CBSE

NCERT Class 12 Chemistry Solutions for Chapter 8  d-and f-Block Elements provides solutions to the questions provided in the textbook. These solutions are provided by the subject experts and are helpful for the students appearing for boards or competitive exams. The students can refer to these and enhance their conceptual knowledge.

NCERT Solutions are provided in the CBSE, MP, UP, and Gujarat boards. The students appearing for these boards can practice through the NCERT Solutions to score well in the examination.

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

The chapter d and f block elements is very important from the examination perspective. It explains the elements in the groups 3-12. A detailed understanding of this chapter will help the students to differentiate between the characteristics of d and f block elements. It also explains a comparative account of lanthanoids and actinoids.

The NCERT Solutions for Class 12 Chemistry Chapter 8 provides in-depth details of d and f blocks elements. The students are advised to refer to these solutions for better results.

NCERT IN-TEXT QUESTIONS

Question 1.
Silver has completely filled d-orbitals (4d10) in its ground state. How can you say that it is a transition metal ?
Answer:
Silver (Z = 47) belongs to group 11 of (7-block (Cu, Ag, Au) and its outer electronic configuration is 4d¹⁰5s¹. It shows + 1 oxidation state (4d10 configuration) in silver halides (e.g. AgCl). However, it can also exhibit + 2 oxidation state (4d⁹ configuration) in compounds like AgF₂ and AgO. Due to the presence of half filled d-orbital, silver is a transition metal.

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?
Answer:
In 3d series from Sc to Zn, all elements have one or more unpaired e^-1 s except Zn which has no unpaired electron as its outer EC is 3d¹⁰4s². Hence, the intermetallic bonding is weakest in zinc. Therefore, enthalpy of atomisation is lowest.

Question 3.
Which of the 3d series of transitional metals exhibits largest oxidation states and why?
Answer:
Mn (Z = 25) with electronic configuration [Ar]3d⁵4s² shows maximum oxidation state (+ 7) in its compounds since it has the maximum number of unpaired five i.e., seven. It shows largest variable oxidation state from + 2 to + 7 ( + 2, + 3, + 4, + 5, + 6, + 7) in its compounds.

Question 4.
The E°(M²⁺/M) value for copper is positive (+ 0·34 V). What is possibly the reason for this ? (C.B.S.E. Outside Delhi 2012, Sample Paper 2012)
Answer:
E°(M²⁺/M) for any metal is based upon three factors which have been discussed in the text part.
M(s) + ∆_aH → M(g) ; (∆_aH = Enthalpy of atomisation)
M(g) + ∆_fH → M²⁺(g) ; (∆_fH = Ionisation enthalpy)
M²⁺(g) + aq → M²⁺(aq); (∆_hydH = Hydration enthalpy)
Copper has very high enthalpy of atomisation (energy required) and low enthalpy of hydration (energy released). In nut shell, the ∆_fH i.e. ionisation enthalpy needed is not compensated by the energy released. Therefore E°(Cu²⁺/Cu) is positive.

Question 5.
How would you account for the irregular variation in ionisation enthalpies (first and second) in first series of transition elements ?
Answer:
The ionisation enthalpies of the transition metals are higher than those of s-block elements and less than the elements of p-block. Thus, these are less electropositive than the elements of s-block and at the same time more electropositive than the elements belonging to p-block present in the same period. In a transition series, the ionisation enthalpies increase from left to right. However, the gaps in the values of the two successive elements are not regular.

Question 6.
Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size and high electronegativity. Hence, they can oxidise the metal to the highest O.S.

Question 7.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺ and why ? (SamplePaper 2011, C.B.S.E. Outside Delhi 2010, 2014)
Answer:
Cr²⁺ is a stronger reducing agent than Fe²⁺. This is quite evident from the E° values ;
E°Cr³⁺/Cr²⁺ = – 0-41 V and E°Fe³⁺/Fe²⁺ = 0-77 V.
Reason : d⁴ → d³ occurs when Cr²⁺ changes to Cr³⁺ ion while d⁶ → d⁵ takes place when Fe²⁺ gets converted to Fe³⁺ ion.

Now, d⁴ → d³ transition is easier as compared to d⁶ → d⁵ transition because in the second case, an electron is removed from a paired orbital which is rather difficult. Therefore, Cr²⁺ is a stronger reducing agent than Fe²⁺.

Question 8.
Calculate the spin magnetic moment of M²⁺(aq) ion (Z = 27).
Answer:
Electronic configuration of element M(Z = 27) : [Ar] 3d⁷4s²
Electronic configuration M²⁺ (aq) ion : 3d⁷ or
Magnetic moment of M²⁺ (aq) ion with n = 3 ; \(\mu =\sqrt { n\left( n+2 \right) } \)
\(=\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87BM.\)

Question 9.
Explain why Cu⁺ ion is not stable in an aqueous solution. (C.B.S.E. Delhi 2011)
Answer:
In aqueous solution Cu⁺ (aq) undergoes disproportionation to form Cu²⁺ (aq) ion and Cu.
2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)
The higher stability of Cu²⁺ (aq) in an aqueous solution may be attributed to its greater negative ∆_hydH^⊝ than that of Cu⁺ (aq). It compensates for the second ionisation enthalpy of Cu involved in the formation of Cl²⁺ ion. Thus, Cu⁺ (aq) ion changes to Cu²⁺ (aq) ion which is more stable.

Question 10.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why?
Answer:
This is due to poor shielding by 5f-electrons in the actinoids than that by 4f e^-1s in lanthanoids.

NCERT EXERCISE

Question 1.
Write down the electronic configuration of :
(a) Cr³⁺
(b) Cu⁺
(c) Co²⁺
(d) Mn²⁺
(e) Pm³⁺
(f) Ce⁴⁺
(g) Lu²⁺
(h) Th⁴⁺
Answer:

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ towards oxidation to their+3 state?
Answer:
A half-filled d orbital or a completely filled d orbital is more stable than any other state. Mn²⁺ already has a half-filled stable state hence would not undergo oxidation to form Mn⁺³. On the other hand, Fe²⁺ on oxidation to Fe³⁺ will have half-filled d orbitals which are more stable.

Question 2.
Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to their +3 state?
Answer:
Electronic configuration of Mn²⁺ is 3d⁵ while that of Fe²⁺ is 3d6. This shows that the Fe²⁺ ion has an urge to change to Fe³⁺ ion by losing an electron whereas the Mn²⁺ ion has no such tendency. Thus, the + 2 oxidation state of Mn is more stable as compared to the + 2 oxidation state of Fe.

Question 3.
Explain briefly how + 2 oxidation state becomes more and more stable in the first half of the first-row transition elements with increasing atomic number.
Answer:
In all the elements listed, with the removal of valence 45 electrons (+2 oxidation state), the 3d-orbitals get gradually occupied. Since the number of empty d-orbitals decreases or the number of unpaired electrons in 3d orbitals increases, the stability of the cations (M²⁺) increases from Sc²⁺to Mn²⁺.

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 an example.
Answer:
In the transition series, the oxidation states which lead to exactly half-filled or completely filled d-orbitals are more stable. For example, the electronic configuration of Fe(Z = 26) is [Ar] 3d⁶4s². It shows various oxidation states but Fe (III) is most stable because it has the configuration [Ar]3d⁵.

Question 5.
What must be the stable oxidation state of the transition elements with the following electronic configuration in the ground states of their atoms: 3d³, 3d⁵, 3d⁸, 3d⁴?
Answer:
The maximum oxidation states of reasonable stability in the transition metals of 3d series correspond to the sum of s and d-electrons upto Mn. However, after Mn there is an abrupt decrease in oxidation states. In the light of this, most stable oxidation states of the elements are:
3d³ : 3d³s² (+ 5);3 : 3d⁵4s¹ (+ 6) and 3d⁵4s² (+ 7)
3d⁸ : 3d⁸4s² (+ 2); 3d⁴ : 3d⁴4s² or 3d⁵4s¹ (+6)

Question 6.
Name the oxometal anions in the first transition series of transition metals in which the metal exhibits an oxidation state equal to its group number.
Answer:
Vanadate: VO₃^–
chromate: CrO₄^2-
permanganate: MnO₄^–

Question 7.
What is lanthanoid contraction? What are the consequences of lanthanoid contraction? (C.B.S.E. Delhi 2013)
Answer:
One common property associated with the elements in the periodic table is the variation in their atomic and ionic radii down the group and along a period. In general, these increase down the group due to the increase in the number of shells and decrease along a period considerably because of the increase in the magnitude of the effective nuclear charge.
Consequences of Lanthanoid Contraction
(a) Separation of Lanthanoids: Separation of lanthanoids is possible only due to lanthanoid contraction. All the lanthanoids have quite similar properties and due to this reason, they are difficult to separate. However, because of lanthanoid contraction, their properties (such as ionic size, ability to form complexes etc.,) vary slightly.

