Important Notes for Class 12 – Solid State

Revision Notes for Solid State 

  1. HCP and CCP have equal ef ciency i.e., 74% of space is occupied and coordination number is 12. CCP arrangement has FCC lattice.
  2. Coordination number is the number of nearest neighbouring points surrounding a particular lattice point (point may be atom, ions or molecules).
  3. Packing ef ciency in simple cubic unit cell is 52.4%, in bcc arrangement is 68% and in fcc is 74%.
  4.  Unoccupied spaces in solids are called interstitial voids or interstitial sites.
  5. Two types of interstitial voids are :
    (i) tetrahedral void
    (ii) octahedral void
    * No. of tetrahedral voids = 2  N (where N is number of closed packed
    particles).
    * No. of octahedral voids = N.
  6. Valency defect lowers the density of a crystal.
  7. Interstitial defect increases the density of a crystal.
  8. Point defects in the ionic crystal may be classified as :
    (i) Stoichiometric defect also known as intrinsic or thermodynamic defect. Ratio of cations and anions is the same in defective crystal as in ideal crystal.
    (ii) In non-stoichiometric defect ratio of cations to anions is the difference in
    defective crystal from the ideal crystal.
    (iii) Impurity defect (due to presence of some other ions at the lattice sites).
  9. Schottky defect arises due to missing of equal number of cations and anions from lattice sites in the crystalline solid of the type A+B- and it lowers and density of alkali metal halides, e.g., NaCl, KCl etc.
  10. Frenkel defect is the combination of vacancy and interstitial defects. Cations leave their actual lattice sites and occupy the interstitial space in the solid. Density remains the same in Frenkel defect.
  11. Non-stoichiometric defect
    (i) Metal excess defect due to anion vacancies.
    (ii) Metal excess defect due to presence of extra cations.
    (iii) Metal deciency due to absence of cations.
  12. F-Center : In metal excess defect, electrons are trapped in the anion vacancies which act as colour centres, e.g., NaCl gives yellow colour when heated in sodium vapour.
  13. Doping is the process of increasing the conductivity of intrinsic semiconductors  by adding an appropriate amount of suitable impurity in Si or Ge.
    * n-type semiconductors : Silicon or Germanium (group 14) doped with electron rich impurity (group 15 element like P or As). Here, conductivity is due to the extra electrons or delocalized electrons.
    * p-type semiconductors : Silicon or Germanium (group 14) doped with group 13 elements like B or Al. Here, conductivity is due to positively charged electron holes.
    * 13-15 group compounds, e.g., InSb, AlP, GaAs.
    * 12-16 group compounds, e.g., ZnS, CdS, CdSe, HgTe.
    * These compounds have average valence of four and are used in semiconductor devices.
  14. Magnetic Properties
    * Ferromagnetic substances : A few substances like iron, cobalt, nickel and CrO2 etc. are attracted very strongly by a magnetic eld. Such substances are called ferromagnetic substances. All molecular domains are arranged permanently in the same direction under inuence of magnetic field.
    * Antiferromagnetism : Substances like MnO showing anti ferromagnetism have domain structure similar to ferromagnetism substances, but their domains are oppositely oriented and cancel out each other ís magnetic moment and so cannot be attracted towards magnet.
    * Ferrimagnetism : When the magnetic moments of the domains in the substances are aligned in parallel and antiparallel directions in unequal number.

Important Questions – Solid State for Board Exams

Level 1 ( Short Answer type questions)

1. Define ‘forbidden zone’ of an insulator. ( 2010)
Ans. The large energy gap between valence band and conduction band in an insulator is called forbidden zone.

