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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