What is a semiconductor? Describe two main types of semiconductors, giving examples and their distinctive features.

Semiconductors : These are the solids with conductivities in the intermediate range from 10–6 to 104 ohm–1m–1.

In case of semiconductors, the gap between the valence band and conduction band is small Therefore, some electrons may jump to conduction band and show some conductivity. Electrical conductivity of semiconductors increases with rise in temperature, since
more electrons can jump to the conduction band. Substances like silicon and germanium show this type of behaviour and are called intrinsic semiconductors.
The conductivity of these intrinsic semiconductors is too low to be of practical use. Their conductivity is increased by adding an appropriate amount of suitable impurity. This process is called doping.

There are two types of semiconductor
i) n- type semiconductor
ii) p- type semiconductor

Element of group14 of the periodic table, when doped with a group 15 element like P or As, Four out of five electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron is extra and becomes delocalised.

These delocalised electrons increase the conductivity of doped silicon (or germanium). Here the increase in conductivity is due to the negatively charged electron, hence silicon doped with electron-rich impurity is called n-type semiconductor.

p-type : 

Silicon or germanium can also be doped with a group 13 element like B, Al or Ga which contains only three valence electrons. The place where the fourth valence electron is missing is called electron hole or electron vacancy. This hole can be act as charge carrier in the lattice. This known as p - type semiconductor.



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Assign reasons the following:
(i) Phosphorus doped silicon is a semi-conductor.
(ii) Schottky defect lowers the density of a solid.
(iii) Some of the very old glass objects appear slightly milky instead of being transparent.

(i) When silicon is doped with phosphorus four valence electrons of phosphorus are involved in bond formation with the neighbouring silicon atoms, while the fifth valence electron is left free to conduct electricity. This type of conduction which arises due to the availability of excess electrons is called n-type conduction.

(ii) Schottky defect arises when equal number of cations and anions are missing from their lattice sites. As the number of ions decreases due to this defect In order to maintain electrical neutrality, the number of missing cations and anions are equal, the mass decreases, but the volume remains the same. As a result, the density of the solid decreases.

(iii) Some of the very old glass objects appear slightly milky instead of being transparentbecause of some crystallisation at that point.

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How the crystalline solids are classified on the basis of the nature of bonding? Give suitable examples and nature of the forces present in different types of solids.

i) A crystalline solid usually consists of a large number of small crystals, each of them having a definite Characteristic geometrical shape. In a crystal, the arrangement of constituent particles (atoms, molecules or ions) is ordered. It has long range order which means that there is a regular pattern of arrangement of particles which repeats itself periodically over the entire crystal. Sodium chloride and quartz are typical examples of crystalline solids.

ii) Crystalline solid have a sharp melting point.

Crystalline solids can be classified on the basis of nature of intermolecular forces operating in them into four categories.
i) molecular
ii)ionic
iii)metallic
iv)covalent solids.

i) molecular: Molecules are the constituent particles of molecular solids. These are further sub divided into the following categories:

a) Non polar Molecular Solids: They comprise of either atoms, for example, argon and helium or the molecules formed by non polar covalent bonds for example H2, Cl2.

b) Polar Molecular Solids: The molecules in which solids are held together by relatively stronger dipole-dipole interactions.for example HCl, SO2, etc.

c) Hydrogen Bonded Molecular Solids: The molecules of such solids contain polar covalent bonds between H and F, O or N atoms. Strong hydrogen bonding binds molecules of such solids like H2O (ice).

ii) Ionic solid : Ions are the constituent particles of ionic solids. In ionic solid cations and anions bound by strong coulombic (electrostatic) forces. for example NaCl, KCl etc.

iii) metallic solid: These solids contain metal atoms as constituent particles. As metals have a good tendency to lose their valence electron and change in to positively charged metal ions (kernel). These electrons can easily move throughout the whole crystal and form the sea of free electrons. for example iron, calcium etc.

iv) covalent solids: A wide variety of crystalline solids of non-metals result from the formation of covalent bonds between adjacent atoms throughout the crystal. They are also called giant molecules. for example diamond, silicon carbide, etc.



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An element E crystallizes in body centred cubic structure. If the edge length of the cell is 1.469 x 10–10 m and the density is 19.3 g cm–3, calculate the atomic mass of this element. Also calculate the radius of this element.

Solution:
We have given that 
Z= 2 since it is bcc structure
edge length, a = 1.469 x 10-10
Density, d= 19.3 g cm-3
Mass = ?

d=2×3a3×NA19.3 = 2×M(1.469×10-8)3 × 6.023 × 1023or M = 19.3×3.17×10-34×6.023×10232         = 19.3×3.17×6.023×102          = 18.4 g mol-1.

mass of element is 18.4g mol-1


(ii) Cell edge,

         a=1.469×10-10m

Body diagonal = 4R

Then 4R = 39
             = 3×1.469×10-10m
or     R = 3×1.469×10-104m    = 0.64 × 10-10m    = 6.4 × 10-11m.

Radius of element is 6.4 x 10-11m.
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Analysis shows that a metal oxide has the empirical formula M0.98 O1.00. Calculate the percentage of M2+ and M3+ ions in this crystal.

Or

In an ionic compound the anion (N) form cubic close type of packing. While the cation (M+) ions occupy one third of the tetrahedal voids. Deduce the empirical formula of the compound and the coordination number of (M) ions.


Solution:

M0.98 O1.00 is non-stoichiometric compound and is a mixture of M2+ and M3+ ions.
Let x atoms of M3+ are present in the compound. This means that x M2+ have been replaced by M3+ ions.
∴  Number of M2+ ions = 0.96 – x.
For electrical neutrality, positive charge on compound

= negative charge on compound.
∴ 2(0.96 – x) + 3x = 2
1.92 – 2x + 3x = 2
or x = 2 – 1.92 = 0.08

∴     % of M3+ ions  = 0.080.96×100 = 8.33%

Thus, M2+ is 92% and M3+ is 8% in given sample having formula, M0.96 O1.00.

Or

The number of tetrahedral voids formed is equal

to twice the number of atoms of element N and only 13rd of these are occupied by the element M. Hence the ratio of number of atoms of M and is 2×13:1  or  2:3.  So, the formula of the compound is M2N3.

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