Chemistry Class 12 NCERT Solutions: Chapter 1 The Solid State Part 9

Q: 17. What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Ans:

Semiconductors are substances having conductance in the intermediate range of Equation Equation .

The two main types of semiconductors are:

(i) n-type semiconductor

(ii) p-type semiconductor

n-type semiconductor:

The semiconductor whose increased conductivity is a result of negatively-charged electrons is called an n-type semiconductor. When the crystal of a group 14 element such as Equation or Equation is doped with a group 15 element such as P or As, an n-type semiconductor is generated

Equation have four valence electrons each. In their crystals, each atom forms four covalent bonds. On the other hand, P and As contain five valence electrons each. When Equation is doped with P or As, the latter occupies some of the lattice sites in the crystal. Four out of five electrons are used in the formation of four covalent bonds with four neighbouring Equation atoms. The remaining fifth electron becomes delocalised and increases the conductivity of the doped Equation .

p-type semiconductor:

The semiconductor whose increased in conductivity is a result of electron hole is called a p- type semiconductor. When a crystal of group 14 elements such as Equation is doped with a group 13 element such as Equation (which contains only three valence electrons), a p-type of semiconductor is generated.

When a crystal of Equation is doped with B, the three electrons of B are used in the formation of three covalent bonds and an electron hole is created. An electron from the neighbouring atom can come and fill this electron hole, but in doing so, it would leave an electron hole at its original position. The process appears as if the electron hole has moved in the direction opposite to that of the electron that filled it. Therefore, when an electric field is applied, electrons will move toward the positively-charged plate through electron holes. However, it will appear as if the electron holes are positively-charged and are moving toward the negatively- charged plate.

Q: 18. Non-stoichiometric cuprous oxide, Equation can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than Equation . Can you account for the fact that this substance is a p-type semiconductor?

Answer :

In the cuprous oxide Equation prepared in the laboratory, copper to oxygen ratio is slightly less than 2:1. This means that the number of Equation is slightly less than twice the number of Equation ions. This is because some Equation have been replaced by Equation ions. Every Equation ion replaces two Equation ions, thereby creating holes. As a result, the substance conducts electricity with the help of these positive holes. Hence, the substance is a p-type semiconductor.

Q: 19. Ferric oxide crystallises in a hexagonal close-packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.

Ans:

Let the number of oxide Equation ions be x.

So, number of octahedral voids Equation

It is given that two out of every three octahedral holes are occupied by ferric ions.

So, number of ferric Equation

Therefore, ratio of the number of Equation ions to the number of Equation ions

Equation

Equation

Equation

Hence, the formula of the ferric oxide is Equation .

Q: 20. Classify each of the following as being either a n – type semiconductor

(I) Ge doped with In

(II) B doped with Si

Ans:

(I) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.

(II) B (a group 13 element) is doped with Si (a group 14 element). Thus, a hole will be created and the semiconductor generated will be a p-type semiconductor.

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