Class XII  - ELECTROCHEMISTRY

Class XII - SOLID STATE

Class XII Sample papers

Class XII unit wise Q.Bank

Distinction test

Name reactions

p- Block - At a Glance

Organic - (Conversions, Named Reactions, Mechanisms and Word Problems)


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Assignments Class 12 (Chapter 1 to 3)

                                                         ASSIGNMENTS

                                       CHAPTER 1- SOLID STATE

                                                    Level one

Q.1. Why is glass considered a super cooled liquid?

Q.2 .Why do solids have a definite volume?

Q.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic

Acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiber glass, copper.

Q.4. what type of solids are electrical conductors, malleable and ductile?

Q.5. Name the parameters that characterize a unit cell.

Q.6. What is meant by the term 'coordination number'?

Q.7. What is the coordination number of atoms:

a) In a cubic close-packed structure?

b) In a body-centred cubic structure?

                                                 Level Two

Q.1. 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?

Q.2. Ionic solids conduct electricity in molten state but not in solid state. Explain.

Q.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fiber glass, copper.

Q.4. Define the term amorphous. Give a few examples of amorphous solids.

                                               Level three

Q.5. What makes a glass different from a solid such as quartz? Under what conditions quartz              could be converted into glass?

                                   CHAPTER 2-SOLUTIONS

                                              Level One

Q1.What is meant by semi-molar and deci-molar solutions?

Q2.What will be the mole fraction of water in ethanol solution containing equal number of moles of water and ethanol?

Q3.State Roult’s Law in its general form in reference to solutions?

Q4.How Is colligative property of the solution when a solute in a solution undergoes:

(i)association (ii)dissociation ?

Q5.Give one practical application of depression of freezing point in automobiles.

Q6.Mention one large scale use of the phenomenon called “Reverse Osmosis”.

Q7.Which one will have greater boiling point 0.1 M NaCl or 0.1 BaCl2 solutions in water?

Q8.What will happen to the vapour pressure of water, if a tablespoon of sugar is added to it?

Q9.Why does the use of pressure cooker reduce the cooking time?

Q10. The dissolution of ammonium chloride in water is an endothermic process .What is the effect of temperature on its solubility?

Q11. Why is liquid ammonia bottle first cooled in ice before opening it?

                                              LevelTwo

Q1. Out of three solutions (i) 1M cane sugar solution (ii) 1M NaCl solution (iii) 1M BaCl2 solution,

which one will have (a)highest boiling point (b)highest freezing point (c)lowest osmotic pressure?

Q2.(a)For which of the following, the van Hoff factor is not greater than 1?

NaNO3 , BaCl2 , K3Fe(CN)6 , C6H12O6

(b)If glycerine and methyl alcohol are sold at the same price per kg, which one would be preferred for use, as an anti-freeze in the car radiators?

Q3.Define Osmotic Pressure. Arrange the following solutions in increasing order of their osmotic pressure. urea,sucrose,NaCl,glucose.(if 1g substance is added).

Q4.What is the effect of the following changes on the osmotic pressure of a solution containing a non-volatile solute:

(i)Solvent is added to the solution

(ii)Solute is added to the solution

(iii)Temperature is raised ?

Q5.Carbon Tetrachloride and Water are immiscible whereas, ethanol and water are miscible in all proportions. Co-relate this behavior with molecular structures of these compounds.

Q6.Give reason, why at higher altitudes, people suffer from a disease called Anoxia.

In this disease, they become weak and cannot think clearly.

                                                            Level Three

Q1.Why do aquatic species feel more comfortable in lakes in winters than in summers?

Q2. Why is a person suffering from high blood pressure is adviced to take minimum quantity of

common salt?

Q3. If Kf for water is 1.86˚C/m, explain why 1 m NaCl in water does not have a freezing point equal

to : (i) -1.86˚C (ii) -3.72˚C

Q4.Derive a relationship between relative lowering of the vapour pressure and the mole fraction of the volatile liquid.

Q5. Why dissolution of some solid compounds is exothermic while that of some others is endothermic?

