NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State

 

NCERT solutions for class 12 chemistry chapter 1 The Solid State - As we know that there are three states of matter which are solid, liquid, and gas but in the NCERT solutions for class 12 chemistry chapter 1, you will only get to know the solutions to Solid State. In this chapter of NCERT book, you will be introduced to different types of structure of solids which exist in nature. For eg., NaCl has a cubic structure and Graphite has Hexagonal structure. At the end of the chapter, you will also get to know the defects in these solid structures. These defects in the solids also introduce some magnetic properties. NCERT solutions for class 12 chemistry chapter 1 The Solid State also contains questions based on topics like the correlation between the nature of interactions within the constituent particles and several properties of solids. How these properties get changed due to the structural imperfections or by the presence of impurities would also be explored. It is an important chapter for CBSE Board exam because it carries 5 marks in the chemistry examination, as well as it is important for competitive exams like JEE, NEET, BITSAT, and KVPY, etc. There are 26 questions in exercise and 24 intext questions in this chapter. As this chapter involves the conceptual knowledge rather than just learning hence it is recommended to clear your doubts to get the good hold on this chapter. You can refer to CBSE NCERT solutions for class 12 chemistry chapter 1 The Solid State for clearing your doubts. The solutions of NCERT for class 12 chemistry chapter 1 The Solid State are prepared in a very comprehensive manner. These NCERT solutions help students to grasp the topic thoroughly.

After completing NCERT class 12 chemistry chapter 1 The Solid State you will be able to describe characteristics of solid-state, differentiate between amorphous solids and crystalline solids, define crystal lattice and unit cell and also able to explain close packing of particles, describe different types of voids and close-packed structure etc. The NCERT class 12 chemistry chapter 1 talks about only solid-state. Some of the properties of solids are -

  • Solids have a definite shape and volume,
  • Inetmolecularforce is strong
  • Intermolecular distance is short
  • Solids are rigid so it cannot be compressed etc

In chemistry chapter 1 The Solid State of class 12, solids are classified into two types: Crystalline and Amorphous.

Property

Crystalline solids

Amorphous solids

Shape

Definite geometric shape

Indefinite or irregular shape

Melting Point

Sharp M.P

Diffused M.P

Anisotropy Nature

Anisotropic

Isotropic

Topics and sub-topics of NCERT class 12 chemistry chapter 1 The Solid State-

1.1 General Characteristics of Solid State

1.2 Amorphous and Crystalline Solids

1.3 Classification of Crystalline Solids

1.4 Crystal Lattices and Unit Cells

1.5 Number of Atoms in a Unit Cell

1.6 Close-Packed Structures

1.7 Packing Efficiency

1.8 Calculations Involving Unit Cell Dimensions

1.9 Imperfections in Solids

1.10 Electrical Properties

1.11 Magnetic Properties

 

NCERT solutions for class 12 chemistry chapter 1 The Solid State

Solutions to In-Text questions Ex 1.1  to 1.24

Q 1.1 Why are solids rigid?

Answer:

Solids have good intermolecular force of attraction, which tend to keep molecules of solid closer. And thus make solid rigid.

 

Q 1.2 Why do solids have a definite volume?

Answer:

At room temperature, intermolecular forces bring molecules so close that they cling to one another and occupy fixed positions. They oscillate about their mean position but they have fixed volume. So, due to the fixed position of particles and strong intermolecular force, solids have a definite volume.

Q 1.3(i) Classify the following as amorphous or crystalline solids: Polyurethane

Answer:

         Polyurethane is an amorphous solid.

Q 1.3(ii)  Classify the following as amorphous or crystalline solids:

naphthalene

Answer:

Naphthalene is a crystalline solids.

Q 1.3(iii) Classify the following as amorphous or crystalline solids: benzoic acid,

Answer:

Benzoic acid is a crystalline solid.

Q 1.3(iv) Classify the following as amorphous or crystalline solids:

teflon

Answer:

Teflon is an amorphous solid.

Q 1.3(v) Classify the following as amorphous or crystalline solids:

potassium nitrate

Answer:

Potassium nitrate is a crystalline solid.

Q 1.3(vi) Classify the following as amorphous or crystalline solids:

cellophane,

Answer:

Cellophane is an amorphous solid.

Q 1.3(vii) Classify the following as amorphous or crystalline solids:

polyvinyl chloride

Answer:

Polyvinyl chloride is an amorphous solid.

Q 1.3(viii) Classify the following as amorphous or crystalline solids:

Fibre glass,

Answer:

Fibre glass is an amorphous solid.

Q 1.3(ix) Classify the following as amorphous or crystalline solids:

copper.

Answer:

Copper is a crystalline solid.

Q 1.4 Why is glass considered a super cooled liquid?

