unit cell,
Let us calculate the density of an element crystallising in face-centred cubic lattice.
Questions asked in previous years’ board examinations
Ques. The density of copper metal is 8.95 g cm−3. If the radius of copper atom is 127.8 pm, is the copper unit cell a simple cubic, a body-centred cubic or a face centred cubic structure? (Given: At. Mass of Cu = 63.54 g mol−1 and NA = 6.02 × 1023 mol−1)
(3 marks)
−2010 CBSE Delhi
Sol: We know that density,
For SCC: z = 1 and a = 2r
For BCC: z = 2 and
For FCC: z = 4 and
Substituting the values of z and a in equation (i) we can calculate the value of density.
It is given that M = 63.54 g mol−1
NA = 6.02 × 1023 mol−1
r = 127.8 × 10−10 cm
The given value of density is 8.95 g/cm3. Hence, from the given data, we can conclude that copper unit cell is face centred cubic or fcc.
Ques. 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)
(3 marks)
−2009 CBSE Delhi
Sol: In a body-centred cubic unit cell, number of atoms present = 2
At mass of iron = 56 g mol−1
Density of iron = 7.87 g cm−3
Mass of iron = 7.87 × Volume
Volume in BCC = (a)3
= (286.65)3 pm
= 2.34 × 10−23 cm
Mass = 7.87 × 2.34 × 10−23 g
∴ Avogadro’s number = 6.022 × 1023
Ques. An element has a body-centred cubic structure with a cell edge of 288 pm. The density of the element is 7.2 g cm−3 Calculate the number of atoms presents in 208 g of the element.
(3 marks)
−2006 CBSE Delhi
Sol: Cell edge (a) = 288 pm
Volume of unit cell = a3
= (288 pm)3
= (288 × 10−10 cm)3
= 2.389 × 10−23 cm3
Volume of 208 g of the element =
= 28.89 cm3
Number of unit cells =
= 12.09 × 1023
In a bcc structure, the number of atoms per unit cell = 2
∴ Number of atoms in 208 g of the given element = 2 × 12.09 × 1023
= 24.18 × 1023
Ques. Aluminium metal forms a cubic close-packed crystal structure. Its atomic radius is 125 x 10 -12 m.
(a) Calculate the length of the side of the unit cell.
(b) How many such unit cells are there in 1.00 m3 of aluminium?
(3 marks)
−2005 CBSE Delhi
Soli:
(a) For a cubic close-packed crystal structure,
Radius of an atom,
Where, a → Edge length
Therefore, a =
Hence, the length of the side of the unit cell is 354 × 10−12 m.
(b) Volume of the unit cell = a3
Ques. Calculate the density of silver which crystallises in the face-centred cubic structure. The distance between the nearest silver atoms in this structure is 287 pm.
(3 marks)
−2004 CBSE Delhi
Sol: Given, molar mass of Ag, M = 107.87 g mol−1
NA = 6.02 × 1023 mol−1
In case of fcc lattice, number of atoms per unit cell, z = 4
Distance between two nearest Ag atoms = 287 pm
Thus, edge length = 406 pm (approx)
= 406 × 10−12 m
Therefore, density of silver is given by
Q1)
The edge length of a unit cell of a metal having molecular mass 23 g/mol is 422.5 pm. If the metal crystallises in a cubic lattice, having a density of 1.51g/cc, then the radius of the metal atom is
182 nm
1.82 nm
18.2 pm
182 pm
Ans :
Number of atoms per unit cell is 2, which indicates that the metal has a body centred cubic lattice.
Therefore, the correct answer is D.
Q2)
Number of unit cells present in a 1.8 g cube shaped ideal crystal of NaCl is
4.6 × 1021
5.8 × 1022
2.3× 1023
4.9× 1022
Ans :
Gram formula mass of NaCl = 58.5 g
Number of formula units of NaCl in 1.8 g of NaCl
There are 4 formula units per unit cell in NaCl.
∴Number of unit cells
Q3)
A metal crystallises into simple, body centred and face centred cubic structure, whose unit cell lengths are A1, A2 and A3 respectively. The correct order of lengths of the unit cells is
A1 < A2 < A3
A2 < A3 < A1
A1 < A3 < A2
A3 < A1 < A2
Ans :
Let the radius of the atom of the metal be r.
