NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements

NCERT Solutions for Class 12 Chemistry Chapter 8 – Free PDF Download

NCERT Solutions for Class 12 Chemistry Chapter 8 The d and f Block Elements is a powerful study material that has answers to textbook exercises and important questions from the previous year and sample papers. These NCERT Solutions for Class 12 Chemistry are updated to the latest term – II CBSE Syllabus for 2021-22 and are given in a simple way for ease of understanding. Important tips and tricks have also been laid out along with these solutions.

The NCERT Solution for Class 12 Chemistry for the d and f block elements provided here can help the students clear all their doubts and understand the fundamentals better. Further, these solutions can be used for preparing chapter notes and during revisions. Download the free PDF of the NCERT Solutions for Class 12 Chemistry Chapter 8 from the link given below.

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Class 12 Chemistry NCERT Solutions Chapter 8 The d and f Block Elements – Important Questions


Q 8.1:

Write down the electronic configuration of:

(i) Cr3+ (ii) Pm3+ (iii) Cu+ (iv) Ce4+ (v) Co2+ (vi) Lu2+ (vii) Mn2+  (viii) Th4+

Ans :

(i) Cr3+: 1s2 2s2 2p6 3s2 3p6 3d3

Or, [Ar]18 3d3

(ii) Pm3+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f4

Or, [Xe]54 4f4

(iii) Cu+: 1s2 2s2 2p6 3s2 3p6 3d10

Or, [Ar]18 3d10

(iv) Ce4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6

Or, [Xe]54

(v) Co2+: 1s2 2s2 2p6 3s2 3p6 3d7

Or, [Ar]18 3d7

(vi) Lu2+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d1

Or, [Xe]54 4f14 5d1

(vii) Mn2+: 1s2 2s2 2p6 3s2 3p6 3d5

Or, [Ar]18 3d5

(viii) Th4+: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 4f14 5s2 5p6 5d10 6s2 6p6

Or, [Rn]86

Q 8.2:

Why are Mn2+ compounds more stable than Fe towards oxidation to their +3 state?

Ans : 

Electronic configuration:               Fe2+ is [Ar]18 3d6.

Mn2+ is [Ar]18 3d5.

We know that half and completely filled orbitals are more stable. Hence, Mn with (+2) state has a stable d5 configuration, which is why Mn2+ shows resistance to oxidation to Mn3+. We know that, Fe2+ has 3d6 configuration and by losing one electron, its configuration changes to a more stable 3d5 configuration. Therefore, Fe2+ easily gets oxidized to Fe+3 oxidation state.

Q 8.3:

Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Ans :

In the table below, the oxidation states of the first half of the first row of transition metals are displayed.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.3

We can see that except Sc all other metals have a +2 oxidation state. As moving in the increasing order of atomic number from 21 to 25, that is from Sc to Mn, the number of electrons in 3d orbitals increases from 1 to 5.

Sc (+2) = d1

Ti (+2) = d2

V (+2) = d3

Cr (+2) = d4

Mn (+2) = d5

+2 oxidation state is attained by the loss of the two 4s electrons by these metals. Since the number of d electrons in (+2) state also increases from Ti(+2) to Mn(+ 2), the stability of +2 state increases (as d-orbital is becoming more and more half-filled). Mn (+2) has d5 electrons (that is half-filled d shell, which is highly stable).

Q 8.4:

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.

Ans :

The oxidation states with the maximum number (+2 to +7) are exhibited by the elements of Mn oxidation states that are in the first half of the transition series. Ti  With the increase in atomic number, the stability of +2 oxidation state increases. Once more electrons are filled in the d-orbital, this happens to a great extent.  The +2 oxidation state is not shown by the Sc. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+, +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar], Ti (+ 4) and V(+5) are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its d-orbital is exactly half-filled, [Ar] 3d5

Q 8.5:

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?

Ans :

Electronic configuration in ground state Stable oxidation states
(i) 3d3 (Vanadium) +2, +3, +4 and +5
(ii) 3d5 (Chromium) +3, +4, +6
(iii) 3d5 (Manganese) +2, +4, +6, +7
(iv) 3d8(Cobalt) +2, +3
(v) 3d4 There is no 3d4 configuration in ground state.

Q 8.6:

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.

Ans :

(i) Chromate, CrO42CrO^{2-}_{4}

Oxidation state of Cr is + 6.

(ii) Permanganate, MnO4MnO^{-}_{4}

Oxidation state of Mn is + 7.

(iii) Vanadate,  VO3VO^-_3

Oxidation state of V is + 5.

Q 8.7:

What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Ans:

On moving along the lanthanoid series, the atomic number gradually increases by one. With the increase in atomic number, the number of protons and electrons present in the atom also increases by one. The effective nuclear charge increases as electrons are being added to the same shell. Owing to the proton addition being pronounced more above the interelectronic repulsions resulting from electron addition, nuclear attraction increase happens. Also, with the increase in the atomic number, the number of electrons in the 4f orbital also increases. The 4f electrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with an increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction

(i) It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from La(OH)3 to Lu(OH)3.)

(ii) There is a similarity in the properties of second and third transition series.