Answer: Glass is an amorphous solid. Just like liquids, amorphous solids also have a tendency to flow, though very slowly. Therefore, these are called pseudo solids or supercooled liquids.


 Q 1.5 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?

Answer:

We know that amorphous solid are isotropic in nature i.e., their properties are same in all directions. According to given description, the nature of given solid is amorphous.

Clevage property :- When we cut the solid with a sharp edged tool, they cut into two pieces with irregular surfaces.

Q 1.6(iii) Classify the following solids in different categories based on the nature of intermolecular forces operating in them:

benzene

Answer:

Benzene is a covalent molecule but a molecular solid because its molecules are together by the intermolecular force of attraction.  

Q 1.6(viii) Classify the following solids in different categories based on the nature of intermolecular forces operating in them:

graphite,

Answer:

Graphite is made up of carbon atoms covalently bonded with each other.

So graphite is a covalent solid or network solid.

Q 1.7 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?

Answer:

As per the given properties, Solid A shows the absence of ionic properties in it.  A is a covalent or network solid (exception graphite, which is soft and conductor of electricity).

Q 1.8 Ionic solids conduct electricity in molten state but not in solid-state. Explain.

Answer:

In the solid state, the ions are not free to move about so they are electrical insulators. But, in the molten state or when ionic solids are dissolved in water, the ions become free to move about so they conduct electricity.

Q 1.9 What type of solids are electrical conductors, malleable and ductile?

Answer:

Metallic solids are hard but malleable and ductile. Also, they are conductors in solid as well as in the molten state.

Q 1.10 Give the significance of a ‘lattice point’.

Answer:

A crystal lattice is the pattern of points representing the locations of motifs. And each point in a lattice is called lattice points. These points are significant as they represent one constituent particle which may be an atom, a molecule (a group of atoms) or an ion.

Q 1.11 Name the parameters that characterise a unit cell.

Answer:

A unit cell is characterised by:- 

  (i) its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular.

 (ii) angles between the edges, α (between b and c), β (between a and c) and γ (between a and b). Thus, a unit cell is characterised by six parameters

      a, b, c, α, β and γ. 

Q 1.12 Distinguish between

(i) Hexagonal and monoclinic unit cells

(ii) Face-centred and end-centred unit cells.

Answer:

(i) Hexagonal and monoclinic unit cells:-   

         Properties               Hexagonal               Monoclinic
    Possible Variation               Primitive

             Primitive,  End centered

     Axial distance           a=b\neq c             a\neq b\neq c
      Axial angles

        \alpha =\beta =90^{\circ}

         \gamma =120^{\circ}

       \alpha =\gamma =90^{\circ}

           \beta \neq 90^{\circ}

     Examples       Graphite, ZnO       Monoclinic sulphur

(ii) Face-centred and end-centred unit cells:-

  • A face-centred unit cell contains atoms at all the corners and at the centre of all the faces, whereas in end - centred unit cell one pair of opposite faces contains atoms apart from atoms at all the corners.
  • The total number of atoms in a unit cell differs by 1.

Q 1.13 Explain how much portion of an atom located at

(i) corner and

(ii) bodycentre of a cubic unit cell is part of its neighbouring unit cell.

Answer:

(i) At corner:-    Each atom at a corner is shared between eight adjacent unit cells, four unit cells in the same layer and four unit cells of the upper (or lower) layer. Therefore, only \frac{1}{8}thof an atom (or molecule or ion) actually belongs to a particular unit cell. 

(ii) At body centre of cubic cell:-   Since body centre atom completely belongs to the unit cell in which it is present, thus it is not part of any neighbouring unit cell. So, one atom belongs to a one unit cell.

Q 1.14 What is the two dimensional coordination number of a molecule in square close-packed layer?

Answer:

In two dimensional square closed packed layer, one atom is in contact with 4 of its neighbouring atoms. So its cordination number is 4.

 

Q 1.15  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?

Answer:

Amount of compound given = 0.5 mol

We know that,                         

                                                 Moles = \frac{No.\ of\ atoms}{6.022 \times 10^{23}}

                                 So, No. of atoms  =0.5\times 6.022 \times 10^{23} = 3.011 \times10^{23} atoms

 

We also know that, No. of tetrahedral voids =   2(No. of atoms in closed packing)

                                                                       =2\times 3.011 \times10^{23} = 6.011\times10^{23}

                                      No. of octahedral voids = No. of atoms present in closed packing.

                                                                             = 3.011 \times10^{23}

So,   Total number of voids =   No. of tetrahedral voids + No. of octahedral voids.

                                           =6.022 \times10^{23} +\ 3.011 \times10^{23}

                                           =9.033 \times10^{23}\ voids 

Q 1.16  A compound is formed by two elementsM and N. The element N forms ccp and atoms of M occupy \frac{1}{3}rd  of tetrahedral voids. What is the formula of the compound?

Answer:

It is given that element N forms CCP.

