NCERT Solutions for Class 12 Chemistry Chemistry Part 1 Chapter 1
The Solid State Class 12
Chapter 1 The Solid State Exercise Solutions
In text : Solutions of Questions on Page Number : 4
Why are solids rigid?
Why do solids have a definite volume?
Classify the following as amorphous or crystalline solids:
Answer :
Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass
Why is glass considered a super cooled liquid?
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
An isotropic solid has the same value of physical properties when measured along different directions. Therefore, the given solid, having the same value of refractive index along all directions, is isotropic in nature. Hence, the solid is an amorphous solid.
Classify the following solids in different categories based on the nature of intermolecular forces operating in them:
Answer :
Potassium sulphate → Ionic solid Tin → Metallic solid
Benzene → Molecular (non-polar) solid Urea → Polar molecular solid Ammonia → Polar molecular solid
Water → Hydrogen bonded molecular solid Zinc sulphide → Ionic solid
Graphite → Covalent or network solid Rubidium → Metallic solid
Argon → Non-polar molecular solid
Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Whattype of solids are electrical conductors, malleable and ductile?
Give the significance of a 'lattice point'.
Name the parameters that characterize a unit cell.
The six parameters that characterise a unit cell are as follows.
(i) Its dimensions along the three edges, a, b, and c
These edges may or may not be equal.
(ii) Angles between the edges
Distinguish between
(ii) Face-centred and end-centred unit cells.
(i) Hexagonal unit cell
In a face-centred unit cell, the constituent particles are present at the corners and one at the centre of each face.
Explain how much portion of an atom located at (i)corner and (ii)body-centre of a cubic unit cell is part of its neighbouring unit cell.
(i) An atom located at the corner of a cubic unit cell is shared by eight adjacent unit cells. Therefore, portion of the atom is shared by one unit cell.
(ii)
What is the two dimensional coordination number of a molecule in square close packed layer?
A compound forms hexagonal close-packed structure. What is the total number of voids in 0.5 mol of it? How many of these are tetrahedral voids?
Number of close-packed particles = 0.5 x 6.022 x 1023 = 3.011 x 1023 Therefore, number of octahedral voids = 3.011 x 1023
And, number of tetrahedral voids = 2 x 3.011 x 1023 = 6.022 x 1023
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rdof tetrahedral voids. What is the formula of the compound?
The ccplattice is formed by the atoms of the element N.
Here, the number of tetrahedral voids generated is equal to twice the number of atoms of the element N.
According to the question, the atoms of element M occupy of the tetrahedral voids.
Therefore, the number of atoms of M is equal to of the number of atoms of N.
Therefore, ratio of the number of atoms of M to that of N is M: N
Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centred cubic and
Answer :
An element with molar mass 2.7 x 10-2kg mol-1forms a cubic unit cell with edge length 405 pm. If its density is
Answer :
It is given that density of the element, d= 2.7 ×103 kg m - 3
Molar mass, M = 2.7 ×10 - 2kg mol - 1
Edge length, a= 405 pm = 405 ×10 - 12m
= 4.05 ×10 - 10m
It is known that, Avogadro's number, NA= 6.022 ×1023 mol - 1
This implies that four atoms of the element are present per unit cell. Hence, the unit cell is face-centred cubic (fcc) or cubic close-packed (ccp).
What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
When a solid is heated, vacancy defect can arise. A solid crystal is said to have vacancy defect when some of the lattice sites are vacant.
What type of stoichiometric defect is shown by:
Answer :
(i) ZnS shows Frenkel defect.
(ii)
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
The colour develops because of the presence of electrons in the anionic sites. These electrons absorb energy from the visible part of radiation and get excited.
For example, when crystals of NaCl are heated in an atmosphere of sodium vapours, the sodium atoms get deposited on the surface of the crystal and the chloride ions from the crystal diffuse to the surface to form NaCl with the deposited Na atoms. During this process, the Na atoms on the surface lose electrons to form Na+ions and the
A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ferromagnetic substances would make better permanent magnets.
In solid state, the metal ions of ferromagnetic substances are grouped together into small regions. These regions are called domains and each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance, the domains are randomly oriented. As a result, the magnetic moments of the domains get cancelled. However, 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.
Exercise : Solutions of Questions on Page Number : 30
Define the term 'amorphous'. Give a few examples of amorphous solids.
Amorphous solids are the solids whose constituent particles are of irregular shapes and have short range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.
What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
The arrangement of the constituent particles makes glass different from quartz. In glass, the constituent particles have short range order, but in quartz, the constituent particles have both long range and short range orders.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(ii) Ammonium phosphate (NH4)3PO4 (viii) Brass
(iv) I2 (x) LiBr
Answer :
Ionic → (ii)Ammonium phosphate (NH4)3PO4, (x) LiBr Metallic → (viii)Brass, (ix)Rb
Molecular → (i) Tetra phosphorus decoxide (P4O10), (iv) I2, (v) P4. Covalent (network) → (iii) SiC, (vii)Graphite, (xi) Si
(i) What is meant by the term 'coordination number'?
