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CLASS 12TH CHAPTER -9 Coordination Compounds |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry : The NCERT solutions provided here will enhance the concepts of the students, as well as suggest alternative methods to solve particular problems to the teachers. The target is to direct individuals towards problem solving strategies, rather than solving problems in one prescribed format. The links below provide the detailed solutions for all the Class 12 Chemistry problems.

Chemistry is much more than the language of Science. We have made sure that our solutions reflect the same. We aim to aid the students with answering the questions correctly, using logical approach and methodology. The NCERT Solutions provide ample material to enable students to form a good base with the fundamentals of the subject.

NCERT Solutions for Class 12 Chemistry Chapter :9 Coordination Compounds

INTEXT Questions

Question 1.
Write the formulas for the following coordination compounds :

Tetraamminediaquacobalt (lll) chloride
Potassium tetracyanonickelate (ll)
Tris(ethane-1,2-diamine) chromium (lll) chloride
Amminebromidochloridonitrito-N- platinate (ll)
Dichloridobis(ethane-1,2-diamine) platinum (IV) nitrate
Iron (lll) hexacyanoferrate (ll)

Solution:

Question 2.
Write the IUPAC names of the following coordination compounds :

Solution:

Hexaamminecobalt (III) chloride
Pentaamminechloridocobalt (III) chloride (HO₃)₂
Potassium hexacyanoferrate (III)
Potassium trioxalatoferrate (III)
Potassium tetrachloridopalladate (II)
Diamminechlorido (methylamine) platinum (II) chloride

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers :

Solution:

Question 4.
Give evidence that [CO(NH₃)5CI] SO₄ and [CO(NH₃)₅SO₄]Cl are ionisation isomers.
Solution:
The ionisation isomers dissolve in water to yield different ions and thus react differently to various reagents :

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCI₄]^2- ion with tetrahedral geometry is paramagnetic.
Solution:

Question 6.
[NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Solution:
In Ni(CO)₄, Ni is in zero oxidation state whereas in NiCl₄^2-, it is in +2 oxidation state. In the presence of CO ligand, the unpaired d-electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Solution:
In presence of CN^–, (a strong ligand), the 3d electrons pair up leaving only one unpaired electron. The hybridisation is d²sp³ forming inner orbital complex. In the presence of H₂O, (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³ d² forming an outer orbital complex containing five unpaired electrons hence, it is strongly paramagnetic.

Question 8.
Explain [CO(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Solution:
in the presence of NH₃, the 3d electrons pair up leaving two d orbitals empty to be involved in d² sp³ hybridisation forming inner orbital complex in case of [CO(NH₃)₆]³⁺.
In Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d³ configuration, the hybridisation involved is sp³ d² forming outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planer [Pt(CN)₄]^2- ion.
Solution:
For square planer shape, the hybridisation is dsp² . Hence, the unpaired electrons in 5d orbital pair up to make one f-orbital empty for dsp² hybridisation. Thus there is no unpaired electron.

Question 10.
The hexaaquomanganese (ll) ion contains five unpaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using Crystal Field Theory.
Solution:
Mn(II) has 3d⁵ electronic configuration. Water is a weak field ligand and therefore ∆₀ is small. Therefore, the hexaaqua complex will be high spin complex containing 5 unpaired electrons. On the other hand, CN^– is a strong field ligand and therefore, ∆₀ is large. Therefore, in its cyano complex, the electrons pair up and have only one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the Cu(NH₃)₄²⁺ion, given that β₄ for this complex is 2.1 × 10¹³.
Solution:

NCERT Exercises

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Solution:
The main postulates are :

In coordination compounds metals show two types of linkages (valencies)-primary and secondary.
The primary valencies are normally ionisable and are satisfied by negative ions.
The secondary valencies are non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valency is equal to the coordination number and is fixed for a metal.
The ions/groups bound by the secondary linkages to the metal have characteristic spatial .arrangements corresponding to different coordination numbers.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu²⁺ion. Explain why?
Solution:
FeSO₄ solution when mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio forms a double salt FeSO₄(NH₄)₂SO₄-6H₂O which when dissolved in water dissociates into simple ions to give tests for its constituent ions. When CuSO₄ is mixed with aqueous ammonia a complex ion [Cu(NH₃)₄]²⁺ is formed which does not give Cu²⁺ in the solution.

Question 3.
Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Solution:
(i) Coordination entity : A coordination entity constitutes a central metal atom or ion bonded loa fixed number of ions or molecules. For example, [COCl₃(NH₃)₃] is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are [Ni(CO)₄], [PtCl₂(NH₃)₂], [Fe(CN)₆]^4-, [CO(NH₃)₆]³⁺.

(ii) Ligand : The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. These may be simple ions such as Cl^–, small molecules such as H₂O or NH₃, larger molecules such as H₂NCH₂CH₂NH₂ or N(CH₂CH₂NH₂)₃ or even macromolecules, such as proteins.

(iii) Coordination number : The coordination number [C.N.] of a metal ion in a complex can be defined as the number of ligand or donor atoms to which the metal is directly bonded. For example, in the complex ions, [PtCl₆]^2- and [Ni(NH₃)₄]²⁺, the coordination number of Pt and Ni are 6 and 4 respectively. Similarly, in the complex ions, [Fe(C₂O₄)₃]^3- and [CO(en)₃]³⁺,the coordination number of both, Fe and Co, is 6 because C₂O₄^2-and en (ethane-1,2-diamine) are bidentate ligands.

(iv) Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom. The most common coordination polyhedra are octahedral, square planar and tetrahedral. For example, [CO(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄]^2- is square planar.

(v) Homoleptic and heteroleptic complexe : Complexes in which a metal is bound to only one kind of donor groups, e.g., [CO(NH₃)₆]³⁺, are known as homoleptic. Complexes in which a metal is bound to more than one kind of donor groups, e.g., [CO(NH₃)₄Cl₂]⁺, are known as heteroleptic.

Question 4.
What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each.
Solution:
Unidentate ligands are those which bind to the metal ion through a single donor atom, e.g., Cl^– , H₂O.

Bidentate ligands are those which bind to the metal ion through two donor atoms. e.g., ethane-1,2-diamine (H₂NCH₂CH₂NH₂), oxalate (C₂O₄^2-) ion.

Ambidentate ligands are those which can bind to metal ion through two different donor atoms, e.g., NO₂^– and SCN^– ion.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities :

Solution:

Question 6.
Using IUPAC norms write the formulas for the following :

Tetrahydroxozincate (ll)
Potassium tetrachloridopalladate (ll)
Diamminedichloridoplatinum (ll)
Potassium tetracyanonickelate (ll)
Pentaamminenitritio-O-cobalt (lll)
Hexaamminecobalt (lll) sulphate
Potassium tri(oxalato)chromate (lll)
Hexaammineplatinum (IV)
Tetrabromidocuprate (ll)
Pentaamminenitrito-N-cobalt (lll)

Solution:

Question 7.
Using IUPAC norms write the systematic names of the following :

Solution:

Hexaamminecobalt (III) chloride
Diamminechloridomethylamine platinum (II) chloride
Hexaaquatitanium (III) ion
Tetraamminechloridonitrito-N-cobalt (III) chloride
Hexaaquamanganese (II) ion
Tetrachloridonickelate (II) ion
Hexaammine nickel (II) chloride
Tris (ethane-1,2-diamine) cobalt(III) ion
Tetracarbonylnickel (0)

Question 8.
List various types of isomerism possible for coordination compounds, giving an example of each.
Solution:
(i) Geometrical isomerism : This type of isomerism arises in heteroleptic complexes due to different possible geometric arrangements of the ligands. Important examples of this behaviour are found with coordination numbers 4 and 6. In a square planar complex of formula [MX₂L₂] (X and L are unidentate), the two identical ligands may be arranged adjacent to each other in a cis- isomer, or opposite to each other in a trans isomer.

Another type of geometrical isomerism occurs in octahedral coordination entities of the type [Ma₃b₃] like [CO(NH₃)₃(NO₂)₃]. If three donor atoms of the same ligands occupy adjacent positions at the corners of an octahedral face, we have the facial (fac) isomer. When the positions are around the meridian of the octahedron, we get the meridional (mer) isomer.

(ii) Optical isomerism : Optical isomerism is common in octahedral complexes involving bidentate ligands. In a coordination entity of the [PtCl₂(en)₂]²⁺, only the ds-isomer shows optical activity.

(iii) Linkage isomerism : Linkage isomerism arises in a coordination compound containing ambidentate ligand. A simple example is provided by complexes containing the thiocyanate ligand, NCS^–, which may bind through the nitrogen to give M-N CS or through sulphur to give M-SCN. This behaviour was seen in the complex [CO(NH₃)₅(NO₂)]Cl₂, which is obtained as the red form, in which the nitrite ligand is bound through oxygen ( ONO), and as the yellow form, in which the nitrite ligand is bound through nitrogen (-NO₂).

(iv) Coordination isomerism : This type of isomerism arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. An example is provided by [Co(NH₃)₆] [Cr(CN)₆], in which the NH₃ ligands are bound to CO³⁺ and the CN^– ligands to Cr³⁺. In its coordination isomer [Cr(NH₃)₆][CO(CN)₆], the NH₃ ligands are bound to Cr³⁺ and the CN^– ligands to CO³⁺.

(v) Ionisation isomerism : This form of isomerism arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. An example is provided by the ionisation isomers [CO(NH₃)₅SO₄]Br and [CO(NH₃)₅Br]SO₄.

(vi) Solvate isomerism : This form of isomerism is known as ‘hydrate isomerism in case where water is involved as a solvent. This is similar to ionisation isomerism. Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. An example is provided by the aqua complex [Cr(H₂O)₆]Cl₃ (violet) and its solvate isomer [Cr(H₂O)₅Cl]Cl₂ H₂O (grey green).

Question 9.
How many geometrical isomers are possible in the following coordination entities?

[Cr(C₂O₄)₃]^3-
[CO(NH₃)₃CI₃]

Solution:

Zero
Two – Facial and meridional.

Question 10.
Solution:

Question 11.
Draw all the isomers (geometrical and optical) of :

Solution:

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(CI)(py)] and how many of these will exhibit optical isomerism?
Solution:

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :

a green precipitate with aqueous potassium fluoride and
a bright green solution with aqueous potassium chloride. Explain these experimental results.

Solution:

Question 14.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S_(g) is passed through this solution?
Solution:

CN^– is a strong field ligand, therefore, [Cu(CN)₄]^2- is highly stable and has large value of stability constant. On passing H₂S, CuS is not formed because this coordination entity does not give Cu²⁺ ion.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory :

Solution:

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Solution:
Let us assume that the six ligands are positioned symmetrically along the Cartesian axes, with metal atom at the origin.

As the ligands approach, first there is an increase in energy of d-orbitals relative to that of the free ion just as would be the case in a spherical field.

The orbitals lying along the axes (d_z2 and d_x2 – y₂) get repelled more strongly than d_xy, d_yz and d_zx orbitals which have lobes directed between the axes.

The d_z2 and d_x2-y2 orbitals get raised in energy and d_xy d_yz, d_xz orbitals are lowered in energy relative to the average energy in the spherical crystal field. Thus, the degenerate set of d-orbitals get split into two sets : the lower energy orbitals set, t_2g and the higher energy orbitals set, e_g. The energy is separated by ∆₀

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Solution:
The crystal field splitting, ∆₀, depends upon the field produced by the ligand and charge on the metal ion. Some ligands are able to produce strong fields in which, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below :

Such a series is termed as spectrochemical series.

If ∆₀ < P, the fourth electron enters one of the eg orbitals giving the configuration t₂³_g g e¹_g. Ligands for which ∆₀ < P are known as weak field ligands and form high spin complexes. If ∆₀ > P, it becomes more energetically favourable for the fourth electron to occupy a t₂_g orbital with configuration t₂_g ⁴e⁰_g. Ligands which produce this effect are known as strong field ligands and form low spin complexes.

Question 18.
What is crystal field splitting energy? How does the magnitude of ∆₀ decide the actual configuration of d-orbitals in a coordination entity?
Solution:

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Solution:

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Solution:
[Ni(H₂O)₆]2+has unpaired electrons due to weak H₂O ligands which absorb light from visible region and radiate complementary colour i.e., green whereas [Ni(CN)₄]^2- does not have any unpaired electron due to strong CN^– ligand, therefore, does not absorb light from visible region hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Solution:

Both ligands show different magnitude of crystal field splitting energy due to different nature hence, absorb different wavelengths ans show different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Solution:

The metal-carbon bond in metal carbonyls possess both s and p character. The M-C σ bond is formed by the donation of lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M-C π bond is formed by the donation of a pair of electrons from a filled d-orbital of metal into the vacant antibonding π* orbital of carbon monoxide. The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes :

Solution:

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state,electronicconfigurationandcoordination number. Also give stereochemistry and magnetic moment of the complex :

Solution:

Question 25.
What is meant by stability of the coordination compound in solution? State the factors which govern stability of complexes.
Solution:
The stability of a complex in solution refers to the degree of association between the two species involved in the state of equilibrium. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type :
then, larger the stability constant, the higher is the proportion of ML₄ that exists in solution. Free metal ions rarely exist in the solution so that M will usually be surrounded by solvent molecules which will compete with the ligand molecules, L and be successively replaced by them. For simplicity, we generally ignore these solvent molecules and write four stability constants as follows :

Factors affecting stability of complexes :

Smaller the size of cation, greater will be the stability of complex e.g., Fe³⁺ forms more stable complex than Fe²⁺.
Greater the charge on central metal ion, more stable will be the complex e.g., Pt⁴⁺ forms more stable complex than Pt²⁺.
Stronger the ligand, more stable will be the complex formed e.g., CN forms more stable complex then NH₃.

Question 26.
What is meant by the chelate effect1 Give an example.
Solution:
When a di-or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it is said to be a chelate ligand. The number of such ligating groups is called the denticity of the ligand. Such complexes, called chelate complexes tend to be more stable than similar complexes containing unidentate ligands. Example : EDTA, DMG, etc.

Question 27.
Discuss briefly giving an example in each case the role of coordination compound in :

biological systems
medicinal chemistry
analytical chemistry
extraction/metallurgy of metals.

Solution:
(i) Coordination compounds are of great importance in biological systems. The pigment responsible for photosynthesis, chlorophyll, is a coordination compound of magnesium. Haemoglobin, the red pigment 1 of blood which acts as oxygen carrier is a coordination compound of iron. Vitamin B₁₂, cyanocobalamine, the anti- pernicious anaemia factor, is a coordination compound of cobalt. Among the other compounds of biological importance with coordinated metal ions are the enzymes like, carboxypeptidase A and carbonic anhydrase (catalysts of biological systems).

(ii) There is growing Interest in the use of chelate therapy in medicinal chemistry. An example is the treatment of problems caused by the presence of metals in toxic proportions in plant/animal systems. Thus, excess of copper and iron are removed by the chelating ligands D-penicillamine and desferrioxime B via the formation of coordination compounds.

EDTA is used in the treatment of lead poisoning. Some coordination compounds of platinum effectively inhibit the growth of tumours. Examples are: ds-platin and related compounds.

(iii) Coordination compounds find use in many qualitative and quantitative chemical analysis. The familiar colour reactions given by metal ions with a number of ligands (especially chelating ligands), as a result of formation of coordination entities, form the basis for their detection and estimation by classical and instrumental methods of analysis. Examples of such reagents include EDTA, DMG (dimethylglyoxime), a-nitroso- β-naphthol, cupron, etc.

(iv) Some important extraction processes of metals, like those of silver and gold, make use of complex formation. Gold, for example, combines with cyanide in the presence of oxygen and water to form the coordination entity [Au(CN)2]- in aqueous solution. Gold can be separated in metallic form from this solution by the addition of zinc.