Let us assume, the number of atoms of element N (which forms ccp) is x.

Then no. of tetrahedral voids = 2x.

It is also given that M occupies  \frac{1}{3}rd of tetrahedral voids.

So the number of atoms of element M : 

            =\frac{2x}{3}

So the molecular formula bocomes : M2N3.

Q 1.17 Which of the following lattices has the highest packing efficiency

(i) simple cubic

(ii) body-centred cubic and

(iii) hexagonal close-packed lattice?

Answer:

(i) Simple cubic:-  In a simple cubic lattice the atoms are located only on the corners of the cube.

                              Thus, the edge length or side of the cube ‘a’, and the radius of each particle, r are related as a = 2r

                              Volume of cubic unit cell = (2r)^3 = 8r^3

                            And  Volume of 1 atom : 

                                                                =\frac{4}{3}\Pi r^3

                              Packing\ efficiency = \frac{Volume\ of\ one\ atom}{Volume\ of\ cubic\ unit\ cell}\times100 \%  

                                                                          = \frac{\frac{4}{3}\Pi r^3}{8\Pi r^3}\times100 \% = \frac{\Pi }{6}\times100\%

                                                                          = 52.4\%

(ii) Body centred cubic:-  In body centred cubic, we have atoms at all corners and at body centre.

                                        Clearly, the atom at the centre will be in touch with the other two atoms diagonally arranged.

                                         b = \sqrt{2}a ;                      and           c = \sqrt{3}a           

      Also, the length of body diagonal is equal to 4r.

                                            \sqrt{3}a = 4r

                                                a = \frac{4r}{ \sqrt{3}}

The volume of the cube              :             

                                                       = a^3 = \left ( \frac{4r}{ \sqrt{3}} \right )^3                                                                                                                                         

In BCC, a total number of atoms is 2. 

                                               

                                                 Packing\ efficiency = \frac{Volume\ of\ one\ atom}{Volume\ of\ cubic\ unit\ cell}\times100 \%

                                                                                             = \frac{2\times(\frac{4}{3}\Pi r^3)}{(\frac{4}{\sqrt{3}}r)^3}\times100 \%   

                                                                                              = 68\%

(iii) Hexagonal close-packed:-    We know that both types of (hcp and ccp) are equally efficient. We also know that the packing efficiency of ccp is 74 percent.

(i) Simple cubic = 52.4%

(ii) Body centred cubic=68%

(iii) Hexagonal close-packed=74%

Thus among all, packing efficiency of hcp is the highest.

Q 1.18.   An element with molar mass  2.7\times 10^{-2}kg\; mol^{-1}  forms a cubic unit cell with edge length 405\; pm. If its density is 2.7\times 103\; kg\; m^{-3}, what is the nature of the cubic unit cell?

Answer:

We know that :

                           Density\ of\ unit\ cell = \frac{z.M}{a^3.N_A}

We are given :        Density = 2.7\times10^3\ Kg\ m^{-3}

and                        Molar\ mass = 2.7\times10^{-2}\ Kg\ mol^{-1}

and                      Edge\ length = 405 pm =4.05\times10^{-10}\ m

  So we need to find z (no. of atoms present in unit cell) by putting all values in formula of density.

 We get,                          z \approx 4.

It is known that face centred cubic also has 4 atoms in its unit cell, So given cubic unit cell is face centred.

Q 1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?

Answer:

When a solid is heated some of the atoms moves out due to increase in its thermal energy. This leads to a vacancy defect. Affected physical property:- Since some of the atoms moved out so its density decreases.

Q 1.20(i)  What type of stoichiometric defect is shown by:

          ZnS

Answer:

ZnS shows Frenkel defect because there is a large difference in the size of Zn and S.

Q 1.20(ii) What type of stoichiometric defect is shown by:

   AgBr

Answer:

AgBr shows both, Frenkel as well as Schottky defects.

Q 1.21.  Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.

Answer:

In impurity defect, the higher valency cation replaces the lower valency cation. An important point to be noted is that electrical neutrality is maintained.

For e.g. In case of NaCl and SrCl2 , some of the sites of Na+ ions are occupied by Sr2+ . Two Na+ ions are replaced by Sr+2 ions. It occupies the site of one ion and the other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions.

Q 1.22.  Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.

Answer:

In metal excess defect F-centres are generated. F-centres are the anionic sites occupied by unpaired electrons. These F-centres are responsible for the developing of colour.

For e.g. when crystals of KCl are heated in an atmosphere of sodium vapour, the potassium atoms comes on the surface of the KCl. Then the Cl– ions go to the surface of the KCl and get combined with K atoms to give KCl. This happens by loss of an electron by potassium atoms to form K+ ions. The released electrons during the process go into the KCl and occupy anionic sites. Thus the crystal now has an excess of potassium. Generation of F-centres takes place. They impart violet colour to the crystals of KCl. 