(a) in a cubic close-packed structure?
Answer :
(i) The number of nearest neighbours of any constituent particle present in the crystal lattice is called its coordination number.
(ii) The coordination number of atoms
(a) in a cubic close-packed structure is 12, and
(b)
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.
By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.
Let 'a' be the edge length of a unit cell of a crystal, 'd' be the density of the metal, 'm' be the mass of one atom of the metal and 'z' be the number of atoms in the unit cell.
[Since mass of the unit cell = Number of atoms in the unit cell × mass of one atom] [Volume of the unit cell = (Edge length of the cubic unit cell)3]
Now, mass of one atom of metal (m) Therefore,
If the edge lengths are different (say a, b and c), then equation (ii) becomes:
'Stability of a crystal is reflected in themagnitude of its melting point'. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
|
Higher the melting point, greater is the intermolecular force of attraction and greater is the stability. A substance with higher melting point is more stable than a substance with lower melting point.
The melting points of the given substances are:
Solid water → 273 K Ethyl alcohol → 158.8 K Diethyl ether → 156.85 K Methane → 89.34 K
How will you distinguish between the following pairs of terms:
(ii) Crystal lattice and unit cell?
Answer :
i.
|
|
Figure 2 |
Figure 3 |
Now, the next layer can be placed over layer B in 2 ways.
In this case we get hexagonal close-packing. This is shown in figure 4. In figure 4.1, layer B is present over the voids a and layer C is present over the voids c. In figure 4.2, layer B is present over the voids b and layer C is present over the voids c. It can be observed from the figure that in this arrangement, the spheres present in layer C are present directly above the spheres of layer A. Hence, we can say that the layers in hexagonal close-packing are arranged in an ABAB….. pattern.
|
|
Figure 4.1 |
Figure 4.2 |
In this case we get cubic close-packing. In figure 5.1, layer B is present over the voids a and layer C is present over the voids d. In figure 5.2, layer B is present over the voids b and layer C is present over the voids d. It can be observed from the figure that the arrangement of particles in layer C is completely different from that in layers A or B.
How many lattice points are there in one unit cell of each of the following lattice?
(ii) Face-centred tetragonal
Answer :
(i) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred cubic.
(ii) There are 14 (8 from the corners + 6 from the faces) lattice points in face-centred tetragonal.
(iii)
Explain
(ii) Ionic solids are hard and brittle.
(i) The basis of similarities between metallic and ionic crystals is that both these crystal types are held by the electrostatic force of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points.
The basis of differences between metallic and ionic crystals is that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot conduct electricity. However, in molten state or in aqueous solution, they do conduct electricity.
(ii)
Calculate the efficiency of packing in case of a metal crystal for
(ii) body-centred cubic
Answer :
Let the edge length of the cube be 'a' and the radius of each particle be r. So, we can write:
a= 2r
Now, volume of the cubic unit cell = a3
= (2r)3
= 8r3
We know that the number of particles per unit cell is 1.
Therefore, volume of the occupied unit cell
Hence, packing efficiency
Itcan be observed from the above figure that the atom at the centre is in contact with the other two atoms diagonally arranged.
Again, from ΔAFD, we have:
Let the radius of the atom be r. Length of the body diagonal, c= 4π
or,
Volume of the cube,
A body-centred cubic lattice contains 2 atoms.
From ΔABC, we have:
Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10-8cm and density is 10.5 g cm-3, calculate the atomic mass of silver.
It is given that the edge length, a= 4.077 × 10 - 8cm Density,d = 10.5 g cm - 3
As the lattice is fcc type, the number of atoms per unit cell, z= 4 We also know that, NA= 6.022 × 1023mol - 1
= 107.13 gmol - 1
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body- centre. What is the formula of the compound? What are the coordination numbers of P and Q?
It is given that the atoms of Q are present at the corners of the cube.
Therefore, number of atoms of Q in one unit cell
It is also given that the atoms of P are present at the body-centre. Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:Q = 1:1 Hence, the formula of the compound is PQ.
Niobium crystallises in body-centred cubic structure. If density is 8.55 g cm-3, calculate atomic radius of niobium using its atomic mass 93 u.
It is given that the density of niobium, d= 8.55 g cm - 3
Atomic mass, M = 93 gmol - 1
As the lattice is bcc type, the number of atoms per unit cell, z= 2 We also know that, NA= 6.022 × 1023mol - 1
= 3.612 × 10 - 23cm3
So, a = 3.306 × 10 - 8cm
= 1.432 × 10 - 8cm
= 14.32 × 10 - 9cm
If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
A sphere with centre O, is fitted into the octahedral void as shown in the above figure. It can be observed from the figure that ΔPOQ is right-angled
∠POQ = 900
Copper crystallises into a fcc lattice with edge length 3.61 x 10-8cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm-3.