                                      CHAPTER 3 ELCTROCHEMISTRY

                                                 Level One

Q 1 What is the effect of temperature on molar conductivity?

Q 2 What are the units of molar conductivity?

Q 3 Name the factors on which emf of a cell depends.

Q 4 What is the emf of the cell when the cell reaction attains equilibrium?

Q 5 What is the electrolyte used in dry cell?

Q 6 How is the cell constant calculated from conductance value?

Q 7 What flows in the internal circuit of a galvanic cell?

Q 8 Define electrochemical series.

                                                           Level Two

Q 1 How can you increase reduction potential of an electrode for the reaction :

Mn+ aq + ne− ⟶ M(s) ?

Q 2 What happens when a piece of copper is added to (a) an aq solution of FeSO4

(b) aq solution of FeCl3

Q 3 Define corrosion. Write chemical formula of rust.

Q 4 State Kohlrausch Law. How do you determine molar conductivity of weak electrolyte at infinite dilution?

Q 5 What is fuel cell? Write its electrode reaction.

Q 6 Electrolysis OF KBr give bromine at anode, but KF does not give F2 .

                                                          Level Three

Q 1 Write three differences between potential difference and emf.

Q 2 How many grams of sodium will be deposited at cathode by passing 10 A current for 10 s

through the electrolysis of NaCl.

Q 3 The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCl solution at 298 K is 0.146 × 10-3 S cm-1.

Q 4 Calculate maximum work done that can be obtained from the following cell under standard

conditions at 250C

𝑍𝑛 𝑠 | 𝑍𝑛2+ ∥ 𝐶𝑢2+ 𝑎𝑞 |𝐶(𝑠)

𝐸0𝐶𝑢2+/𝐶𝑢= 0.34 𝑣, 𝐸0𝑍𝑛2+//𝑍𝑛= −0.76 𝑣

                                   ASSIGNMENTS (Answers)

                                      CHAPTER 1- SOLID STATE

                                                   Level one

Ans.1. Glass is an amorphous solids, it has tendency to flow but very slowly. This is the cause that glass is considered as super cooled liquid.

Ans.2. The intermolecular force of attraction make the particles of solid closely packed and force them

To only oscillate at their fixed positions. These give solids a definite volume.

Ans.3.Polyurethane, Teflon, cellophane, polyvinyl chloride, fiber glass – Amorphous solids

Naphthalene, benzoic acid, potassium nitrate, copper – Crystalline solids.

Ans.4. Metallic solids are conductor of electricity, malleable and ductile.

Ans.5.Unit cells are characterize on six parameters – dimensions along three edges and three   angles between their edges, i.e. a, b, c which are edges and α, β and γ which are angles between the edges.

Ans.6. Coordination number: - it is the total number of surrounding atoms of any constituent particle present in the crystal lattice.

Ans.7. The coordination number of atoms

a) In a cubic close-packed structure is 12

b) In a body-centred cubic structure is 8

                                                   Level Two

Ans.1.Given solid ‘A’ is a covalent solids, such as diamond.

Ans.2. Ionic solids conduct electricity because of movement of their ions. In solid state ions present in ionic solids do not move hence do not conduct electricity while in molten state ions can move and thus conduct electricity.

Ans.3.Polyurethane, Teflon, cellophane, polyvinyl chloride, fiber glass – Amorphous solids

Naphthalene, benzoic acid, potassium nitrate, copper – Crystalline solids.

Ans.4. Amorphous solids exhibits following properties: -

     • Their constituent particles are of irregular shapes and have short range order.

     • They are isotropic in nature and melt over a range of temperature. Because of which they

          are also known as pseudo solids or super cooled liquids.

     • When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces.

     • They do not have definite heat of fusion.

     • Examples - glass, rubber, plastic etc.

                                                  Level three

Ans.5. The arrangement of the constituent particles is different in glass & quartz. In glass, the

Constituent particles have short range order whereas in quartz, the constituent particles                have both long range as well as short range orders. We can convert Quartz into glass by heating    and then cooling it rapidly.