Relationship between unit cell length and radius of the atom in:
Simple cubic
A1 = 2r = 2.00r
Body centred cubic
Face centred cubic
Therefore, the correct answer is 1.
Q4)
The interionic distance for caesium iodide crystal is
a
a/2
Ans :
Radius of Cs+ =
Radius of Cl- =
These radius are related with edge length, ‘a’ of unit cell as
Here, is interionic distance.
Q5)
The unit cube length of LiCl (NaCl structure) is 5.14 Å. Assuming anion-anion contacts, what is the radius of chloride ion?
1.24 Å
1.81 Å
2.97 Å
4.85 Å
Ans :
Inter-ionic distance of LiCl = 5.14/2 = 2.57 Å
Therefore,
The correct answer is B.
Defects
Irregularities or deviations from the ideal arrangement of constituent particles Two types:
Point defects − Irregularities in the arrangement of constituent particles around a point or an atom in a crystalline substance.
Line defects − Irregularities in the arrangement of constituent particles in entire rows of lattice points.
Do not disturb stoichiometry of the solid Also called intrinsic or thermodynamic defects Two types − (i) Vacancy defect
(ii) Interstitial defect
Vacancy defect
When some of the lattice sites are vacant Shown by non-ionic solids Created when a substance is heated Results in the decrease in density of the substance
Interstitial defect
Shown by non-ionic solids Created when some constituent particles (atoms or molecules) occupy an interstitial site
of the crystal.
Ionic solids show these two defects as Frenkel defect and Schottky defect. Frenkel defect
Shown by ionic solids containing large differences in the sizes of ions Created when the smaller ion (usually cation) is dislocated from its normal site to an
interstitial site Creates a vacancy defect as well as an interstitial defect Also known as dislocation defect Ionic solids such as AgCl, AgBr, AgI and ZnS show this type of defect.
Schottky defect
Basically a vacancy defect shown by ionic solids An equal number of cations and anions are missing to maintain electrical neutrality Results in the decrease in the density of the substance Significant number of Schottky defect is present in ionic solids. For example, in NaCl,
there are approximately 106 Schottky pairs per cm3, at room temperature. Shown by ionic substances containing similar-sized cations and anions; for example,
NaCl, KCl CsCl, AgBr
Impurity Defect
Point defect due to the presence of foreign atoms For example, if molten NaCl containing a little amount of SrCl2 is crystallised, some of
the sites of Na+ ions are occupied by Sr2+ ions. Each Sr2+ ion replaces two Na+ ions, occupying the site of one ion, leaving the other site vacant. The cationic vacancies thus produced are equal in number to those of Sr2+ ions.
Solid solution of CdCl2 and AgCl also shows this defect
Non-Stoichiometric Defects
Result in non-stoichiometric ratio of the constituent elements Two types −
Metal excess defect Metal deficiency defect
Metal excess defect
Metal excess defect due to anionic vacancies:
Alkali metals like NaCl and KCl show this type of defect.
When crystals of NaCl are heated in an atmosphere of sodium vapour, the sodium atoms are deposited on the surface of the crystal. The Cl− ions diffuse from the crystal to its surface and combine with Na atoms, forming NaCl. During this process, the Na atoms on the surface of the crystal lose electrons. These released electrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.
When the ionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F-centres.
Metal excess defect due to the presence of extra cations at interstitial sites:
When white zinc oxide is heated, it loses oxygen and turns yellow.
Then, zinc becomes excess in the crystal, leading the formula of the oxide to . The excess Zn2+ ions move to the interstitial sites, and the electrons move to the neighbouring interstitial sites.
Metal deficiency defect
Arises when a solid contains lesser number of cations compared to the stoichiometric proportion.
For example, FeO is mostly found with a composition of . In crystals of FeO, some Fe2+ ions are missing, and the loss of positive charge is made up by the presence of the required number of Fe3+ ions.
Questions asked in previous years’ board examinations
Ques. Which point defect in crystals of a solid decreases the density of the solid?
(1 mark)
−2010 CBSE Delhi
Sol: Vacancy defect decreases the density of a substance. Vacancy defect in ionic solids is known as Schottky defect.
Ques. Which point defect in crystals does not affect the density of the relevant solid?