Question 28.
How many ions are produced from the complex CO(NH₃)₆CI₂ in solution?

(a) 6
(b) 4
(c) 3
(d) 2

Solution:

Question 29.
Amongst the following ions which one has the highest magnetic moment value?

(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [Zn(H₂O)₆]²⁺

Solution:

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(a) +1
(b) +3
(c) -1
(d) -3
Solution:
(c) : + 1 + x + 4(0) = 0
∴ x = – 1.

Question 31.
Amongst the following, the most stable complex is
(a) [Fe(H₂O)₆]³⁺
(b) [Fe(NH₃)₆]³⁺
(c) [Fe(C₂O₄)₃]^3-
(d) [FeCI₆]^3-
Solution:
(c) : Since C₂O₄^2-is a bidentate ligand, it forms the most stable complex.

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region for the following :

Solution:
In spectrochemical series the order of the given ligands is H₂O < NH₃ < NO₂– . Hence, the wave length of light will be absorbed in the opposite order since E = $\frac { hc }{ \lambda }$

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CLASS 12TH CHAPTER -8 d-and f-Block Elements |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter :8 d-and f-Block Elements

INTEXT Questions

Question 1.
Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
Solution:
Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence it is a transition element.

Question 2.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 kJ mol^-1. Why?
Solution:
In the formation of metallic bonds, no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d-series, electrons from the d-orbitals are always involved in the formation of metallic bonds.

Question 3.
Which of the 3d-series of the transition metals exhibits the largest number of oxidation states and why?
Solution:
Manganese (Z = 25), as its atom has the maximum number of unpaired electrons.

Question 4.
The E°(M²⁺/M) value for copper is positive (+0.34 V). What is possibly the reason for this? (Hint: consider its high ∆_aH° and low ∆_hydH°)
Solution:
The high energy to transform Cu_(s) to Cu²⁺_(aq)is not found balanced by its hydration energy. Hydration energy and lattice energy of Cu²⁺ is more than Cu.

Question 5.
How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Solution:
Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d – configurations (e.g., d°, d⁵, d¹⁰ are exceptionally stable).

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Solution:
Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state.

Question 7.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why?
Solution:
Cr²⁺ is stronger reducing agent than Fe²⁺ Reason: d⁴ → d³ occurs in case of Cr²⁺ to Cr²⁺. But d⁶ → d⁵ occurs in case of Fe²⁺ to Fe²⁺.
In a medium (like water) d’1 is more stable as compared to d⁵.

Question 8.
Calculate the ‘spin only’ magnetic moment of M²⁺_(aq)ion (Z = 27).
Solution:

Question 9.
Explain why Cu⁺ ion is not stable in aqueous solutions?
Solution:
Cu⁺ in aqueous solution undergoes disproportionation, i.e.,
2Cu⁺_(aq) → Cu²⁺_(aq) + Cu_(s)
The stability of Cu²⁺_(aq) rather than Cu⁺_(aq) is due to the much more negative ∆_hyd H°of Cu²⁺_(aq) than Cu⁺, which more than compensates for the second ionisation enthalpy of Cu.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Solution:
The 5f electrons are more effectively shielded from nuclear charge. In other words the 5f electrons themselves provide poor shielding from element to element in the seriest.

NCERT Exercises

Question 1.
Write down the electronic configuration of

Cr³⁺
Pm³⁺
Cu⁺
Ce⁴⁺
CO²⁺
Lu²⁺
Mn²⁺
Th⁴⁺

Solution:

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation of their +3 state?
Solution:
Mn²⁺ has an electronic configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵ 4s⁰. To go to Mn³⁺ state the electron has to be taken out from stable d⁵ orbital which is half filled and requires very high ionisation energy. In case of Fe²⁺ ion, the third electron is taken out from 3d⁶ configuration which results in more stable 3d⁵ configuration.

Question 3.
Explain briefly how +2 state becomes more and more stable in the first half of the first row With increasing atomic number the transition elements with increasing atomic numbers?
Solution:
With increasing atomic number the effective nuclear charge increases after losing two electrons from s-orbital. The ionic size decreases which results in more stability. The slability is less in the beginning due to too few electrons to lose or share.

Question 4.
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
Solution:
Sc³⁺ has stable electronic configuration (vacant d-orbital), therefore Sc³⁺ is more stable than Sc⁺.

Fe³⁺ is more stable than Fe²⁺ due to half filled d-orbitals.

Mn²⁺ is more stable than Mn³⁺ due to half filled d-orbitals.

V⁵⁺ is more stable (due to vacant d-orbital) than V³⁺.

Question 5.
What may be the stable oxidation state of the transition element with the following d-electron configurations in the ground state of their atoms: 3d³, 3d⁵, 3d⁸ and 3d⁴ ?
Solution:

3d³ : Stable oxidation state will be +5 due to outer electronic configuration 3d³4s² (+2, +3, +4, +5).
3d⁵: Stable oxidation state will be +2 and +7 due to outer electronic configuration 3d⁵4s² (+2, +3, +4, +6, +7).
3d⁸: Stable oxidation state will be +2 due to outer electronic configuration 3d⁸ 4s² (+2, +3, +4).
3d⁴ : Stable oxidation state will be +3 and +6 due to outer electronic configuration 3d⁴4s¹. There is no d⁴ configuration in ground state, as it becomes 3d⁵4s¹

Question 6.
Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
Solution:
Manganese shows oxidation state of +7 in its oxometal anion MnO₄^– which is equivalent to its group number 7. Cr in Cr₂O₇^2- and CrO₄^2- show oxidation state +6 which is equivalent to its group number 6.

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Solution:
The overall decrease in atomic and ionic radii from lanthanum to lutetium is a unique feature in the chemistry of the lanthanoids. It has far reaching consequences in the chemistry of the third transition series of the elements.

This contraction is attributed to the imperfect shielding of one electron by another in the same sub-shell. However, the shielding of one 4f electron by another is less than one d electron by another with the increase in nuclear charge along the series. There is fairly regular decrease in the sizes with increasing atomic number.

The cumulative effect of the contraction of the lanthanoid series, known as lanthanoid contraction, causes the radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. The almost identical radii of Zr (160 pm) and Hf (159 pm), a consequence of the lanthanoid contraction, account for their occurrence together in nature and for the difficulty faced in their separation.

Question 8.
What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?
Solution:
General characteristics of transition elements.

Electronic configuration – (n – 1) d^1-10 ns^1-2
Metallic character – With the exceptions of Zn, Cd and Hg, they have typical metallic structures.
Atomic and ionic size – Ions of same charge in a given series show progressive decrease in radius with increasing atomic number.
Oxidation state – Variable ; ranging from + 2 to + 7.
Paramagnetism – The ions with unpaired electrons are paramagnetic.
Ionisation enthalpy – Increases due to increase in molecular charge.
Formation of coloured ions – Due to unpaired electrons.
Formation of complex compounds – Due to small size and high charge density of metal ions.
They possess catalytic properties – Due to their ability to adopt multiple oxidation states.
Formation of interstitial compounds.
Alloy formation.

They are called transition elements due to their incompletely filled d-orbitals in ground state or any stable oxidation state and they are placed between s and p-block elements. Zn, Cd and Hg have fully filled d° configuration in their ground state hence may not be regarded as the transition elements.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?.
Solution:
The electronic configuration of the transition elements is (n – 1 )d^1-10 ns^1-2. (n-1) stands for penultimate shell and d-orbitals may have one to ten d electrons and n denotes valence s or the outermost shell which can have one or two electrons. Hence the basic difference in electronic configuration of transition metals is that their penultimate shell is incomplete and progressively filled and not the valence shell.

Question 10.
What are the different oxidation states exhibited by the lanthanoids?
Solution:
In lanthanoids +3 oxidation state is predominant. However occasionally +2 and +4 ions in the solution or in solid compounds are also obtained, e.g. Ce⁴⁺, Tb⁴⁺, Eu²⁺, Yb²⁺, etc. +2 and +4 oxidation states are exhibited due to extra stability of empty, half – filled or fully filled f – subshells.

Question 11.
Explain giving reasons :

Transition metals and many of their compounds show paramagnetic behaviour.
The enthalpies of atomisation of the transition metals are high.
The transition metals generally form coloured compounds.
Transition metals and their many compounds act as good catalysts.

Solution:
(i) Paramagnetism arises from the presence of unpaired electrons, each such electron has magnetic moment associated with its spin angular momentum and orbital angular momentum. For the compounds of the first series of transition metals, the contribution of the orbital angular momentum is effectively quenched and hence is of no significance. For these, the magnetic moment,is determined by the number of unpaired electrons and is calculated by using the ‘spin’ only’ formula, i.e., µ = $\sqrt { n(n+2) }$ B.M

Where n is the number of unpaired electrons and µ is the magnetic moment in units of Bohr magneton (BM). A single unpaired electron has a magnetic moment of 1.73 Bohr magneton (BM).

(ii) Because of large number of unpaired electrons in their atoms they have stronger interatomic interactions and hence stronger metallic bonding between atoms resulting in higher enthalpies of atomisation.

(iii) Due to presence of unpaired electrons and d-d transitions, the transition metals are generally coloured. When an electron from a lower energy d orbital is excited to a higher energy d-orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. The colour observed corresponds to the complementary colour of the light absorbed. The frequency of the light absorbed is determined by the nature of the ligand.

(iv) The transition metals and their compounds are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Vanadium(V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in r Catalytic Hydrogenation) are some of the examples.Catalysts at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst.

Question 12.
What are interstitial compounds? Why are such compounds well known for transitions metals ?
Solution:
Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of metals. They are usually non stoichiometric and are neither typically ionic nor covalent, for example, TiC, Mn₄N, Fe₃H, VH_0.56 and TiH1.7, etc.

These do not correspond to any normal oxidation state of the metal. Because of the nature of their composition, these compounds are referred to as interstitial compounds.The principal physical and chemical characteristics of these compounds are as follows :

They have high melting points, higher than those of pure metals.
They are very hard, some borides approach diamond in hardness.
They retain metallic conductivity
They are chemically inert.

Question 13.
How is the variability in oxidation states of transition metals different from that of the non-transition metals ? Illustrate with examples.
Solution:
The variability of oxidation states, a characteristic of transition elements, arises out of incomplete filling of d-orbitals in such a way that their oxidation states differ from each other by unity, e.g., V^II, V^III, V^IV, V^V. This is in contrast with the variability of oxidation sates of non transition elements where oxidation states normally differ by a unit of two.

In the p-block the lower oxidation states are favoured by the heavier members (due to inert pair effect), the opposite is true in the groups of d-block. For example, in group 6, Mo(VI) and W(VI) are found to be more stable than Cr(VI). Thus Cr(VI) in the form of dichromate in acidic medium is a strong oxidising agent, whereas MoO₃ and WO₃ are not.

Question 14.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Solution:
Potassium dichromate is prepared from chromate, which in turn is obtained by the fusion of chromite ore (FeCr2O4) with sodium or potassium carbonate in free access of air. The reaction with sodium carbonate occurs as follows :
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7-2H2O can be crystallised.
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Sodium dichromate is more soluble than potassium dichromate. The latter is therefore, prepared by treating the solution of sodium dichromate with potassium chloride.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Orange crystals of potassium dichromate crystallise out. The chromates and dichromates are interconvertible in aqueous solution depending upon pH of the solution. If pH of potassium dichromate is increased it is converted to yellow potassium chromate.
2CrO₄^2- + 2H⁺ → Cr₂O₇^2- + H₂O
Cr₂O₇^2- + 2OH^– → 2CrO₄^2- + H₂O

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with

iodide
iron(ll) solution and
H₂S.

Solution:
K₂Cr₂O₇ is a powerful oxidising agent. In dilute sulphuric acid medium the oxidation state of Cr changes from +6 to +3. The oxidising action can be represented as follows :

Question 16.
Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron (ii) ions (ii) SO₂ and (iii) oxalic acid ? Write the ionic equations for the reactions.
Solution:
Potassium permanganate (KMnO₄) is prepared by the fusion of a mixture of pyrolusite (MnO2), potassium hydroxide and oxygen, first green coloured potassium manganate is formed.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
The potassium manganate is extracted by water, which then undergoes disproportionation in neutral or acidic solution to give potassium permanganate.
Electrolytically :

Question 17.
For M²⁺/M and M³⁺/M²⁺ systems the P values for some metals are as follows
Use this data to comment upon

the stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺ and
the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Solution:

From the E° values for M³⁺/M²⁺ for Cr, Mn and Fe, it is very clear that Fe³⁺ is more stable than Mn³⁺ in an acidic medium, whereas less stable than Cr³⁺ because of the higher reduction potential in comparison to Cr³⁺/Cr²⁺ and lower reduction potential than Mn³⁺/Mn²⁺.
Reduction potential for Mn²⁺/Mn is most negative and therefore, it will be most easily oxidised and ease of getting oxidised will be Mn > Cr > Fe.

Question 18.
Predict which of the following will be coloured in aqueous solution? Ti³⁺, V³⁺, Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and CO³²⁺. Give reasons for each.
Solution:
The configuration of the given metal ions can be given as

Question 19.
Compare the stability of +2 oxidation state for the elements of the first transition series.
Solution:
The common oxidation state of 3d series elements is + 2 which arises due to participation of only 4s electrons. The tendency to show highest oxidation state increases from Sc to Mn, then decreases due to pairing of electrons in 3d subshell. Thus in the series Sc(II) does not exist, Ti(II) is less stable than Ti(IV). At the other end of the series, oxidation state of Zn is +2 only.

Question 20.
Compare the chemistry of actinoids with that of the lanthanoids with special reference to
(i) electronic configuration
(ii) atomic and ionic sizes
(iii) oxidation state and
(iv) chemical reactivity.
Solution:
(i)Electronicconfiguration : Lanthanoids have general electronic configuration of [Xe] 4f^1-14 5d^0-1 6s² and actinoids have general electronic configuration of [Rn]5f^1-14 6d^0-1 7s2. Thus, lanthanoids belong to 4 f-series whereas actinoids belong to 5 f-series.

(ii) Atomic and ionic sizes : The atomic size of lanthanoids decreases from lanthanum to lutetium. Though the decrease is not regular, in case of atomic radii, the decrease in the ionic size (M³⁺) is regular. Decrease in size between two successive elements is higher in actinoids due to poor screening by 5f electrons.

(iii) Oxidation state : The most common oxidation state of lanthanoids is +3 while actinoids show more variable oxidation states than lanthanoids ranging from +3 to +7. The tendency of showing greater range of oxidation states can be attributed to the fact that the 5f 6d and 7s levels are of comparable energies and larger distance of 5fas compared to 4f from the nucleus.

(iv) Chemical reactivity : Actinoids are far more reactive than lanthanoids. They react with non-metals at moderate temperatures whereas lanthanoids react at high temperatures. Most actinoids are attacked by HCl but are slightly affected by HNO₃ due to formation of a protective layer of oxide and alkalies give no reaction.

Lanthanoids liberate hydrogen from dilute acids and burn in halogens to form halides.

Question 21.
How would you account for the following?
(i) Of the d⁴ species, Cr²⁺ is strongly reducing while manganese(lll) is strongly oxidising.
(ii) Cobalt(ll) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised.
(iii) The d¹ configuration is very unstable in ions.
Solution:
(i) Both Cr2+ and Mn3+ have d4 configuration, Cr2+ is reducing since its configuration is converted to d3 from d4. d3 has half filled t28, configuration with higher stability. Mn3+ is oxidising since in changing from d4 to d5 the configuration becomes half filled which has extra stability,

(ii) Co(II) gets oxidised to Co(III) in presence of complexing agent because Co(III) is more stable than Co(II). Most of the strong field ligands cause pairing of electrons forming diamagnetic octahedral complexes which are very stable due to very large crystal field stabilization energy.

