Solid state previous year questions CBSE with answers( Two and Three Mark )

Solid state previous year questions CBSE with answers( Two and Three Mark )
Solid state previous year questions CBSE with answers( Two and Three Mark )

These are Solid State previous year questions for CBSE .
Two Marks Question:

1.       An element having bcc structure has atomic mass 50u and density 6.81 gcm-3 Calculate the edge length of the unit cell. (NA=6.02 * 1023 mol-1)                                                                          (AI 2001 C)

2.       Unit cell of an element (atomic mass 108u and density 10.5g cm-3)has an edge length 409pm Deduce the type of crystal lattice . (NA=6.02 * 1023 mol-1)

3.       Nacl crystals are doped with 2*10-3 mol percent of SrCl2 , calculate the cation vacancies per mole. (NA  =6.02 * 1023 mol-1) (Delhi 2001C)

4.       Lithium metal crystal has body centred  cubic structures  density is 0 53 g cm-3 and  its atomic mass is 6.94 g mol -1 .Calculate the volume  of a unit cell of lithium metal . (NA  =6.02 * 1023 mol-1)  (Al 2001)

5.                Sodium crystallizes in the cubic lattice and the edge of the unit cell is 430 pm .Calculate the number of atoms in the unit cell (Atomic mass of Na = 23.0u, density of sodium =0.963g cm-3) )  (NA  =6.02 * 1023 mol-1) (Al 2001)

6.       An element crystallizes in fcc structure  .200 g of this element has 4.12*1024 atoms. The density of A is  7.2 g cm-3 . Calculate the edge length of the unit cell.  (AI. 2001)

7.       State the difference between Schottky and Frenkel defects.  Which of the two changes the density of the solid? (A1 2002)

8.       The density of chromium metal is 7.2 g cm-3. If the unit cell has edge length of 289pm , determine  the type of unit cell .(atomic mass of Cr =52 u, NA  =6.02 * 1023 mol-1) (Al 2003 C)



9.       An element occurs in bcc  structure with cell edge 300pm. The density of the element is 5.2 g cm-3. How many atoms of the element does 200 g of the element contain? (Delhi 2003 C)

10.   Explain the following with suitable examples

                       An n type semiconductor (Delhi 2006 C)

11.   Copper crystallizes into an fcc lattice with edge length of3.61*10 -8cm . Calculate the density of copper (Given Cu  =63.5g mol-1 , NA  =6.02 * 1023 mol-1 ) (Delhi 2009 C)

12.   A. In reference to crystal structure explain the meaning of coordination number?

(Delhi 2009 C)

                 B.What is the number of atoms in a unit cell of
                   a.) face centred cubic structure

                  b.) body centred cubic structure                                                                                         (Delhi 2009 C)


13.   How is crystals affected by Schkotty and Frenkel defects? (Delhi 2009 C)

14.   The well known mineral Fluorite is chemically calcium chloride .It is known that in one unit cell of this mineral there are 4 Ca2+ ions and 8 F- ions and that Ca2+are arranged in a fcc lattice .The F - ions fill all the Tetrahedral holes in the face centred cubic lattice of Ca2+ ions .The edge of the unit cell is 5.46 *10-8 cm in length. The density of the solids is 3 .18 g cm-3. Use the information to calculate Avogadro's number (Molar mass of CaF2, 78.08 g mol-1) (Delhi 2010)

15.   In terms of band theory explain the difference between a conductor and a semiconductor and give one example of each (AI.2011 C)

16.   Explain the following terms with suitable example of each

i.)Ferromagnetism                                            ii.)Antiferromagnetism           (Delhi 2011 C)

17.   Calculate the packing efficiency of a metal crystal for a simple cubic lattice (AI.2011)

18.   Calculate the number of unit cells in 8.1g of aluminium if it crystallizes in a face-centred  structure. (Atomic mass of Al = 27 g mol-1)                   (2017)