(b) Variation in basic strength of hydroxides: The basic strength of oxides and hydroxides decreases from La(OH)₃ to Lu(OH)₂. Due to lanthanoid contraction, size of M³⁺ ions decreases and thus there is a corresponding increase in the covalent character in M—OH bond. The acidic strength which involves the cleavage of the O—H bond follows the reverse trend i.e. it increases along the series.

(c) Similarly in the atomic sizes of the elements of the second and third transition series present in the same group: We know that the atomic sizes of the elements generally increase appreciably down a group. Similar trend is also expected in the elements present in the different groups of d-block.

(d) Variation in standard reduction potential: Due to lanthanoid contraction there is a small but steady increase in the standard reduction potential (E°) for the reduction process.
M³⁺ (aq) + 3e^– → M (aq)

(e) Variation in physical properties like melting point, boiling point, hardness etc: Various physical properties like m.pt., b.pt., hardness etc., increase with the increase in atomic number. This is because the attraction forces between the atoms increase as the size decreases.

Question 8.
What are the characteristics of transition elements and why are they called transition elements? Which of the d- block elements may not be regarded as the transition elements?
Answer:
In the transition elements, d-orbitais are successively filled. The general electronic configuration of transition elements is (n – 1) d^1-10 ns^1-2. There are three transition series. The first transition series involves the filling of 3d-orbitais. It starts from scandium (Z = 21) and goes upto zinc (Z = 30).

The second transition series involves the filling of 4 d-orbitais and includes 10 elements from yttrium (Z = 39) to cadmium (Z = 48). The third transition series invokes the filling of 5d-orbitals. The first element of this series lanthanum (Z = 57). It is followed by fourteen elements called lanthanides which involve the filling of 4f-orbitais. The net nine elements from hafnium (Z = 72) to mercury (Z = 80) belong to the third transition series.

There is an incomplete fourth transition series. it involves the filling of 6d- subshell starting from actinium (Z = 89) followed by elements with atomic number 104 onwards.

Question 9.
In what way is the electronic configuration of the transition elements different from that of the non-transition elements?
Answer:
Transition elements contain partially filled d-orbitals whereas non-transition elements have no d-orbitals or have completely filled d -orbitals.

Question 10.
What are the different oxidation states exhibited by lanthanoids?
Answer:
+3 is the common oxidation state of the lanthanoids.
In addition to +3, oxidation states +2 and +4 are also exhibited by some of the lanthanoids.

Question 11.
Explain giving the reason:
(a) Transition metals and many of their compounds show paramagnetic behaviour. (H.P. Board 2014)
(b) The enthalpies of atomisation of transition metals are high. (Jharkhand Board 2010, C.B.S.E. Outside Delhi 2008, 2012, H.P. Board 2014)
(c) The transition metals generally form coloured compounds. (C.B.S.E. 2010. 2012)
(d) The transition metals and their compounds act as good catalysts. (C.B.S.E.Outside Delhi 2010) (C.B.S.E. Delhi 2008, Sample Paper 2010, H.P. Board 2017)
Answer:
(i) Paramagnetic arises due to the presence of unpaired electrons, each such electron has a magnetic moment associated with it due to its spin angular momentum. Transition metals have in its ground state or ionized state has a number of unpaired d-electrons which gives them a paramagnetic behaviour.

(ii) Transition metals have very high interatomic metallic interaction due to the involvement of greater number of electrons from (n – l)d in addition to the ns electrons. The greater the number of valence electrons, the stronger is the resultant bonding due to the greater overlapping of half-filled orbitals. Hence, more amount of energy is required to break these metallic bonds. Thus enthalpy of atomisation of transition metal is very high.

(iii) Colour of transition metal – compounds is due to the excitation of an electron from a lower energy d-orbital to a higher energy d orbital. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary colour of the light absorbed (whose frequency lies on the visible region). The frequency of the light absorbed depends on the nature of ligand. Transition metals form coloured compounds due to the presence of vacant d-orbitals for the d-d transition of e’ which causes the colour.

(iv) The catalytic activity of transition metals is ascribed to their ability to adopt multiple oxidation states and to form complexes. Catalyst at a solid surface involve the formation of bonds between reactant molecules and atoms of the surface of the catalyst this has the effect of increasing the concentration of the reactants at the catalyst surface and also weakening of the bonds in the reacting molecules (activation energy is lowered. Transition metals have d and s orbitals to from these bonds.

Question 12.
What are interstitial compounds? Why are such compounds well known for transition metals?
Answer:
Interstitial compounds are the compounds formed as a result of the trapping of atoms of small elements like H, N, C, B etc. in the crystal lattices of certain metals. These are non-stoichiometric in nature and are neither ionic nor covalent. In fact, no proper bonds exist in the atoms of metals and non-metals involved in these compounds. Transition metals have a tendency to form such compounds. A few examples are: TiC,
M_n4N, Fe₃H, VH_0·56, V_se0·98 and Fe_0·94O etc.

1. These are generally non-stoichiometric in nature. Therefore, they cannot be represented by a definite structure or formula.
2. The compounds are neither covalent nor ionic and they don’t represent the normal oxidation states of the metals.
3. Since the strengths of the metallic bonds in these compounds increase due to greater electronic interactions, they show high melting points and high metallic conductivity. However, these compounds are chemically inert.
4. The conductivity of the metals remains unaffected in the corresponding interstitial compounds.

Question 13.
How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate with examples.
Answer:
The variability of oxidation states, characteristics 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 This is in contrast with the variability of oxidation states of non-transition elements where oxidation states normally differ by a unit of two.
eg: Vanadium: V⁺², V⁺³, V⁺⁴, V⁺⁵
Chromium : Cr⁺², Cr⁺³, Cr⁺⁴, Cr⁺⁵, C⁺⁶
Nitrogen: +5, +3, +1, -1, -3.

Question 14.
Describe the preparation of potassium dichromate from chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?
Answer:
Preparation from chromite:
Potassium dichromate is generally prepared from chromite ore (FeCr₂O₄). It is in fact, a mixed oxide Fe0.Cr₂O₃ of iron and chrome also called ferrochrome or chrome iron.
(i) Conversion of chromite ore into sodium chromate: Chromite ore is fused with sodium hydroxide or sodium carbonate in the presence of air.

(ii) Conversion of sodium chromate into sodium dichromate : The fused mass obtained above is extracted with water. Sodium chromate which is soluble in water goes into the solution leaving behind the insoluble ferric oxide (Fe₂O₃). The yellow solution of sodium chromate obtained above is treated with concentrated H₂SO₄ to form sodium dichromate which has an orange colour.

(iii) Conversion of sodium dichromate into potassium dichromate: Sodium dichromate is more soluble and less stable than potassium dichromate.

Effect of increasing pH : The solution of potassium dichromate (K₂Cr₂O₇) in water is orange in colour. On increasing the pH i.e. on adding the base, the potassium dichromate changes to potassium chromate (K₂CrO₄) which is yellow in colour. Thus, on increasing the pH, the colour of the solution changes from orange to yellow.

Question 15.
Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(a) iodide,
(b) iron (II) solution
(c) H₂S.
Answer:

Question 16.
Describe the preparation of potassium permanganate. How does acidified permanganate solution react with:
(a) iron (II) solution
(b) SO₂
(c) oxalic acid?
Write the ionic equations for the reactions.
Answer:
Potassium permanganate is prepared on a large scale from the mineral pyrolusite, MnO₂.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

Question 17.
For M²⁺/M and M³⁺/M²⁺ systems, the E° values of some metals are given :

Use this data to comment upon :
(a) The stability of Fe³⁺ in acid solution as compared to that of Cr³⁺ or Mn³⁺.
(b) The ease with which iron can be oxidised as compared to the similar process for either chromium or manganese metal.
Answer:
As \({ E }_{ { Cr }^{ 3+ }/{ Cr }^{ 2+ } }^{ \circ }\) is negative (-0·4 V), this means that Cr³⁺ ions in solution cannot be reduced to Cr²⁺ ions or Cr³⁺ ions are very stable. As farther comparison of E° values shows that Mn³⁺ ions can be reduced to Mn²⁺ ion more readily than Fe³⁺ ions. Thus, in the light of this, the order of relative stabilities of different ions is Mn³⁺ < Fe³⁺ < Cr³⁺.
(b) From the E° values, the order of oxidation of the metal to the divalent cation is Mn > Cr > Fe.