Q 1.23  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?

Answer:

The 14 group elements have four valence electrons, whereas group 15 elements like P have 5 valence electrons. In this, 4 out of 5 are used in making a bond with group 14 element and 1 is extra and becomes delocalised. This result in extra conductivity. So group 14 elements are required to be doped with the group 15 elements for forming n-type conductors.

Q 1.24.  What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.

Answer:

Substances like iron, cobalt (Ferromagnetic) can make better permanent magnets because in the solid state these substances are grouped together in small regions called domains. Each domain acts like a tiny magnet. They are randomly arranged when placed in the unmagnetised state. But when they are placed in magnetic field domains align in direction of the magnetic field. But in case of ferrimagnetic materials domains are aligned in parallel and anti-parallel direction, which lead to loss of total magnetic effect.

NCERT Solutions for Class 12 Chemistry Chapter 1 The Solid State Exercises

Q 1.1 Define the term 'amorphous'. Give a few examples of amorphous solids.

Answer:

Amorphous solids have the constituent particles arranged only in short-range order and consequently, they behave like supercooled liquids. So do not have sharp melting points. Also, they are isotropic in nature i.e., their physical properties are the same in all directions.

E.g.  quartz glass, polymers, gels etc.

Q 1.2 What makes a glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?

Answer:

The arrangement of constituent particles makes glass different from quartz. In glass short-range order of particles exists whereas in quartz long range of particles is seen. Glass is an amorphous solid while quartz is a crystalline form of silica.

Quartz can be converted into glass by heating it strongly till it comes to a molten state and then cooling it rapidly so that it does not get time to crystallize.

Q 1.3(iii) Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

 SiC

Answer:

(iii) SiC:-   It is a covalent or network solid.

Q 1.3(vii) Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

Graphite

Answer:

(vii) Graphite :-  It is a covalent or network solid.

Q 1.3(xi) Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.

 Si

Answer:

(xi) Si :-  It is a network or covalent solid.

Q 1.4(i) What is meant by the term 'coordination number'?

Answer:

Coordination number is defined as the number of the nearest neighbours of a particle.

Q 1.4(ii) What is the coordination number of atoms:

(a) in a cubic close-packed structure?

(b) in a body-centred cubic structure?

Answer:

(i) In ccp coordination number is 12.

(ii) Coordination number of atoms in bcc is 8.

Q 1.5 How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell? Explain.

Answer:

We know that density is related to molar mass and edge length by the formula :

                                                      Density (d)= \frac{z.M}{a^3.N_A}

                                                                       M=\frac{d.a^3.N_A}{Z}

where, d=Density
            a3 = The volume of the unit cell
            M= Atomic mass
          Z = No. of atoms in unit cell    
         Na= Avogadro’s number.

Q 1.6 'Stability of a crystal is reflected in the magnitude of its melting points'. 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?

Answer:

If a crystal has a high melting point implies that it will require high temperature or say energy to break the intermolecular bonds. Also if the intermolecular forces are strong then the crystal is more stable. Thus it can be said that higher the melting point high is the stability.

Melting points of compounds are given below:- 

(i) Water:- 273 K

(ii) Ethyl alcohol:- 155.7 K

(iii) Diethyl ether:- 156.8 K

(iv) Methane:-  90.5 K

It can be concluded from the above data that among all, the intermolecular force of attraction is highest in water and lowest in methane. The possible reason for this is the hydrogen bonding present in water whereas the Vanderwall force of attraction is present in case of methane.

Q 1.7(i) How will you distinguish between the following pairs of terms:

Hexagonal close-packing and cubic close-packing?

Answer:

(i) Hexagonal close-packing:- In this, tetrahedral voids of the second layer are covered by the spheres of the third layer. In this case, the atoms (spheres) of the third layer are aligned with those of the first layer. This results in the pattern of spheres to be repeated in alternate layers. We get a pattern as ABAB ....... . This structure is known as the hexagonal close-packed (hcp) structure.

(ii) Cubic close-packing:-  In this case, the third layer is placed above the second layer such that its spheres cover the octahedral voids. In this manner, the atoms (spheres) of the third layer are not aligned with those of either the first or the second layer. This pattern of layers is written as ABCABC ........... . This structure is called cubic close-packed (ccp)

Q 1.7(ii) How will you distinguish between the following pairs of terms:

Crystal lattice and unit cell?

Answer:

(i) Crystal lattice:-   Crystal lattice is a three-dimensional array of points. The crystal structure is generated by using structural motifs with lattice points.Each point in a crystal lattice denotes one constituent particle which can be either an atom, a molecule (a group of atoms).

(ii) Unit cell:- It is the smallest unit of a crystal lattice which when repeated gives the crystal structure.