Three Marks Questions:

1.       Determine the type of cubic lattices to which the iron crystal belongs if it has edge length of 290 pm and the density is 7.87 g cm-3 (At mass of Fe =56 g mol-1, , NA  =6.02 * 1023 mol-1 )                                                                         (Delhi 2005)


2.       Aluminium crystals in a cubic close-packed structure is metallic radius is 125 pm .
a.)    What is the length of the edge of the unit cell?
b.)    How many such unit cells are there in a 100 cm3 of aluminium                 (Delhi 2005)

3.       What is a semiconductor? Describe two main types of semiconductors, giving examples and their distinctive features .                                                                                  (AI.2006 )

4.       Assign reasons for the following
i.)Phosphorus doped silicon is a semiconductor
ii.)Schottky defect lower the density of a solid
iii.) Some of the very old glass objects appear slightly milky instead of being transparent

5.       Silver crystallises in an fcc lattice. The edge length of its unit cell is 4. 077 10-8 cm and is 10.5 g cm-1. Calculate on this basis the atomic mass of silver (N=6.023x1023 mol-1 )                                         (A.1.2008)

6.       What is semiconductor? Describe two main types of semiconductors and explain mechanism  of their conduction.                                                                          (Delhi 2008)

7.       Explain the following giving a suitable example in each case:
i.) Paramagnetism
ii.)Frenkel defect
iii.) Ferrimagnetism                                                                                                 (Foreign 2008)

8.       How would you account for the following?
i.) Frenkel defects are not found in ionic solids of nearly equal sizes of cations and anions ii.)Impurity doped silicon is a semiconductor                                                   (Foreign 2008)

9.       Silver crystallizes with face-centred cubic until cells. Each side of the unit cell has a length of 409 pm. What is the radius of an atom of silver? (Assume that each face atom is touching corner atoms)                                                                                           (AI.2009)

10.   Iron has a body centred cubic unit cell with a cell edge of 286.65 pm. The density of iron is7.87 g cm-1 . Use this information to calculate Avogadro's number (At mass of Fe = 56 g mol-1 )(Delhi 2009)
11.   The density of lead is 11.35g cm-3 and the metal crystallizes with fcc unit cell. Estimate t

radius of lead atom .                                                                                                   (Delhi 2011)

12.   Calculate the distance between Na+ and Cl-  ions in NaCl crystal  if its density is 2. 165 g cm-3[Molar mass of Nacl = 58.5 g mole-1 ,NA =6.02*1023 mol-1  ]                            (2006)

13.   The density of Copper metal is 8.95 g cm-1 .If the radius of copper atom be 127.8 pm , is the copper unit cell simple cubic, body-centred or face centred cubic?(Given Atomic mass of  Cu =63. 54 g mol-1   , NA =6.02*1023 mol-1)                                                   (2010)

14.   An element with molar mass 27 g mol-1   forms a cubic unit cell with edge length 4.05*10-8 cm its density is 2.7g cm-3, what is the nature of the cubic unit cell?
 (Delhi 2015)

15.   An element crystallizes in a fcc lattice with cell edge of 400 pm. The density of the element is 7 g cm-3 . How many atoms are present in 280g of the element?  (AI. 2016)                                                                            

16.   An element crystallizes in a f.c.c lattice with cell edge of 250 pm. Calculate the density of 300 g of this element containing  2*1024 atoms.                                          (Delhi 2016)

17.   a.)Based on the nature of intermolecular forces classify the following solids:
 Silicon carbide.Argon
 (b) ZnO turns yellow on heating. Why?
(c) What is meant by groups 12-16 compounds? Give an example .                         (2017)
HINTS & SOLUTIONS
SOLID STATE
Two Marks Questions:

1.       Z =2, M=50 gm mol-1 , d=6.81g cm -3
a3  = ZxM
        d x NA
a3   =2x 50
     6.18 x6.02x1023
a=289.6 cm

2.       Z=?, M=108 u, d=10.5 g cm, a = 409 pm, NA= 6.02x1023 mol-1
d =ZxM
    a3   x NA
Z= d xa3   x NA
                        M
              Z= 10.5gcm-3 x(409)3 x 10-30 cm x 6.02x1023 Cm
                                      108
              Z=4, FCC structure

3.       Cation vacancies = Number of Sr2+ ions ,add =2x10-5 mol = 2x10-5 x 6.023x1023 = 1.2046x1019 mol

4.       Z=2, d= ZxM
            a3   x NA
a3  = ZxM    =  2x 6.94                    
        d x NA      0.53x 6.23 x1023

= 4.348x 10-23 cm

5.       Z=d x NA x a3
                                   M
                 =0.9623 cm3 x6.023x1023 x (430)3 x (10-10 )3cm 3 
                                                            23
                Z=2


    6. Z=4, M=200 g, N=4.12x1027, d=7.2 g cm3

     a3= Z x M
            d x N
      a3=    4    x   200
            7.2x4.12x1027
a=299.8 pm
7.       Schottky defect arises due equal number of cation and anion missing - density decreases  Frenkel defect  arises due to equal no of cation & anion leaves its normal site and occupies interstitial site - density remains constant

8.       Z=d x a3 x NA   =  7.2 g cm-3x (289)3 x 10-30 cm3 x 6.02 x 1023
        M                                               52

Z =2 (body centre)

9.       Z=2

N=2 x  M
      a3 x d

N= 2x 200
(300)3 x 10-10 x 5.2 cm 3

N=          400                                           
           27 x 10 27 x5.2

=2.85x1024 atoms

10.   When the Group - 14 element doped with upon -15, element n-type semiconductor is able

e.g. silicon doped with phosphorous

11.    d= Z x M
                    a3  x NA  

=   4 x 63.5
(3.61 x 10-8 )3 x 6.022 x 1023

= 8.965 g cm-3

12.   a) Coordination no: the no of atom surrounded by nearest  neighbour

b) (i) Four (ii) Two

        14.The No of molecule per unit cell is 4
                      N = ZxM
                            a3  xd
Z=4. M=78.08 g mol-1 ,a=5.46x10-8, d=3.18 g cm -3
NA =4x78.08
(5.46x 10-8)3 X3.18

=6.03x1023cm

15.Conductor According to band theory the energy gap is very small in conductor Semiconductor: According to band theory the energy gap is more than conductor but less than insulator



17.Let the edge length is a, radius of atom = r ,V of cube = a3
      AB = r+r
       A=2r
       Volume of cube = (2r)3 = 8r3
       No of atom in simple cube = 1
       Volume of one atom = 4/3 Ï€r3
Solid state previous year questions CBSE


 

                                       
Packing fraction = 4/3 Ï€r3  x100  = 52.4%
                                   8r3
18. Moles of aluminium = Mass/ molecular mass
We know that one unit f.c.c., No. of atoms = 4                                              nAl =8.1/2.7 =0.3moles
4 - atoms are found in unit cell = 1
1- atoms are found in unit cell =1/4
 (1 mole) NA atoms are found in unit cell = NA/4
0.3 moles atoms are found in unit cell = NA/4x0.3= 0.075xNA

Three Marks Questions:

1.       d= ZxM
    a3 x NA

7.87=Z x 56
     (290 x 10-10)3 x 6.02 x 102

Z=2 bcc structure

2.       a.) In cubic close packing structure
 4r=√2a
r = 125x10-12 m
 a=4r/√2 = r2√2 = 125x 10-12 x 2√2 = 354pm

b) a3 = (354x10-12 m)3
 a= 44.21x10-30 m3
No. of unit cell in 1 cm3 =  total value/V. of one unit cell = 10-6  m3 2.261x1023  unit cell             
                                                                                                   44.21 x 10-30 m3

3.       The solids with intermediate conductivities between insulators and conductors are termed semiconductors
(i) n-type semiconductor : It is obtained by doping Si or Ge with a group 15 element like P. Out of 5 valence electrons  only 4 are involved in bond formation and the fifth electron is delocalized and can be easily provided to the conduction band. The conduction is thus mainly caused by the movement of electron.

