Chemistry Class Twelve

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Solid State

NCERT Exercise Solution (part 3)

Question – 1.19 - 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.

Solution:

Given,

Ferric oxide crystallizes in a hexagonal close packed array of oxide ions.

Two out of every three octahedral holes are occupied by ferric ions.

Let the number of oxide ions = x

Therefore number of octahedral voids = x

Since, two out of every three octahedral holes are filled by ferric ions,

Thus, voids filled by ferric ions

class 12 solid state ncert exercise solution24

Therefore, number of ferric ions

class 12 solid state ncert exercise solution24

Now, ratio of ferric ions to the oxide ions

class 12 solid state ncert exercise solution25

Thus, formula of ferric oxide is (Fe)22 O3

Question – 1.20 – Classify each of the following as being either a p-type or a n-type semiconductor:

(i) Ge doped with In (ii) Si doped with B.

Solution:

(i) Ge doped with In – ‘Ge’ belongs to group 14 in periodic table and ‘In’ belongs to group 13. Thus, when ‘Ge’ is doped with ‘In’, it makes hole or electron vacancy and acts as p-type of conductor.

(ii) Si doped with B – ‘Si’ belongs to group 14 in periodic table and ‘B’ belongs to group 13. Thus, when ‘Si’ is doped with ‘B’, it makes hole or electron vacancy and acts as p-type of conductor.

Question – 1.21 – Gold (atomic radius = 0.144 nm) crystallises in a face-centred unit cell. What is the length of a side of the cell?

Solution:

Given, atomic radius = 0.144 nm

Type of unit cell = face centered

class 12 solid state ncert exercise solution26

Thus side of the given cell = 407 nm

Question – 1.22 - In terms of band theory, what is the difference

(i) between a conductor and an insulator

(ii) between a conductor and a semiconductor?

Solution:

Molecular orbirals of metals are formed by atomic orbitals. These orbitals are so close to each other as they form band or valence band.

(i) Difference between conductor and insulator - In conductors there is no energy gap between the valence band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

While in insulators there is large energy gap between the valence band and electrons cannot jump to it i.e. large energy gap prevents the flow of electricity.

(ii)Difference between conductors and semiconductor - In conductors there is no energy gap between the valence band and conduction band, which facilitates the flow of electrons easily under an applied electric field and metals show conductivity.

While in semi conductors, there is small energy gap between valence bond and conduction band. The small gap between band facilitates some electrons to jump to the conduction band by acquiring extra energy.

Question – 1.23 - Explain the following terms with suitable examples:

(i) Schottky defect (ii) Frenkel defect (iii) Interstitials and (iv) F-centres.

Solution:

(i) Schottky defects : When cations and anions both are missing from regular sites, the defect is called Schottky Defect. In Schottky Defects, the number of missing cations is equal to the number of missing anions in order to maintain the electrical neutrality of the ionic compound.

Schottky Defect is type of simple vacancy defect and shown by ionic solids having cations and anions; almost similar in size, such as NaCl, KCl, CsCl, etc. AgBr shows both types of defects, i.e. Schottky and Frenkel Defects.

Since, Schottky Defects arises because of mission of constituent particles, thus it decreases the density of ionic compound.

(ii) It is a type of vacancy defect. In ionic compounds, some of the ions (usually smaller in size) get dislocated from their original site and create defect. This defect is known as Frenkel Defects. Since this defect arises because of dislocation of ions, thus it is also known as Dislocation Defects. As there are a number of cations and anions (which remain equal even because of defect); the density of the substance does not increase or decrease.

Ionic compounds; having large difference in the size between their cations and anions; show Frenkel Defects, such as ZnS, AgCl, AgBr, AgI, etc. These compounds have smaller size of cations compared to anions.

(iii) Interstitials – Sometime in the formation of lattice structure some of the atoms or ions occupy vacant interstitial site, and are known as interstitials. These interstitials are generally small size non-metals, such as H, B, C, etc. Defect arises because of interstitials is called interstitial defect.

(iv) F-centres – This is type of defect and called metal excess defect. These type of defects seen because of missing of anions from regular site leaving a hole which is occupied by electron to maintain the neutrality of the compound. Hole occupied by electron is called F-centre and responsible for showing colour by the compound.