                                        CHAPTER 2-SOLUTIONS

                                                    Level One

Ans.1.Semi molar means M/2,i.e.,half mole of the solute is present in 1 liter of the solution.

  Dec molar means M/10,i.e.,one-tenth mole of the solute is present in 1 liter of the solution.

Ans.2. 0.5

Ans.3.  The relative lowering of the vapour pressure is equal to the mole fraction of the solute.

Ans.4. (i) Value of colligative property decreases.

           (ii)Value of colligative property increases.

Ans.5. The use of anti-freeze (ethylene glycol) in car radiators.

Ans.6. In the Desalination of Sea water.

Ans.7. 0.1 M BaCl2.

Ans.8. The vapour pressure of water will decrease as the addition of a non-volatile solute   lowers the vapour pressure of the solvent.

Ans.9.At higher pressure over the liquid (due to increase in pressure of cooker ,the liquid boils at higher temperature. Therefore, cooking occurs faster.

Ans.10. Since the dissolution of NH4Cl in water is an endothermic process, its solubility increases with the rise in temperature (Le-Chatelier principle)

Ans.11. At room temperature the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Therefore, the liquid ammonia will not splash out.

                                                     Level Two

Ans.1. (a) highest boiling point -1M BaCl2 solution

           (b) Highest freezing point 1M cane sugar solution

           (c) Lowest osmotic pressure -1M cane sugar solution

Ans.2. (a) C6H12O6

            (b) Methyl alcohol

Ans.3. Osmotic Pressure is defined as the hydrostatic pressure built up on the solution which, just, stops the flow of solvent.

             Sucrose < Glucose < Urea < NaCl

Ans.4. (i) Decreases

          (ii)Increases

          (iii) Increases

Ans.5. CCl4 does not form hydrogen bond with water, while ethyl alcohol forms hydrogen bond with water.

Ans.6. At higher altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentration of oxygen in the blood and the tissues of the people living in the higher altitudes. As a result of low oxygen in blood, the people become weak and are unable to think clearly. These are the symptoms of the condition, known as Anoxia.

                                               Level Three

Ans.1. Aquatic species require dissolved O2 for breathing. As solubility of gases decreases with

increase in temperature, less oxygen is available in summer than winter. Hence, they feel more

comfortable in winters (low temperature) when the solubility of O2 is higher.

Ans.2. Osmotic pressure is directly proportional to the concentration of the solutes. Our body fluids

contain a number of solutes. If a person takes more salt, the concentration of Na+ and Cl- ions

entering the body fluid , raises the concentration of the solutes. As a result, osmotic pressure

increases which may rupture the blood cells.

Ans.3.

(i) There are two moles if ions per mol of NaCl. Therefore, ΔTf cannot be equal to 1.86 and freezing point cannot be equal to 0 - 1.86 = -1.86.

(ii) ΔTf cannot be equal to 2×1.86 = 3.72 and freezing point = -3.72 because there are strong

interionic attractions for 1m solution so that the degree of ionization is not 100% at the freezing

point . The value of I is somewhat less than 2.

Ans.4. According to Roult’s Law , (po -ps )/ po =x2 (mole fraction of the non-volatile solute)

Mole fraction of the volatile liquid, x1 =1- x2 or x2 =1- x1  therefore, (po -ps )/ po=1- x1 or x1=1- (po -ps) / po = ps/ po

Ans.5. If solute-solvent interactions are stronger than the solute- solute or solvent- solvent interactions, energy

is released and the dissolution is exothermic. And if solute-solvent interactions are weaker than the solutesolute

or solvent- solvent interactions, energy is absorbed and the dissolution is endothermic.

                                       CHAPTER 3 ELCTROCHEMISTRY

                                                     Level One

Ans.1. Molar conductivity of an electrolyte increases with increase in temperature.

Ans.2. cm2ohm-1mol-1 or S Cm2mol-1

Ans.3. nature of reactants, concentration of solution, temperature, pressure of gas.