Q 1.7(iii) How will you distinguish between the following pairs of terms:

Tetrahedral void and octahedral void?

Answer:

(iii) Tetrahedral void:- When a sphere (atom) of the second layer is above the void of the first layer (or vice versa) production of a tetrahedral void takes place. These voids are known as tetrahedral voids as, when the centres of these four spheres are joined a tetrahedron is formed.

(iv) Octahedral void:-  The voids having a triangular in shape (or triangular voids) in the second layer are above the triangular voids in the first layer, also the triangular shapes of these do not overlap. One of them has the apex of the triangle pointing upwards and the other downwards. These voids are surrounded by six spheres/atoms and called as octahedral voids.

 

                                                                   

Q 1.8(i) How many lattice points are there in one unit cell of each of the following lattice?

Face-centred cubic

Answer:

In face centred cubic atoms are present at all 8 corners and at the centre of each face (6 faces in one unit cell). So total no. of lattice points are 14. 

Q 1.8(ii) How many lattice points are there in one unit cell of each of the following lattice? 

Face-centred tetragonal

Answer:

In face centred tetragonal atoms are present at all 8 corners and at centre of each face (total 6 faces).

So total lattice points are 14.

Q 1.8(iii) How many lattice points are there in one unit cell of each of the following lattice? 

Body-centred

Answer:

In body centred atoms are present at all 8 corners and 1 atom is present at body centre. So total number of lattice points are 9 in bcc.

Q 1.9(i) Explain,

The basis of similarities and differences between metallic and ionic crystals.

Answer:

Similarities:- Both ionic and metallic crystals are hard in nature due to the good force of attraction between molecules. Both have fairly high melting points. In both ionic and metallic bond is non-directional.

Differences:-  In ionic solids, attractive forces are coulombic or electrostatic whereas in case of metallic solids forces attractive forces are metallic bonding.

Moreover, ionic solids are insulators in the solid state whereas metallic solids are a very good conductor of electricity in both solid and molten state.

Q 1.10(i) Calculate the efficiency of packing in case of a metal crystal for

simple cubic

Answer:

(i) Simple cubic :-  In a simple cubic lattice the atoms are located only on the corners of the cube.

                               Thus, the side of the cube ‘a’, and the radius of each particle, r are related as:

                                                     a = 2r

                               Volume of cubic unit cell = (2r)^3 = 8r^3                               

                           And Volume of 1 atom : 

                                                                =\frac{4}{3}\Pi r^3

                              Packing\ efficiency = \frac{Volume\ of\ one\ atom}{Volume\ of\ cubic\ unit\ cell}\times100 \%

                                                                          

                                                                          = \frac{\frac{4}{3}\Pi r^3}{8\Pi r^3}\times100 \% = \frac{\Pi }{6}\times100\%

                                                                          = 52.4\%

Thus packing efficiency of simple cubic is 52.4\%

Q 1.10(ii) Calculate the efficiency of packing in case of a metal crystal for

body-centred cubic

 

Answer:

(ii) Body centred cubic:-  In body centred cubic, we have atoms at all corners and at body centre.

                                        Clearly, the atom at the centre will be in touch with the other two atoms diagonally arranged.

                                         b = \sqrt{2}a ;  and  c = \sqrt{3}a           

      Also the length of body diagonal is equal to 4r.

     Thus,                              \sqrt{3}a = 4r

                                            a = \frac{4r}{ \sqrt{3}}

The volume of a cube :             

                                                       = a^3 = \left ( \frac{4r}{ \sqrt{3}} \right )^3                                                                                                                                         

In BCC, the total number of atoms is 2.

                                                 Packing\ efficiency = \frac{Volume\ of\ one\ atom}{Volume\ of\ cubic\ unit\ cell}\times100 \%

                                                                                             = \frac{2\times(\frac{4}{3}\Pi r^3)}{(\frac{4}{\sqrt{3}}r)^3}\times100 \%

                                                                                              = 68\%

Q 1.10(iii) Calculate the efficiency of packing in case of a metal crystal for

face-centred cubic (with the assumptions that atoms are touching each other).

Answer:

In fcc, we know that we have a total of 4 efficient atoms present.

Also, radius and edge length of the sphere can be related by the following relation :         

                                               r = \frac{a}{2\sqrt{2}}

The above relation can be found out by equating diagonal of 1 face to 4r. (2r from the atom at face centre and r from each atom at a corner.)

Thus packing efficiency becomes:- 

                        Packing\ efficiency = \frac{Volume\ occupied\ by\ atoms}{Total\ volume\ of\ unit\ cell}\times100\%

                                                                    = \frac{4\times\frac{4}{3}\Pi r^3}{(2\sqrt{2}r)^3}\times100\%

                                                                     = 74\%

Q 1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07\times 10^{-8}cm and density is 10.5\; g\; cm^{-3} , calculate the atomic mass of silver.