(ii) p-type semiconductor: It is obtained by doping Si or Ge with a group 13 element like Gallium which contains only 3 valence electrons. Due to missing of 4th valence electron, electron hole or electron vacancy is created. The movement of these  positively charged hole is responsible for the conduction

4.       i.)Because after doping the phosphorus as the conductance is intermediate between conductor & Insulator

ii) Because in Schottky defect equal no of cation & anion is missing
iii) Because in day ,glass is heating but in night, glass is cooling e.g. annealing after some year glass acquired crystalline structure.

5.       M = d xa3 x NA
               Z
=10.5 g cm3 x(4.077 x 10-8 )3x6.023 x 1023
                                     4
=107.143

6.       See solution of question No. 3 (Three Marks)

7.       (i) Paramagnetism : Materials which are weakly attracted by magnetic fields are called paramagnetic materials and the property that exhibited is called paramagnetism. Paramagnetic substances contain unpaired electrons, e.g. TiO,CuO,O2 and VO2  etc.
(ii) Piezoelectric effect: When the electricity is produced by applying mechanical stress polar crystals, is known as piezoelectric effect. Quartz show this property

(iii) Frenkel defect in crystals: Frenkel defect occurs when an atom or ion (generally cation) leaves the normal site in the crystal -lattice (creating a vacancy) and occupies on interstitial site. This defect is generally found is silver halides because of the small size of Ag+  ions.

8.       i.)Because in Frenkel defect size of cation should be smaller than the size of anion so cation goes to void.

(ii) When silicon doped with impurities like boron positive hole is formed when is doped with arsenic free electron present.

9.       145.03 pm
In fcc d = a/√2

r=a/2√2  = 409/ 2√2 =144.6 pm

10.   NA   =    ZxM
 1      a3 x NA

NA  = 2x56g mol-1
          (286.65x 10-10)3 x7.87
NA = 6 x 10-23

11.   1.7494,10-8 cm.
a3= ZxM
     d x NA
a3  =  4x207
       6.02 x1023 x11.35

r=a/2√2  ,  r = 4.948x10-8  cm                                        a = 4.948 x 10-8cm
                          2 x 1.4142


12.    d= ZxM
    a3 x NA

 a=281 pm

13.   4 (fcc)

14.    Molar mass of the given element, 27 g mol-1 = 0.027 kg mol-1
Edge length, =4.05x 10-8 cm = 4.05x10-10 m
Density, =2.7 g cm-3 = 2.7x103
 Applying the relation
d= ZxM
    a3 x NA

Where, Z is number of atoms in the unit cell and NA is the Avogadro number
 Z=d xa3 xNA 
             M
= 2.7x103 (4.05 x 10-10)3 x6.022x1023/ 0.027 = 4
 Since the number of atoms in the unit cell is four, the given cubic unit cell has face-centred cubic (fcc) or cubic-close packed (ccp) structure.

15.   Volume of unit cell = a3 [ a= edge length ] = 400 pm
= (400 x 10-12m)3 =(400 x 10-10cm)3  = 64*10-24 cm3

Volume of 208 g of the Element = mass/ density
=208g/7g cm -3  = 29.71 cm3
No. of unit cells in this volume = volume of 208 g of element
                                                            volume of one unit cell
=29.71/64 x 10 -24 = 0.46 x 1024
Since each fcc unit cell contains 4 atoms therefore ,total no of atoms in 208 g
 =4 x 0.46 x1024
=1.84 x 1024 atoms

16.   Molecular mass of element = 300 x 6 x 1023  = 90 amu
                                                          2x1024

Density = Z M
                    N0xa3

=4x90
6 x 1023 x ( 250 x 10-10)3

=4x90   g/cc
6x 2.5x 2.5 x 2.5 x 10-1

17.    On the basis of intermolecular forces: - (i) Silicon carbide : - Covalent network solid (Covalent Bonding)
 (ii) Argon : - Non-polar molecular solid which posses dispersion or london forces.