Ans.4. zero

Ans.5. a paste of NH4Cl, MnO2 and Carbon

Ans.6. cell constant= Specific conductance/ Observed conductance

                                

Ans.8.The arrangement of various electrodes in the decreasing or increasing order of their standard reduction potential.

                                           Level Two

Ans.1. (i) Increase in concentration of Mn+ ions in solution.

            (ii) By increasing the temperature.

Ans.2. (a) nothing will happen when the piece of copper is added to FeSO4 solution because reduction potential of copper is more than reduction potential of iron.

(b) Copper will dissolve in aq. Solution of FeCl3 because reduction potential of iron is more than

     reduction potential of copper.

Ans.3. Corrosion is a process of determination of metal as a result of its reaction with air and water and surroundings. It is due to formation of sulphide, oxide, carbonate and hydroxide. Formula of rust is Fe2O3.xH2O

Ans.4.Molar conductivity of strong electrolytes at infinite dilution is the sum of two values one

depending upon cation and other on the anion. Using Kohlraush law we can determine molar

conductivity of weak electrolyte at infinite dilution.

Ans.5. The device which converts chemical energy of fuel into electrical energy is called fuel cell.

2𝐻2 + 4𝑂𝐻− ⟶ 4𝐻2𝑂 + 4𝑒

2𝐻2𝑂 + 2𝐻2 + 4𝑒 ⟶ 4𝑂𝐻−

__________________________

2𝐻2 + 𝑂2 ⟶ 22𝐻2𝑂

Ans.6. Oxidation takes place at anode. Now higher the oxidation potential easier to oxidise. Oxidation

potential of 𝐵𝑟−, 𝐻2𝑂, 𝐹− are in the following order.

𝐵𝑟− > 𝐻2𝑂 > 𝐹−

Therefore in aq. Solution of KBr gives Br2 at anode, but KF give oxygen O2 instead of F2.

                                        Level Three

Ans.1. E.M.F POTENTIAL DIFFERENCE

(i)Potential difference of two electrodes when no current is flowing through

circuit. Potential difference between electrode in closed circuit.

(ii)It is maximum voltage obtained from a cell. It is less than maximum voltage obtained

      from cell.

(iii)It is responsible for steady flow of current. It is not responsible for steady flow of

      current.

Ans.2. m = zct

                 = 23×10×10/96500

                 = 23/965 g sodium

                 = 0.024 g sodium (approx.)

Ans.3. Conductivity = 1/R × cell constant

             cell constant = R × conductivity

                                     = 1500 × 0.146 × 10-3 cm-1

Ans.4  0𝑐𝑒𝑙𝑙 = 𝐸0𝐶𝑢 2+𝐶𝑢− 𝐸0𝐶𝑢 2+𝐶𝑢

                  = 0.34 – ( – 0.76)

                       = 0.34 + 0.76

                       = 1.10 v

             △ 𝐺0 = −𝑛𝐹𝐸0𝑐𝑒𝑙𝑙

                       = −2 × 96500 × 1.10 = −212300 𝐽

                       = −212.3 𝑘𝐽

Solid State Question Bank

QUESTION BANK OF SOLID STATE

LEVEL 1

1.       How does amorphous silica differ from quartz?

2.       Why glass is called supper cooled liquids?

3.       Some of the very old glass objects appear slightly milky instead of being transparent why?

4.       What is anisotropy?

5.       What is the coordination number of atoms?a) in fcc structure b) in bcc structure.

6.       How many lattice points are there in one cell of a) fcc b) bcc c) simple cubic

7.       What are the co‐ordination numbers of octahedral voids and tetrahedral voids?.

8.       Why common salt is sometimes yellow instead of being of being pure white?

9.       A compound is formed by two elements X and Y. The element Y forms ccp and atoms of X occupyoctahedral voids. What is formula of the compound?

LEVEL 2

1.       Explain how electrical neutrality is maintained in compounds showing Frenkel and Schottkydefect.

2.       Calculate the number of atoms in a cubic unit cell having one atom on each corner and twoatoms on each body diagonal.