(iii) d¹ configuration is very unstable in ions because after losing one more electron it will become more stable due to vacant d-orbital. All elements with d¹ configuration are either reduced or undergo disproportionation, e.g.,

Question 22.
What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.
Solution:
Disproportionation reaction involves the oxidation and reduction of the same substance. The two examples of disproportionation reaction are

2Cu⁺ → Cu²⁺ + Cu
3MnO₄^2- + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O

Question 23.
Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?
Solution:
Copper exhibits +1 oxidation state in the first series of transition metals because when one electron is lost, the configuration becomes stable due to fully filled d¹⁰ configuration.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions : Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which ono of these is the most stable in aqueous solution ?
Solution:

Question 25.
Give examples and suggest reasons for the following features of the transition metal chemistry :
(i) The lowest, oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Solution:
(i) Lowest oxidation compounds of transition metals are basic due to their ability to get oxidised to higher oxidation states. Whereas the higher oxidation state of metal and compounds gets reduced to lower ones and hence acts as acidic in nature.
e.g. MnO is basic whereas Mn₂O₇ is acidic.

(ii) Due to high electronegativities of oxygen and fluorine, the oxides and fluorides of transition metals exhibit highest oxidation state.
e.g. OSF₆,V₂O₅

(iii) In oxoanions of metals, the metals from bonds with oxygen and hence are present in their highest oxidation states. For example : Cr forms CrO₄^2- and Cr2O₇^2-, both contain chromium in +6 oxidation state.

Permanganate ion, MnO₄^– contains Mn in its highest oxidation state of +7.

Question 26.
Indicate the steps in the preparation of

K₂Cr₂O₇ from chromite ore
KMnO₄ from pyrolusite ore.

Solution:

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Solution:
An allov is a blend of metals prepared by mixing the components. Alloys are homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of other. Misch metal is an alloy which contains some of the lanthanoid metals. It contains 95% lanthanoid metals, 5% iron and traces of S, C, Ca and Al. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint.

Question 28.
What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers oftheinnertransition element: 29, 59, 74, 95,102,104.
Solution:
Lanthanoids and actinoids are called inner transition elements because inner f-orbitals are progressively filled and the last electron goes to anti penultimate f orbital.

Elements with atomic number 59, 95, 102 are inner transition metals because they belong to lanthanoids and actinoids.

Question 29.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Solution:
The actinoids are radioactive elements and the earlier members have relatively long half-lives, the latter ones have half-life values ranging from a day to 3 minutes for lawrencium (Z = 103). The latter members could be prepared only in nanogram quantities. These facts render their study more difficult.

There is a greater range of oxidation states, which is attributed to the fact that the 5f, 6d and 7s levels are of comparable energies. The actinoids show in general +3 oxidation state. The elements, in the first half of the series frequently exhibit higher oxidation states. For example, the maximum oxidation state increases from +4 in Th to +5, +6 and +7 respectively in Pa, U and Np but decreases in succeeding elements, The actinoids resemble the lanthanoids in having more compounds in +3 state than in the +4 state. However, +3 and +4 ions tend to hydrolyse. Because the distribution of oxidation states among the actinoids is so uneven and so different for the earlier and latter elements. It is unsatisfactory to review their chemistry in terms of oxidation states.

Question 30.
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.
Solution:
Lawrencium (Lr) is the last element of actinoids. Its outer electronic configuration is 5f¹⁴ 6d¹ 7s² and its possible oxidation state is +3.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce³⁺ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.
Solution:
The electronic configuration of Ce³⁺ is 4f¹

Question 32.
Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behaviour with the electronic configurations of these elements.
Solution:
All lanthanoids show Ln³⁺ oxidation state. But some of them show +2 and +4 states also in solution and solid form like Ce⁴⁺, Eu²⁺, Yb²⁺, Tb⁴⁺, etc. The variable oxidation state is related to electronic configuration due to extra stability of half filled, fully filled or empty orbitals, e.g., Ce⁴⁺ has 4f⁰, Eu²⁺ has 4f⁷ Tb⁴⁺ has 4f⁷ and Yb²⁺ has 4f¹⁴ configuration.

Question 33.
Compare the chemistry of the actinoids with that of lanthanoids with reference to (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity.
Solution:
Refer answer number 20.

Question 34.
Write the electronic configurations of the elements with the atomic numbers 61,91,101, and 109.
Solution:

Question 35.
Compare the general characteristics of the first series of transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points
(i) electronic configurations
(ii) oxidation states
(iii) ionisation enthalpies and
(iv) atomic sizes.
Solution:
(i) Electronic configurations : In 1^st transition series, 3d-orbitals are progressively filled whereas in 2^nd transition series, 4d-orbitals are progressively filled and in 3^rd transition series, 5d-orbitals are progressively filled.

(ii) Oxidations states : Elements show variable oxidation states in both the series. The highest oxidation state is equal to total number of electrons in ‘s’ as well as ‘d’ orbitals. The number of oxidation states shown are less in 5d transition series than 4d series. In 3d series +2 and +3 oxidation states are common and they form stable complexes in these oxidation states. In other series OsO₄ and PtF₆ are formed which are quite stable in higher oxidation state.

(iii) Ionisation enthalpies : The ionisation enthalpies in each series generally increases gradually from left to right. The ionisation enthalpy of 5d series do not differ appreciably due to lanthanoid contraction.

(iv) Atomic sizes : The atomic sizes of 4d and 5d-series do not differ appreciably due to lanthanoid contraction. The atomic radii of second and third series are larger than 3d series.

Question 36.
Write down the number of 3d electrons in each of the following ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, CO²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Solution:

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Solution:

In first transition series lower oxidation state is more stable whereas in heavier transition elements higher oxidation states are more stable.
The ionisation enthalpy of 5d transition series is higher than 3d and 4d transition series.
M-M bonding is most common in heavier transition metals but less in first series.
The elements of first transition series do not form complexes with higher coordination number of 7 and 8.
The elements of first series can form high spin or low spin complexes depending upon strength of ligands but elements of other series form low spin complexes irrespective of strength of ligands.

Question 38.
What can be inferred from the magnetic moment values of the following complex species?

Solution:

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CLASS 12TH CHAPTER -7 The p-Block Elements |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter :7 The p-Block Elements

INTEXT Questions

Question 1.
Which of the ores mentioned in Table 6.1 (NCERT Textbook) can be concentrated by magnetic separation method?
Solution:
Ores in which one of the components (either the impurity or the actual ore) is magnetic can be concentrated by magnetic separation, e.g., ores containing iron (haematite, magnetite, siderite and iron pyrites).

Question 2.
What is the significance of leaching in the extraction of aluminium?
Solution:
Leaching is significant as it helps in removing the impurities like SiO₂, Fe₂O₃, etc. from the bauxite ore.

Question 3.
The reaction, Cr₂O₃ + 2Al → Al₂O₃ + 2cr (∆G° = – 421 kJ) is thermodynamically feasible as is apparent from the Gibbs energy value. Why does it not take place at room temperature?
Solution:
Certain amount of activation energy is essential even for such reactions which are thermodynamically feasible, therefore heating is required.

Question 4.
Is it true that under certain conditions. Mg can reduce Al₂O₃ and Al can reduce MgO? What are those conditions?
Solution:
Yes, below 1350°C, Mg can reduce Al₂O₃ and above 1350°C, Al can reduce MgO. This can be inferred from ∆G° vs T plots.

NCERT Exercises

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Solution:
The E° of zinc (Zn²⁺/Zn = – 0.76 V) is lower luau that of copper (Cu²⁺/Cu = +0.34 V). Hence, zinc can displace Cu from solutions of Cu²⁺ ions.

But all these metals react with water forming their corresponding ions with the evolution of H₂ gas. Hence, Al, Mg, etc. cannot be used to displace zinc from the solution of Zn²⁺ ions. Thus, copper can be extracted by hydrometal-lurgy but not zinc.

Question 2.
What is the role of depressant in froth-floatation process?
Solution:
In froth -floatation process, the role of the depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. For example, NaCN is used as a depressant to separate lead sulphide (PbS) ore from zinc sulphide (ZnS) ore. The reason is that NaCN forms a zinc complex, Na₂[Zn(CN)₄] on the surface of ZnS preventing it from the formation of the froth. Under these conditions, only PbS forms froth and therefore, it can be separated from ZnS ore.

Question 3.
Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction?
Solution:
In the Ellingham diagram, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (as can be seen in graph that the lines of C, CO and C, CO₂ are at much lower positions). But most of the ores are sulphide and some may also contain iron. So, the sulphide ores are roasted/smelted to give oxides :

Question 4.
Explain : (i) Zone refining (ii) Column chroma-tography.
Solution:
(i) This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. A circular mobile heater is fixed at one end of a rod of the impure metal. The molten zone moves along with the heater which is moved forward. As the heater moves forward, the pure metal crystallises out of the melt and the impurities pass on into the adjacent molten zone. The process is repeated several times and the heater is moved in the same direction. At one end, impurities get concentrated. This end is cut off. This method is very useful for producing semiconductor and other metals of very high purity, c.g., germanium, silicon, boron, gallium and indium.

(ii) Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography etc.

Column chromatography : Column chroma-tography involves separation of a mixture over a column of adsorbent (stationary phase) packed in a glass tube. The column is fitted with a stopcock at its lower end (fig). The mixture adsorbed on adsorbent is placed on the top of the adsorbent column packed in a glass tube.

An appropriate eluant which is a liquid or a mixture of liquids is allowed to flow the column slowly. Depending upon the degree to which the compounds are adsored, complete separation takes place. The most readily adsorbed substances are retained near the top and others come down to various distance in the column.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Solution:
When carbon acts as a reducing agent, it is either converted into CO or CO₂ or both.
2C + O₂ → 2CO
C + O₂ → CO₂
CO is oxidised to CO₂ when it is used as a reducing agent.
2CO + O₂ → 2CO₂
From the Ellingham diagram (refer answer number 3), it is clear that at the temperature 673 K, the AG° of the formation of CO₂ from CO is more negative than the formation of CO or CO₂ from carbon. Hence, at temperature 673 K, CO is a better reducing agent than C.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Solution:
Many of the metals such as copper, silver, gold, aluminium, lead, etc., are purified by this method. This is perhaps the most important method. The impure metal is made anode while a thin sheet of pure metal acts as a cathode. The electrolytic solution consists of generally an aqueous solution of a salt or a complex of the metal. On passing the current, the pure metal is deposited on the cathode and equivalent amount of the metal gets dissolved from the anode. Thus, the metal is transferred from anode to cathode through solution. The soluble impurities pass into the solution while the insoluble one, especially less electropositive impurities collect below the anode as anodic mud or anode sludge. Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum.

Question 7.
Write down the reactions taking place in different zones in the blast furnace during the extraction of iron.
Solution:
Near the bottom of the furnace (zone of combustion, 2170 K), coke first combines with air to form CO₂ which then combines with more coke (zone of heat absorption, 1423 K) to form CO. The CO thus produced acts as the reducing agent and reduces iron oxide to spongy iron near the top of the furnace (zone of reduction, 823 K).

At the lower part of the furnace (zone of fusion, 1423-1673 K) the spongy iron melts and dissolves some carbon, S, P, SiO₂, Mn, etc.

The molten slag being less dense floats over the surface of the molten iron. The molten iron is then tapped off from the furnace and is then solidified to give blocks of iron called cast iron or pig iron.

Question 8.
Write chemical reactions taking place in the extraction of zinc from zinc blende.
Solution:
Concentration : When zinc blende is used, the powdered ore is concentrated by froth-floatation process.

Roasting : The concentrated ore is heated in excess of oxygen at about 900°C. Zinc sulphide is oxidised to zinc oxide. If some of the ore is oxidised to zinc sulphate, it also decomposes at 900°C into ZnO.

For roasting, a reverberatory furnace may be used.

Reduction : The principal reaction that takes place during reduction is the conversion of the oxide into the metal with the help of carbon.

Electrolytic refining : Purification of zinc is done by electrolytic refining using pure Zn as cathode and impure Zn as anode. The electrolyte is ZnSO₄.

Reaction at cathode : Zn2⁺_(aq)+ 2e^– → Zn_(s)
Reaction at anode : 2H₂O_(l) → O_2(g)+ 4H⁺_(aq)+ 4e^–
ZnSO₄ electrolyte is added from time to time.

Question 9.
State the role of silica in the metallurgy of copper.
Solution:
Iron present in pyrites has greater affinity for oxygen than copper. The copper oxide formed reacts with unchanged iron sulphide to form iron oxide so, most of the iron sulphide is oxidised to ferrous oxide.
2FeS + 3O₂ → 2FeO + 2sO₂
Ferrous oxide combines with silica which acts as flux and forms ferrous silicate. By this reaction most of the iron is removed as slag.

Question 10.
What is meant by the term “chromatography”?
Solution:
Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Solution:
The stationary phase is selected in such a way that the impurities are more strongly adsorbed or are more soluble in the stationary phase than element to be purified. Under these conditions, when the column is extracted, the impurities will be retained by the stationary phase whereas the pure component is easily removed.

Question 12.
Describe a method for refining nickel.
Solution:
Nickel is purified by Mond’s process. Impure nickel is treated with carbon monoxide at 60-80°C when volatile compound nickel carbonyl is formed. Nickel carbonyl decomposes at 180°C to form pure nickel and carbon monoxide.

Question 13.
How can you separate alumina from silica in a bauxite ore associated with silica? Give equations, if any.
Solution:
Serpeck’s process is used when silica is present in considerable amounts in bauxite ore. The ore is mixed with coke and heated at 1800°C in presence of nitrogen, where AIN is formed.
Al₂O₃ + 3C + N₂ → 2AlN + 3CO
Silica is reduced to silicon which volatilises off at this temperature.
SiO₂ + 2C → Si + 2CO

Question 14.
Giving examples, differentiate between ‘roasting’ and ‘calcination’.
Solution:
Calcination : It is the process of converting an ore into its oxide by heating it strongly below its melting point either in absence or limited supply of air.

Roasting : In roasting, the ore is heated in a regular supply of air in a furnace at a temperature below the melting point of the metal.
2 ZnS + 3O₂ → 2ZnO + 2SO₂
2 PbS + 3O₂ → 2PbO + 2SO₂

Question 15.
How is’cast iron’different from ‘pig iron’?
Solution:
The iron obtained from Blast furnace contains about 4% carbon and many impurities in smaller amount (e.g., S, P, Si, Mn). This is known as pig iron and cast into variety of shapes.

Cast iron is different from pig iron and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.

Question 16.
Differentiate between “minerals”and “ores’.
Solution:
The natural substances in which the metal or their compounds occur in the earth are called minerals. The mineral has a definite composition. It may be a single compound or complex mixture. The minerals from which the metals can be conveniently and economically extracted are known as ores. All the ores arc minerals but all minerals cannot be ores, e.g., both bauxite (Al₂O₃.xH2O) and clay (Al₂O₃.2SiO₂.2H₂O) are minerals of aluminium. It is bauxite which is used for extraction of aluminium and not clay. Thus bauxite is an ore of aluminium.

Question 17.
Why copper matte is put in silica lined converter ?
Solution:
Copper matte consists of Cu2S and FeS. Wheia a blast of hot air is passed through molten matte taken in a silica lined converter, FeS present in matte is oxidised to FeO which combines with silica (SiO2) to form FeSiO3 (slag).