Answer:

The relation between density, molar mass and edge length is given by:-

                                                 Density = \frac{z.M}{a^3.N_A}

It is given that silver crystallises in fcc lattice, so the value of z = 4. (since it is known that 4 atoms are completely efficient in fcc lattice)                 

                                                          M = \frac{Density.a^3.N_A}{z} 

                                                              = \frac{10.5\ g\ cm^{-3} \times (4.07\times10^{-8}cm)^3\times N_A}{4}

 Convert these into SI units,                = 107.09\ u

Q 1.12 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?

Answer:

It is given that element Q is at corners of the cube.

So total atoms of element Q per unit cell = 1                                       (8\times\frac{1}{8} = 1)

And it is also given that element P is present at body centre.

So the total number of atoms of element P per unit cell = 1.

Therefore formula of the compound becomes: PQ

It is clear that it is a bcc lattice so coordination number of P and Q is 8.

Q 1.13 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.

Answer:

Here we will use the relation between density and edge length.

                                          Density = \frac{z.M}{a^3.N_A}

It is given that Niobium crystallises in body-centred cubic structure so the value of z = 2.

So we will put the value of density, molar mass and z.

We get,                            

                                          8.55 = \frac{2\times93}{a^3\times6.022 \times10^{23}}

or                                       a^3 = 36.124 \times10^{-24}

or                                       a = 3.3057 \times10^{-8}\ cm

We know the relation between the radius of the atom and edge length in bcc lattice.

                                          r = \frac{\sqrt{3}a}{4}

we get the radius of niobium atom    = 143.1 pm.

Q 1.14 If the radius of the octahedral void is r and radius of the atoms in closepacking is R, derive relation between r and R.

Answer:

From the figure, it is clear that we can use Pythagoras theorem and find the relation between R and r.

Using pythagoreas theorem :  

                                            (2R)^2 = (R+r)^2 + (R+r)^2

or                                         4R^2 = R^2+2Rr +r^2 + R^2+2Rr+r^2

or                                         2R^2 = 4Rr +2r^2

or                                            r = 0.414R

 

Q 1.15 Copper crystallises into a fcc lattice with edge length 3.61\times 10^{-8}cm. Show that the calculated density is in agreement with its measured value of 8.92\; g\; cm^{-3}.

Answer:

The relation between density and edge lenght gives : 

                                                  Density = \frac{z.M}{a^3.N_A}

Since it crystallises in fcc lattice, thus z = 4.

Molar mass of copper = 63.546 u.

So,          

                                                              Density = \frac{4\times63.546}{(3.61\times10^{-8})^3\times6.022\times10^{23}}\ cm

                                                                                 = 8.97\ g\ cm^{-3}

Q 1.16 Analysis shows that nickel oxide has the formula  Ni_{0.98}O_{1.00}. What fractions of nickel exist as Ni^{2+} and Ni^{3+}\; ions \; ? 

Answer:

Given formula is Ni0.98O1.00.

Ratio of Ni to O is 0.98.

This means 98 Ni are required for 100 atoms of O.

Let us assume that Ni+2 ions are x.

So the number of Ni+3 ions will be 98 - x.

Using the charge neutrality principle, the net charge of Ni should be equal to a net charge of O.

So the equation becomes :

              x(2) + (98-x)3 =  2(100)

or          2x  +  294 - 3x = 200

or                x = 94.

Thus the fraction of Ni+2 ions =        

                                                      \frac{94}{98}\times100 = 96\%

and fraction of Ni+3 ions =   4%

fraction of Ni+2 ions = 100-4 =96%

Q 1.17 What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism. 

Answer:

Semiconductors are the solids with have conductivities in the range from 10^{-6} to  10^{4}  ohm^{-1}\ m^{-1}. In the case of semiconductors, the gap between the valence band and the conduction band is small. Thus, some electrons may jump to the conduction band and may show some conductivity.

Based on the doping process we have two types of semiconductors.

(i) n-type semiconductors:- In the case of n-type semiconductor doping is done by using group 15 elements like P(have 5 valence electrons).

 Group 14 elements have 4 valence electrons; these all 4 valence electrons are bonded with 4 valence electron of group 15 element. One left out electron delocalised and increases the conductivity. Due to the presence of one extra electron, it is also called as electron-rich impurities.

(ii) p-type semiconductors:- In the case of p-type semiconductors we use group 13 impurities such as  Al (having 3 valence electrons). This results in the generation of electron hole due to the missing electron in 4th place. From a neighbouring atom, an electron can come and fill the electron-hole, but it will then create a new electron-hole at the position from where electron moved. In this circumstance, it would appear as if the electron-hole is moving in the direction opposite to that of the electron that filled it. 

Q 1.18 Non-stoichiometric cuprous oxide, Cu_{2}O can be prepared in a 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?    