3.       Gold crystallizes in an FCC unit cell. What is the length of a side of the cell(r=0.144mm)?

4. Classify each of the following as either a p‐type or n‐type semi‐conductor.

a) Ge doped with In (b) B doped with Si

5.       In terms of band theory what is the difference between a conductor, an insulator and a semiconductor?

6.       The electrical conductivity of a metal decreases with rise in temperature while that of a semiconductorincreases. Explain.

7.       What type of substances would make better permanent magnets, ferromagnetic orferromagnetic, why?

8.       Zinc oxide is white but it turns yellow on heating. Explain?

9.       In compound atoms of element Y forms ccp lattice and those of element X occupy 2/3rd oftetrahedral voids. What is the formula of the compound?

10.    AgCl is doped with 10^–2mol% of CdCl₂, find the concentration of cationvacancies.

Level Three Hots

Q-1   CsCl have bcc arrangement and its unit cell edge length is 400 pm.

Calculate interionic distance.

Q-2    Analysis show that Ni Oxide has formula Ni _0.98 O _1.8 What fractions

of Ni exist as Ni²⁺ and Ni³⁺ ions?

Q-3    on terms of band theory what is the difference

(i)   between conductor and insulators.

(ii)  between conductor and semi-conductors.

Q-4.     KF has NaCl structure calculate the edge length of its unit cell if its

dencity is 2.48 cm^-3. molar mass of  KF  =  58.1 g mol^-1,

NA = 6.02 x 10²³ mol^-1.

Q5.    In the cubic crystal of CSCl d = 3.97g/cm³. the eight corners are

occupied by Cl^- with Cs^- at centre and vice versa .Calculate the distance

 between Cs⁺ and Cl^- ions. What is the radius ratio of two

ions^-. (atmass of Cs = 132.9 and Cl = 35.46.

VALUE-BASED QUESTIONS

Q.1 A goldsmith under training was asked by his boss to make bangles of gold. He used a bar of 24carat for this purpose , The bangles made by him were so soft that they could be easily bent. The boss told him not to use pure gold but rather 22 carat gold. The bangles were now quite hard and could not be bent on applying some force?

(a) What was the mistake commited by goldsmith?

(b) Why did boss ask him to use 22 carat gold?

(c) In what way is 22 carat gold better than 24 carat gold?

(d) What is the value associated with this?

Q.2  A student working in the laboratory passed electric current through a rod of copper as well as solid copper sulphate crystals . To his surprise, the current could pass through copper rod connected in a cell and not through the crystals .

(a) was his observation correct?

(b) what was the reson behind this observation?

(c) how is this behaviour helpful in explaning conducting of electrolytes?

Q.3 In India , there is shortage of electricity . people have to face big cuts particulary when the summer is in peak . Industry has to suffer a lot . conventional sources of electricity such as hydel power, thermal power and even nuclear power etc.fail to meet the ever growing demand of electricity . now a days , there is emphasis on the use  of solar energy . photovoltaic material can convert solar energy into electrical energy.

(a) what are the drawbacks of traditional source of energy?

(b) how is solar energy better than conventional source of energy

(c) why is solar power not very popular in india?

(d) what is the value associated with the use of solar energy?

Q.4  Daimond and graphite are two crystalline allotropic forms of carbon . however , they differ in most of their physical characteristis. for example , diamond is very hard while graphite is quite soft. Daimond is a poor conductor of e.lectricity . on the other hand , graphite is a good conductor . Answer the following questions.

(a)why is diamond hard while graphite is soft?

(b)why is diamond a poor conductor of  electricity while graphite is a good conductor?

(c) why are certain diamonds very shining?

(d) what is the value associated with the diamonds?

Q.5 Sudanshu made a model of the unit cell of diamond. It resembled the unit cell of ZnS. If the unit cell of ZnS has 4 units of ZnS per unit cell. It has the same packing efficiency as ZnS. But diamond is the hardest known substance.

(a)What is the number of atoms of carbon per unit cell of diamond?