Question 18.
Predict which of the following will be coloured in an aqueous solution? Ti³⁺, V³⁺,Cu⁺, Sc³⁺, Mn²⁺, Fe³⁺ and Co²⁺ Give reasons for each.
Answer:
Among the above mentioned, ions Ti³⁺, V³⁺ Mn²⁺, Fe³⁺, CO²⁺ will be coloured in its aqueous solution due to the ability of e’ to jump from a lower energy d orbital to a higher energy d orbital. In case of the ions Cu⁺, Sc³⁺ this d-d transition cannot take place either due to the absence of any e- in 3d orbital or due to complete filling of d orbital.

Question 19.
Compare the stability of the +2 oxidation state for the elements of the first transition series.
Answer:
The common oxidation state of 3d series elements is + 2 which arises due to the participation of only 4s electrons. The tendency to show the highest oxidation state increases from Sc to Mn then decreases due to the 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, the 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 and
(iii) oxidation state
(iv) chemical reactivity.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] ^0-14 5d ^0-1 6s² and that of actinoids is [Rn] 5f ^0-14 6d^0-1 7s². Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical activity
In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 21.
How would you account for the following :
(a) Of the d⁴ species, Cr²⁺ is strongly reducing while Mn³⁺ is strongly oxidising in nature.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.
(c) d¹ configuration is very unstable in ions.
Answer:
(a) E° value of Cr³⁺/Cr²⁺ is negative (-0·41 V) while that of Mn³⁺/Mn²⁺ is positive (+ 1·57 V). This means Cr²⁺ ions can lose electrons to form Cr³⁺ ions and act as a reducing agent while Mn³⁺ ions can accept electrons and can act as oxidising agent.
(b) Cobalt (II) is stable in aqueous solution but in the presence of complexing agent, it undergoes change in oxidation state from +2 to +3 and is easily oxidised.
(c) The ion with d¹ configuration is expected to be extremely unstable and has a great urge to acquire d° configuration (very stable) by losing the only electron present in the d-subshell.

Question 22.
What is meant by ‘disproportionation’ ? Give two examples of disproportionation reaction in aqueous solution.
Answer:
In a disproportionation reaction, an element undergoes an increase as well as decrease in its oxidation state forming
two different compounds. In other words, we can say that it can act both as reducing agent as well as oxidising agent.

Question 23.
Which metal in the first series of transition metals exhibits + 1 oxidation state most frequently and why?
Answer:
Copper, because with +1 oxidation state an extra stable configuration, 3d¹⁰ results.

Question 24.
Calculate the number of unpaired electrons in the following gaseous ions :
Mn³⁺, Cr³⁺, V³⁺.
Answer:

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 the highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.
Answer:
(i) In case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atoms are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding. As a result, it can accept electrons and behave as an acid. For example, Mn^IIO is basic and Mn^VIIO is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits oxidation states in oxides and fluorides. For example, in OsF₆ and V₂O₅, the oxidation states of Os and V are + 6 and respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size so, oxoanions of metal have the highest oxidation state. For example, in MnO₄^–, the oxidation state of Mn is +7.

Question 26.
Give the steps in the preparation of (C.B.S.E. Delhi 2009 comptt.)
(a) K₂Cr₂O₇ from chromite ore
(b) KMnO₄ from pyrolusite ore.
Answer:
(a)

(b)

Question 27.
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer:
An alloy is a homogeneous mixture of two or more metals or metals and non-metals. An important alloy that contains lanthanoid metal is mischmetal which contains 50% Cerium and 25 % Lanthanum, with small amounts of Nd (Neodymium) and Pr (Praseody-mium). It is used in Mg-based alloy to produce bullets, shells, and lighter flints.

Question 28.
What are inner transition elements? Decide which of the following atomic numbers belong to inner transition elements :
29, 59, 74, 95, 102, 104.
Answer:
The inner transition elements also called/-block elements include the series of lanthanoids (Z = 58 to 71) and actinoids (Z = 90 to 103). This means that the elements with atomic numbers 59, 95, and 102 belong to inner transition elements.

Question 29.
The chemistry of actinoid elements is not so smooth as that of lanthanoids. Justify this statement by giving some examples of the oxidation states of these elements.
Answer:
Lanthanoids show a limited number of oxidation states, such as +2, +3 and +4 (+3 is the principal oxidation state). This is because of large energy gap between 5d and 4f- subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation state also. For example, Uranium (Z = 92) exhibits oxidation states of +3, +4, +5, +6 and Neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of small energy difference between 5f and 6d orbitals.

Question 30.
Which is the last element in the series of actinoids? Write the electronic configuration of the element. Comment upon its possible oxidation state.
Answer:
Lawrencium (Lr = 103); [Rn] 5f¹⁴6d¹7s² oxidation state = +3.

Question 31.
Use Hund’s rule to derive the electronic configuration of Ce3+ ion and calculate its magnetic moment on the basis of the spin-spin formula.
Answer:
Cerium electronic configuration = [Xe]4f¹ 5d¹ 16s²
Ce⁺³ ion = [Xe]4f¹
i.e., one unpaired electron is present
Magnetic moment, µ = \(\sqrt{n(n+2)}\) = \(\sqrt { 3 }\) = 1.73BM.

Question 32.
Name the members of the lanthanoid series which exhibit+4oxidatk>nstatesandthosewhichexhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.
Answer:
+ 4 oxidation state in Ce (Z = 58), Pr (Z = 59), Tb (Z = 65).
+ 2 oxidation state in Nd (Z = 60), Sm(Z=62), Eu (Z = 63), Tm (Z=69), Yb (Z = 70).
+ 2 oxidation state is exhibited when the lanthanoid has the configuration 5cf 6s2 so that two electrons are-easily lost.
+ 4 oxidation state is exhibited by the elements which after losing four electrons acquire configuration 4f° or 4f¹

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.
Answer:
(i) Electronic configuration:
The general electronic configuration for lanthanoids is [Xe] ^0-14 5d ^0-1 6s² and that of actinoids is [Rn] 5f ^0-14 6d^0-1 7s². Unlike 4f orbitals, 5 f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states:
The principal oxidation states of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4. This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the 5f, 6d and 7s levels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in the +3 state than in the +4 state.

(iii) Atomic and ionic size Similar to lanthanoids, actinoids also exhibit actinoid contraction overall decrease in atomic and ionic radii. The contraction is greater due to the poor shielding effect of 5f orbitals. Hence there is an unexpected in the atomic and ionic sizes of actinoids.

(iv) Chemical reactivity
In the lanthanoid series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In the case of acids, they are slightly affected by nitric acid (because of the -formation of a protective oxide layer).

Question 34.
Write the electronic configuration of the elements with atomic numbers 61, 91, 101, 109.
Answer:
Promethium or Pm (Z = 61) [Xe]⁵⁴4f⁵5d⁰6s²
Protactinium or Pa (Z = 91) [Rn] 4f²6d¹7s²
Mendelevium or Md (Z = 101) [Rn] 5f¹6d⁰7s²
Meitnerium or Mt (Z = 109) [Rn] 5f¹⁴6d⁷7s²

Question 35.
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in the respective vertical columns. Give special emphasis on the following points :
(i) electronic configuration
(ii) oxidation states
(iii) ionisation enthalpies
(iv) atomic sizes.
Answer:
(i) Electronic configuration. There are some exceptions in the electronic configurations in all the three series.

(ii) Oxidation state. The elements belonging to the different series but present in the same group have similar electronic configurations and therefore, exhibit almost same variable oxidation states. In general, these are maximum in the middle of the series while minimum towards the end. Transition elements show variable oxidation states due to the participation of ns and (n – 1) d electrons in bonding because the energies of ns and (n – 1) d-subshells are quite close. The stability of a particular oxidation state depends upon the nature of the element with which the transition metal forms the compound.