When complete iron has been removed as slag, some of the Cu2S undergoes oxidation to form Cu2O which then reacts with more Cu2S to form copper metal.
2CU₂S + 3O₂ → 2CU₂O + 2SO₂
2CU₂O + Cu₂S → 6Cu + SO₂
Thus, copper matte is heated in silica lined converter to remove FeS present in matte as FeSiO₃ (slag).

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Solution:
The role of cryolite is as follows :

It makes alumina a good conductor of electricity.
It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Question 19.
How is leaching carried out in case of low grade copper ores?
Solution:
The leaching of the low grade copper ores is carried out with acids in the presence of air when copper goes into solution as Cu²⁺ ions. Therefore,

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Solution:
The standard free energy of formation (∆_fG°) of CO₂ from CO is higher than that of the formation of ZnO from Zn. Hence, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Solution:

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Solution:
The free energy of formation (∆_f G°) of CO from C becomes lower at temperatures above 1120 K whereas that of CO₂ from C becomes lower above 1323 K than ∆_fG° of ZnO. However, ∆_fG° of CO₂ from CO is always higher than that of ZnO. Therefore, C can reduce ZnO to Zn but not CO. Therefore, out of C and CO, C is a better reducing agent than CO for ZnO.
Reduction of ZnO is usually carried out around

Question 23.
The choice of a reducing agent in a particular case depends on thermodynamic factor. How far do you agree with this statement? Support your opinion with two examples.
Solution:
Some basic concepts of thermodynamics lielp us in understanding the theory of metallurgical transformations. Gibb’s energy is the most significant term.
The graphical representation of Gibb’s energy was first used by H.J.T. Ellingham. This
provides a sound basis for considering the k choice of reducing agent in the reduction of oxides. This is known as Ellingham diagram. (Tor diagram refer answer number 3) Such diagrams help us in predicting the feasibility of thermal reduction of an ore. The criterion of feasibility is that at a given temperature Gibb’s energy of the reaction must be negative.
Examples :
(i) Thermodynamics helps us to understand how coke reduces the oxide and why blast furnace is chosen. One of the main reduction steps in this process is :
FeO_(s) + C_(s) → Fe_(s/l) + CO_(g)… (1)
It can be seen as a couple of two simpler reactions. In one, the reduction of FeO is taking place and in the other, C is being oxidised to CO :

When both the reactions take place to yield the equation (1), the net Gibb’s energy change becomes :

Naturally, the resultant reaction will take place when the right hand side in equation (3) is negative. In ∆G° vs T plot representing reaction 2, the plot goes upward and that representing the change C → CO (C, CO) goes downward. At temperatures above 1073 K (approx), the C, CO line comes below the Fe, FeO line [∆G_(c,co) < ∆G_(Fe,FeO)] So in this range, coke will be reducing the FeO and will itself be oxidised to CO. In a similar way the reduction of Fe₃O₄ and Fe₂O₃ at relatively lower temperatures by CO can be explained on the basis of lower lying points of intersection of their curves with the CO, CO₂ curve in Ellinghan diagram.

(ii) In the graph of ∆r.G° vs T for formation of oxides, the Cu₂O line is almost at the top. So it is quite easy to reduce oxide ores of copper directly to the metal by heating with coke (both the lines of C, CO and C, CO₂ are at much lower position in the graph particularly after 500 – 600 K). However most of the ores are sulphide and some may also contain iron. The sulphide ores are roasted/smelted to give oxides:
2CU₂S + 3O₂ → 2Cu₂O + 2SO₂
The oxide can then be easily reduced to metallic copper using coke.
Cu₂O + C → 2Cu + CO

Question 24.
Name the processes from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCI is subjected to electrolysis?
Solution:
Sodium metal is prepared by Down’s process. This process involves the electrolysis of a fused mixture of NaCI and CaCl₂ at 873 K. During electrolysis, sodium is discharged at the cathode and Cl₂ is obtained at the anode as a by-product.

Question 25.
What is the role of graphite rod in the electrometallurgy of aluminium?
Solution:
In this process, a fused mixture of alumina, cryolite and fluorspar (CaF₂) is electrolysed using graphite as anode and steel as cathode. During electrolysis, Al is liberated at the cathode whereas CO and CO₂ are liberated at the anode.

If some other metal is used as the anode other than graphite, then O₂ liberated will not only oxidise the metal of the electrode but would also convert some of the Al liberated at the cathode back to Al₂O₃. Since graphite is much cheaper than any metal, graphite is used as the anode. So, the role of graphite in electrometallurgy of Al is to prevent the liberation of O₂ at the anode which may otherwise oxidise some of the liberated Al back to Al₂O₃.

Question 26.
Outline the principles of refining of metals by the following methods :

Zone refining
Electrolytic refining
Vapour phase refining

Solution:
(i) Refer answer number 4(i).

(ii) In this method, the impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud.
Anode : M → Mⁿ⁺ ne^–
Cathode : Mⁿ⁺ + ne^– → M
Copper is refined using an electrolytic method. Anodes are of impure copper and pure copper strips are taken as cathode. The electrolyte is acidified solution of copper sulphate and the net result of electrolysis is the transfer of copper in pure form from the anode to the cathode :
Anode : Cu → Cu²⁺ + 2e^–
Cathode : Cu²⁺ + 2e^– → Cu
Impurities from the blister copper deposit as anode mud which contains antimony, selenium, tellurium, silver, gold and platinum; recovery of these elements may meet the cost of refining.
Zinc may also be refined this way.

(iii) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. So, the two requirements are :

(a) The metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, so that the recovery is easy.

Following example will illustrate this technique.
Mond process for Refining Nickel : In this process, nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl :

The carbonyl is subjected to higher temperature so that it is decomposed giving the pure metal :

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Solution:
The two equations are :

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CLASS 12TH CHAPTER -6 General Principles and Processes of Isolation of Elements |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter :6 General Principles and Processes of Isolation of Elements

INTEXT Questions

NCERT Exercises

At the lower part of the furnace (zone of fusion, 1423-1673 K) the spongy iron melts and dissolves some carbon, S, P, SiO₂, Mn, etc.

Reduction : The principal reaction that takes place during reduction is the conversion of the oxide into the metal with the help of carbon.

Electrolytic refining : Purification of zinc is done by electrolytic refining using pure Zn as cathode and impure Zn as anode. The electrolyte is ZnSO₄.

Reaction at cathode : Zn2⁺_(aq)+ 2e^– → Zn_(s)
Reaction at anode : 2H₂O_(l) → O_2(g)+ 4H⁺_(aq)+ 4e^–
ZnSO₄ electrolyte is added from time to time.

Question 10.
What is meant by the term “chromatography”?
Solution:
Chromatographic method is based on the principle that different components of a mixture are differently adsorbed on an adsorbent. The adsorbed components are removed (eluted) by using suitable (eluant). There are several chromatogrpahic techniques such as paper chromatography, column chromatography, gas chromatography, etc.

Cast iron is different from pig iron and is made by melting pig iron with scrap iron and coke using hot air blast. It has slightly lower carbon content (about 3%) and is extremely hard and brittle.

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Solution:
The role of cryolite is as follows :

It makes alumina a good conductor of electricity.
It lowers the fusion temperature of the bath from 2323 K to about 1140 K.

Question 21.
The value of ∆_fG° for formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ possible with Al?
Solution:

Question 26.
Outline the principles of refining of metals by the following methods :

Zone refining
Electrolytic refining
Vapour phase refining

Solution:
(i) Refer answer number 4(i).

(iii) In this method, the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal. So, the two requirements are :

(a) The metal should form a volatile compound with an available reagent.
(b) The volatile compound should be easily decomposable, so that the recovery is easy.

Question 27.
Predict conditions under which Al might be expected to reduce MgO.
Solution:
The two equations are :

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CLASS 12TH CHAPTER -5 Surface Chemistry |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter :5 Surface Chemistry

INTEXT Questions

Question 1.
Why are substances like platinum and palladium often used for carrying out electrolysis of aqueous solutions?
Solution:
Due to their inert nature, these metals do not affect the products of electrolysis. They have good adsorbing capacity for hydrogen.

Question 2.
Why does physisorption decrease with the increase of temperature?
Solution:
Physical adsorption of a gas by a solid is generally reversible. Thus,
Solid + Gas $\rightleftharpoons$ Gas / Solid + Heat
Since the adsorption process is exothermic, the physical adsorption occurs readily at low temperature and decreases with increasing temperature (Le Chatelier’s principle).

Question 3.
Why are powdered substances more effective adsorbents than their crystalline forms?
Solution:
The extent of adsorption increases with increase in surface area of the adsorbent. Finely powdered substances have large, porous areas and act as good adsorbents.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process?
Solution:
CO is a catalytic poison. It reacts with iron to form iron carbonyl thus inhibiting the activity of catalyst.

Question 5.
Why is the ester hydrolysis slow in the beginning and becomes faster after sometime?
Solution:
The acid formed during the reaction provides hydrogen ions which act as catalyst for the reaction and it becomes faster.

Question 6.
What is the role of desorption in the process of catalysis?
Solution:
In catalysis the products formed are desorbed and detached from the surface so that, more reactants can get adsorbed on the surface of catalyst.

Question 7.
What modification can you suggest in the Hardy Schulze law?
Solution:
The Hardy Schulze law considers the coagulation of sols because of neutralisation of their charges. Since coagulation can also occur by mixing two oppositely charged sols, it should also include “when oppositely charged sols are mixed in proper proportions to neutralize the charges of each other, the coagulation of both the sols occurs.”

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Solution:
Few impurities which are soluble in water and are adsorbed on the surface of the precipitate are removed by washing them with water.

NCERT Exercises

Question 1.
Distinguish between the meaning of the terms adsorption and absorption. Give one example of each.
Solution:
The phenomenon of accumulation of the molecules of a substance on a solid or liquid surface resulting in the increased concentration of the molecules on the surface is called adsorption. In absorption, the substance is uniformly distributed throughout the bulk of the solution. A distinction can be made by taking an example of water vapours. Water vapours are absorbed by anhydrous calcium chloride but adsorbed by silica gel.

	Adsorption	Absorption
1.	It is a surface phenomenon, i.e., it occurs only on the adsorbent surface.	It occurs throughout the body of the material. It is called bulk phenomenon.
2.	The concentration on the adsorbent’s surface is different from that in the bulk.	The concentration is same throughout.
3.	The rate varies throughout the process.	The rate remains the same.

Question 2.
What is the difference between physisorption and chemisorption?
Solution:

Physisorption		Chemisorption
1.	The adsorbate and adsorbent are held by weak van der Waals forces.	The adsorbate and adsorbent are held by forces similar to a chemical bond.
2.	Heat of adsorption is of the order of 20 kj/mol.	Heat of adsorption is of the order of 200 kj/mol.
3.	It is reversible.	It is irreversible.
4.	It decreases with increase in temperature and occurs at lower temperatures.	It increases with temperature and occurs at high temperature.
5.	It is not specific in nature, i.e., all gases are adsorbed on all solids to some extent.	It is specific in nature and occurs only when a chemical bond is formed between the adsorbate and adsorbent.
6.	Multimolecular layers may be formed on the adsorbent.	Usually unimolecular layer is formed on the adsorbent.

Question 3.
Give reason why a finely divided substance is more effective as an adsorbent.
Solution:
The extent of adsorption increases with increase in surface area of the adsorbent. Thus, finely divided metals and porous substances having large surface areas are good adsorbents.

Question 4.
What are the factors which influence the adsorption of a gas on a solid?
Solution:
Factors affecting adsorption of a gas on solids are :

Nature of the adsorbent : The same gas is adsorbed to different extents by different solids at the same temperature. Also, greater the surface area of the adsorbent, more is the gas adsorbed.

Nature of the adsorbate : Different gases are adsorbed to different extents by different solids at the same temperature. Higher the critical temperature of the gas, greater is its amount adsorbed.

Temperature : Since adsorption is an exothermic process, applying Le Chatelier’s principle, we can find out that adsorption decreases with an increase in temperature.

Specific area of the adsorbent : Surface area available for adsorption per gram of the adsorbent increases the extent of adsorption. Greater the surface area, higher would be the adsorption therefore, porous or powdered adsorbents are used.

Pressure : At constant temperature, the adsorption of gas increases with pressure.

Activation of adsorbent : It means increasing the adsorbing power of an adsorbent by increasing its surface area. It is done by :

making the adsorbent’s surface rough
removing gases already adsorbed
subdividing the adsorbent into smaller pieces.

Question 5.
What is an adsorption isotherm? Describe Freundlich adsorption isotherm.
Solution:
Adsorption isotherm is a graph between the amount of the gas adsorbed by an adsorbent and equilibrium pressure of the adsorbate at constant temperature. Freundlich obtained an empirical relationship between the quantity of gas adsorbed by unit mass of solid adsorbent and pressure at a particular temperature. It is mathematically represented in the following way :

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Solution:
Activation of an adsorbent means increasing its adsorbing power by increasing the surface area of the adsorbent by making its surface rough, and removing adsorbed gases from it. With an increase in surface area the adsorption increases.

Question 7.
What role does adsorption play in heterogenous catalysis?
Solution:
Adsorption of reactants on solid surface of the catalysts increases the rate of reaction. There are many gaseous reactions of industrial importance involving solid catalysts. Manufacture of ammonia using iron as a catalyst, manufacture of H₂SO₄ by Contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis.

Question 8.
Why is adsorption always exothermic?
Solution:
During adsorption, there is always a decrease in residual forces of the surface, i.e., there is decrease in surface energy which appears as heat. Adsorption therefore, is invariably an exothermic process. In other words, ∆H of adsorption is always negative to keep the value of ∆G negative for the reaction to be spontaneous as ∆S decreases during adsorption.

Question 9.
How are the colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Solution:
Colloids can be classified into eight types depending upon the physical state of the dispersed phase and the dispersion medium.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Solution:

Adsorption decreases with an increase in, temperature because it is an exothermic process and according to Le Chatelier’s principle the reaction will proceed in backward direction with increase in temperature.
At a constant temperature, adsorption increases with pressure.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Solution:
There are two types of colloidal sols :

(i) Lyophilic sols : The word lyophilic means solvent loving. They are obtained by directly mixing the dispersed phase and the dispersion medium, e.g., sols of gum, gelatin, starch, etc. They are solvent attracting hence quite stable and cannot be coagulated easily.

(ii) Lyophobic sols : They cannot be prepared by directly mixing the dispersed phase and dispersion medium but are prepared by special methods, e.g., sols of metals. They are solvent repelling. Hydrophobic sols are easily coagulated due to repulsion between water and dispersed phase.

Question 12.
What is the difference between multimolecular and macromoiecular colloids? Give one example of each. How are associated colloids different from these two types of colloids?
Solution:
Depending upon the type of particles of the dispersed phase, colloids are classified as : multimolecular, macromoiecular and associated colloids.

(i) Multimolecular colloids : On dissolution, a large number of atoms or smaller molecules of a substance aggregate together to form species having size in the colloidal range (diameter < 1 nm). The species thus formed are called multimolecular colloids. For example, a gold sol may contain particles of various sizes having many atoms. Sulphur sol consists of particles containing a thousand or more of S8 sulphur molecules.

(ii) Macromoiecular colloids : Macromolecules in suitable solvents form solutions in which the size of the macromolecules may be in the colloidal range. Such systems are called macromoiecular colloids. These colloids are quite stable and resemble true solutions in many respects. Examples of naturally occurring macromolecules are starch, cellulose, proteins and enzymes; and those of man-made macromolecules are polythene, nylon, polystyrene, synthetic rubber, etc.