(b)What is the value that Sudanshu can derive from these facts?

(c)What is the value that Sudanshu can derive from these facts?

QUESTION BANK - SOLID STATE (ANSWERS)

LEVEL 1.

Ans1. In amorphous silica, SiO₄ tetrahedral are randomly joined to each other whereas in quartz they arelinked in a regular manner.

Ans2.Schottky defect.

Ans3. It has tendency to flow like liquid.

Ans4. Due to crystallization

Ans5. Physical properties show different values when measured along different directions in crystalline solids

Ans6. a) 12 b) 8

Ans7. a) 14 b) 9 c) 8

Ans8.  6 and 4 respectively

Ans9. Due to the presence of electrons in some lattice sites in place of anions these sites act as F‐centers.These electrons when excited impart color to the crystal.

Ans10. No. of Y atoms be N No. of octahedral voids N

No. of X atoms be =N Formula XY

LEVEL 2.

Ans1. In compound showing Frenkel defect, ions just get displaced within the lattice. While incompounds showing Schottky defect, equal number of anions and Cations are removed fromthe lattice. Thus, electrical neutrality is maintained in both cases.

Ans2.  8 corner *1/8 atom per unit cell = 1atom

There are four body diagonals in a cubic unit cell and each has two body centre atoms.

So 4*2=8 atoms therefore total number of atoms per unit cell =1+8=9

Ans3.  r=0.144nm

a=2*√2r

=2*1.414*0.144nm

=0.407nm

Ans4. (a) Ge is group 14 elements and In is group 13 element. Therefore, an electron deficit holeiscreated. Thus semi‐conductor is p‐type.

(b) Since B is group 13 element and Si is group 14 elements, there will be a free electron, thus it is n type semi conductor.

Ans5. The energy gap between the valence band and conduction band in an insulator is very large while in a conductor, the energy gap is very small or there is overlapping between valence band andconductionband.

Ans6. In metals with increase of temperature, the kernels start vibrating and thus offer resistance to the flow of electrons. Hence conductivity decreases. In case of semi‐conductors, with increase oftemperature, more electrons can shift from valence band to conduction band. Hence conductivity increases.

Ans7.. Ferromagnetic substances make better permanent magnets. This is because the metal ions of a ferromagnetic substance are grouped into small regions called domains. Each domain acts as tiny magnet and get oriented in the direction of magnetic field in which it is placed. This persists even in the absence of magnetic field.

Ans8. When ZnO is heated it looses electron and which occupies the  interstitial sites . This electron absorbs the radiation from the visible region and transmitted radiations and reflects the colour.

Ans9.

No. of Y atoms per unit cell in ccp lattice=4

No. of tetrahedral voids= 2*4=8

No. of tetrahedral voids occupied by X= 2/3*8=16/3

Therefore formula of the compound =X_16/3 Y₄

=X₁₆ Y1₂

=X₄ Y₃

Ans10. Doped atom = 10^–2mol% of CdCl₂

= 10^–2     x 6.022 X 10²³ atoms.

One crystal of CdCl₂ create one cation vacancy when doped with AgCl

No of cation vacancies = 6.022 X 10²¹

LEVEL 3.

Ans1 Face Diagonal = √a² +a²  = √ 2   .a

Body Diagonal   =  √(√2a)² + a²

Body Diagonal =  √3    . a .

Body Diagonal=  √3 x 400  = 1.73x400

Body Diagonal =  692

Interior Distance  = ½ x Body Diagonal = ½ x 692 = 346pm.

 Ans 2. Ni ²⁺ ions present the ratio of Ni : O is 1:1  i.e, 100Ni²⁺ ion should

have 100 O^2-. Now replace 3Ni²⁺ ions 2Ni³⁺ ions are required it

means presence of two Ni³⁺  would reduce one No. of Ni out of 100

numbers.In Ni _0.98 O _1.00  ratio 98:100. It means two Ni ions are lesser

than oxide ions ( out of 100 atoms) Then four Ni³⁺ ions are present

and 96 Ni²⁺ are present for each 100 oxide ions. Thus % of Ni²⁺ is

96% and Ni³⁺ is 4%.

