Solid State Chemistry Class 12 Important All Questions with Answers pdf

Solid State Chemistry Class 12 Important Questions and Answers with Free pdf

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SOLID STATE

1. Write two features by which you can distinguish metallic solid from ionic solid.

Ans. 

Metallic solids are malleable and ductile while ionic solids are hard and brittle. Metallic solids conduct electricity in solid state due to movement of electrons, while ionic solids conduct electricity in molten state or aqueous solution due to the movement of ions.

2. The radius of Nat is 95 pm and CF is 181 pm. Find co-ordination, number of Na+ ion.

Ans. Radius ratio=r( Na+)/r(cl-)= 95/181 =0.524. Since

radius ratio lies between 0.414- 0.732, so co-ordination number of Na+ ion is 6.

3. Silver crystallises in fcc unit cell. The edge length The unit cell is 409 pm. Find the radius of the silver atom.

Ans. a= 409 pm

for a Fcc unit cell

r=a/2√3= 409/2×1.414 = 144.5pm

4. What is co-ordination number in solids ? Write the co-ordination number in hep, ccp and bcc

crystal structure.

Ans. Co-ordination number is defined as the number of nearest neighbours with which a given sphere is in contact.

Coordination number in

hcp packed structure = 12

ccp packed structure =12

bec packed structure = 8

5. A non-stoichiometric oxide of iron has the formula Fe⁰.⁹³ 0¹.⁰⁰What % of iron is present in

the form of Fe (III)?

Ans. Iron has both +3 and +2 oxidation states. Let

x atoms of Fe Ions have been replaced by Feion

Number of Felons 0.93-x

Total +ve charge = 2(0.93 – x) +3x

Also total +ve charge =2 (charge of oxide)

For electrically neutrality

2(0.93 x) + 3x = 2

=> 1.86+x=2

=> x=0.14

:• % of iron as present in the form of Fe³+

0.14

  —-     x100 15.05

0.93

6. Iron crystallises in a body centered cubic lattice, whose edge length is 2.93 A”. Calculate the radius of the sphere and distance between centres of neighbouring spheres.

Ans. Given a = 2.93A°

since for bcc, arrangement

r= a√3/4

=> r= 2.93x √3/4 =1.268A°

Distance between the centres of two neighbouring

spheres 2r 2 x 1.268 2.536A⁰

7. Copper metal crystallises in a fec lattice with cell edge, a=361.6 pm. Find the density of Cu crystals. (At mass of Cu = 63.5)

Ans. M =63.5, a 361.6 pm 361.6 x 10–¹⁰ cm

N= 6.023 x103, z= 4 atom per unit cell

d= ZM/a³N=  4×63.5/(361.6x 10-¹⁰) x6.023 x 10²³

=8.94gm–³

8. Chromium forms a body centred cubic lattice of unit cell edge length 288 pm. Calculate the density of chromium (At mass of Cr = 52)

Ans. For bcc lattice the number of atoms per unit cell- 2

However mass of each ‘Cr’ atom= 52g/6.023×10

=> Mass of 2 ‘Cr’ atoms = 2x52g / 6.023 x10²³

 = mass of unit cells.

Volume of unit cell = a³ =(288×10–¹⁰cm)³

Now density of chromium =mass/ volume

=> 6.023 x10²³x(2.88 x10–⁸cm /2x52g )³

= 7.23gcm–³

9. The edge length of the unit cell of a metallic crystal is 3.624. The density of metal is 8.93gcm–³ Calculate the number of unit cells in 100g of the metal.

Ans. Volume of unit cell =

a³= (3.62A⁰) =(3.62×10–⁸ cm)³

Volume of 100 g metal= mass/density

= 100g /8.93gcm–³  = 11.2cm³

Therefore number of unit cells in above volume

volume of 100g metal/volume of unit cel

 = 11.2cm³/(3.62x 10–⁸cm)³

= 2.36x 10²³

10. A cubic solid 1S made oI two elements A and B. Atoms A are at the corners of the cube and B at the edge centres and the body centre. Find the formula of the compound.