(iii) Ionisation enthalpies. In general, the ionisation enthalpies in all three transition series increase from left to the right. However, the gaps in the two successive elements in a particular series are small and are also not regular. The first three ionisation enthalpies of the elements present in the first transition series are given in the text part. The ∆_iH₁ [ values of the elements belonging to the 5d series and higher as compared to those belonging to 3d and Ad series in the same group because of poor shielding by intervening 4f electrons present.

(iv) Atomic size. In all three transition series, the atomic, as well as ionic radii of the elements, increase from left to right. The values for 3d series are given in the text part. However, the increase in their values is not as much as expected since the shielding by (n – 1 )d electrons is not as much as expected. In a particular group, the atomic radius of the elements belonging to Ad series is more than the elements in the 3d series. However, the gaps in the elements belonging to Ad and 5d series are negligible on account of lanthanoid contraction which the elements of the 5d experience.

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).
Answer:
The number of 3d electrons in the ions are :

For the explanation of the involvement of 3d orbitals in the hydrated ions (octahedral in nature) consult the next unit on coordination compounds.

Question 37.
Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.
Answer:
The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.
(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements of the 2^nd and 3_rd transition series). However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series this is due to lanthanoid configuration.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomization of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of strong metallic bonding (M – M bonding)

(v) The dements of the first transition series from low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

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

Answer:
The magnetic moment of a compound is given by the relation (µ) = \(\sqrt { n\left( n+2 \right) } \) B.M, where n is the number of unpaired electrons.
For one unpaired electron (n = 1) ; µ = \(\sqrt { 1\left( 1+2 \right) } =\sqrt { 3 } =1\cdot 73\quad B.M.\)
For two unpaired electrons (n – 2) ; µ = \(\sqrt { 2\left( 2+2 \right) } =\sqrt { 8 } =2\cdot 83\quad B.M.\)
For three unpaired electrons (n = 3); µ = \(\sqrt { 3\left( 3+2 \right) } =\sqrt { 15 } =3\cdot 87\quad B.M.\)
For four unpaired electrons (n = 4) ; µ = \(\sqrt { 4\left( 4+2 \right) } =\sqrt { 24 } =4\cdot 9\quad B.M.\)
For five unpaired electrons (n = 5) ; µ = \(\sqrt { 5\left( 5+2 \right) } =\sqrt { 35 } =5\cdot 92\quad B.M.\)
*In the light of the above value, let us gather the desired information about the complex species that are mentioned
(i) K₄[Mn(CN)₆]
Oxidation state of Mn : [Mn(CN)₆]^4- , x + 6(-l) = -4 or x = -4 + 6 = + 2
The magnetic value of 1·73 B.M. indicates the presence of one unpaired electron in the complex. When six, CN^– ions (or ligands) approach Mn²⁺ ion, electrons in 3d orbitals pair up to make available six vacant orbitals involving d²sp³ hybridisation.

The complex is octahedral and is paramagnetic due to one unpaired electron.
(ii) [Fe(H₂O)₆]²⁺
Oxidation state of Fe : [Fe(H₂O)₆]²⁺ ; r + 6 (0) = +2
The magnetic moment value of 5·3 B.M. indicates that there are four unpaired electrons in the complex. This means that the electrons in Fe²⁺ ion do not pair up when six H20 molecules (or ligands) approach it. Since the desired number of vacant orbitals (six) are available, die complex formed is sp³d² hybridised.

The complex is octahedral and is paramagnetic due to four unpaired electrons. It is also called outer orbital complex because 4d (n = 4) orbitals are involved.
(iii) K₂[MnCl₄]
Oxidation state of Mn : [MnCl₄]^2-, x + 4(-1) = -2 or x = -2 + 4= + 2
The magnetic moment value of 5·9 B.M. indicates that there are five unpaired electrons in the complex. This means that all the five 3d orbitals in Mn²⁺ ion are involved in the bond formation. The complex is sp³ hybridized in which one vacant 4s and three vacant 4p orbitals participate.

The complex is therefore, tetrahedral in nature.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 8 d-and f-Block Elements, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 5 are the practice guide for all the students. It contains solved questions provided by the subject matter experts. The solutions are accurate and the answers can be written in the exams.

NCERT Solutions are provided for reference in various boards (CBSE, UP board, Gujarat board, MP board). These help the students practice well before the examination and know their shortcomings. NCERT Solutions not only help the students score well in the board exams but also help them get through competitive exams.

NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry

Class 12 Chemistry Chapter 5 Surface Chemistry is an important chapter and explains the types of chemical reactions, its pocesses and mechanisms that take place at the surface of the material. The chapter also contains the details on various important concepts such as Tyndall effect, Brownian motion, and catalyst reactions. Notes on various topics like colloids, emulsions, electrodialysis, enzyme reactions, electrophoresis, etc. are also provided here for reference.

NCERT IN-TEXT QUESTIONS

Question 1.
Why are substances like platinum and palladium often used for carrying out the electrolysis of aqueous solutions ?
Answer:
The metals like platinum and palladium are used as inert electrodes for carrying out the process of electrolysis because these are not attacked by the ions involved in the process.

Question 2.
Why does physisorption decrease with increase in temperature ?
Answer:
Physisorption or physical adsorption of a gas on the surface of a solid is exothermic in nature.

When temperature is increased, the equilibrium gets shifted in the backward direction to neutralise the effect of increase in temperature. Consequently, physisorption decreases with the increase in temperature.

Question 3.
Why are powdered substances more effective . adsorbents than their crystalline forms?
Answer:
Powdered substances have greater surface area as compared to their crystalline forms. Greater the surface area, greater is the adsorption.

Question 4.
Why is it necessary to remove CO when ammonia is obtained by Haber’s process ?
Answer:
Carbon monoxide (CO) acts as a poison for the catalyst iron as well as promoter molybdenum which are used in the Haber’s process. Moreover, it is likely to combine with iron to form iron carbonyl Fe(CO)₅. Therefore, it is necessary to remove it from the reaction mixture by suitable means.

Question 5.
Why is ester hydrolysis slow in the beginning and becomes fast after sometime ?
Answer:
In ester hydrolysis, an acid and alcohol are formed as the products. For example,

Acid will release H⁺ ions in solution which act as catalyst (auto-catalysis) for the reaction. That is why, the hydrolysis is slow in the beginning and becomes faster later on.

Question 6.
What is the role of desorption in the process of catalysis?
Answer:
Desorption makes the surface of the solid- catalyst-free for fresh adsorption of the reactants on the surface.

Question 7.
What modification can you suggest for Hardy-Schulze Law?
Answer:
According to Hardy-Schulze Law, the ions carrying charge opposite to the charge on sol particles neutralise their charge and thus cause their coagulation or precipitation. The law takes into account the charge carried by the ion and not its size. Smaller the size of the ion more will be its polarising power. Thus, the law should be modified in terms of the polarising power of the flocculating ion or the ion causing the precipitation. The modified form of the law states that “Greater the polarising power of the flocculating ion added, greater is its power to cause precipitation.”

Question 8.
Why is it essential to wash the precipitate with water before estimating it quantitatively?
Answer:
Some amounts of the electrolyte are mixed to form the ppt. Some of these electrolytes remain adsorbed on the surface of the particles of the ppt. Hence, it is essential to wash the ppt with water to remove the sticking electrolytes (or any other impurities) before estimating it quantitatively.

NCERT EXERCISE

Question 1.
Distinguish between the meaning of terms adsorption and absorption. Give one example in each case.
Answer:
Differences between Adsorption and Absorption:

Adsorption:

1. It is a process as a result of which one substance gets concentrated only on the surface of the other.
2. The concentration of adsorbate on the surface of the adsorbent is different than in the bulk.
3. It is a surface phenomenon.
4. Example: Adsorption of water vapour on silica gel.

Absorption:

1. It is a process as a result of which one substance gets uniformly distributed in the volume of the other.
2. Concentration is uniform in the entire solid system.
3. It is a bulk phenomenon.
4. Example: Adsorption of water vapour by dry calcium chloride.

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

Question 3.
Why is a finely divided substance more effective as an adsorbent ?
Answer:
With the increase in surface area of adsorbent adsorption increases. Thus, in the powdered state (finely divided substance) or in porous state surface area of metals is more. Therefore, adsorption is more in these states.