Edge length, a= 3.61 × 10 - 8cm
As the lattice is fcc type, the number of atoms per unit cell, z= 4 Atomic mass, M = 63.5 g mol - 1
We also know that, NA= 6.022 × 1023mol - 1
Applying the relation:
= 8.97 g cm - 3
Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+and Ni3+ions?
The formula of nickel oxide is Ni0.98O1.00.
Therefore, the ratio of the number of Ni atoms to the number of O atoms, Ni : O = 0.98 : 1.00 = 98 : 100
Now, total charge on 100 O2 - ions = 100 × ( - 2)
= - 200
Let the number of Ni2+ions be x. So, the number of Ni3+ions is 98 - x.
Now, total charge on Ni2+ions = x(+2)
= +2x
And, total charge on Ni3+ions = (98 - x)(+3)
= 294 - 3x
Since, the compound is neutral, we can write: 2x+ (294 - 3x) + ( - 200) = 0
⇒ - x+ 94 = 0
⇒ x= 94
Therefore, number of Ni2+ions = 94 And, number of Ni3+ions = 98 - 94 = 4
Hence, fraction of nickel that exists as Ni2+
= 0.959
And, fraction of nickel that exists as
= 0.041
Alternatively, fraction of nickel that exists as Ni3+= 1 - 0.959
What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.
Semiconductors are substances having conductance in the intermediate range of 10-6to 104ohm-1m-1. The two main types of semiconductors are:
(i) n-type semiconductor
(ii) p-type semiconductor
n-type semiconductor: The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. When the crystal of a group 14 element such as Si or Ge is doped with a group 15 element such as P or As, an n-type semiconductor is generated.
p-type semiconductor: The semiconductor whose increased in conductivity is a result of electron hole is called a p- type semiconductor. When a crystal of group 14 elements such as Si or Ge is doped with a group 13 element such as B, Al, or Ga (which contains only three valence electrons), a p-type of semiconductor is generated.
When a crystal of Si is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate.
Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?
Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Let the number of oxide (O2 - ) ions be x. So, number of octahedral voids = x
It is given that two out of every three octahedral holes are occupied by ferric ions.
So, number of ferric (Fe3+) ions
= 2 : 3
Classify each of the following as being either a p-type or an n-type semiconductor:
Answer :
(i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.
(ii)
Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?
It is given that the atomic radius, r= 0.144 nm So,
= 0.407 nm
In terms of band theory, what is the difference
(ii) Between a conductor and a semiconductor
(i) The valence band of a conductor is partially-filled or it overlaps with a higher energy, unoccupied conduction band.
On the other hand, in the case of an insulator, the valence band is fully- filled and there is a large gap between the valence band and the conduction band.
(ii)
Explain the following terms with suitable examples:
(ii) Frenkel defect
(iv) F-centres
(i)
(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions show this type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well as an interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids such as AgCl, AgBr, AgI, and ZnS show this type of defect.
(iii)
(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons, the ionic sites are called F- centres. These unpaired electrons impart colour to the crystals. For example, 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.
Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
(ii) How many unit cells are there in 1.00 cm3of aluminium?
(i)
= 353.55 pm
= 354 pm (approximately)
(ii) Volume of one unit cell = (354 pm)3
= 4.4 × 107pm3
= 4.4 × 107× 10 - 30cm3
= 4.4 × 10 - 23cm3
Therefore, number of unit cells in 1.00 cm3=
If NaCl is doped with 10-3mol % of SrCl2, what is the concentration of cation vacancies?
It is given that NaCl is doped with 10 - 3mol% of SrCl2.
This means that 100 mol of NaCl is doped with 10 - 3mol of SrCl2.
Therefore, 1 mol of NaCl is doped with mol of SrCl2
= 10 - 5mol of SrCl2
Hence, the concentration of cation vacancies created by SrCl2is 6.022 × 108per mol of NaCl.
Explain the following with suitable examples:
(v)12-16 and 13-15 group compounds.
(i) 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, and CrO2.
(ii) Paramagnetism:The substances that are 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.
(iii) 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 MgFe2O4and ZnFe2O4.
(iv)
(v) 12-16 and 13-15 group compounds:The 12-16 group compounds are prepared by combining group 12 and group 16 elements and the 13-15 group compounds are prepared by combining group 13 and group15 elements. These compounds are prepared to stimulate average valence of four as in Ge or Si. Indium (III) antimonide (IrSb), aluminium phosphide (AlP), and gallium arsenide (GaAS) are typical compounds of groups 13-15. GaAs semiconductors have a very fast response time and have revolutionised the designing of semiconductor devices. Examples of group 12-16 compounds include zinc sulphide (ZnS), cadmium sulphide (CdS), cadmium selenide (CdSe), and mercury (II) telluride (HgTe). The bonds in these compounds are not perfectly covalent. The ionic character of the bonds depends on the electronegativities of the two elements.