(b) Žinc oxide is white in colour at room temperature. On heating it loses oxygen & turns yellow

Zno - Zn2+ + ½ O2 + 2e-
 the excess Zn2+ ions move to interstitial sites and the electron to neighbouring interstitial sites

(c) Some of the compound like Zns, CdSe and HgTe are example of group 12 - 16 compound.In these compound bonds are having same ionic character along with covalent.

Check out : Electrostatics Five mark questions with answers for CBSE Class 12 (previous year )

Please Share and Help Others

Comments

Post a Comment

Feel free to ask if you have any queries and if want any information . I feel happy to interact with you all and help you .

Popular posts from this blog

Notice Writing Format with Examples ( Notice writing )

Image
 Notice Writing Format with Examples ( Notice writing ) Notices are frequent question which are asked in both CBSE class 12 and 10  board papers.  The  format of notice writing with the rules to be followed is described here . Notice Writing : A notice is written form of formal message or a type of written announcement made to everyone . As notice is a written form of formal message so it is made to inform all about various activities like the activities may refer to football tournament or debate competition to be held in school , notice for  a holiday , a new scheme implemented by the college for the welfare of girls , inter - school competitions and many more .  The notices simple and passive form language to deliver a direct message to the readers of the notice . It is written in 50 words .  Image by      Rules to be followed : Notices are always written in a box . There should be usuage of passive voice  while writing a notice . It means you have to
Read more

CSAB (Fee Refund)

Image
CSAB ( Central Seat Allocation Board)   Fee Refund You will get full information along with personal experience on CSAB Fee Refund . CSAB every year holds counselling for the students who have appeared in Jee Mains . It is after JOSAA counselling only that it holds counselling in the form of CSAB Special Rounds and CSAB NEUT Counselling . There are mostly 2 rounds conducted every year . But you might be confused whether CSAB refunds the money back or not?  So, here I am to tell you all about it as I also took part in it . CSAB  FEE REFUND As the counselling of CSAB starts it is just a matter of  3 to 4 days that it ends. This counselling is done very fast as at this time all the NIT and IIIT colleges get started. As we take part in counselling the first thing one has to do is to pay counselling fees which was 36,000 for General and OBC students and around 16,500 for SC and ST. Image by  rainesUMD  from  Pixabay I also took part in CSAB Special Round couns
Read more

All Derivations of Ray Optics Class 12 ( Ray Optics )

Image
All Derivations of Ray Optics Class 12 These important derivations are all from previous years question papers : All Derivations of Ray Optics Class 12 ( Important derivations) Five Marks a)With the help of a suitable ray diagram , derive the mirror formula for a concave mirror . b)The near point of a hypermetropic person is 50 cm from the eye . What is the power of the lens required to enable the person to read clearly a book held at 25cm from the eye ? (CBSE 2009) Trace the ray of light showing the formation of image due to a point object placed on the axis of a spherical surface separating the two media of refractive indices n1 and n2 . Establish the relation between the distances of the object , the image and the radius of curvature from the central point of the spherical surface . Hence , derive the expression of lens maker's formula . Or Draw the labelled ray diagram for the formation of image by a compound microscope . Derive the expression for th
Read more

Solid state previous year questions CBSE with answers( One mark)