Question 4.
What are the factors which influence the adsorption of a gas on a solid ?
Answer:
There are various factors that affect the rate of adsorption of a gas on a solid surface.

• Nature of the gas: Easily liquefiable gases such as NH₃, HCl, etc. are adsorbed to a great extent in comparison to gases such as H₂, O_2, etc. This is because Van der Waal’s forces are stronger in easily liquefiable gases.
• The surface area of the solid: The greater the surface area of the adsorbent, the greater is the adsorption of a gas on the solid surface.
• Effect of pressure: Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore adsorption increases with an increase in pressure.
• Effect of temperature: Adsorption is an exothermic process. Thus in accordance with Leehatelie’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 5.
What is adsorption isotherm ? Distinguish between Freundlich adsorption isotherm and Langmuir adsorption isotherm.
Answer:
Adsorption isotherm represents the variation of the amount of the gas adsorbed and the corresponding pressure at a certain temperature. The mathematical forms of the two adsorption isotherms are :

The main points of distinction in the two adsorption isotherms are:

• Freundlich Adsorption isotherm is applicable to all types of adsorption whereas Langmuir Adsorption isotherm is applicable mainly to chemical adsorption or chemisorption.
• Freundlich adsorption isotherm fails at high pressure of the gas whereas Langmuir Adsorption isotherm can be applied under all pressures.

Question 6.
What do you understand by activation of adsorbent? How is it achieved?
Answer:
By activating an absorbent, we tend to increase the adsorbing power of the adsorbent. Some ways to activate an adsorbent are:

1. By increasing the surface area of the adsorbent. This can be done by breaking it into smaller pieces or powdering it.
2. Some specific treatments can also lead to the activation of the adsorbent.

For example, wood charcoal is activated by heating it between 650K and 1330K in vacuum pr air. It expels all the gases absorbed or adsorbed and thus, creates a space for the adsorption of gases.

Question 7.
What role does adsorption play in heterogeneous catalysis?
Answer:
Heterogeneous catalysis is generally carried on the surface of the finely divided metals of the transition series. Due to the availability of large surface area, the reacting species get adsorbed on the surface either by physical adsorption or by chemisorption. The adsorbed species get opportunity to mutually combine to form the products which are released or desorbed from the surface so as to accommodate more reacting species.

• Diffusion of the reactants on the surface of the catalyst.
• Some association between the catalyst surface and the reactants i. e., adsorption.
• The occurrence of the chemical reactions on the catalyst surface.
• Dissociation of the reaction products from the catalyst surface i.e., desorption.
• Diffusion of the products from the catalyst surface.

Question 8.
Give two chemical methods for the preparation of colloids.
Answer:
These are formed in two ways:

• Condensation methods
• Dispersion methods.

Condensation methods: The particles of the dispersed phase are very small in size. They have to be condensed suitably to be of colloidal size.
A colloidal solution of sulphur is obtained when H₂S gas is bubbled through the solution of oxidising agent like bromine water, sulphur dioxide, dilute HNO₃ etc.
Dispersion methods: In these methods, bigger particles of a substance (suspension) are broken into smaller particles of colloidal dimensions. The substance whose colloidal solution is to be prepared, is first ground to coarse particles. It is then mixed with the dispersion medium to get a suspension

Question 9.
How are colloidal solutions classified on the basis of physical states of the dispersed phase and dispersion medium?
Answer:
There are in all eight types of colloidal solutions.

Question 10.
Discuss the effect of pressure and temperature on the adsorption of gases on solids.
Answer:
Effect of pressure:
Adsorption is a reversible process and is accompanied by a decrease in pressure. Therefore adsorption increases with an increase in pressure.

Effect of temperature:
Adsorption is an exothermic process. Thus, in accordance with Le-chatelier’s principle, the magnitude of adsorption decreases with an increase in temperature.

Question 11.
What are lyophilic and lyophobic sols? Give one example of each type. Why are hydrophobic sols easily coagulated?
Answer:
Lyophillic colloids (solvent loving) are those substances that directly pass into the colloidal state when brought- in contact with the solvent, e.g., proteins, starch, rubber, etc.
These sols are quite stable because of the strong attractive forces between the particles of disperse phase and the dispersion medium.
Lyophobic colloids (solvent hating) are those substances that do not form the colloidal sol readily when mixed with the dispersion medium.These sols are less stable than the lyophilic sols. Examples of lyophobic sols include sols of metals and their insoluble compounds like sulphides and hydroxides.
The stability of hydrophobic sol is only due to the presence of charge on the colloidal parties. If charge is removed, e.g., by addition of suitable electrolytes, the particles will come nearer to each other to form aggregate, i.e., they will coagulate and settle down. On the other hand, the stability of hydrophilic sol is due to charge as well as solvation of the colloidal particles. Thuf, for coagulation to occur easily both the mentioned factors have to be removed.

Question 12.
What is the difference between muitimolecular and macromolecular colloids ? Give one example of each. How are associated colloids different from these two types of colloids ? (C.B.S.E. 2008, 2009, 2010)
Answer:
Difference between multimolecular and macromolecular colloids
The main points of distinction are listed.

Colloidal sol of sulphur (Sg) is an example of multimolecular colloid while colloidal sol of starch represents macromolecular colloid.

Associated colloids also called micelles, are generally electrolytes. They exist as ions at low concentrations. However, above a particular concentration called critical micellear concentration (CMC) and above a particular temperature called Kraft temperature (T_k), these get associated and exhibit colloidal behaviour. Soap is a common example of associated colloids.

Multimolecular colloids: In these colloids, the individual particles consist of an aggregate of atoms or small molecules with molecular size less than 10³ pm. For example, gold sol consists of particles of various sizes having several atoms. Similarly, a sulphur sol consists of particles each having eight sulphur atoms (S_g). In these colloids, the particles are held by van der Waals’ forces.

Macromolecular colloids: In this type, the particles of the dispersed phase are sufficiently big in size (macro) to be of colloidal dimensions. These are normally polymers. A few naturally occurring macromolecules are starch, cellulose and proteins. The examples of artificial macromolecules are those of polythene, nylon, polystyrene, plastics etc.

Question 13.
What are enzymes? Write in brief the mechanism of enzyme catalysis.
Answer:
Enzymes are complex nitrogenous compounds which are produced by living plants and animals. In fact, these are proteins produced by living systems and catalyse certain biological reactions. These are, therefore, often known as bio-chemical catalysts and this phenomenon is known as bio-chemical catalysis.

The rate of enzyme-catalyzed reaction which is initially of first-order changes to zero-order as the concentration of substrate species on the catalyst surface increases.
Two models have been proposed by bio-chemists to explain the mechanism of enzyme catalyzed reactions. These are briefly discussed.

Question 14.
How are colloids classified on the basis of :
(a) physical states of components
(b) nature of dispersion medium
(c) interaction between the dispersed phase and dispersion medium?
Answer:
(a) Based on physical states of components. Based on the physical states of components i.e., dispersed phase and dispersion medium, there are eight types of colloidal solutions.

(b) Nature of dispersion medium. The dispersion medium can be either gas, liquid or solid. Based upon its nature, the colloids or colloidal solutions are of three types.

• Aerosols: Air or gases act as the dispersion medium
• Liquid sols: Liquids like water, alcohol or benzene act as the dispersion medium.
• Solid sols: Solid acts as the dispersion medium.

(c) Interaction between the dispersed phase and dispersion medium. Colloidal solutions are classified into two types. These are lyophilic and lyophobic sols.
(i) Lyophilic colloids: The colloidal solution in which the particles of the dispersed phase have a great affinity (or love) for the dispersion medium, are called lyophilic colloids. Such solutions are easily formed the moment the dispersed phase and the dispersion medium come in direct contact. e.g., sols of gum, gelatin, starch, etc.

(ii) Lyophobic colloids: The colloidal solutions in which the particles of the dispersed phase have no affinity or love, rather have hatred for the dispersion medium, are called lyophobic colloids. The solutions of metals like Ag and Au, hydroxides like Al(OH)₃ and Fe(OH)₃ and metal sulphides like As₂S₃ are examples of lyophobic colloids.