2. Which point defect in crystal units alters the density of a solid? ( 2009, 2010)
Ans.  Interstitial defect ( density increases)
Vacancy defect ( density decreases) e.g. Scottky defect

3. Crystalline solids are anisotropic in nature. Explain.
4. Frenkel defects are not found in alkali metal halides. Why?
5. How many lattice points are there in one unit cell of
a) fcc
b) bcc
c) simple cubic
Ans a) 14
b) 9
c) 8
6. What are the co-ordination numbers of octahedral voids and tetrahedral voids?
Ans. 6 and 4 respectively

Level 2 (2 MARKS QUESTIONS)

1. Explain how electrical neutrality is maintained in compounds showing Frenkel and Schottky defect.
Ans. In compound showing Frenkel defect, ions just get displaced within the lattice. While in compounds showing Schottky defect, equal number of anions and cations are removed from the lattice. Thus, electrical neutrality is maintained in both cases.

2. Calculate the number of atoms in a cubic unit cell having one atom on each corner and two atoms on each body diagonal.
Ans. 8 corner atoms × 1/8 atom per unit cell = 1 atom
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 atoms

3.The electrical conductivity of a metal decreases with rise in temperature while that of a semi-conductor increases.Explain.
Ans. 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 semiconductors, with increase of temperature, more electrons can shift from valence band to conduction band. Hence conductivity increases.

4. What type of substances would make better permanent magnets, ferromagnetic or ferromagnetic,Why?
Ans. 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.

5. In a crystalline solid, the atoms A and B are arranged as follows:
a. atoms A are arranged in ccp array
b. atoms B occupy all the octahedral voids and half of the tetrahedral voids.
What is the formula of the compound?

Ans: Let no. of atoms of A be N, No. of octahedral voids = N, No. of tetrahedral voids = 2N

i)There will be one atom of B in the octahedral void
ii)There will be one atom of B in tetrahedral void (1/2 X 2N) Therefore, total 2 atoms of B for each atom of A Therefore formula of the compound = AB2

LEVEL 3 (3 Marks questions)

7. An element E crystallizes in body centered cubic structure. If the edge length of the cell is 1.49 X 10-10 m and the density is 19.3g cubic cm , calculate the atomic mass of this element. Also calculate the radius of an atom of the element.

Ans ρ = Z x M /a3 x No

Common Mistake – Interconversión of units of edge length 1 pm = 10-12m = 10-10 cm
1Å =10-10m=10-8cm

2XM
19.3gcm-3 = (2xM) /  (1.469 X 10-8 cm)3 X ( 6.022 X 1023)

M = 18.42gmol-1

For a bcc structure , the radius of an atom is = √ 3/4 .a

= (1.732 X 1.489 X 10-10m)/4

= 6.36 X 10-11 m

8. Analysis shows that a metal oxide has the empirical formula M0.96O1.00. Calculate the percentage of M2+ and M3+ ions in this crystal.

OR

i) What type of substances show antiferromagnetism?
ii) Assign reason for the fact that silicon doped with phosphorus is a semiconductor.

Ans.  Suppose M2+ = X M3+= 96-X

Total charge on M2+ and M3+ is 2X + 3(96-X) = 2

X= 88
% of M2+ = 88/96 = 91.7%

% of M3+ = 100-91.7 = 8.3%

OR

i) Substances which are expected to possess paramagnetism or ferromagnetism on the basis of magnetic moments of the domains but actually possess zero net magnetic moment. e.g. MnO. It is due to the presence of equal number of domains in the opposite directions.

Si doped with P is an n-type semi-conductor because Si is tetravalent and P is pentavalent and one electron is left free and is not involved in bonding.

9. Calculate the density of silver which crystallizes in face centred cubic form. The distance between nearest metal atoms is 287pm.(molar mass of Ag=107.87gmol-1, No=6.022 x 1023)

Ans:d=Z×M/a3 ×NA
distance between nearest neighbor, 2r = 287pm therefore ,

a= √2x2r = 405.87pm, Z= 4 M=107.87gmol-1 No=6.022 x 1023 d=4×107.87gmol-1 / (405.87×10-10cm)3×6.022 x 1023mol-1

Density = 10.72g /cm-3

 

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