(iii) Separation of lanthanoids is possible due to lanthanide contraction.

Q 8.8:

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?

Ans:

Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled d-orbital. These elements are found in the d-block and show a transition of properties between s-block and p-block. Hence, these are called transition elements.

Elements such as Cd, Hg and Zn cannot be called transition elements because they have completely filled d-sub shell.

Question 8.9:

In what way is the electronic configuration of the transition elements different from that of the non transition elements?

Ans :

Transition metals have a partially filled d−orbital. Hence, the electronic configuration of transition elements is (n − 1)d1-10 ns0-2.

The non-transition elements either do not have a d−orbital or have a fully filled d−orbital.

Therefore, the electronic configuration of non-transition elements is ns1-2 or ns2 np1-6.

Q 8.10:

What are the different oxidation states exhibited by the lanthanoids?

Ans :

In the case of lanthanoids , +3 oxidation state is most common.

That is, Ln(III) compounds are predominant.

However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Q 8.11:

Explain giving reasons:
(i) Transition metals and many of their compounds show paramagnetic behaviour.
(ii) The enthalpies of atomisation of the transition metals are high.
(iii) The transition metals generally form coloured compounds.
(iv) Transition metals and their many compounds act as good catalyst.

Ans :

(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

(ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the d−orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(iv) The catalytic activity of the transition elements can be explained by two basic facts.

(a) Transition metals provide a suitable surface for the reactions to occur.

(b) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, Ea, for the reaction.

Q 8.12:

What are interstitial compounds? Why are such compounds well known for transition metals?

Ans :

Transition metals are large in size and they also contain lots of interstitial sites.

These interstitial sites can be used to trap atoms of other elements (that have small atomic size), such as H, C, N. The compounds resulting are called interstitial compounds.

Q 8.13:

How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.

Ans :

In the case of transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc.

Q 8.14:

Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Ans :

From iron chromite ore, Potassium dichromate is prepared as in the following steps.

Step (1): Preparation of sodium chromate

4FeCr2O4+16NaOH+7O28Na2CrO4+2Fe2O3+8H2O4Fe Cr_2 O_4 +16 Na OH+7 O_2\rightarrow 8 Na_2 Cr O_4 + 2Fe_2 O_3 + 8 H_2 O

Step (2): Conversion of sodium chromate into sodium dichromate

2Na2CrO4+conc.H2SO4Na2Cr2O7+Na2SO4+H2O2Na_{2} Cr O_4 + conc. H_2 SO_4 \rightarrow Na_2 Cr_{2} O_7 + Na_2 SO_4 + H_2 O

Step (3): Conversion of sodium dichromate to potassium dichromate

Na2Cr2O7+2KClK2Cr2O7+2NaClNa_{2} Cr_2 O_7 + 2KCl \rightarrow K_2 Cr_{2} O_7 + 2Na Cl

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

The dichromate ion (Cr2O72)(Cr_2 O_7^{2-}) exists in equilibrium with chromate (CrO42)(Cr O_4^{2-}) ion at pH 4.

However, by changing the pH, they can be inter converted.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.14

Q 8.15:

Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:
(i) iodide (ii) iron(II) solution and (iii) H2S

Ans :

k2Cr2O7k_2 Cr_2 O_7 acts as a very strong oxidising agent in the acidic medium.

k2Cr2O7+4H2SO4K2SO4+Cr2(SO4)3+4H2O+3[O]k_2 Cr_2 O_7 + 4 H_2 SO_4\rightarrow K_2 SO_4 + Cr_2(SO_4)_3 + 4 H_2O+3[O] k2Cr2O7k_2 Cr_2 O_7 takes up electrons to get reduced and acts as an oxidizing agent. The reaction of k2Cr2O7k_2 Cr_2 O_7 with H2S, iron (II) solution and other iodide are given below.

(i) k2Cr2O7k_2 Cr_2 O_7 oxidizes iodide to iodine.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.15(i)

(ii) k2Cr2O7k_2 Cr_2 O_7 oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferricions.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.15(ii)

(iii) k2Cr2O7k_2 Cr_2 O_7 oxidizes H2S to sulphur.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.15(iii)

Q 8.16:

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.

Ans :

Potassium permanganate can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KClO4, to give K2MnO4.

2MnO2+4KOH+O2heat2K2MnO4+2H2O2 Mn O_2 + 4 KOH + O_2 \overset{heat}{\rightarrow} 2K_2 MnO_4 + 2 H_2 O

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution. Electrolytic oxidation

K2MnO42k++MnO42K_2 Mn O_4\leftrightarrow 2k^+ +MnO^{2-}_4 H2OH++OHH_2 O\leftrightarrow H^+ + OH^{-}

At anode, manganate ions are oxidized to permanganate ions.