Answer:

In the given compound Cu2O, the charge on Cu is +1. But the charge on copper in normal compounds is +2.

So Cu+2 will try to replace Cu+1 in atmospheric conditions.

This will lead to a generation of positive charge holes which are the cause of conductivity in p-type semiconductors.

So given compound can act as p-type semiconductors.

Q 1.19 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.

Answer:

Let the number of oxide ions be x.

Then, the number of octahedral voids = number of oxide ions = x.

According to question, number of ferric ions 

                                                                  = \frac{2}{3}x

Ratio of ferric ion: oxide ion = 2:3

So the required formula of the compound is Fe2O3.

Q 1.20(i) Classify each of the following as being either a p-type or a n-type semiconductor:

 Ge doped with In.

Answer:

We know that Ge is a group 14 element (having 4 valence electrons) and In is a group 13 element  (having 3 valence electrons).

So this is a p-type semiconductor.

Q 1.20(ii) Classify each of the following as being either ap-type or a n-type semiconductor: 

 Si doped with B.

Answer:

Si belongs to group 14 and B belongs to group 13 so it is a p-type semiconductor.

Q 1.21 Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Answer:

Atomic radius = 0.144 nm

 for  a face-centred unit cell

a=2 \sqrt{2} r

a=2 *1414*0.144 nm

a=0.407 nm

the length of a side of the cell is 0.407.

Q 1.22(i) In terms of band theory, what is the difference
between a conductor and an insulator

Answer:

Conductor: A conductor may conduct electricity through the movement of electrons or ions. The conductivity of metals depends upon the number of valence electrons available per atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other as to form a band. If this band is partially filled or it overlaps with a higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal shows conductivity.

Insulator: If the gap between the filled valence band and the next higher unoccupied band is large then electrons cannot jump to it and such a substance behaves as an insulator.

Q 1.22(ii) In terms of band theory, what is the difference
between a conductor and a semiconductor?

Answer:

Conductor: A conductor may conduct electricity through the movement of electrons or ions. The conductivity of metals depends upon the number of valence electrons available per atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other as to form a band. If this band is partially filled or it overlaps with a higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal shows conductivity.

Semi-conductor: The gap between the valence band and the conduction band is small. Therefore, some electrons may jump to the conduction band and show some conductivity. The electrical conductivity of semiconductors increases with rising in temperature since more electrons can jump to the conduction band.

 

Q 1.23(i) Explain the following terms with suitable examples:
Schottky defect

Answer:

 (i) Schottky defect

The Schottky defect is basically a vacancy defect in ionic solids. To maintain electrical neutrality, the number of missing cations and anions are equal as shown in the figure. Schottky defect decreases the density of the substance. The number of such defects in ionic solids is quite significant. For example, in NaCl, there are approximately 106 Schottky pairs per cm3 at room temperature. In 1 cm3 there are about 1022 ions. Thus, there is one Schottky defect per 1016 ions. The Schottky defect is shown by ionic substances in which the cation and anion are of almost similar sizes. For example, NaCl, KCl, CsCl and AgBr. 

The figure showing a Schottky defect is as shown :

Q 1.23(ii)  Explain the following terms with suitable examples

Frenkel defect

Answer:

(ii)Frenkel defect  :

Frenkel defect is shown be ionic solids. The smaller ion (usually cation) is dislocated from its normal site to an interstitial site as shown in the figure . It creates a vacancy defect at its original site and an interstitial defect at its new site. Frenkel defect is also called a dislocation defect. Frenkel defect does not affect the density of the solid. Frenkel defect is shown by the ionic substance in which there is a large difference in the size of cations and anions, for example, ZnS, AgCl, AgBr and AgI due to the small size of cations and large size of anions, Frenkel defect can be observed in these.

The figure showing the Frenkel defect is as shown :

Q 1.23(iii) Explain the following terms with suitable examples:

Interstitials

Answer:

Interstitial Defect: When some constituent particles (atoms or molecules) occupy an interstitial site as shown in the figure, the crystal is said to have the interstitial defect. This defect increases the density of the substance. Interstitial defects can be shown by non-ionic solids. 

The figure representing the interstitial defect is as shown :

Q 1.23(iv) Explain the following terms with suitable examples 

 F-centres

Answer:

When any negative ion is absent from lattice site then result the crystal now has an excess of cations. To maintain electrical neutrality the vacant anionic site is occupied by an electron as shown in the figure.

The anionic sites occupied by unpaired electrons are called F-centres. The F- centre is responsible for most interstitial properties of the compound.

This defect can be observed in  NaCl.

Q 1.24(i) Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
 What is the length of the side of the unit cell?

Answer:

For  cubic close-packed structure,

we have a=2\sqrt{2} . r

Given : r=125 pm

           a =2\times 1.414\times 125

           a=354 pm

Q 1.24(ii)  Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.
How many unit cells are there in 1.00 cm3 of aluminium?