(1 mark)
2009 CBSE Delhi
Sol: Frenkel defect in crystals does not affect the density of the relevant solid.
Ques. Which point defect in its crystal units alters the density of a solid?
(1 mark)
2008 CBSE Delhi
Sol: Schottky defect in crystals units alters the density of a solid.
Ques. How would you account for the following?
(i) Frenkel defects are not found in alkali metal halides.
(ii) Schottky defects lower the density of related solids.
(iii) Impurity doped silicon is a semiconductor.
(3 marks)
2008 CBSE Delhi
Sol: (i) Frenkel defects are shown by ionic solids having large differences in the sizes of ions. Solids such as ZnS, AgCl show these defects due to the small size of Zn2+ and Ag+ ions, and the large size of anions. Alkali metals are not so small so as to show these defects. Hence, Frenkel defects are not found in alkali metal halides.
(ii) Schottky defects are basically vacancy defects in ionic solids. In these defects, lattice sites become vacant. As a result, the density of the substance decreases.
(iii) Silicon is an intrinsic semi-conductor in which conductivity is very low. To increase its conductivity, silicon is doped with an appropriate amount of suitable impurity. When doped with electron-rich impurities such as P or As, n-type semi-conductor is obtained, and when doped with electron-deficient impurities, p-type semi-conductor is obtained. In n-type semiconductor, negatively charged electron is responsible for increasing conductivity, and in p-type
semiconductor, electron hole is responsible for increasing conductivity.
Ques. What are the types of lattice imperfections found in crystals?
(1 mark)
2004 CBSE Delhi
Sol: Two types of lattice imperfections are found in crystals.
(i) Point defects (irregularities in arrangement around a point or an atom)
(ii) Line defects (irregularities in arrangement in entire rows of lattice points)
Ques. Explain interstitial defect with examples.
(3 marks)
2004 CBSE Delhi
Sol:Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial site of the crystal. The density of a substance increases because of this defect.
Q1)
Consider the diagram given below:
Which type of crystal defect is shown in the given diagram?
Frenkel defect
Schottky defect
Impurity defect
Interstitial defect
Ans :
Schottky defect arises when lattice sites are vacant. This defect decreases the density of the crystal.
Q2)
The ionic solid defect in which the density of ionic solid decreases is
Dislocation defect
Frenkel defect
Schottky defect
Interstitial defect
Ans :
In Schottky defect, equal number of cations and anions are missing from their lattice sites and hence results in decrease in density of solid.
Q3)
Which of the following does not exhibit Frenkel defect?
AgBr
AgCl
KBr
ZnS
Ans :
KBr does not show frenkel defect. Frenkel defect is shown by compounds which have large difference between size of cation and anion. KBr does not have such large difference in size. Actually KBr shows schottky defect in which there is small difference between size of cation and anion.
Q4)
When NaCl is doped with MgCl2, the nature of defect produced is
Schottky defect
Frenkel defect
Interstitial defect
Impurity defect
Ans :
In impurity defect, some of the sites of cations of crystal are occupied by cations of another element (termed as impurity) having higher charge. Here, one Mg2+ replaces two Na+ ions leaving one vacant site. Thus, cationic vacancies are produced equal to the number of Mg2+ ions.Hence, the correct option is D.
Q5)
Which of the following statements is correct with respect to defects in solids?
Frenkel defect is usually favoured by a very small difference in the sizes of cations and anions.
Schottky defect is a dislocation defect.
Trapping of an electron in the lattice leads to the formation of F-centres.
Schottky defect has no effect on the physical properties of solids.
Ans :
Frenkel defect is favoured when the difference in the size of one cation and anion is quite large. It is a dislocation defect. Schottky defect decreases the density of the solid.
The correct answer is C.
Study MaterialChange Chapter
The Solid State 0% Topic: Electrical PropertiesTake a Chapter Test View NCERT Solutions View Revision Notes
Conduction of Electricity in Metals
Metals conduct electricity in molten state. The conductivity of metals depends upon the number of valence electrons.
In metals, the valence shell is partially filled, so this valence band overlaps with a higher energy unoccupied conduction band so that electrons can flow easily under an applied electric field.
In the case of insulators, the gap between filled valence shell and the next higher unoccupied band is large so that electrons cannot jump from the valence band to the conduction band.