Question 15.
Explain what is observed
(i) when a beam of light is passed through a colloidal sol.
(ii) an electrolyte, NaCI is added to hydrated ferric oxide sol.
(iii) electric current is passed through a colloidal sol.
Answer:
(i) Scattering of light by colloidal particles takes place and the path of the light becomes visible (Tyndall effect).
(ii) The positively charged colloidal particles of Fe(OH)₃ get coagulated by the oppositely charged Cl^– ions provided by NaCl.
(iii) On passing electric current, the colloidal particles move towards the oppositely charged electrode where they lose their charge and get coagulated. This is the electrophoresis process.

Question 16.
What are emulsions? What are their different types? Give an example of each type.
Answer:
(a) Oil-in-water emulsion (O/W type). In this case, the dispersed phase is oil while the dispersion medium is water. Milk is a common example in which liquid fats are dispersed in water. Similarly, if a few drops of nitrobenzene (oil) is added to water, an emulsion results. Vanishing cream is another example of this type.

(b) Water-in-oil emulsion (W/O type). In this type of emulsions, the dispersed phase is water while the dispersion medium is oil. Butter is an example of water in oil emulsion in which water is dispersed in oil. Cod liver oil and cold cream are the other examples of these emulsions.

Question 17.
What is demulsification? Name two demulsifiers.
Answer:
The process of separation of constituent liquids of an emulsion is called demulsification. Demulsification can be done by centrifuging or boiling.

Question 18.
The action of soap is due to demulsification and micelle formation. Comment.
Answer:
Soap is sodium or potassium salt of a higher fatty acid and may be represented as RCOO^– Na⁺ (e.g., sodium stearate CH₃(CH₂)]₁₆ COO^– 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 ionichead’), 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 the hydrophobic part of the stearate ions is in the oil droplet and the hydrophilic part projects out of the grease droplet like the bristles

Question 19.
Give four examples of heterogeneous catalysts.
Answer:
(i) The combination between nitrogen and hydrogen to form ammonia in the presence of finely divided iron acting as catalyst. This is known as Haber’s process.

(ii) Formation of sulphur trioxide by the oxidation of sulphur dioxide in the presence of platinum catalyst is the basis of the manufacture of sulphuric acid in Contact process.

(iii) Oxidation of ammonia into nitric oxide in the presence of platinum catalyst is employed for the commercial preparation of nitric acid in Ostwald process.

(iv) In the hydrogenation of vegetable oils (unsaturated in nature) resulting in solid fats (saturated in nature), hydrogen gas is passed through the oil in the presence of nickel catalyst at about 473 K.

Question 20.
What do you mean by activity and selectivity of catalysts?
Answer:
(a) Activity: The activity of a catalyst depends upon the strength of chemisorption to a large extent. The reactants must get adsorbed reasonably strongly onto 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 catalyst 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 catalyze another reaction.

Question 21.
Describe some features of catalysis by zeolites.
Answer:

1. Zeolites are hydrated alumino-silicates. They have a three-dimensional network structure. They contain water molecules in their pores,
2. Zeolites are heated to remove the water from hydration. The pores become vacant and zeolites are ready to act as catalysts.
3. The size of the pores varies from 260 pm to 760 pm. This shows that only those molecules can be adsorbed in these pores whose size is small enough to enter these pores. Thus, zeolites a molecular sieve and the shape-selective catalysts.

Question 22.
What are shape-selective catalysts?
Answer:
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 a three-dimensional network of silicates in which some silicon atoms are replaced by aluminium atoms giving an 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 synthesized for catalytic selectivity.

Question 23.
Explain the following terms:
(i) Electrophoresis
(ii) Coagulation
(iii) Dialysis
(iv) Tyndall effect.
Answer:
(i) Electrophoresis:
The movement of colloidal particles under the influence of an electric field is known as electrophoresis. Positively charged particles move to the cathode, while negatively charged particles move towards the anode. As the particles reach oppositely charged electrodes, they become neutral and get coagulated.

(ii) Coagulation:
The process of setting down colloidal particles i.e., conversion of a colloid into a precipitate is called coagulation.

(iii) Dialysis: The process of removing dissolved substances from a colloidal solution by the means of diffusion through a membrane is known as dialysis. This process is based on the principle that ions and small molecules can pass through animal membranes, unlike colloidal particles.

(iv) Tyndall effect:
When a beam of light is allowed to pass through a colloidal solution, it becomes visible like a column of light. This is known as the Tyndall effect. This phenomenon takes place as particles of colloidal dimension scatter light in all directions.

Question 24.
Give four uses of emulsions.
Answer:
Four uses of emulsions:

1. the cleansing action of soaps is based on the formation of emulsions
2. digestion of fats in the intestines takes place by the process of emulsification.
3. Antiseptics and disinfectants when added to water form emulsions.
4. The process of emulsification is used to make medicines.

Question 25.
What are micelles? Give an example of the micelles system.
Answer:
Micelles are substances that behave as normal strong electrolytes at low concentration but at high concentrations behave as colloids due to the 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:

1. Alcosol
2. Aerosol
3. Hydrosol.

Answer:

1. Alcosol: It is a colloidal solution in which alcohol is the dispersion medium. For example, colloids which has cellulose nitrate as a dispersed phase and ethyl alcohol as the dispersion medium.
2. Aerosol: It is a colloidal solution in which liquid is a dispersed phase and gas is a dispersion medium e.g., fog, mist, cloud, etc.
3. Hydrosol: It is a colloidal solution in which solid is a dispersed phase and water is a dispersion, e.g., gold sol, arsenious sulphide sol, ferric oxide sol, etc.

Question 27.
Comment on the statement that colloid is not a substance but a state of substance’.
Answer:
Common salt (atypical crystalloid in an aqueous medium) behaves as a colloid in a benzene medium. Hence, we can say that a colloidal substance does not represent a separate class of substances. When the size of the solute particle lies between 1 nm and 1000 nm, it behaves as a colloid.

Hence, we can say that colloid is not a substance but a state of the substance which is dependent on the size of the particle.
A colloidal state is intermediate between a true solution and a suspension.

We hope the NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 5 Surface Chemistry, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 Chemistry Chapter 6 contains concept wise details of the chapter. The solutions help the students prepare well for the examination. The diagrammatic representations and step wise solutions make it easy for the students to understand. Also the answers are accurate and are provided by the subject matter experts.

NCERT Solutions are provided for reference in various boards (CBSE, MP board, UP board, Gujarat board). The students appearing for these boards and competitive exams can score well by referring to the NCERT Solutions.

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

Class 12 Chemistry chapter 6 explains the applications of various metals in our day-to-day life. Different terminologies such as refining, calcination, concentration, benefaction, roasting, etc. are very well explained here. The concepts of thermodynamics for the extraction of copper, aluminium and zinc are also mentioned here.

This chapter contains the details of various chemical processes associated with metallurgy. Being an important chapter, it will help the students score well during the examinations.

NCERT IN-TEXT QUESTIONS

Question 1.
Name some ores which can be concentrated by magnetic separation method.
Answer:
Only those ores can be concentrated by magnetic separation method in which either the ore particles or the impurities associated with it are of magnetic nature. For example, ores of iron such haematite (Fe₂O₃), magnetite (Fe₃O₄), siderite (FeCO₃) are magnetic and can be concentrated by this method. Similarly, casseterite (SnO₂) an ore of tin is non-magnetic while the impurities of tungstates of iron and chromium are of magnetic nature. Magnetic separation is effective in this case also.

Question 2.
What is the significance of leaching in the extraction of aluminium?
Answer:
Aluminium contains silica (SiO₂), iron oxide (Fe₂O₃) and titanium oxide (TiO₄) as impurities. These impurities can be removed by the process of leaching. During leaching, the powdered bauxite ore is heated with a concentrated (45%) solution of NaOH at 473-523 K, where alumina dissolves as sodium meta-aluminate and silica as sodium silicate leaving Fe₂O₃, TiO₂ and other impurities behind:

The impurities are filtered off and solution of sodium meta-aluminate is neutralised by passing CO₂ when hydrated alumina separates out while sodium silicate remains in solution. The hydrated aluminathus obtained is filtered, dried and heated to give back pure alumina.

Thus, by leaching, pure alumina can be obtained from bauxite ore.

Question 3.
The reaction : Cr₂O₂(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s) ; ∆G° = – 421 kJ is thermodynamically feasible as is apparent from the value of ∆G°. Why does not it take place at room temperature ?
Answer:
Though the reaction is feasible, it does not proceed at room temperature because all the reactants and products are solids. At elevated temperature, when chromium starts melting, the reaction becomes feasible.