(iii) Associated colloids (Micelles) : There are some substances which at low concentrations behave as normal strong electrolytes but, at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregated particles thus formed are called micelles. These are also known as associated colloids. The formation of micelles takes place only above a particular temperature called Kraft temperature (Tk) and above a particular concentration called critical micelle concentration (CMC). On dilution, these colloids revert back to individual ions. Surface active agents such as soaps and synthetic detergents belong to this class. For soaps, the CMC is 10^-4 to 10^-3 mol L^-1. These colloids have both lyophobic and lyophilic parts. Micelles may contain as many as 100 molecules or more.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Solution:
Enzymes are complex nitrogenous organic compounds which act as biological catalysts and increase the rate of cellular processes. According to the lock and key model, like every lock has a specific key, similarly every enzyme acts at a specific substrate.

When the substrate fits the active site (lock) of the enzyme, the chemical change begins. But it has also been noticed that enzyme changes shape, when substrate lands at the active site. This induced-fit model of enzyme action pictures the substrate inducing the 1 active site to adopt a perfect fit, rather than a rigid shaped lock and key. Therefore, the new model for enzyme action is called induced fit model.

Question 14.
How are colloids classified on the basis of

physical states of components
nature of dispersion medium and
interaction between dispersed phase and dispersion medium?

Solution:

Refer answer number 9.
Depending upon the nature of dispersion medium colloids can be classified as sol if the dispersion medium is liquid, gel if the dispersion medium is solid. If the dispersion medium is water, the sol is called hydrosol and if the dispersion medium is alcohol, it is called alcosol. A colloid in which the dispersion medium as well as dispersed phase are liquids, is called emulsion.
Refer answer number 11.

Question 15.
Explain what is observed

when a beam of light is passed through a colloidal sol.
an electrolyte, NaCI is added to hydrated ferric oxide sol.
electric current is passed through a colloidal sol?

Solution:
(i) When a beam of light is passed through colloidal particles, its path becomes clearly visible and is known as Tyndall effect. It is due to scattering of light by colloidal particles. The bright cone of the light is called Tyndall cone.

(ii) When NaCI is added to hydrated ferric oxide sol coagulation takes place. Since ferric oxide is a positive sol, it is coagulated by the negative chloride ions.

(iii) When electric potential is applied across two platinum electrodes dipped in a colloidal solution, the colloidal particles move towards one or the other electrode. The movement of colloidal particles under an applied electric potential is called electrophoresis.

Positively charged particles move towards the cathode while negatively charged particles move towards the anode. Since all the colloidal particles in a given colloidal solution carry the same charge, the particles move to one or the other electrode depending on the charge.

Question 16.
What are emulsions? What are their different types? Give example of each type.
Solution:
These are liquid-liquid colloidal systems, i.e., the dispersion of finely divided droplets in another liquid. If a mixture of two immiscible or partially miscible liquids is shaken, a coarse dispersion of one liquid in the other is obtained which is called emulsion. Generally, one of the two liquids is water. There are two types of emulsions :

Oil dispersed in water (O/W type) and
Water dispersed in oil (W/O type).

In the first system, water acts as dispersion medium. Examples of this type of emulsion are milk and vanishing cream. In milk, liquid fat is dispersed in water. In the second system, oil acts as dispersion medium. Common examples of this type are butter and cream.

Question 17.
What is demulsification? Name two demulsifiers.
Solution:
The process of converting the emulsion back into two distinct components, oil and water is called demulsification. This can be done by

boiling
freezing
changing pH
electrostatic precipitation.

Question 18.
Action of soap is due to emulsification and micelle formation. Comment.
Solution:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na⁺ (e.g., sodium stearate CH₃(CH₂)]₁₆COO^– Na⁺, which is a major component of many bar soaps). When dissolved in water, it dissociates into RCOO^– and Na+ ions. The RCOO^– ions, however, consist of two parts – a long hydrocarbon chain R (also called non-polar ‘tail’) which is hydrophobic (water repelling), and a polar group COO^– (also called polar-ionic ‘head’), which is hydrophilic (water loving).

The RCOO^– ions are, therefore, present on the surface with their COO^– groups in water and the hydrocarbon chains R staying away from it and remain at the surface. But at critical micelle concentration, the anions are pulled into the bulk of the solution and aggregate to form a spherical shape with their hydrocarbon chains pointing towards the centre of the sphere with COO^– part remaining outward on the surface of the sphere. An aggregate thus formed is known as ‘ionic micelle’.

The cleansing action of soap is due to the fact that soap molecules form micelle around the oil droplet in such a way that hydrophobic part of the stearate ions is in the oil droplet and hydrophilic part projects out of the grease droplet like the bristles

(a) Grease on cloth
(b) Stearate ions (from soap) arranging around the grease droplets
(c) Micelle formed

Since the polar groups can interact with water, the oil droplet surrounded by stearate ions is now pulled in water and removed from the dirty surface. Thus soap helps in emulsification and washing away of oils and fats. Tne negatively charged sheath around the globules prevents them from coming together and forming aggregates.

Question 19.
Give four examples of heterogeneous catalysis.
Solution:

Question 20.
What do you mean by activity and selectivity of catalysts?
Solution:
(a) Activity : The activity of a catalyst depends upon the strength of chemisorption to a large extent. The reactants must get adsorbed reasonably strongly on to the catalyst to become active. But adsorption must not be so strong that they are immobilised. It is observed that maximum activity is shown by elements of groups 7 – 9 of the periodic table
2H₂ + O₂ $\underrightarrow { Pt }$ 2H₂O

(b) Selectivity : The selectivity of a calatyst is its ability to yield a particular product in the reaction e.g.,

Thus, a selective catalyst can act as a catalyst in one reaction and may fail to catalyse another reaction.

Question 21.
Describe some features of catalysis by zeolites.
Solution:
(a) Zeolites are hydrated aluminosilicates which have a three dimensional network structure containing water molecules in their pores.
(b) The pores are made vacant by heating before catalysis.
(c) The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules and also on the pores and cavities in them, e.g., ZSM-5 converts alcohols to hydrocarbons by dehydrating them.

Alcohols $\underrightarrow { ZSM-5 }$ Hydrocarbons

Question 22.
What is shape selective catalysis?
Solution:
The catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules is called shape-selective catalysis. Zeolites are good shape-selective catalysts because of their honeycomb-like structures. They are microporous aluminosilicates with three dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving Al-O-Si framework. The reactions taking place in zeolites depend upon the size and shape of reactant and product molecules as well as upon the pores and cavities of the zeolites. They are found in nature as well as synthesised for catalytic selectivity.

Question 23.
Explain the following terms :

Electrophoresis
Coagulation
Dialysis
Tyndall effect

Solution:
(i) Electrophoresis : Refer answer number 15 (iii)

(ii) Coagulation or precipitation : The stability of the lyophobic sols is due to the presence of charge on colloidal particles. If somehow, the charge is removed, the particles will come nearer to each other to form aggregates (or coagulate) and settle down under the force of gravity. The process of settling down of colloidal particles is called coagulation.

(iii) Dialysis : It is the process of removing dissolved substances from a colloidal solution by means of diffusion through a suitable membrane. Since particles (ions or smaller molecules) in a true solution can pass through animal membrane (bladder) or parchment paper or cellophane sheet but not the colloidal particles, the membrane can be used for dialysis. The apparatus used for this purpose is called dialyser. A bag of suitable membrane containing the colloidal solution is suspended in a vessel through which fresh water is continuously flowing. The molecules and ions diffuse through membrane into the outer water and pure colloidal solution is left behind.

(iv) Tyndall effect : Refer answer number 15 (i)

Question 24.
Give four uses of emulsions.
Solution:

Some of the medicines are effective as emulsions.
Paints are emulsions which are used in our daily life.
Soaps and detergents act as cleansing agents, action of which is based on emulsification.
Photographic films are coated with emulsion of AgBr on its surface.

Question 25.
What are micelles? Give an example of a micelle system.
Solution:
Micelles are substances that behave as normal strong electrolytes at low concentration but at high concentrations behave as colloids due to formation of aggregates. They are also called associated colloids, e.g., soaps and detergents. They can form ions and may contain 100 or more molecules to form a micelle.

Question 26.
Explain the terms with suitable examples :

Alcosol
Aerosol
Hydrosol

Solution:
(i) Alcosol : The sol in which alcohol is used as dispersion medium is called alcosol e.g., sol of cellulose nitrate in ethyl alcohol.

(ii) Aerosol : The sol in which dispersion medium is gas and dispersed phase is either solid or liquid, the colloidal system is called aerosol e.g., fog, insecticides, sprays, etc.

(iii) Hydrosol : The sol in which dispersion medium is water is called hydrosol e.g., starch sol.

Question 27.
Comment on the statement that’colloid is not a substance but a state of substance’.
Solution:
Colloid is not a substance, but a state of substance because the same substance may exist as a colloid or crystalloid under different conditions e.g., sulphur. Colloidal solution of sulphur consists of sulphur molecules dispersed in water. In this state, sulphur atoms combine to form multimolecules whose size lies between 1 nm to 1000 nm and form colloidal state. Sulphur forms true solution in carbon disulphide. Similarly soap is a solution at low concentration but a colloid at higher concentration.

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CLASS 12TH CHAPTER -4 Chemical Kinetics |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter : 4 Chemical Kinetics

INTEXT Questions

Question 1.
For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Solution:

Question 2.
In a reaction, 2A → products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval.
Solution:

Question 3.
For a reaction, A + B → product; the rate law is given by, r = k[A]^1/2 [B]2. What is the order of the reaction?
Solution:

Question 4.
The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?
Solution:
The reaction, X → Y, follows second order kinetics hence the rate law equation will be
Rate = kC², where C = [X]
If concentration of X increases three times, now, [X] = 3C mol L^-1
∴ Rate = k(3C)² = 9kC²
Thus the rate of reaction will become 9 times. Hence, the rate of formation of Y will increase 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5 g of this reactant take to reduce to 3 g?
Solution:

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:

Question 7.
What will be the effect of temperature on rate constant?
Solution:
It has been found that for a chemical reaction, with rise in temperature by 10°, the rate constant is nearly doubled.

The temperature dependence of the rate of a chemical reaction can be accurately explained by Arrhenius equation.
K = Ae^-Ea/RT
Where A is the Arrhenius factor or the frequency factor. It is also called pre exponential factor. It is a constant specific to a particular reaction. R is gas constant and E_a is activation energy measured in joules/mole (J mol^-1).

Question 8.
The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_a.
Solution:

Question 9.
The activation energy for the reaction 2Hl_(g) → H₂_(g)+ l_2(g) is 209.5 kJ mol^-1 at 581 K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy.
Solution:
Fraction of molecules having energy equal to or greater than activation energy is given as

NCERT Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solution:

Question 2.
For the reaction : 2A + B → A2B the rate = k[A][B]2 with k = 2.0 × 10^-6 mol^-2 L² s^-1 Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1 [B] = 0.2 mol L^-1 Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
Solution:

Question 3.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol L^-1 s^-1 ?
Solution:

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by
Rate = k[CH₃OCH₃]^3/2
The rate of reaction is followed by increase in pressure in closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = K(pCH₃OCH₃)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants ?
Solution:
In terms of pressure,
Units of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Solution:
Following are the factors on which rate of reaction depends.
(i) Nature of the reactant : Rate of reaction depends on nature of reactant.
Example : Reactions of ionic compounds are faster than that of covalent compounds.

(ii) State of reactants : Solid reactions are slow, reactions of liquids are fast whereas that of gases are very fast.

(iii) Temperature : Rate of reaction largely depends on temperature. It has been observed that every 10°C rise in temperature increases rate of reaction by 2-3 times.
$\frac { { r }_{ t }+10 }{ { r }_{ t } }$ = 2 – 3 . This ratio is called temperature coefficient.
There are two reasons for increasing rate of reaction with increasing temperature.

(a) Increase in temperature increases average kinetic energy of reactant molecules. Hence, rate of collision increases.
(b) With increase in temperature number of molecules having threshold energy also increases i.e. number of active molecules increases. As a result, number of effective collisions increases. Hence, rate of reaction increases.

(iv) Concentration : Rate of reaction also depends on concentration of reactants.
Rate = k × C^”, where n = order of reaction, C = concentration of reactant.

(v) Presence of catalyst : Rate of reaction also depends on presence of catalyst. Catalyst increases rate of reaction by any of the following ways:
(a) Increasing surface area of reaction.
(b) Adsorbing the reactants on its surface and thus increasing chance of collision.
(c) By forming unstable intermediate with the substrate.
(d) By providing alternate path of lower activation energy.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is

doubled
reduced to half ?

Solution:

Question 7.
What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively?
Solution:
The rate constant increases with increase in temperature and becomes almost double for every 10° increase in temperature. Swedish chemist, Arrhenius derived a quantitative relation between rate of reaction and temperature. According to Arrhenius,

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained :

Calculate the average rate of reaction between the time interval 30 to 60 seconds.
Calculate the pseudo first order rate constant for the hydrolysis of ester.

Solution:

Question 9.
A reaction is first order in A and second order in B.

Write the differential rate equation.
How is the rate affected on increasing the concentration of 6 three times?
How is the rate affected when the concentrations of both A and B are doubled?

Solution:
(i) Reaction is first order in A and second order in B, hence differential rate equation is

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below :

What is the order of the reaction with respect to A and B ?
Solution:

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2 A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:

Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table :

Solution:

Question 13.
Calculate the half-life of the first order reaction from their rate constants given below:

200 s^-1
2 min^-1
4 year^-1

Solution:

Question 14.
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first order kinetics. Therefore,

Question 15.
The experimental data for the decomposition of N₂O₅
[2N₂O₅ → 4NO₂ + O₂]
in gas phase at 318K are given below :

Plot [N₂O₅] against t.
Find the half-life period for the reaction.
Draw a graph between log [N₂O₅] and t.
What is the rate law?
Calculate the rate constant.
Calculate the half-life period from k and compare it with (ii).

Solution:

Question 16.
The rate constant for a first order reaction is 60 s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16 th value ?
Solution:

Question 17.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If 1 µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically ?
Solution:

Question 18.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:

Question 19.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained.

Calculate the rate constant
Solution:

Question 21.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.

Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution:

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below :

Draw a graph between In k and 1/7 and calculate the value of A and Ea. Predict the rate constant at 30°C and 50°C.
Solution:
The values of rate constants for the decomposition of N₂O₅ at various temperatures are given below :

Question 23.
The rate constant for the decomposition of a hydrocarbon is 2.418 × 10^-5 s^-1 at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor.
Solution:

Question 24.
Consider a certain reaction A → Products with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours ?
Solution:
Sucrose decomposes according to first order rate law, hence

Question 26.
The decomposition of a hydrocarbon follows the equation

Solution:

Question 27.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 104 K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:

Question 28.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kJ mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1 ?
Solution:

Question 29.
The time required for 10% completion of the first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1, calculated at 318 K and E_a.
Solution:

Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:

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CLASS 12TH CHAPTER -3 Electrochemistry |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter : 3 Electrochemistry

INTEXT Questions

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg?
Solution:
Set up an electrochemical cell consisting of MglMgSO₄(1 M) as one electrode by dipping a magnesium wire in 1 M MgSO₄ solution and standard hydrogen electrode Pt, H₂ (1 atm) | H⁺(1 M) as the second electrode. Measure the EMF of the cell and also note the direction of deflection in the voltmeter. The direction of deflection shows that the electrons flow from magnesium electrode to hydrogen electrode, i.e., oxidation takes place on magnesium electrode and reduction on hydrogen electrode.