Conduction of Electricity in Semiconductors
The gap between the valence band and conduction band is so small that some electrons may jump to the conduction band.
Electrical conductivity of semiconductors increases with increase in temperature. Substances like Si, Ge show this type of behaviour, and are called intrinsic
semiconductors. Doping − Process of adding an appropriate amount of suitable impurity to increase
conductivity
Doping is done with either electron-rich or electron-deficient impurity as compared to the intrinsic semiconductor Si or Ge.
There are two types of semiconductors:
i. n − type semiconductorii. p − type semiconductor
n − type semiconductor
Conductivity increases due to negatively charged electrons Generated due to the doping of the crystal of a group 14 element such as Si or Ge, with a
group 15 element such as P or As
p − type semiconductor
Conductivity increases as a result of electron hole Generated due to the doping of the crystal of a group 14 element such as Si or Ge, with a
group 13 element such as B, Al or Ga
Applications of n − type and p − type semiconductors
In making a diode, which is used as a rectifier
In making transistors, which are used for detecting or amplifying radio or audio signals In making a solar cell, which is a photo diode used for converting light energy into
electrical energy
A large number of compounds (solid) have been prepared by the combination of groups 13 and 15 or 12 and 16 to stimulate average valence of four as in Si or Ge.
Examples of compounds of groups 13 − 15 are InSb, AlP, GaAs Examples of compounds of groups 12 − 16 are ZnS, CdS, CdSe, HgTe
Some transition metal oxides like TiO, CrO2, ReO3 behave like metals.
For example, ReO3 resembles metallic copper in its conductivity and appearance Some oxides like VO, VO2, VO3, TiO3 show metallic or insulating properties depending
on temperature.
Do you know?
Polyacetylene, an organic compound shows conductivity when exposed to iodine vapours.
Questions asked in previous years’ board examinations
Ques. What is semiconductor? Describe the two main types of semiconductors and explain mechanisms for their conduction.
(3 marks)
−2008 CBSE Delhi
Sol: Solids having intermediate conductivities (from 10−6 to 104 Ω−1 m−1) are calledsemiconductors. Germanium and silicon are two examples of semi-conductors. These substances act as insulators at low temperatures and as conductors at high temperatures. There are two types of semiconductors:
i. n-type semiconductor:
When the crystal of a semiconductor is doped with group-15 elements (P, As, Sb or Bi), only four of the five valence electrons of the doped atoms participate in forming covalent bonds with the atoms of the semiconductors. The fifth electron is free to conduct electricity. As these crystals contain extra electrons, these are known as n-type semiconductors.
ii. p-type semiconductor:
When the crystal of a semiconductor is doped with group-13 elements (Al, Ga or In), only three covalent bonds are formed by the atoms of the doped atoms as they contain three valence electrons. A hole is created at the place where the electron is absent. The presence of such holes increases the conductivity of the semiconductor as the neighbouring electrons can move into these holes, thereby creating newer holes. As these crystals contain lesser electrons than un-doped crystals, they are known as p-type semiconductors.
Q1)
Electrical conductivity of semiconductors
increases with increase in temperature
decreases with increase in temperature
first increases and then decreases with increase in temperature
first decreases and then increases with increase in temperature
Ans :
Electrical conductivity of semiconductors increases with increase in temperature. This is because in case of semiconductors, conduction is due to impurities and defects, which increase with increase in temperature.Therefore, the correct answer is A.
Q2)
Silicon doped with arsenic is an example of which type of semiconductor?
p-type
n-type
n,p-type
Intrinsic
Ans :
Arsenic belongs to group-15 elements having 5 valence electrons and thus, it forms n-type semiconductor with silicon (group-14). Four out of five electrons are used in the formation of four covalent bonds with the four silicon atoms. Thus, the fifth electron remains unused which increases the conductivity of doped silicon.
Q3)
A semiconductor of Ge can be made p-type by adding
trivalent impurity
divalent impurity
tetravalent impurity
pentavalent impurity
Ans :
Ge is group-14 element. Positive holes can be created by adding group-13 element i.e. trivalent impurity.
Q4)
Use the following information to answer the question.
When a crystal of Si is doped with , it leads to the formation of semi-conductor.