Question 4
Is it true that under certain conditions, Mg can reduce Al₂O₃ and Al can reduce MgO ? What are those conditions ?
Answer:
If we look at the Ellingham diagram, it becomes evident that the plots for Al and Mg cross each other at 1350 °C (1623 K). Below this temperature, Mg can reduce Al₂O₃ and above this temperature, Al can reduce MgO.

NCERT EXERCISE

Question 1.
Copper can be extracted by hydrometallurgy but not zinc. Explain.
Answer:
Copper is a comparatively less active metal as its reduction potential i.e. E° (Cu²⁺/Cu) is high (+0-34V). It can be displaced from a solution of Cu²⁺ ions by more active metals which have E° value lower than copper. For example, E° of zinc (Zn²⁺/Zn) is -0.76V and thus, zinc can displace copper from the solution of Cu²⁺ ions. In contrast, to displace zinc from a solution of Zn²⁺ ions a more reactive metal than zinc is required like, Na, K, Mg, Ca, etc. But, the more active metals readily react with water forming their corresponding ions and evolve hydrogen gas.
[2Na + 2H₂O → 2NaOH + H₂],
Thus, it is difficult to displace zinc from a solution of Zn²⁺ ions. Elance, copper can be extracted by hydrometallurgy but not zinc.

Question 2.
What is the role of the depressant in the froth floatation process?
Answer:
A depressant suppresses the formation of froth with a particular compound in the froth floatation process by reacting chemically with it. Thus, it helps in the separation of two metal sulphides present together in a particular ore.

In actual process, the sulphide ore is finely powdered and is mixed with water to form a slurry in a tank as shown in Fig. 6.3. To this oily component of the sulphide ore particles by water. As a result, ore and oil constitute hydrophobic or water-repelling components while gangue and water form a lyophilic or water-attracting component.

Question 3.
Why is the extraction of copper from its sulphide ore difficult than that from its oxide through reduction?
Answer:
Sulphide ore of copper (Cu₂S) cannot be directly reduced by either coke or hydrogen because ∆_fG° of Cu₂S is more than those of CS₂ and H₂S that will be formed as a result of the reaction.
These reactions are, therefore, not feasible. However, the ∆_fG° of Cu₂O is lower than that of CO₂. Therefore, the sulphide ore is first roasted to Cu₂O which is then reduced.

Question 4.
Explain
(i) zone refining
(ii) column chromatography. (C.B.S.E. Sample Paper 2015)
Answer:
(i) Zone refining (Fractional Crystallisation). This method is used only if a metal in almost pure state is required. Metals like germanium and gallium which are used in semiconductors are purified by this method. The principle of zone refining is based on the fact that impurities are more soluble in molten metal than in solid metal.

(ii) Chromatographic method. This method can also be employed on small scale for the purification of certain metals. Adsorption chromatography is normally used for this purpose. Different components present in a given sample are adsorbed to a different extent on the surface of the adsorbent.

Question 5.
Out of C and CO, which is a better reducing agent at 673 K?
Answer:
When carbon reacts with dioxygen, two reactions are possible
C_(s) + O_2(g) → CO_2(g) ……(i)
2C_(s) + O_2(g) → 2CO_(g) …..(ii)
When CO is used as a reducing agent, it gets oxidized to CO₂
2CO + O₂ → 2CO₂ …(iii)
It is clear from the Ellingham diagram that at 673K the ∆G° for the oxidation of CO to CO₂ is more negative than the reaction (i) and reaction (ii). Therefore, CO is a better reducing agent than C. It is supported by the fact that the curve for the reaction (iii) lies below the curve for the reaction (i) and reaction (ii) at 673K. An element below in the Ellingham diagram reduces the oxide of other metal which lies above it.

Question 6.
Name the common elements present in the anode mud in electrolytic refining of copper. Why are they so present?
Answer:
The common elements present in the anode mud are antimony, selenium, tellurium, silver, gold, and platinum. These elements settle down under anode as anode mud because they are less reactive and are not effected by CuSO₄ – H₂SO₄solution.

Question 7.
Write the chemical reactions which take place in different zones in the blast furnace during the extraction of iron.
Answer:

The chemical reactions which take place in the blast furnace are briefly discussed.
Zone of combustion. At the bottom of the furnace, the blast of hot air causes the combustion of coke into carbon dioxide. The reaction is highly exothermic and a temperature of nearly 2170 K develops. It supplies most of the heat required for the process.

Zone of heat absorption. As CO₂ gas rises up, it combines with more coke to form carbon monoxide. Since the reaction is endothermic, the temperature in the middle of the furnace is nearly 1570 K. It further decreases as the reaction proceeds.

Zone of slag formation. In the middle of the furnace, the temperature is nearly 1123 K. Here limestone (CaCO₃) decomposes to form CaO and CO₂. The former (CaO) combines with silica (SiO₂) which is an impurity in the haematite ore to form calcium silicate (CaSiO₃) that is fusible. This implies that CaO has acted as a basic flux. It has combined with the acidic impurity of Si02 to form slag.

Zone of reduction: This is the upper part of the furnace. The temperature ranges between 500K to 900K. Here haematite (Fe₂O₃) is reduced to FeO.

Further reduction of FeO to Fe occurs at higher temperatures (1123 K) by CO gas.

The direct reduction of iron ore left unreacted also occurs with carbon above 1123 K

Question 8.
Write the chemical reactions which take place in the extraction of zinc from zinc blende.
Answer:
Zinc blende is chemically zinc sulphide (ZnS). It undergoes the following reactions:

Zinc metal is in crude or impure form. It can be refined with the help of electro-refining. In this method, impure zinc is made anode while a plate of pure metal acts as the cathode. The electrolyte is aqueous zinc sulphate containing a small amount of dilute H₂SO₄. On passing electric current, Zn²⁺ ions from the electrolyte migrate towards the cathode and are reduced to zinc metal which gets deposited on the cathode.

From anode, an equivalent amount of zinc gets oxidised to Zn²⁺ ions which migrate to the electrolyte

Question 9.
State the role of silica in the metallurgy of copper.
Answer:
During roasting, copper pyrites are converted into a mixture of FeO and Cu₂0. Thus, acidic flux silica is added during smelting to remove FeO (basic). FeO combines with SiO₂to form famous silicate (FeSiO₃) slag which floats over molten matte.

Question 10.
What is meant by the term chromatography?
Answer:
The term chromatography was originally derived from the Greek word chroma meaning colour and graphy meaning writing because the method was first used for the separation of coloured substances (plant pigments) into individual components. Now the term chromatography has lost its original meaning and the method is widely used for separation, purification and characterization of the components of a mixture whether coloured or colourless.

Question 11.
What criterion is followed for the selection of the stationary phase in chromatography?
Answer:
The stationary phase is selected in such a way that it is capable of adsorbing the impurities more strongly than the elements to be purified. Under this condition, the impurities are retained by stationary phase i. e„ these cannot be eluted easily while the pure component, which is weakly adsorbed is easily eluted.

Question 12.
Describe a method for the refining of nickel.
Answer:
Nickel is refined by Mond’s process.
Mond’s process. Mond’s process is used for the refining of nickel metal. In the process, impure metal is heated in a current of carbon monoxide (CO) at 330 to 350 K to form nickel carbonyl which is of volatile nature. The vapours of the metal carbonyl escape leaving behind impurities. Upon heating to about 450 K, nickel carbonyl decomposes to give pure nickel.

From the above discussion, we may conclude that the basic requirements for the refining of metal by Vand Alkel process and Mond’s process are:

• The metal should form a volatile compound with the available reagent.
• The volatile compound should be easily decomposable so that metal in pure form may be regenerated.

Question 13.
How can you separate alumina from silica in bauxite ore associated with silica? Give equations if any.
Answer:
The purification of bauxite containing silica as the main impurity is done by Serpeck’s process. The powdered ore is mixed with coke and heated to about 2073 K in an atmosphere of nitrogen. Silica (SiO₂) is reduced to silicon which being volatile escapes. Alumina (Al₂O₃) is converted into aluminium nitride (AIN) by reacting with nitrogen. It is hydrolysed upon heating with water to get the precipitate of Al(OH)₃. From the precipitate, Al₂O₃ is recovered upon strong heating.

