Question

. Account for the following:

(i) Schottky defects lower the density of related solids.

(ii) Conductivity of silicon increases on doping it with phosphorus.

Answer 1

(i) Schottky defect is a vacancy defect which is shown by ionic solids. In this defect, an equal number of cations and anions are missing from their sites to maintain electrical neutrality. This leads to decrease in the density of a substance.

(ii) Silicon contains four valence electrons in its valence shell. When doped with phosphorus which is a group 15 element. Phosphorus contains five valence electrons which occupy some of the lattice sites in silicon. Four out of five electrons are used in the formation of four covalent bonds with the four silicon atoms. The fifth electron is extra and becomes delocalised. These delocalised electrons increase the conductivity of doped silicon

Answer 2

Answer:

[ i.] If in an ionic crystal of type A+B-, an equal number of cations and anions are missing from their lattice sites so that electrical neutrality as well as stoichiometry is maintained this is called a Schottky Defect.

[ ii.] When numerous phosphorus atoms are substituted for silicon in crystal, many electrons are free which lead to increase in conductivity of silicon crystal.