Ans3.  

(i)   in conductor the energy bands are very close overlapped by the

conduction bands. Due to this reason e^- can flow very easily from

valency band to conductor band under an electric field, thereby

showing conductivity.

 ii)               In semiconductor the gap between valence band and conduction bond  is  small which make some of the e^- enable to jump from valences band to conduction band. Therefore conductance is observed with the increace in temp. more e^- jump. Hence the electrical conductivity of semiconductor increase with the increase in temperature.

 Ans4.   For Nacl

d = 2xM/a³ x NA

a³ = 2 x M /d x NA

a³ = 4 x 581/2.48 x 6.02 x 10^-3

= 155.659 x 10^-24

a  =  5.379 x 10^-8 cm

=   537.9 p.m.

Ans5.  In one unit cell there is one CsCl 1/8 x 8 Cl^- + 1 Cs ⁺

d = z x M/a³ x N_oa³ = 2 x M / d x N_o

a³ = volume ,  Z = 1m = 132.9 + 35.46   =  168.36

f  =  3.97 g cm^-3No = 6.02 x 10²³.

a³=  1 x 168.36/6.02 x 1023 x 3.97      =    7.04 x 10^-23 cm³

a  =  (7.04 x 10^-23)^1/3a  =  4.13  x  10^-8 cm  = 413 pm.

Hence the diagonals  = √3 x 413  = 715.34 pm.

As the structure with Cs⁺ at centre and Cl^- at corner

2r⁺ + 2r-  =  715.34 or

r⁺ + r^-      =   715.34 / 2    =   357.62 pm

We  assume 2Cl^- touch with each other than length of cell is

2r^-  =  413 pm

r^-  =   206.5 pm

r⁺  =357.67  -  206 .50   =151.17 pm

r+/r^-  =  151.17/206.50      =  0.73

VALUE-BASED ANSWERS

 Ans.1 (a) ornaments are not made from 24 carat gold which is completely pure.

      (b) 22 carat gold contains traces of copper or silver as impurity.

      (c) It is better than puregold since it is  comparatively hard, malleable and ductile.

        Infact, 22 carat is an alloy.

     (d) The boss guided the junior worker about the technique for making gold jewellery.As stated above , pure gold is extremely soft and is also brittle . That is why bangles could not be made out of that. the information is extremley     valuable for all new comers in the field.

Ans.2 (a) yes , his observation was correct.

     (b) Copper metal is conducting by the movement of electrons while copper sulphate crystals are expected to be conducting by the movement of ions. Since ions do not move in the solid state therfore , copper sulphate is non-conducting in the solid state.

     (c) Electrolytes are either taken in molten state or aqueous solution in the electrolyte cells(e.g. electrolysis,electroplating).They conduct due to the movement of ions.

Ans.3 (a) Traditional source of electricity ;both thermal and nuclear energy lead to emissionof smoke,fly ash and extremely harmful radiations.

(b) Sun is the only source of solar energy of energy in a country like India where the span of summer is very long solar energy is very good option .

(c) Solar power is not very popular in India mainly due to lack of awareness.

(d) Use of solar energy in houese,hotelsand in small industrial unitscan solve our electricity problems.

Ans.4 (a) Structure of diamond is very compact on the other hand graphite has a layer structure.

(b) In diamond all the four valence electrons of carbon are involved in bond formation in graphite three valence electrons of carbon are involved while one electron is delocalised to conduct electricity.

(c) Diamond has a high refractive index as a result certain diamonds produce very high total internal refraction responsible for bluish tinge that is why these are very shiny and expensive.

Ans.5  a) The number of atoms of Carbon per unit cell is 8 in diamond.

       b) The C—C bond is very strong in diamond (due to small size of Carbon) unlike the Zn—S bond in ZnS.

       c) Though from the same background ie with the same structure the property can be different, thus, with a little effort, we can do same things differently and bring about major change

 

                 

 

                                                                                     

 

           

  

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