Question 14.
Giving examples, differentiate between calcination and roasting.
Answer:

Question 15.
How is ‘cast iron’ different from ‘pig iron’?
Answer:
Iron obtained from the blast furnace is called pig iron. It contains about 4% carbon and other impurities of S, P, Si, Mn etc. When pig iron is mixed with scrap iron and coke and then heated in a blast of hot air, some impurities are removed. Cast iron is obtained. It contains 3% carbon and some other impurities. It is hard and brittle.

Question 16.
Differentiate between mineral and ore.
Answer:
The naturally occurring chemical substances in form of which the metal occurs in the earth along with impurities are called minerals. The minerals from which the metal can be extracted conveniently and economically are called ore. Thus, all ores are minerals but all minerals are not ores. For example, iron is found in the earth’s crust oxides, carbonates and sulphides. Out of these minerals of iron, the oxides of iron are employed for the extraction of the metal. Therefore, oxides of iron are the ores of iron. Similarly, aluminium occurs in the earth’s crust in form of two minerals, i.e, bauxite (Al₂O₃. x H₂O) and clay (Al₂O₃.2SiO. 2H₂O). Out of these two minerals, Al can be conveniently and economically extracted from bauxite the ore of alluminium.

Question 17.
Why is copper matte put in the silicon-lined converter?
Answer:
Copper matte consists of a mixture of Cu₂S and Cu₂O. Along with that, it also contains a small amount of FeS and FeO. It is put in a silicon-lined converter known as a Bessemer converter. Some silica (SiO₂) is also added and a blast of hot air is blown. As a result, a number of reactions take place.

Question 18.
What is the role of cryolite in the metallurgy of aluminium?
Answer:
The role of cryolite is two-fold

• It makes alumina a good conductor of electricity.
• It lowers the fusion (melting point) temperature of both from 2323K to about 1140K.

Question 19.
How is leaching carried out in case of low-grade copper?
Answer:
Leaching in case of low-grade copper is carried out by reacting with an acid like H₂SO₄ in the presence of air when copper is oxidised to Cu²⁺ ions which pass into the solution. For example.
2 Cu(s) + 2 H₂SO₄ (aq) + O₂ (g) → 2CuSO₄ (aq) + 2 H₂O(l)
or Cu + 2H⁺ (aq) + 1/2 O₂(g) → Cu²⁺ (aq) + H₂O(l)

Question 20.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
The standard free energy of formation (Δ_f G°) of CO₂ from CO (fig 6.8) is higher than that of the formation-of ZnO from Zn. Therefore, CO cannot be used to reduce ZnO to Zn.

Question 21.
The value of ∆_fG° for the formation of Cr₂O₃ is – 540 kJ mol^-1 and that of Al₂O₃ is – 827 kJ mol^-1. Is the reduction of Cr₂O₃ with Al possible? (Pb. Board2009)
Answer:
The two thermochemical equations may be written as follows :

Since ∆G° comes out to be negative, the reaction is feasible.

Question 22.
Out of C and CO, which is a better reducing agent for ZnO?
Answer:
The Δ_f G° of CO2 from CO is always a higher reduction of ZnO to Zn. In contrast, Δ_f G° of CO from C is lower at a temperature above 1180K while that of CO₂ from C is lower at temperatures above 1270K than Δ_f G° of ZnO. Thus above 1270K, ZnO can be reduced to Zn by C. In actual practice, the reduction is usually carried out around 1673K. Thus out of C and CO, C is a better reducing agent than CO for ZnO.

Question 23.
The choice of a reducing agent in a particular case depends on the thermodynamic factors. How do you agree with this statement? Support your opinion with two examples.
Answer:
Thermodynamic factors have a major role in selecting the reducing agent for a particular reaction.
(i) Only that reagent will be preferred which will lead to decrease in free energy (∆G°) at a certain specific temperature.
(ii) A metal oxide placed lower in the Ellingham diagram cannot be reduced by the metal involved in the formation of the oxide placed higher in the diagram.
Examples:
(0 Al₂O₃ cannot be reduced by Cr present in Cr₂O₃ since the curve for Al₂O₃ is placed below that of Cr₂O₃ in the Ellingham diagram
(ii) CO cannot reduce ZnO because there is hardly any change in free energy (∆G°) as a result of the reaction.

Question 24.
Name the processes from which chlorine is obtained as the by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis?
Answer:
Chlorine is obtained as the by-product in the manufacture of sodium by Down’s process in which molten sodium chloride is subjected to electrolysis.

Sodium obtained by this method is almost pure while chlorine is the by-product.
Chlorine can also be obtained by carrying out the electrolysis of an aqueous solution of sodium chloride. The process is carried in Nelson’s cell. The various reactions which take place are as follows :

At cathode. Both Na⁺ and H⁺ ions migrate towards the cathode but H⁺ ions are discharged in preference to Na⁺ ions since their discharge potential is less. Na⁺ ions remain in the solution.

At anode: Both Cl^– and OH^– ions migrate towards the anode but Cl^– ions are discharged in preference to OH” since their discharge potential is less. OH^– ions remain in the solution.

Thus, in the electrolysis of an aqueous NaCl solution, H₂ gas is evolved at the cathode and chlorine at the anode. The solution contains NaOH and is, therefore, basic in nature.
It is always better to prepare chlorine by Down’s process.

Question 25.
What is the role of a graphite rod in the electrometallurgy of aluminium?
Answer:
In the electrometallurgy of aluminium, oxygen gas is evolved at the anode. It reacts with graphite or carbon (graphite electrodes) to form carbon monoxide and carbon dioxide. In case some other metal electrodes act as an anode, then oxygen will react with aluminium formed during the process to form aluminium oxide (Al₂O₃) which will pass into the reaction mixture. Since graphite is cheaper than aluminium, its wastage or consumption can be tolerated.

Question 26.
Outline the principles of refining of metals by the following methods:
(i) Zone refining
(ii) Electrolytic refining
(iii) Vapour phase refining
Answer:
(i) 1. This method is based on the principle that the impurities are more soluble in the molten state of metal (the melt) than in the solid-state. In the process of zone refining, a circular mobile heater is fixed at one end of a rod of impure metal. As the heater moves the molten zone of the rod also moves with it. As a result, pure metal crystallizes out of the melt and the impurities pass onto the adjacent molten zone. This process is repeated several times, which leads to the segregation of impurities at one end of the rod. Then the end with impurities is cut off. Silicon, boron, gallium, indium, etc can be purified by this process.

2. Column chromatography is a technique used to separate components of a mixture where components are in minute quantities. In chromatography, there are two phases: the mobile phase and the stationary phase. The stationary phase is immobile and immiscible. Al₂O₃ column is usually used as the stationary phase in column chromatography. The mobile phase may be a gas, liquid, or supercritical fluid in which the sample extracts dissolve. Then the mobile phase is forced to wave through the stationary phase. The component that is more strongly absorbed on the column takes a long time to travel than the component weakly, absorbed.

(ii) Electrolytic refining is the process of refining impure metal by using electricity. In this process, impure metal is made the anode and a strip of pure metal is made the cathode. A solution of a soluble salt of the same metal is taken as the electrolyte. When an electric current is passed, metal ions from the electrolyte are deposited at the cathode as pure metal and the impure metal from the anode dissolves into the electrolyte in the form options. The impurities present in the impure metal gets collected below the anode. This is called anode mud.

Anode : M → Mⁿ⁺+ ne^–
Cathode : Mⁿ⁺ + ne^– → M.

(iii) Vapour phase refining is the process of refining metal by converting it into its volatile compound and then, decomposing is to obtain a pure metal to carry out this process.
(a) The metal should form a volatile compound with available reagent, and
(b) The volatile compound should be easily decomposable so that the metal can be easily recovered.
Nickel, Zirconium, and titanium are refined using this method.

Question 27.
Predict the conditions under which aluminium can be expected to reduce magnesium oxide.
Answer:
The equations for the formation of the two oxides are :

If we look at the plots for the formation of the two oxides on the Ellingham diagram, we find that they intersect at a certain point. The corresponding value of ∆G° becomes zero for the reduction of MgO by Al metal.

This means that the reduction of MgO by A1 metal can occur below this temperature.
Aluminium (Al) metal can reduce MgO to Mg above this temperature because ∆_fG for Al₂O₃ is less as compared to that of MgO.
NCERT Solutions for Class 12 Chemistry Chapter6 General Principles and Processes of Isolation of Elements 30
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