Answer:

For cubic close-packed structure,

we have a=2\sqrt{2} . r

Given : r=125 pm

           a =2\times 1.414\times 125

           a=354 pm

the volume of one unit cell =a^{3}=354^{3}

                               =(354\times 10^{-10})^{3}                    

                             =4.4 \times 10^{-23} cm^{3}

Total unit cells in 1.00 cm3=\frac{total \, \, volume }{size \, \, of\, \, unit\, \, cell}

                                            =\frac{1 cm^3}{4.4\times 10^{-23} cm^3}

                                            =2.27 \times 1022 unit\, \, cell

Q 1.25 If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation vacancies?

Answer:

NaCl is doped with 10–3 mol % of SrCl_2

Concentration in % so that take a total of 100 mol of solution.

Moles of NaCl = 100 - moles of SrCl_2

Moles of SrCl_2 is very less , so we can neglect them.

Moles of NaCl =100

1 mole of NaCl is dipped with = 10^{-5}  mol of SrCl_2.

So cation vacanties per mole of NaCl =10^{-5} mol 

1mol=6.022\times 10^{23} particles 

So cation vacancies per mol of NaCl = 10-5 \times 6.022\times 10^{23}

                                                          =6.02\times 10^{18}

Hence, the concentration of cation vacancies.=6.02\times 10^{18}

Q 1.26(i) Explain the following with suitable examples:

Ferromagnetism

Answer:

(i) Ferromagnetism

Substances that are attracted very strongly by a magnetic field are called ferromagnetic substances. Example: iron, cobalt, nickel, gadolinium and CrO2. Besides strong attractions, these substances can be permanently magnetised. The metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus, each domain acts as a tiny magnet. In an unmagnetised piece of a ferromagnetic substance, the domains are randomly oriented and their magnetic moments get cancelled. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field as shown in the figure and a strong magnetic effect is produced. This ordering of domains persists even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet.

The figure is as shown :

Q 1.26(ii) Explain the following with suitable examples:

Paramagnetism

Answer:

(ii) Paramagnetism : 

The substances that are weakly attracted by a magnetic field are called paramagnetic substances. They are magnetised in a magnetic field in the same direction. They lose their magnetism in the absence of a magnetic field. Paramagnetism is due to the presence of one or more unpaired electrons which are attracted by the magnetic field. Example:O_2,Cu_2^+ , Fe_3^+ , Cr_3^+.

Q 1.26(iii) Explain the following with suitable examples:

Ferrimagnetism

Answer:

(iii)    Ferrimagnetism : 

When the magnetic moments of the domains in the substance are aligned in parallel and anti-parallel directions in unequal numbers than ferrimagnetism is observed. Refer to the given figure. They are weakly attracted by a magnetic field as compared to ferromagnetic substances. Example:Fe_3O_4 , MgFe_2O_4,ZnFe_2O_4. These substances also lose ferrimagnetism on heating and become paramagnetic.

Q 1.26(iv) Explain the following with suitable examples:

Antiferromagnetism

Answer:

 (iv) Antiferromagnetism : 

Substances like MnO showing antiferromagnetism have domain structure similar to ferromagnetic substance, but their domains are oppositely oriented and cancel out each other's magnetic moment as shown in the figure.

Q 1.26(v) Explain the following with suitable examples:

12-16 and 13-15 group compounds

Answer:

(v)12-16 and 13-15 group compounds

12-16 group compounds: Compounds formed between elements of group 12 and elements of group 16 are called 12-16 group compounds.

Example: ZnS

13-15 group compounds: Compounds formed between elements of group 13 and elements of group 15 are called 13-15 group compounds.

Example: GaAs

NCERT solutions for class 12 chemistry

Chapter 1

CBSE NCERT solutions for class 12 chemistry chapter 1 The Solid State

Chapter 2

NCERT solutions for class 12 chemistry chapter 2 Solutions

Chapter 3

Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry

Chapter 4

CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics

Chapter 5

Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry

Chapter 6

NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements

Chapter 7

CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements

Chapter 8

Solutions of NCERT class 12 chemistry chapter 8 The d and f block elements

Chapter 9

NCERT solutions for class 12 chemistry chapter 9 Coordination compounds

Chapter 10

Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes

Chapter 11

CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers

Chapter 12

Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids

Chapter 13

NCERT solutions for class 12 chemistry chapter 13 Amines

Chapter 14

CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules

Chapter 15

Solutions of NCERT class 12 chemistry chapter 15 Polymers

Chapter 16

NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

 

NCERT solutions for class 12 subject wise

Solutions of NCERT class 12 biology

NCERT solutions for class 12 maths

CBSE NCERT solutions for class 12 chemistry

Solutions of NCERT class 12 physics

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