Question – 1.24 - Aluminium crystallises in a cubic close-packed structure. Its metallic radius is 125 pm.

(i) What is the length of the side of the unit cell?

(ii) How many unit cells are there in 1.00 cm3 of aluminium?

Solution:

Given, radius of atom (r) = 125 pm

(i) For ccp structure, we know that

class 12 solid state ncert exercise solution27

(ii) Volume of 1 unit cell = a3

class 12 solid state ncert exercise solution28

Thus, number of unit cell of aluminium in 1 cm3

class 12 solid state ncert exercise solution29

Question – 1.25 - If NaCl is doped with 10–3 mol % of SrCl2, what is the concentration of cation vacancies?

Solution:

We know that two Na+ ions is replaced by each of the Sr++ ion while SrCl2 is doped with NaCl. But in this case only one lattice point is occupied by each of the Sr++ ion and produce one cation vacancy.

Here 10 – 3 mole of SrCl2 is dopped with 100 moles of NaCl

Thus, cation vacancies produced by NaCl = 10 – 3 mol

Since, 100 mole of NaCl produce cation vacancies after doping = 10 – 3 mol

Therefore, 1 mole of NaCl will produce cation vacancies after doping

class 12 solid state ncert exercise solution30

Question – 1.26 - Explain the following with suitable examples:

(i) Ferromagnetism

(ii) Paramagnetism

(iii) Ferrimagnetism

(iv) Antiferromagnetism

(v) 12-16 and 13-15 group compounds.

Solution:

(i) Ferromagnetism - Substances that are attracted strongly with magnetic field are called ferromagnetic substances, such as cobalt, nickel, iron, gadolinium, chromium oxide, etc. Ferromagnetic substances can be permanently magnetized also.

Metal ions of ferromagnetic substances are randomly oriented in normal condition and substances do not act as a magnet. But when metal ions are grouped together in small regions, called domains, each domains act like a tiny magnet and produce strong magnetic field, in such condition ferromagnetic substance act like a magnet. When the ordering of domains in group persists even after removal of magnetic field a ferromagnetic substance becomes a permanent magnet.

(ii) Paramagnetism - Substances which are attracted slightly by magnetic field and do not retain the magnetic property after removal of magnetic field are called paramagnetic substances. For example O2, Cu2+, Fe3+, Cr3+, Magnesium, molybdenum, lithium, etc.

Substances show paramagnetism because of presence of unpaired electrons. These unpaired electrons are attracted by magnetic field.

(iii) Ferrimagnetism - Substances which are slightly attracted in magnetic field and in which domains are grouped in parallel and anti-parallel direction but in unequal number, are called ferromagnetic substances and this property is called ferrimagnetism.

For example, magnetite (Fe3O4), ferrite (MgFe2O4), ZnFe2O4, etc.

Ferrimagnetic substances lose ferrimagnetism on heating and become paramagnetic.

(iv) Antiferromagnetism - Substances in which domain structure are similar to ferromagnetic substances but are oriented oppositely, which cancel the magnetic property are called antiferromagnetic substances and this property is called antiferromagnetism. For example; MnO.

(v) 12-16 and 13-15 group compounds –

Compounds belong to 12 – 16 group are formed by the combination of elements of 12 and 16 groups. For example – ZnS, Cds, etc.

Compounds belong to 13 – 15 group are formed by the combination of the elements of 13 and 15 groups. For example – InSb, GaAs, etc.

Bonds in these compounds are not perfectly covalent.

The ionic characters of the bonds in these compounds depends on the electronegativities of the two elements.


Previous            Next
Solid State - Introduction
Solid State - Crystal Lattice
Solid State - Close Packing Structure
Solid State - Efficiency of Close Packing Structure - 1
Solid State - Efficiency of Close Packing Structure - 2
Solid State - Imperfections in Solids - Crystal Defects
Solid State - Electrical and Magnetic Properties
Solid State - NCERT In Text Solution 1
Solid State - NCERT In Text Solution 2
Solid State - NCERT In Text Solution 3
Solid State - NCERT Exercise Solution (Part 1)
Solid State - NCERT Exercise Solution (Part 2)
Solid State - NCERT Exercise Solution (Part 3)
Solid State - NCERT Exercise Solution (Part 4)

 
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