Ans. Number of A’ atoms per unit cell =8 (corners)×1/8=1

Number of atom of “B per unit cell

= 12 (edge centres) x1/4+1 (body centr)

3+1=4

So formula of the compound is AB⁴

11. Explain why conductivity decreases for a conductor with increase in temperature, while that of semiconductor increases with increase in temperature.

Ans. Metals conduct electricity due to movement of electrons. When a metal is heated the metal ions start vibrating more vigorously and the motion of valency electrons through the crystal is hindered. Thus electrical conductivity of metals decreases with increase in temperature.

For semiconductors with increase in temperature more electrons can jump from the valence band to the conduction band. Thus electrical conductivity increases with increase in temperature.

12. Explain why the conductivity of silicon increases on doping it with phosphorus.

Ans. Silicon is an insulator. However when it is doped with phosphorus it becomes a n-type

Semiconductor. So conductivity increases due to extra free delocalised electrons.

13. Why does the presénce of excess lithium make licl crystals pink ?

Ans. When a crystal of LiCl is heated in an atmosphere of ‘Li’ vapour, the lithium atoms lose electrons to form Li ions. The released free electrons diffuse into the crystal and occupy anionic sites. These electrons absorb one colour from visible height and undergo electronic transition. This results pink colour to the crystal. This is also known as metal excess defect.

14. Why is glass considered as a super cooled liquid?

Ans. Glass is obtained by rapidly cooling molten silicates, it has the tendency to flow, though very slowly. So it is called a supercooled liquid.

15. Explain why non-stoichiometric sodium chloride is yellow in colour.

Ans. Non-stoichiometric sodium chloride contains excess sodium atoms. These sodium atoms combine with Cl ions by losing electrons. These electrons get entrapped in vacancies created by CI ions. The electrons trapped in anionic vacancies are called F-centres. Now when visible light falls on the crystal the electrons get excited and impart yellow colour.

16. The radius of CIion is 181pm. Calculate the radius of cation that just fits into (a) octahedral

holes of this lattice of anions. (b) tetrahedral holes of this lattice of anions.

Ans. (a) Radius ratio of octahedral voids is,

r+/r- =0.414

radius of cation = radius of anion x 0.414

= 181 x 0.414

=74.93 pm

So cation having radius 74.93 pm will just fit into octahedral voids.

(b) Radius of cation in tetrahedral voids rcl-X0.225pm= 40.725 pm

So cations having radius 40.725 pm will just fit into tetrahedral voids.

17. The radius of Rb+  is 147 pm and Br is 195 pm. Find the most probable co-ordination number of Rb+ in RbBr.

                          147

Ans.= r+/r= —–=0.753 (cubic)

                          195

Since the radius ratio lies between 0.732 to 1, so co-ordination number of Rb is 8.

18. If the radius of Li’ is 60 pm and F– ion is 136 pm. Find the ordination number of Li+ in LiF

  Ans.   rli/rf= 60/136= 0.441

since r+/r– lies between 0.414 to 0.732 therefore it

Will have sodium chloride type structure. Hence co ordination number of Li + ion and F- ion are 6 each.

19. An element crystallises in fcc structure 100g of This element has 2.06 x 104 atoms. The density of the element is 7.2g/cm. Calculate the edge length of the unit cell.

Ans. d 7.2 g/cm³, M-100g, Z-4, N=2.06 x 10-24

Since d=ZM/a³×N => a³= Z×M/d×N

= 4×100/7.2x 2.06×10 024= 27×10-²⁴

=> a=3×10–⁸ cm=3A⁰

So edge length is 3A°

20. Determine the type of cubic lattice to which the iron crystal belongs, if its density is 7.86 g/cm and edge length of the cell is 286 pm (at wt of Fe 56).

Ans. Z ?d= 7.86 g/cm³, a=286 pm

M =56g N= 6.023x 1023

Now Z = dxa³xN/M

= 7.86× (286x 10–¹⁰)³x 6.023 x 10²³/56=2

Hence iron crystal belongs to bcc arrangement.

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