The information in which row completes the given statement?
i ii
B n-type
i ii
Al p-type
i ii
P p-type
i ii
Ga n-type
Ans :
The process of increasing conductivity of intrinsic conductors by adding an appropriate amount of suitable impurity is called doping. Doping can be done with an impurity that may be electron rich or electron deficient as compared to the intrinsic semi-conductor. Silicon and germanium are intrinsic semi-conductors. The doping of crystals of Si or Ga with a group 13 element (electron deficient), such as B, Al, or Ga, leads to the creation of p-type semi-conductor, while that with a group 15 element (electron rich), such as P or As, leads to the creation of n-type semi-conductor.
Therefore, the correct answer is B.
Q5)
Use the following information to answer the next question.
The decomposition reaction of potassium chlorate is represented as
2KClO3 (s) → 2KCl (s) + 3O2 (g)
KCl will conduct electricity in
water
ethanol
benzene
chloroform
Ans :
This question is successfully answered by the student who has knowledge of the properties of KCl. KCl is an ionic solid. This clearly means that it will conduct electricity in water only.
The correct answer is A.
Why alternatives B, C and D are wrong:
Benzene, chloroform, and ethanol are organic solvents, whereas KCl is an inorganic salt. KCl will not conduct electricity in any of these solvents. Hence, alternatives B, C, and D are incorrect.
Each electron in an atom behaves like a tiny magnet. The magnetic moment of an electron originates from its two types of motion.
Orbital motion around the nucleus Spin around its own axis
Thus, an electron has a permanent spin and an orbital magnetic moment associated with it.
An orbiting electron A spinning electron
Based on magnetic properties, substances are classified into five categories −
The substances that are attracted by a magnetic field are called paramagnetic substances.
Some examples of paramagnetic substances are O2, Cu2+, Fe3+ and Cr3+.
Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed.
To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.
Diamagnetism
The substances which are weakly repelled by magnetic field are said to have diamagnetism.
Example − H2O, NaCl, C6H6
Diamagnetic substances are weakly magnetised in a magnetic field in opposite direction. In diamagnetic substances, all the electrons are paired. Magnetic characters of these substances are lost due to the cancellation of moments by
the pairing of electrons.
Ferromagnetism
The substances that are strongly attracted by a magnetic field are called ferromagnetic substances.
Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field.
Some examples of ferromagnetic substances are iron, cobalt, nickel, gadolinium andCrO2.
In solid state, the metal ions of ferromagnetic substances are grouped together into small regions called domains, and each domain acts as a tiny magnet. In an un-magnetised piece of a ferromagnetic substance, the domains are randomly oriented, so their magnetic moments get cancelled. However, when the substance is placed in a magnetic field, all the domains get oriented in the direction of the magnetic field. As a result, a strong magnetic effect is produced. This ordering of domains persists even after the removal of the magnetic field. Thus, the ferromagnetic substance becomes a permanent magnet.
Schematic alignment of magnetic moments in ferromagnetic substances is as follows:
Ferrimagnetism
The substances in which the magnetic moments of the domains are aligned in parallel and anti-parallel directions, in unequal numbers, are said to have ferrimagnetism.
Examples include Fe3O4 (magnetite), ferrites such as MgFe2O4 and ZnFe2O4.
Ferrimagnetic substances are weakly attracted by a magnetic field as compared to ferromagnetic substances.
On heating, these substances become paramagnetic. Schematic alignment of magnetic moments in ferrimagnetic substances is as follows:
Anti-ferromagnetism
Antiferromagnetic substanceshave domain structures similar to ferromagnetic substances, but are oppositely oriented.
The oppositely oriented domains cancel out each other’s magnetic moments. Schematic alignment of magnetic moments in anti-ferromagnetic substances is as
follows:
Do you know?
When a substance offers no resistance to the flow of electric current, it is said to be superconducting. This phenomenon was first discovered by Kammerlingh Onnes.
Questions asked in previous years’ board examinations
Ques. What type of substances exhibits antiferromagnetism?
(1 mark)
−2008 CBSE Delhi
Sol: Antiferromagnetism is exhibited by substances which have domain structure similar to ferromagnetic substances, but are oppositely oriented (thereby cancelling out each other’s magnetic moment), e.g., MnO.
