Class 12 Chemistry

Solid State

NCERT Solution (part 3)

Question – 1.15 - Copper crystallises into a fcc lattice with edge length 3.61 `xx` 10–8 cm. Show that the calculated density is in agreement with its measured value of 8.92 g cm–3.

Answer: Given, edge length (a) = 3.61 `xx` 10-8cm

Given lattice is fcc

Hence, number of atoms per unit cell (z) = 4

Avogadro’s Number (NA = 6.022 `xx`1023 g mol-1

To prove, density (d) = 8.92 g cm-3

We know that density (d) `=(zM)/(a^3N_A)`

Or, `d=(4xx63.5g\text(mol)^(-1))/((3.61xx10^(-8))^3xx6.022xx10^(23)g\text(mol)^(-1))`

Or, `d=(254)/(47.0458xx10^(-24)xx6.022xx10^(23)cm^3)`

Or, `d=(254)/(283.30xx10^(-1)cm^3)`

Or, `d=(2540)/(283.30cm^3)`

Or, `d=8.9657` cm-3

Question – 1.16 - Analysis shows that nickel oxide has the formula Ni0.98O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions?

Answer: Given, formula of nickel oxide = Ni0.98O1.00

So, Ni : O = 0.98 : 1.00 = 98 : 100

Total charge on O-- = 100 `xx` (-2) = -200

Now, let number of Ni2+ `=x`

So, number of Ni3+ `=98-x`

Since compound is neutral

Hence, number of Ni2+ ions + number of Ni3+ + number of O2- ions = 0

Or, `x(-2)+(98-x)xx(-3)+(-200)=0`

Or, `-2x-294+3x-200=0`

Or, `x-94=0`

Or, `x=94`

So, number of Ni2+ ions = 94

Number of Ni3+ ions = 98-94=4

Now, fraction of nickel existing as Ni++ ions `=(94)/(98)=0.959`

Fraction of nickel existing as Ni3+ `=4/(98)=0.041`

Question – 1.17 - What is a semiconductor? Describe the two main types of semiconductors and contrast their conduction mechanism.

Answer: Semiconductor: - Solids having intermediate range of conductivity, i.e. from 10–6 to 104 ohm–1 m–1 are called semiconductors. Semiconductors are of following two types:

  1. n – type of semiconductors
  2. p – type of semiconductors

(i) n – type semiconductors – Semiconductors formed after doping with electron rich impurities to increase their conductivity are called n-type of semiconductors.

Example: Silicon and germanium, each has four valence electrons as they belong to 14th group of periodic table. Arsenic and phosphorous belong to 15th group of periodic table and they have valence electrons equal to 5. When silicon or germanium is doped with phosphorous or arsenic, four electrons of phosphorous or arsenic out of five; make covalent bonds with four electrons of silicon or germanium leaving one electron free; which increases the electrical conductivity of silicon or germanium.

class 12 solid state ncert exercise solution7C

Since the electrical conductivity of silicon or phosphorous is increased because of negatively charged particle (electron), thus this is known as n-type of semiconductor.

(ii) p – type of semiconductor - Semiconductors formed by the doping with electron deficient impurities; to increase their conductivity; are called p-type semiconductors. In p - type of semiconductors, conductivity increase because of formation of electron holes.

Example: Electrical conductivity of silicon or germanium is doped with elements, such as Boron, Aluminium or Gallium having valence electrons equal to 3. Three valence electrons present in these elements make covalent bonds with three electrons present in valence shell out of four of silicon or germanium leaving one electron delocalized. The place from where one electron is missing is called electron hole or electron vacancy.

When the silicon or germanium is placed under electrical field, electron from neighbouring atom fill the electron hole, but in doing so another electron hole is created at the place of movement of electron. In the influence of electrical filed electron moves toward positively charge plate through electron hole as appearing the electron hole as positively charged and are moving towards negatively charged plate.

Question – 1.18 - Non-stoichiometric cuprous oxide, Cu2O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor?

Answer: When cuprous oxide is prepared in laboratory; the ratio of copper to the oxygen in the compound becomes slightly less than 2:1. This happens because some of the Cu+ ions are replaced by the Cu2+ ions. In this process, one Cu2+ ions replaces two Cu+ ions. As two Cu+ ions are replaced by one Cu2+ ion, this creates defects because of creating vacant space, i.e. positive holes.

Because of creation of holes due to this defect; this compound conducts electricity through these positive holes.

As semiconductors which are formed by electron deficient impurities are called p-type of semiconductors; thus, cuprous oxides so formed in laboratory are the p – type of semiconductors.


NCERT Exercise Solution (Part 2)

NCERT Exercise Solution (Part 3)

NCERT Exercise Solution (Part 4)

NCERT Exercise Solution (Part 5)