Image
SOLID STATE  Solid state previous year questions CBSE with answers( One mark) These are solid state previous year questions for CBSE . One Mark Questions: 1.        Which point defect lowers the density of ionic crystals? (AI. 2000 C) 2.        What is the difference between ferromagnetic and ferrimagnetic substances? (Delhi 2001 C) 3.        Why does Frenkel defect not change the density of AgCl crystal? (AL. 2001 C) 4.        What other elememt may be added to silicon to make electrons available for the conduction of electricity ? (Delhi 2001 C) 5.        What is Curie temperature? (AI. 2002 C) 6.        What is the effect of Frenkel defect on electrical conductivity of the solid? (AL. 2003 C) 7.        Name one solid in which both Frenkel and Schottky defects occur .(AI. 2003 C) 8.        What is difference between glass and quartz though both contain SiO 4 4- units? (Delhi 2003 C) 9.        In an alloy of gold and cadmium, gold crystallises in cubi
Read more

JEE Mains -Mole Concept and Stoichiometry Previous Year Questions (with Answers)

Image
     JEE Mains -Mole Concept and Stoichiometry Previous Year Questions (with Answers)   JEE Mains - Mole Concept and Stoichiometry Previous Year Questions (with Answers)                                                             JEE Mains -Mole Concept and Stoichiometry Previous Year Questions (with Answers) These are Jee Mains previous year questions for chemistry - MOLE CONCEPT, BALANCING OF REDOX REACTION and STOICHIOMETRY 1.        25 ml of a solution of barium hydroxide on titration with 0 .1 molar solution of hydrochloric acid gave a litre value of 35 ml. The molarity of barium hydroxide solution was  (1) 0.14   (2) 0.28 (3) 0.35 (4) 0.07                                                                                    (MAINS-2003) Ans :   (4.) 2.        The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium h
Read more

CSAB 2020 ( CSAB Fee Refund )

Image
 CSAB (Central Seat Allocation Board) 2020 ( CSAB Fee Refund ) CSAB 2020 is the Special Round Counselling which is conducted to fill the seats left after Jossa counselling in the Academic Year of 2020-21. CSAB 2020 is being conducted by Motilal Nehru National Institute of Technology Allahabad. CSAB 2020 (CSAB Fee Refund) Image by  Picography  from  Pixabay  
Read more

Message Writing,Format for message writing

Image
 Message Writing,Format for message writing You might think how to use message writing in out daily life . How should you write a message. Every thing is discussed below: There are many circumstances when we may confront lack of time like we are in a hurry to go out and do some work . But before going out we need to inform someone like our family , friend or the person who is incharge of the shop like shop owner that you are leaving from that place as you have some work . But what if the person is not present at that time. So how will you inform ? This is really a matter of concern .
Read more

Electrostatics Important Five mark questions with answers for CBSE Class 12 (previous year)

Image
Electrostatics Five mark questions for CBSE Class 12 ( previous year ) Do subscribe the blog and for any queries or study material comment in the comment box . Electrostatics Important Five mark questions for CBSE Class 12 ( previous year ) a. Derive an expression for the energy stored in a parallel plate capacitor C , charged to a potential difference V. b. Obtain the capacitance of the network given below . For a supply of 300 V , determine the charge and voltage across C₄ . Electrostatics Important Five mark questions with answers for CBSE Class 12 (previous year) [2008] Explain the principle of a device that build up high voltages of the order of a few million volts . Draw a schematic diagram and explain the working of this device . Is there any restriction on the upper limit of high voltage set up in this machine ? Explain . [CBSE 2012] Or a. Define electric flux. Write the S.I. units . b. Using Gauss law , prove that the electric field at a point due
Read more

Menu Driven Program For Circular Linked List: creation, traversing, insertion and deletion(Circular Linked List)

Image
 Menu Driven Program For Circular Linked : creation, traversing, insertion and deletion(Circular Linked List) Circular Linked List is a very useful data structure and it also has various daily life applications . The below C++ program shows various operations for Circular Linked List like creation of circular linked list, traversing or displaying it, insertion and deletion operations in circular linked list at various locations that is at the end , in the beginning and at any given location. 
Read more