Question 2.
Can you store copper sulphate solution in a zinc pot ?
Solution:

Question 3.
Consult the table of standard electrode potentials and suggest three substances that can oxidise ferrous ions under suitable conditions.
Solution:
Oxidation of ferrous ions means :

Only those substances can oxidise Fe²⁺ to Fe³⁺ which are stronger oxidising agents and have positive reduction potentials greater than 0.77 V so that EMF of the cell reaction is positive. This is for elements lying below Fe³⁺/Fe²⁺ in the electrochemical series, for example, Br₂, Cl₂ and F₂.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution:

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni_(S) + 2Ag⁺ (0.002 M) → Ni²⁺(0.160 M) + 2Ag_(S)
Given that E°cell = 1.05 V
Solution:

Question 6.
The cell in which the following reaction occurs: 2Fe³⁺_(aq) + 2l^–_(aq) → 2Fe²⁺_(aq) + l_2(s) has E°cell = 0.236 V at 298 K. Calculate the standard Gibb’s energy and the equilibrium constant of the cell reaction.
Solution:

Question 7.
Why does the conductivity of a solution decrease with dilution?
Solution:
Conductivity of solution decreases with dilution because number of ions per unit volume decreases.

Question 8.
Suggest a way to determine the A°m value of water.
Solution:
Water is a weak electrolyte. Its A°m value can be determined with the help of Kohlrausch’s law.

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1. Calculate its degree of dissociation and dissociation constant. Given λ (H⁺) = 349.6 S cm² mol^-1 and λ (HCOO^–) = 54.6 S cm² mol^-1
Solution:

Question 10.
lf a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow though the wire?
Solution:

Question 11.
Suggest; a list of metals that are extracted electrolytically.
Solution:
Ca, Na, K, Al are extracted electrolytically.

Question 12.
Consider the reaction:
Cr₂O^2-₇ + 14H⁺ + 6e^– → 2Cr³⁺ + 7H₂O,
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂ O^2-₇?
Solution:

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Solution:
Chemical reactions while recharging :
2PbSO₄ + 2H₂O → PbO₂ + Pb + 2H₂SO₄
Electricity is passed through the electrolyte PbSO₄ which is converted into PbO₂ and Pb.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Solution:
CH₄ and CO can be used in fuel cell instead of hydrogen.

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Solution:
The following reactions take place at the surface of iron metal which acts as an electrochemical cell.

The water layer present on the surface of iron dissolves acidic oxides of air like C0₂ to form acids which dissociate to give H⁺ ions. Fe starts losing electrons in presence of H⁺ ions.
H₂O + CO₂ → H₂CO₃ $\rightleftharpoons$ 2H⁺ + C0₃^2-

NCERT Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Solution:
Mg, Al, Zn, Fe, Cu

Question 2.
Given the standard electrode potentials,
K⁺/ K = – 2.93 V, Ag⁺/ Ag = 0.80 V, Hg²⁺/ Hg = 0.79 V, Mg²⁺/ Mg = -2.37 V, Cr³⁺/ Cr = – 0.74 V
Arrange these metals in their increasing order of reducing power.
Solution:
Higher the oxidation potential, more easily it is oxidized and hence greater is the reducing power. Thus, increasing order of reducing power will be
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn_(S) + 2Ag⁺_(aq)Zn²⁺_(aq), + 2Ag_(s) takes place. Further show :

Which of the electrode is negatively charged?
The carriers of the current i n the cell.
Individual reaction at each electrode.

Solution:

Question 4.
Calculate the standard cell potentials of galvanic cell in which the following reactions take place :

Solution:

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :

Solution:

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place :

Solution:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Solution:
The reciprocal of resistivity is known as specific conductance or simply conductivity. It is denoted by K (kappa). Thus, if K is the specific conductance and G is the conductance of the solution, then

Now, if I = 1 cm and A = lsq.cm, then K = G.

Hence, conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross¬section. Alternatively, it may be defined as conductance of one centimetre cube of the solution of the electrolyte.

Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from 1 mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∆_m.

Variation of conductivity and molar conductivity with concentration: Conductivity always decreases with decrease in concentration, for both weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Molar conductivity increases with decrease in concentration. This is because that total volume, V, of solution containing one mole of electrolyte also increases. It has been found that decrease in K on dilution of a solution is more than compensated by increase in its volume.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 S cm^-1. Calculate its molar conductivity.
Solution:

Question 9.
The resistance of a conductivity cell containing 0.001 M KCI solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001 M KCI solution at 298 K is 0.146 × 10^-3 S cm^-1.
Solution:
Cell constant = K × R = 0.146 × 10^-3 × 1500 = 0.219 cm^-1

Question 10.
The conductivity of sodium chloride at 298K has been determined at different concentrations and the results are given below:
Concentration/M 0.001 0.010 0.020 0.050 0.100
10² × K/S m^-1 1.237 11.85 23.15 55.53 106.74
Calculate ∆_m for all concentrations and draw a plot between ∆_m and c^1/2. Find the value of ∆°_m
Solution:

Question 11.
Conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity. If ∆°_m for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant?
Solution:

Question 12.
How much charge is required for the following reductions:

1 mol of Al³⁺to Al ?
1 mol of Cu²⁺ to Cu ?
1 mol of MnO₄^–to Mn²⁺?

Solution:

Question 13.
How much electricity in terms of Faraday is required to produce

20.0 g of Ca from molten CaCl₂?
40.0 g of Al from molten Al₂O₃?

Solution:

Question 14.
How much electricity is required in coulomb for the oxidation of

1 mol of H₂O to O₂?
1 mol of FeO to Fe₂O₃?

Solution:

Question 15.
A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
(At. mass of Ni = 58.7)
Solution:

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄ respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
(At. wt. of Ag = 108, Cu = 63.5, Zn = 65.3)
Solution:

Question 17.
Using the standard electrode potentials predict if the reaction between the following is feasible :

Solution:

Question 18.
Predict the product of electrolysis in each of the following:

An aqueous solution of AgNO₃ with silver electrodes.
An aqueous solution of AgNO₃ with platinum electrodes.
A dilute solution of H₂SO₄ with platinum electrodes.
An aqueous solution of CuCI₂ with platinum electrodes.

Solution:

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CLASS 12TH CHAPTER -2 Solutions |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter : 2 Solutions

NCERT TEXTBOOK QUESTIONS SOLVED

2.1. Calculate the mass percentage of benzene (C₆H₆) and carbon tetrachloride (CCl₄) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride.
Ans: Mass of solution = Mass of C₆H₆ + Mass of CCl₄
= 22 g+122 g= 144 g
Mass % of benzene = 22/144 x 100 =15.28 %
Mass % of CCl₄ = 122/144 x 100 = 84.72 %

2.2. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride.
Ans: 30% by mass of C₆H₆ in CCl₄ => 30 g C₆H₆ in 100 g solution
.’. no. of moles of C₆H₆,(ⁿC₆h₆) = 30/78 = 0.385

2.3. Calculate the molarity of each of the following solutions
(a) 30 g of Co(NO₃)26H₂O in 4·3 L of solution
(b) 30 mL of 0-5 M H₂SO₄ diluted to 500 mL.
Ans:

2.4. Calculate the mass of urea (NH₂CONH₂) required in making 2.5 kg of 0.25 molal aqueous solution.
Ans: 0.25 Molal aqueous solution to urea means that
moles of urea = 0.25 mole
mass of solvent (NH₂CONH₂) = 60 g mol^-1
.’. 0.25 mole of urea = 0.25 x 60=15g
Mass of solution = 1000+15 = 1015g = 1.015 kg
1.015 kg of urea solution contains 15g of urea
.’. 2.5 kg of solution contains urea =15/1.015 x 2.5 = 37 g

2.5. Calculate
(a) molality
(b) molarity and
(c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI solution is 1·202 g mL^-1.
Ans:
Step I. Calculation of molality of solution
Weight of KI in 100 g of the solution = 20 g
Weight of water in the solution = 100 – 20 = 80 g = 0-08 kg
Molar mass of KI = 39 + 127 = 166 g mol^-1.

Step II. Calculation of molarity of solution

Step III. Calculation of mole fraction of Kl

2.6. H₂S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H₂S in water at STP is 0.195 m, calculate Henry’s law constant.
Ans: Solubility of H₂S gas = 0.195 m
= 0.195 mole in 1 kg of solvent
1 kg of solvent = 1000g

2.7. Henry’s law constant for CO₂ in water is 1.67 x 10⁸Pa at 298 K. Calculate the quantity of CO₂ in 500 mL of soda water when packed under 2.5 atm CO₂ pressure at 298 K.
Ans.:

2.8 The vapour pressures of pure liquids A and B are 450 mm and 700 mm of Hg respectively at 350 K. Calculate the composition of the liquid mixture if total vapour pressure is 600 mm of Hg. Also find the composition in the vapour phase.
Ans:
Vapour pressure of pure liquid A (P∘A) = 450 mm
Vapour pressure of pure liquid B (P∘B) = 700 mm
Total vapour pressure of the solution (P) = 600 mm

2.9. Vapour pressure of pure water at 298 K is 23.8 m m Hg. 50 g of urea (NH₂CONH₂) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.
Ans:

2.10. Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.
Ans:

2.11 Calculate the mass of ascorbic acid (vitamin C, C₆H₈O₆) to be dissolved in 75 g of acetic acid to lower its melting point by 1·5°C. (K_f for CH₃COOH) = 3·9 K kg mol^-1)
Ans:

2.12. Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.
Ans:

NCERT EXERCISES

2.1. Define the terra solution. How many types of solutions are formed? Write briefly about each type with an example.
Sol: A solution is a homogeneous mixture of two or more chemically non-reacting substances. Types of solutions: There are nine types of solutions.
Types of Solution Examples
Gaseous solutions
(a) Gas in gas Air, mixture of 0₂ and N₂, etc.
(b) Liquid in gas Water vapour
(c) Solid in gas Camphor vapours in N2 gas, smoke etc.
Liquid solutions
(a) Gas in liquid C02 dissolved in water (aerated water), and 02 dissolved in water, etc.
(b) Liquid in liquid Ethanol dissolved in water, etc.
(c) Solid in liquid Sugar dissolved in water, saline water, etc.
Solid solutions
(a) Gas in solid Solution of hydrogen in palladium
(b) Liquid in solid Amalgams, e.g., Na-Hg
(c) Solid in solid Gold ornaments (Cu/Ag with Au)

2.2. Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What type of solid solution is this likely to be ?
Sol: The solution likely to be formed is interstitial solid solution.

2.3 Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage
Sol: (i) Mole fraction: It is defined as the ratio of the number of moles of the solute to the total number of moles in the solution. If A is the number of moles of solute dissolved in B moles of solvent, then Mole fraction of solute

(ii) Molality: It is defined as die number of moles of a solute present in 1000g (1kg) of a solvent.

NOTE: Molality is considered better way of expressing concentration of solutions, as compared to molarity because molality does not change with change in temperature since the mass of solvent does not vary with temperature,
(iii) Molarity: It is defined as the number of moles of solute present in one litre of solution.

NOTE: Molarity is the most common way of expressing concentration of a solution in laboratory. However, it has one disadvantage. It changes with temperature because volume of a solution alters due to expansion and contraction of the liquid with temperature.
(iv) Mass percentage: It is the amount of solute in grams present in 100g of solution.

2.4. Concentrated nitric acid used in the laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of acid if the density of the solution is 1·504 g mL^-1 ?
Sol: Mass of HNO₃ in solution = 68 g
Molar mass of HNO₃ = 63 g mol^-1
Mass of solution = 100 g
Density of solution = 1·504 g mL^-1

2.5. A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1 .2 g m L^-1, then what shall be the molarity of the solution?
Sol: 10 percent w/w solution of glucose in water means 10g glucose and 90g of water.
Molar mass of glucose = 180g mol^-1 and molar mass of water = 18g mol^-1

2.6. How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na₂C0₃and NaHCO₃containing equimolar amounts of both?
Sol: Calculation of no. of moles of components in the mixture.

2.7. Calculate the percentage composition in terms of mass of a solution obtained by mixing 300 g of a 25% and 400 g of a 40% solution by mass.
Sol:

2.8. An antifreeze solution is prepared from 222.6 g of ethylene glycol, (C₂H₆O₂) and200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^-1, then what shall be the molarity of the solution?
Sol:

2.9. A sample of drinking water was found to be severely contaminated with chloroform (CHCl₃), supposed to be a carcinogen. The level of contamination was 15 ppm (by mass).
(i) express this in percent by mass.
(ii) determine the molality of chloroform in the water sample.
Sol: 15 ppm means 15 parts in million (10⁶) by mass in the solution.

2.10. What role does the molecular interaction play in solution of alcohol in water?
Sol: In case of alcohol as well as water, the molecules are interlinked by intermolecular hydrogen bonding. However, the hydrogen bonding is also present in the molecules of alcohol and water in the solution but it is comparatively less than both alcohol and water. As a result, the magnitude of attractive forces tends to decrease and the solution shows positive deviation from Raoult’s Law. This will lead to increase in vapour pressure of the solution and also decrease in its boiling point.

2.11. Why do gases always tend to be less soluble in liquids as the temperature is raised?
Sol: When gases are dissolved in water, it is accompanied by a release of heat energy, i.e., process is exothermic. When the temperature is increased, according to Lechatlier’s Principle, the equilibrium shifts in backward direction, and thus gases becomes less soluble in liquids.

2.12. State Henry’s law and mention some of its important applications.
Sol:
Henry’s law: The solubility of a gas in a liquid at a particular temperature is directly proportional to the pressure of the gas in equilibrium with the liquid at that temperature.
or
The partial pressure of a gas in vapour phase is proportional to the mole fraction of the gas (x) in the solution. p = KHX
where KH is Henry’s law constant.
Applications of Henry’s law :
(i) In order to increase the solubility of CO₂ gas in soft drinks and soda water, the bottles are normally sealed under high pressure. Increase in pressure increases the solubility of a gas in a solvent according to Henry’s Law. If the bottle is opened by removing the stopper or seal, the pressure on the surface of the gas will suddenly decrease. This will cause a decrease in the solubility of the gas in the liquid i.e. water. As a result, it will rush out of the bottle producing a hissing noise or with a fiz.
(ii) As pointed above, oxygen to be used by deep sea divers is generally diluted with helium inorder to reduce or minimise the painfril effects during decompression.
(iii) As the partial pressure of oxygen in air is high, in lungs it combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is comparatively low. Therefore, oxyhaemoglobin releases oxygen in order to carry out cellular activities.

2.13. The partial pressure of ethane over a solution containing 6.56 × 10^-3 g of ethane is 1 bar. If the solution contains 5.00 × 10^-2 g of ethane, then what shall be the partial pressure of the gas?
Sol:

2.14. According to Raoult’s law, what is meant by positive and negative deviaitions and how is the sign of ∆_solH related to positive and negative deviations from Raoult’s law?
Sol: Solutions having vapour pressures more than that expected from Raoult’s law are said to exhibit positive deviation. In these solutions solvent – solute interactions are weaker and ∆_solH is positive because stronger A – A or B – B interactions are replaced by weaker A – B interactions. Breaking of the stronger interactions requires more energy & less energy is released on formation of weaker interactions. So overall ∆_solH is positive. Similarly ∆_solV is positive i.e. the volume of solution is some what more than sum of volumes of solvent and solute.
So there is expansion in volume on solution formation.
Similarly in case of solutions exhibiting negative deviations, A – B interactions are stronger than A-A&B-B. So weaker interactions are replaced by stronger interactions so , there is release of energy i.e. ∆_sol H is negative.

2.15. An aqueous solution of 2 percent non-volatile solute exerts a pressure of 1·004 bar at the boiling point of the solvent. What is the molecular mass of the solute ?
Sol:
According to Raoult’s Law,

2.16 Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35.0 g of octane?
Sol.