Ques. Explain each of the following with a suitable example:
(i) Paramagnetism
(ii) Piezoelectric effect
(3 marks)
−2007 CBSE Delhi
Sol: (i) Paramagnetism:
The phenomenon due to which a substance gets attracted towards a magnetic field is called paramagnetism. The substances attracted by a magnetic field are called paramagnetic substances. Some examples of paramagnetic substances are O2, Cu2t, Fe3t and Cr3t.
Paramagnetic substances get magnetised in a magnetic field in the same direction, but lose magnetism when the magnetic field is removed. To undergo paramagnetism, a substance must have one or more unpaired electrons. This is because the unpaired electrons are attracted by a magnetic field, thereby causing paramagnetism.
(ii) Piezoelectric effect:
The production of electricity due to the displacement of ions, on the application of mechanical stress, or the production of mechanical stress and/or strain due to atomic displacement, on the application of an electric field is known as piezoelectric effect. Piezoelectric materials are used in transducers − devices that convert electrical energy into mechanical stress/strain or vice-versa. Some piezoelectric materials are lead-zirconate (PbZrO3), ammonium dihydrogen phosphate (NH4H2PO4), quartz, etc.
Ques. What makes alkali metal halides sometimes coloured, which are otherwise colourless?
(1 mark)
−2004 CBSE Delhi
Alkali metal halides have anionic sites occupied by unpaired electrons. These are called F-centres, and impart colour to the crystals of alkali metal halides. For example, the excess of lithium in LiCl makes it pink.
Q1)
Which of the following oxides is ferromagnetic in nature?
Cr2O3
Fe3O4
FeO
CrO2
Ans :
Ferromagnetic substances are those that are strongly attracted by magnetic field. These exhibit permanent magnetism even in the absence of magnetic field.
Therefore, the correct answer is D.
Q2)
The ferromagnetic substances turn paramagnetic on
cooling
heating
washing
quenching
Ans :
In ferromagnetic substances, the parallel and anti-parallel alignment of magnetic dipoles get randomised upon heating. These make them paramagnetic.Therefore, the correct answer is B.
Q3)
Use the following information to answer the question.
substances are strongly attracted, while substances are weakly attracted by a magnetic field.
The information in which row completes the given statement?
i ii
Ferromagnetic paramagnetic
i ii
Paramagnetic anti-ferromagnetic
i ii
Diamagnetic ferromagnetic
i ii
Paramagnetic ferromagnetic
Ans :
Ferromagnetic substances are strongly attracted, while paramagnetic substances are weakly attracted by a magnetic field. Fe, Co, Ni, Gd, CrO2 are some examples of ferromagnetic substances. They can be permanently magnetised.
In a magnetic field, paramagnetic substances are magnetised in the same direction. In the absence of a magnetic field, these substances lose their magnetism. O2, Cu2+, Fe3+, Cr3+ are some examples of paramagnetic substances.
The correct answer is A.
Q4)
Use the following information to answer the next question.
substances are strongly attracted while substances are weakly attracted by magnetic fields.
The information in which alternative completes the given statement?
i ii
Ferromagnetic paramagnetic
i ii
Paramagnetic anti-ferromagnetic
i ii
Diamagnetic ferromagnetic
i ii
Paramagnetic ferromagnetic
Ans :
Ferromagnetic substances are strongly attracted while paramagnetic substances are weakly attracted by a magnetic field. Fe, Co, Ni, Gd, CrO2 are some examples of ferromagnetic substances. They can be permanently magnetized.
Paramagnetic substances are magnetized in a magnetic field in the same direction. In the absence of magnetic field, these substances lose their magnetism. O2, Cu2+, Fe3+, Cr3+ are some examples of paramagnetic substances.
The correct answer is A.
Q5)
Use the following information to answer the next question.
The empirical formula of a metal oxide is .
In this crystal, the respective percentages of M2+ and M3+ ions are
10.53% and 89.47%
12.24% and 87.76%
87.76% and 12.24%
89.47% and 10.53%
Ans :
The formula shows that,
M:O = 0.95:1.00
= 95:100
That is, if there are 100 O atoms, then number of M atoms = 95
Total charge on 100 O2− ions
= 100 × (−2)
= −200
Let number of M atoms as M2+ = x
Therefore, number of M atoms as M3+ = 95 − x
Then, total charge on M2+ and M3+ ions
As the metal oxide is neutral, the sum of the charges on cations and anions is equal to zero.