2.17. The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a non-volatile solute in it
Sol: 1 molal solution of solute means 1 mole of solute in 1000g of the solvent.

2.18. Calculate the mass of a non-volatile solute (molecular mass 40 g mol^-1) that should be dissolved in 114 g of octane to reduce its pressure to 80%. (C.B.S.E. Outside Delhi 2008)
Sol: According to Raoult’s Law,

2.19. A solution containing 30g of non-volatile solute exactly in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18g of water is then added to the solution and the new of vapour pressure becomes 2.9 kPa at 298 K. Calculate
(i) molar mass of the solute.
(ii) vapour pressure of water at 298 K.
Sol: Let the molar mass of solute = Mg mol^-1

2.20. A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Sol: Mass of sugar in 5% (by mass) solution means 5gin 100g of solvent (water)

2.21. Two elements A and B form compounds having formula AB₂ and AB₄. When dissolved in 20g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. The molar depression constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Sol:

2.22. At 300 K, 36 g glucose present per litre in its solution has osmotic pressure of 4·98 bar. If the osmotic pressure of the solution is 1·52 bar at the same temperature, what would be its concentration?
Sol:

2.23. Suggest the most important type of intermolecular attractive interaction in the following pairs:
(i) n-hexane and n-octane
(ii) I₂ and CCl₄.
(iii) NaCl0₄ and water
(iv) methanol and acetone
(v) acetonitrile (CH₃CN) and acetone (C₃H₆0)
Sol: (i) Both w-hexane and n-octane are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(ii) Both I₂ and CCl₄ are non-polar. Thus, the intermolecular interactions will be London dispersion forces.
(iii) NaCl0₄ is an ionic compound and gives Na⁺ and Cl0₄^– ions in the Solution. Water is a polar molecule. Thus, the intermolecular interactions will be ion-dipole interactions.
(iv) Both methanol and acetone are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.
(v) Both CH₃CN and C₃H₆O are polar molecules. Thus, intermolecular interactions will be dipole-dipole interactions.

2.24. Based on solute solvent interactions, arrange the following in order of increasing solubility in n-octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Sol: n-octane (C₈H₁₈) is a non-polar liquid and solubility is governed by the principle that like dissolve like. Keeping this in view, the increasing order of solubility of different solutes is:
KCl < CH₃OH < CH₃C=N < C₆H₁₂ (cyclohexane).

2.25. Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol
Sol: (i) Phenol (having polar – OH group) – Partially soluble.
(ii) Toluene (non-polar) – Insoluble.
(iii) Formic acid (form hydrogen bonds with water molecules) – Highly soluble.
(iv) Ethylene glycol (form hydrogen bonds with water molecules) Highly soluble.
(v) Chloroform (non-polar)- Insoluble.
(vi) Pentanol (having polar -OH) – Partially soluble.

2.26. If the density of lake water is 1·25 g mL-1, and it contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na⁺ ions in the lake. (C.B.S.E. Outside Delhi 2008)
Sol:

2.27. If the solubility product of CuS is 6 x 10^-16, calculate the maximum molarity of CuS in aqueous solution.
Sol:

2.28. Calculate the mass percentage of aspirin (C₉H₈O₄ in acetonitrile (CH₃CN) when 6.5g of CHO is dissolved in 450 g of CH3CN.
Solution:

2.29. Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 x 10^-3 m aqueous solution required for the above dose.
Solution:

2.30. Calculate the amount of benzoic acid (C₅H₅COOH) required for preparing 250 mL of 0· 15 M solution in methanol.
Solution:

2.31. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.
Solution:

Fluorine being more electronegative than chlorine has the highest electron withdrawing inductive effect. Thus, triflouroacetic acid is the strongest trichloroacetic acid is second most and acetic acid is the weakest acid due to absence of any electron withdrawing group. Thus, F₃CCOOH ionizes to the largest extent while CH₃COOH ionizes to minimum extent in water. Greater the extent of ionization greater is the depression in freezing point. Hence, the order of depression in freezing point will be CH₃COOH < Cl₃CCOOH < F₃CCOOH.

2.32. Calculate the depression in the freezing point of water when 10g of CH₃CH₂CHClCOOH is added to 250g of water. Ka = 1.4 x 1o^-3 Kg = 1.86 K kg mol^-1.

Solution:

2.33. 19.5g of CH₂FCOOH is dissolved in 500g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’s Hoff factor and dissociation constant of fluoroacetic acid.

Solution:

2.34. Vapour pressure of water at 293 K is 17·535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Solution:
According to Raoult’s Law,

2.35. Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10⁵ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

Solution:

2.36. 100g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000g of liquid B (molar mass 180g mol^-1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr.

Solution:

2.37. Vapour pressures of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8 mm Hg respectively. Assuming that they form ideal solution over the entire range of composition, plot P_total, P_chlroform and P_acetone as a function of χ_acetone. The experimental data observed for different compositions of mixtures is:

Plot this data also on the same graph paper. Indicate whether it has positive deviation or negative deviation from the ideal solution.

Solution:

2.38. Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80g of benzene is mixed with 100g of toluene.

Solution:

2.39. The air is a mixture of a number of gases. The major components are oxygen and nitrogen with an approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen are 3.30 x 10⁷ mm and 6.51 x 10⁷ mm respectively, calculate the composition of these gases in water.

Solution:

2.40. Determine the amount of CaCl₂ (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Solution:

2.41. Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated. (C.B.S.E. 2013)
Solution:
Step I. Calculation of Van’t Hoff factor (i)
K₂SO₄ dissociates in water as :

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CLASS 12TH CHAPTER -1 The Solid State |NCERT CHEMISTRY SOLUTIONS | EDUGROWN

NCERT Solutions for Class 12 Chemistry Chapter : 1 Solid State

NCERT TEXTBOOK QUESTIONS SOLVED

1.1. Why are solids rigid?
Ans: The constituent particles in solids have fixed positions and can oscillate about their mean positions. Hence, they are rigid.

1.2. Why do solids have definite volume?
Ans: Solids keep their volume because of rigidity in their structure. The interparticle forces are very strong. Moreover, the interparticle spaces are very few and small as well. As a result, their volumes cannot change by applying pressure.

1.3. Classify the following as amorphous or crystalline solids: Polyurethane, naphthalene, benzoic acid, Teflon, potassium nitrate, cellophane, polyvinyl chloride, fibreglass, copper
Ans: Crystalline solids: Benzoic acid, potassium nitrate, copper Amorphous solids: Polyurethane, Teflon, cellophane, polyvinyl chloride, fibreglass

1.4. Why is glass considered as super cooled liquid ? (C.B.S.E. Delhi 2013)
Ans: Glass is considered to be super cooled liquid because it shows some of the characteristics of liquids, though it is an amorphous solid. For example, it is slightly thicker at the bottom. This can be possible only if it has flown like liquid, though very slowly.

1.5. Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property?
Ans: As the solid has same value of refractive index along all directions, it is isotropic in nature and hence amorphous. Being amorphous solid, it will not show a clean cleavage and when cut, it will break into pieces with irregular surfaces.

1.6. Classify the following solids in different categories based on the nature of the intermolecular forces: sodium sulphate, copper, benzene, urea, ammonia, water, zinc sulphide, diamond, rubedium, argon, silicon carbide.
Ans: Ionic, metallic, molecular, molecular, molecular (hydrogen-bonded), molecular (hydrogen-bonded), ionic, covalent, metallic, molecular, covalent (network).

1.7. Solid A is a very hard electrical insulator in. solid as well as in molten state and melts at extremely high temperature. What type of solid is it?
Ans: It is a covalent or network solid.

1.8. Why are ionic solids conducting in the molten state and not in the solid-state?
Ans: In the ionic solids, the electrical conductivity is due to the movement of the ions. Since the ionic mobility is negligible in the solid state, these are non-conducting in this state. Upon melting, the ions present acquire some mobility. Therefore, the ionic solids become conducting

1.9. What type of solids are electrical conductors, malleable and ductile?
Ans: Metallic solids

1.10. Give the significance of a lattice point.
Ans: The lattice point denotes the position of a particular constituent in the crystal lattice. It may be atom, ion or a molecule. The arrangement of the lattice points in space is responsible for the shape of a particular crystalline solid.

1.11. Name the parameters that characterise a unit cell.
Ans: A unit cell is characterised by the following parameters:
(i)the dimensions of unit cell along three edges: a, b and c.
(ii)the angles between the edges: α (between b and c); β (between a and c) and γ (between a and b)

1.12. Distinguish between :
(i) Hexagonal and monoclinic unit cells
(ii) Face-centred and end-centred unit cells.
Ans:
(i) In a hexagonal unit cell :
a = b # c; α = β = 90° and γ = 120°
In a monoclinic unit cell :
a # b # c and α = γ = 90° and β # 90°
(ii) In a face-centered unit cell, constituent particles are located at all the corners as well as at the centres of all the faces.
In end-centered unit cell, constituent particles are located at all the corners as well as at the centres of two opposite faces. (C.B.S.E Foreign 2015)

1.13. Explain how many portions of an atom located at
(i)corner and (ii)body centre of a cubic unit cell is part of its neighbouring unit cell.
Ans: (i) An atom at the comer is shared by eight adjacent unit cells. Hence, portion of the atom at the comer that belongs to one unit cell=1/8.
(ii)An atom at the body centre is not shared by any other unit cell. Hence, it belongs fully to unit cell.

1.14. What is the two-dimensional coordination number of a molecule in a square close-packed layer?
Ans: In the two-dimensional square close-packed layer, a particular molecule is in contact with four molecules. Hence, the coordination number of the molecule is four.

1.15. A compound forms hexagonal close-packed. structure. What is the total number of voids in 0. 5 mol of it? How many of these are tetrahedral voids?
Ans:
No. of atoms in close packings 0.5 mol =0.5 x 6.022 x 10²³ =3.011 x 10²³
No. of octahedral voids = No. of atoms in packing =3.011 x 10²³
No. of tetrahedral voids = 2 x No. of atoms in packing
= 2 x 3.011 x 10²³= 6.022 x 10²³Total no. of voids = 3.011 x 10²³+ 6.022 x 10²³
= 9.033 x 10²³

1.16. A compound is formed by two elements M and N. The element N forms ccp and atoms of the element M occupy 1/3 of the tetrahedral voids. What is the formula of the compound? (C.B.S.E. Foreign 2015)
Ans: Let us suppose that,
the no. of atoms of N present in ccp = x
Since 1/3rd of the tetrahedral voids are occupied by the atoms of M, therefore,
the no. of tetrahedral voids occupied = 2x/3
The ratio of atoms of N and M in the compound = x : 2x/3 or 3 : 2
∴ The formula of the compound = N₃M₂ or M₂N₃

1.17. Wh ich of the following lattices has the highest packing efficiency (i) simple cubic (ii) body-centered cubic and (iii) hexagonal close-packed lattice?
Ans: Packing efficiency of:
Simple cubic = 52.4% bcc = 68% hcp = 74%
hcp lattice has the highest packing efficiency.

1.18. An element with molar mass 2:7 x 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2:7 x 103 kg m^-3, what is the nature of the cubic unit cell ? (C.B.S.E. Delhi 2015)
Ans:

Since there are four atoms per unit cell, the cubic unit cell must be face centred (fcc) or cubic close packed (ccp).

1.19. What type of defect can arise when a solid is heated? Which physical property is affected by it and in what way?
Ans: When a solid is heated, vacancy defect is produced in the crystal. On heating, some atoms or ions leave the lattice site completely, i.e., lattice sites become vacant. As a result-of this defect, density of the substances decreases.

1.20. What types of stoichiometric defects are shown by (C.B.S.E. Delhi 2013)
(i) ZnS
(ii) AgBr?
Ans:
(i) ZnS crystals may show Frenkel defects since the cationic size is smaller as compared to anionic size.
(ii) AgBr crystals may show both Frenkel and Schottky defects.

1.21. Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it.
Ans: Let us take an example NaCl doped with SrCl, impurity when SrCl2 is added to NaCl solid as an impurity, two Na+ ions will be replaced and one of their sites will be occupied by Sr21- while the other will remain vacant. Thus, we can say that when a cation of higher valence is added as an impurity to an ionic solid, two or more cations of lower valency are replaced by a cation of higher valency to maintain electrical neutrality. Hence, some cationic vacancies are created.

1.22. Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example.
Ans: Let us take an example of NaCl. When NaCl crystal is heated in presence of Na vapour, some Cl^–ions leave their lattice sites to combine with Na to form NaCl. The e^-1 s lost by Na to form Na⁺ (Na⁺ + Cl^–—> NaCl) then diffuse into the crystal to occupy the anion vacancies. These sites are called F-centres. These e-s absorb energy from visible light, get excited to higher energy level and when they fall back to ground state, they impart yellow colour to NaCl crystal.

1.23. A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong?
Ans: Impurity from group 15 should be added to get n-type semiconductor.

1.24. What type of substances would make better permanent magnets, ferromagnetic or ferrimagnetic. Justify your answer.
Ans: Ferromagnetic substances make better permanent magnets. This is because when placed in magnetic field, their domains get oriented in the directions of magnetic field and a strong magnetic field is produced. This ordering of domains persists even when external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

NCERT EXERCISES

1.1. Define the term ‘amorphous’. Give a few examples of amorphous solids.
Sol. Amorphous solids are those substances, in which there is no regular arrangement of its constituent particles, (i.e., ions, atoms or molecules). The arrangement of the constituting particles has only short-range order, i.e., a regular and periodically repeating pattern is observed over short distances only, e.g., glass, rubber, and plastics.

1.2. What makes glass different from a solid such as quartz? Under what conditions could quartz be converted into glass?
Sol. Glass is a supercooled liquid and an amorphous substance. Quartz is the crystalline form of silica (SiO₂) in which tetrahedral units SiO₄ are linked with each other in such a way that the oxygen atom of one tetrahedron is shared with another Si atom. Quartz can be converted into glass by melting it and cooling the melt very rapidly. In the glass, SiO₄ tetrahedra are joined in a random manner.

1.3 Classify each of the following solids as ionic, metallic, modular, network (covalent), or amorphous:
(i) Tetra phosphorus decoxide (P₄O₁₀) (ii) Ammonium phosphate, (NH₄)₃PO₄ (iii) SiC (iv) I₂ (v) P₄(vii) Graphite (viii), Brass (ix) Rb (x) LiBr (xi) Si
Sol.

1.4 (i) What is meant by the term ‘coordination number’?
(ii) What is the coordination number of atom
(a) in a cubic close-packed structure?
(b) in a body centred cubic structure?
Sol. (i) The number of nearest neighbours of a particle are called its coordination number.
(ii) (a) 12 (b) 8

1.5. How can you determine the atomic mass of an unknown metal if you know its density and dimensions of its unit cell ? Explain your answer. (C.B.S.E. Outside Delhi 2011)
Sol.

1.6 ‘Stability of a crystal is reflected in the magnitude of its melting points’. Comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules?
Sol. Higher the melting point, greater are the forces holding the constituent particles together and thus greater is the stability of a crystal. Melting points of given substances are following. Water = 273 K, Ethyl alcohol = 155.7 K, Diethylether = 156.8 K, Methane = 90.5 K.
The intermoleoilar forces present in case of water and ethyl alcohol are mainly due to the hydrogen bonding which is responsible for their high melting points. Hydrogen bonding is stronger in case of water than ethyl alcohol and hence water has higher melting point then ethyl alcohol. Dipole-dipole interactions are present in case of diethylether. The only forces present in case of methane is the weak van der Waal’s forces (or London dispersion forces).

1.7. How will you distinguish between the following pairs of terms :
(a) Hexagonal close packing and cubic close packing
(b) Crystal lattice and unit cell
(c) Tetrahedral void and octahedral void.
Sol.
(a) In hexagonal close packing (hcp), the spheres of the third layer are vertically above the spheres of the first layer
(ABABAB……. type). On the other hand, in cubic close packing (ccp), the spheres of the fourth layer are present above the spheres of the first layer (ABCABC…..type).
(b) Crystal lattice: It deplicts the actual shape as well as size of the constituent particles in the crystal. It is therefore, called space lattice or crystal lattice.
Unit cell: Each bricks represents the unit cell while the block is similar to the space or crystal lattice. Thus, a unit cell is the fundamental building block of the space lattice.
(c) Tetrahedral void: A tetrahedral void is formed when triangular void made by three spheres of a particular layer and touching each other.

Octahedral void: An octahedral void or site is formed when three spheres arranged at the corners of an equilateral triangle are placed over anothet set of spheres.

1.8 How many lattice points are there is one unit cell of each of the following lattices?
(i) Face centred cubic (if) Face centred tetragonal (iii) Body centred cubic
Sol.

1.9 Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Sol. (i) Metallic and ionic crystals
Similarities:
(a) There is electrostatic force of attraction in both metallic and ionic crystals.
(b) Both have high melting points.
(c) Bonds are non-directional in both the cases.
Differences:
(a) Ionic crystals are bad conductors of electricity in solids state as ions are not free to move. They can conduct electricity only in die molten state or in aqueous solution. Metallic crystals are good conductors of electricity in solid state as electrons are free to move.
(b) Ionic bond is strong due to strong electrostatic forces of attraction.
Metallic bond may be strong or weak depending upon the number of valence electrons and the size of the kernels.
(ii) Ionic solids are hard and brittle.Ionic solids are hard due to the presence of strong electrostatic forces of attraction. The brittleness in ionic crystals is due to the non- directional bonds in them.

1.10 Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic, (ii) body centred cubic, and (iii) face centred cubic (with the assumptions that atoms are touching each other).
Sol. Packing efficiency: It is the percentage of total space filled by the particles.

1.11 Silver crystallises in fcc lattice. If edge length of the cell is 4.07 x 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Sol.

1.12. A cubic solid is made of two elements P and Q. Atoms Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What is the co-ordination number of P and Q?
Sol. Contribution by atoms Q present at the eight corners of the cube = 18= x 8 = 1
Contribution by atom P present at the body centre = 1
Thus, P and Q are present in the ratio 1:1.
∴ Formula of the compound is PQ.
Co-ordination number of atoms P and Q = 8.

1.13 Niobium crystallises in a body centred cubic structure. If density is 8.55 g cm^-3, calculate atomic radius of niobium, using its atomic mass 93u.
Sol.

1.14 If the radius of the octahedral void is r and radius of the atoms in close-packing is R, derive relation between rand R.
Sol. A sphere is fitted into the octahedral void as shown in the diagram.

1.15 Copper crystallises into a fee lattice with edge length 3.61 x 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Sol.

This calculated value of density is closely in agreement with its measured value of 8.92 g cm³.

Question 16.
Analysis shows that nickel oxide has the formula Ni_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions?
Solution:
98 Ni-atoms are associated with 100 O – atoms. Out of 98 Ni-atoms, suppose Ni present as Ni²⁺ = x
Then Ni present as Ni³⁺ = 98 – x
Total charge on x Ni²⁺ and (98 – x) Ni³⁺ should
be equal to charge on 100 O^2- ions.
Hence, x × 2 + (98 – x) × 3 = 100 × 2 or 2x + 294 – 3x = 200 or x = 94
∴ Fraction of Ni present as Ni²⁺ = 9498 × 100 = 96%
Fraction of Ni present as Ni³⁺ = 498 × 100 = 4%

Question 17.
What are semi-conductors? Describe the two main types of semiconductors and contrast their conduction mechanisms.
Solution:
Semi-conductors are the substances whose conductivity lies in between those of conductors and insulators. The two
main types of semiconductors are n-type and p-type.
(i) n-type semiconductor: When a silicon or germanium crystal is doped with group 15 element like P or As, the dopant atom forms four covalent bonds like Si or Ge atom but the fifth electron, not used in bonding, becomes delocalised and continues its share towards electrical conduction. Thus silicon or germanium doped with P or As is called H-type semiconductor, a-indicative of negative since it is the electron that conducts electricity.

(ii) p-type semiconductor: When a silicon or germanium is doped with group 13 element like B or Al, the dopant is present only with three valence electrons. An electron vacancy or a hole is created at the place of missing fourth electron. Here, this hole moves throughout the crystal like a positive charge giving rise to electrical conductivity. Thus Si or Ge doped with B or Al is called p-type semiconductor, p stands for positive hole, since it is the positive hole that is responsible for conduction.

Question 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?
Solution:
The ratio less than 2 : 1 in Cu₂0 shows cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. For maintaining electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to the presence of these positive holes, hence it is a p-type semiconductor.

Question 19.
Ferric oxide crystallises in a hexagonal dose- 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:
Suppose the number of oxide ions (O^2-) in the packing = 90
∴ Number of octahedral voids = 90
As 2/3^rd of the octahedral voids are occupied by ferric ions, therefore, number of ferric ions 2 present = 23 × 90 = 60
∴ Ratio of Fe³⁺ : O^2- = 60 : 90 = 2 : 3
Hence, the formula of ferric oxide is Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or n-type semiconductor :

Ge doped with In
B doped with Si.

Solution:

Ge is group 14 element and In is group 13 element. Hence, an electron deficient hole is created and therefore, it is a p – type semiconductor.
B is group 13 element and Si is group 14 element, there will be a free electron, So, it is an n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallises in a face centred unit cell. What is the length of the side of the unit cell ?
Solution:
For a face centred cubic unit cell (fcc)
Edge length (a) = 22–√r = 2 x 1.4142 x 0.144 mm = 0.407 nm

Question 22.
In terms of band theory, what is the difference

between a conductor and an insulator
between a conductor and a semiconductor?

Solution:
In most of the solids and in many insulating solids conduction takes place due to migration of electrons under the influence of electric field. However, in ionic solids, it is the ions that are responsible for the conducting behaviour due to their movement.

(i) In metals, conductivity strongly depends upon the number of valence electrons available in an atom. The atomic orbitals of metal atoms form molecular orbitals which are so close in energy to each other, as to form a band. If this band is partially filled or it overlaps with the higher energy unoccupied conduction band, then electrons can flow easily under an applied electric field and the metal behaves as a conductor.

If the gap between valence band and next higher unoccupied conduction band is large, electrons cannot jump into it and such a substance behaves as insulator.

(ii) If the gap between the valence band and conduction band is small, some electrons may jump from valence band to the conduction band. Such a substance shows some conductivity and it behaves as a semiconductor. Electrical conductivity of semiconductors increases with increase in temperature, since more electrons can jump to the conduction band. Silicon and germanium show this type of behaviour and are called intrinsic semiconductors. Conductors have no forbidden band.

Question 23.
Explain the following terms with suitable examples :

Schottky defect
Frenkel defect
Interstitial defect
F-centres.

Solution:
(i) Schottky defect : In Schottky defect a pair of vacancies or holes exist in the crystal lattice due to the absence of equal number of cations and anions from their lattice points. It is a common defect in ionic compounds of high coordination number where both cations and anions are of the same size, e.g., KCl, NaCl, KBr, etc. Due to this defect density of crystal decreases and it begins to conduct electricity to a smaller extent.

(ii) Frenkel defect : This defect arises when some of the ions in the lattice occupy interstitial sites leaving lattice sites vacant. This defect is generally found in ionic crystals where anion is much larger in size than the cation, e.g., AgBr, ZnS, etc. Due to this defect density does not change, electrical conductivity increases to a small extent and there is no change in overall chemical composition of the crystal.

(iii) Interstitial defect : When some constituent particles (atoms or molecules) occupy an interstitial site of the crystal, it is said to have interstitial defect. Due to this defect the density of the substance increases.

(iv) F-Centres : These are the anionic sites occupied by unpaired electrons. F-centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal.

Question 24.
Aluminium crystallises in a cubic close packed structure. Its metallic radius is 125 pm.

What is the length of the side of the unit cell?
How many unit cells are there in 1.00 cm³ of aluminium?

Solution:

Question 25.
If NaCI is doped with 10^-3 mol % SrCl₂, what is the concentration of cation vacancies?
Solution:
Let moles of NaCI = 100
∴ Moles of SrCl₂ doped = 10^-3
Each Sr²⁺ will replace two Na+ ions. To maintain electrical neutrality it occupies one position and thus creates one cation vacancy.
∴ Moles of cation vacancy in 100 moles NaCI = 10^-3
Moles of cation vacancy in one mole
NaCI = 10^-3 × 10^-2 = 10^-5
∴ Number of cation vacancies
= 10^-5 × 6.022 × 10²³ = 6.022 × 10¹⁸ mol^-1

Question 26.
Explain the following with suitable example:

Ferromagnetism
Paramagnetism
Ferrimagnetism
Antiferromagnetism
12-16 and 13-15 group compounds.

Solution:
(i) Ferromagnetic substances : Substances which are attracted very strongly by a magnetic field are called ferromagnetic substances, e.g., Fe, Ni, Co and CrO₂ show ferromagnetism. Such substances remain permanently magnetised, once they have been magnetised. This type of magnetic moments are due to unpaired electrons in the same direction.

The ferromagnetic material, CrO₂, is used to make magnetic tapes used for audio recording.

(ii) Paramagnetic substances : Substances which are weakly attracted by the external magnetic field are called paramagnetic substances. The property thus exhibited is called paramagnetism. They are magnetised in the same direction as that of the applied field. This property is shown by those substances whose atoms, ions or molecules contain unpaired electrons, e.g., O₂, Cu²⁺, Fe³⁺, etc. These substances, however, lose their magnetism in the absence of the magnetic field.

(iii) Ferrimagnetic substances : Substances which are expected to possess large magnetism on the basis of the unpaired electrons but actually have small net magnetic moment are called ferrimagnetic substances, e.g., Fe₃O₄, ferrites of the formula M²⁺Fe₂O₄ where M = Mg, Cu, Zn, etc. Ferrimagnetism arises due to the unequal number of magnetic moments in opposite direction resulting in some net magnetic moment.

(iv) Antiferromagnetic substances : Substances which are expected to possess paramagnetism or ferromagnetism on the basis of unpaired electrons but actually they possess zero net magnetic moment are called antiferromagnetic substances, e.g., MnO. Antiferromagnetism is due to the presence of equal number of magnetic moments in the opposite directions

(v) 13-15 group compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds thus obtained are called 13-15 compounds. For example, InSb, AlP, GaAs, etc.

12-16 group compounds : Combination of elements of groups 12 and 16 yield some solid compounds which are referred to as 12-16 compounds. For example, ZnS, CdS, CdSe, HgTe, etc. In these compounds, the bonds have ionic character.

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Class 12th Chapter -15 Communication Systems |NCERT Physics Solutions | Edugrown

NCERT Solutions for Class 12 Physics Chapter – 15 Communications Systems includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Physics Chapter 15 Communications Systems. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 12 Physics Chapter 15 NCERT Solutions will be most helpful to the students to solve their Home works and Assignments on time.

NCERT Solutions for Class 12 Physics Chapter : 15 Communications Systems

NCERT Exercises

Question 1.
Which of the following frequencies will be suit-able for beyond-the-horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Solution:
(b) : 10 MHz will be suitable frequency for sky waves as lower frequency of 10 kHz will require large radiating antenna and higher frequencies 1 GHz and 1000 GHz will pass through the ionosphere and will not be reflected by it.

Question 2.
Frequencies in the UHF range normally propagate by means of:
(a) Ground waves
(b) Sky waves
(c) Surface waves
(d) Space waves
Solution:
(d) : Frequencies in the UHF range normally propagates by means of space waves. The high frequency space waves are ideal for frequency modulation but do not bend with ground.

Question 3.
Digital signals
(i) do not provide a continuous set of values,
(ii) represent values as discrete steps,
(iii) can utilize binary system, and
(iv) can utilize decimal as well as binary systems.

Which of the above statements are true?
(a) (i) and (ii) only
(b) (ii) and (iii) only
(c) (i), (ii) and (iii) but not (iv)
(d) All of (i), (ii), (iii) and (iv).
Solution:
(c) : Decimal system represents a continuous set of values which cannot be utilized by digital signals.

Question 4.
Is it necessary for a transmitting antenna to be at the same height as that of the receiving antenna for the line of sight communication? A TV transmitting antenna is 81 m tall. How much service area can it cover if the receiving antenna is at the ground level?
Solution:
No, for line of sight communication, the two antenna may not be at the same height. Surface area
$A=\pi { d }^{ 2 }=\pi \left( 2hR \right) =\cfrac { 22 }{ 7 } \times 2\times 81\times 6.4\times { 10 }^{ 6 }$
$=3258.5\times { 10 }^{ 6 }sq.metre=3258.5sq.km$

Question 5.
A carrier wave of peak voltage 12 V is used to transmit a message signal. What should be the peak voltage of the modulating signal in order to have a modulation index of 75%?
Solution:
Modulation index, $\mu =\cfrac { { A }_{ m } }{ { A }_{ c } }$ so, peak voltage
${ A }_{ m }={ \mu A }_{ c }=0.75\times 12=9V$

Question 6.
A modulating signal is a square wave, as shown in Figure

The carrier wave is given by c(t) = 2sin (8πt) volts
(i) Sketch the amplitude modulated wave form
(ii) What is the modulation index?
Solution:
(i) The amplitude modulated wave is shown here’:

(ii) Modulation index, $\mu \cfrac { { A }_{ m } }{ { A }_{ c } } =\cfrac { 1V }{ 2V } =0.5$

Question 7.
For an amplitude modulated wave, the maximum amplitude is found to be 10 V while the minimum amplitude is found to be 2 V. Determine the modulation index, p. What would be the value of p if the minimum amplitude is zero volt?
Solution:
We know
Modulation index, $\mu =\cfrac { { A }_{ m } }{ { A }_{ c } }$
Also, minimum amplitude, ${ A }_{ min }={ A }_{ c }\left( 1-\mu \right)$
Maximum amplitude, ${ A }_{ max }={ A }_{ c }\left( 1+\mu \right)$
So, modulation index, $\mu =\cfrac { { A }_{ max }{ -A }_{ min } }{ { A }_{ max }{ +A }_{ min } }$
or $\mu =\cfrac { 10-2 }{ 10+2 } =\cfrac { 8 }{ 12 } =2/3=0.67$
if A_min = O, then modulation index, $\cfrac { { A }_{ mix }-{ A }_{ min } }{ { A }_{ mix }+{ A }_{ min } } =\cfrac { 10-0 }{ 10+0 } =\cfrac { 10 }{ 10 } =1$

Question 8.
Due to economic reasons, only the upper side band of an AM wave is transmitted, but at the receiving station, there is a facility for generating the carrier. Show that if a device is available which can multiply two signals, then it is possible to recover the modulating signal at the receiver station.
Solution:
Let, the received signal be cos(ω_c + ω_m)t The carrier signal available at the receiving station is A_c cos ω_ct Multiplying the two signals, we get A₁A_c cos (ω_c + ω_m)t cos ω_ct $\cfrac { { A }_{ 1 }{ A }_{ c } }{ 2 } \left[ cos\left( 2{ \omega }_{ c }+{ \omega }_{ m } \right) t+cos{ \omega }_{ m }t \right]$ If this signal is passed through a low pass filter, we can recover the modulating signal $\cfrac { { A }_{ 1 }{ A }_{ c } }{ 2 } cos{ \omega }_{ m }t$

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