MnO42MnO4+eMnO^{2-}_4\leftrightarrow MnO^-_4 + e^-

Oxidation by chlorine

2K2MnO4+Cl22MnO4+2KCl2 K_2 MnO_4 + Cl_2 \rightarrow 2MnO_4 + 2KCl 2MnO42+Cl22MnO4+2Cl2 MnO_4^{2-} + Cl_2 \rightarrow 2MnO_4^- + 2Cl^-

Oxidation by ozone

2KMnO4+O3+H2O2KMnO4+2KOH+O22 KMnO_4 + O_3 + H_2O \rightarrow 2KMnO_4 + 2KOH + O_2 2MnO42+O3+H2O2MnO42+2OH+O22 MnO_4^{2-} + O_3 + H_2O \rightarrow 2MnO_4^{2-} + 2OH^- + O_2

(i) Acidified KMnO4 solution oxidizes Fe (II) ions to Fe (III) ions i.e., ferrous ions to ferric ions.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.16(i)

(ii). Acidified potassium permanganate oxidizes SO2 to sulphuric acid.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.16(ii)

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.16(iii)

Q 8.17:

For M2+/M and M3+/M2+ systems the EV values for some metals are as follows:
Cr2+/Cr -0.9V  Cr3/Cr2+ -0.4 V
Mn2+/Mn -1.2V  Mn3+/Mn2+ +1.5 V
Fe2+/Fe -0.4V Fe3+/Fe2+ +0.8 V

Use this data to comment upon:
(i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and
(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

 Ans :

(i) The reduction potentials for the given pairs increase in the following order.

Mn2+ / Mn < Cr2+ / Cr < Fe2+ /Fe

So, the oxidation of Fe to Fe2+ is not as easy as the oxidation of Cr to Cr2+ and the oxidation of Mn to Mn2+. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: Fe < Cr < Mn

(ii) The EΘE^{\Theta} value for Fe3+/Fe2+ is higher than that for Cr3+/Cr2+ and lower than that for Mn3+/Mn2+. So, the reduction of Fe3+ to Fe2+ is easier than the reduction of Mn3+ to Mn2+, but not as easy as the reduction of Cr3+ to Cr2+. Hence, Fe3+ is more stable than Mn3+, but less stable than Cr3+. These metal ions can be arranged in the increasing order of their stability as: Mn3+ < Fe3+ < Cr3+

Q 8.18:

Predict which of the following will be coloured in aqueous solution? Ti3+, V3+, Cu+, Sc3+, Mn2+, Fe3+ and Co2+. Give reasons for each.

Ans :

Ions with electrons in d-orbital will be the only ones that will be coloured and the ions with empty d-orbital will be colourless.

Element Atomic Number Ionic

State

Electronic configuration in ionic

State

Ti 22 Ti3+ [Ar] 3d1
V 23 V3+ [Ar] 3d2
Cu 29 Cu+ [Ar] 3d10
Sc 21 Sc3+ [Ar]
Mn 25 Mn2+ [Ar] 3d5
Fe 26 Fe3+ [Ar] 3d5
Co 27 Co2+ [Ar] 3d7

All other ions, except Sc3+, will be coloured in aqueous solution because of d−d transitions. Not Sc3+ as it has an empty d-orbital.

Q 8.19:

Compare the stability of +2 oxidation state for the elements of the first transition series.

 Ans:

Sc     +3
Ti +1 +2 +3 +4
V +1 +2 +3 +4 +5
Cr +1 +2 +3 +4 +5 +6
Mn +1 +2 +3 +4 +5 +6 +7
Fe +1 +2 +3 +4 +5 +6
Co +1 +2 +3 +4 +5
Ni +1 +2 +3 +4
Cu +1 +2 +3
Zn   +2

 From the table above:

Mn shows maximum number of oxidation states, changing between +2 to +7.

The number of oxidation states increases as we move on from Sc to Mn.

On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons.

The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving to bottom from the top, it becomes more and more difficult to remove the third electron from the d-orbital.

Question 8.20:

Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
(i) electronic configuration (iii) oxidation state
(ii) atomic and ionic sizes and (iv) chemical reactivity.

Ans :

(i) Electronic configuration

The general electronic configuration for lanthanoids is [Xe]54 4f0-14 5d0-1 6s2 and that for actinoids is [Rn]86 5f1-14 6d0-1 7s2. Unlike 4f orbitals, 5f orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Atomic and lonic sizes :

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.

(iii) 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.

(iv) Chemical reactivity:

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 case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Question 8.21:

How would you account for the following?
(i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising.
(ii) Cobalt(II) 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

 Ans:

(i) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written as configuration, which is a more stable configuration. In the case of Mn3+ (d4), it acts as an oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and is highly stable.

(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.

(iii) The ions in d1 configuration tend to lose one more electron to get into stable d0 configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the d-orbital of these ions. Therefore, they act as reducing agents.

Q 8.22:

What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

Ans :

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation−reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

For example,

(a) 3CrO43+8H+2CrO42+Cr3++4H2O3Cr O^{3-}_4 + 8H^+ \rightarrow 2Cr O^{2-}_4 + Cr^{3+} + 4H_2 O

Cr(V) is oxidized to Cr(VI) and reduced to Cr(III).

(b) 3MnO42+4H+2MnO4+MnO2+2H2O3Mn O^{2-}_4 + 4H^+ \rightarrow 2Mn O^{-}_4 + Mn O_2 + 2H_2 O

Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV).

Question 8.23:

Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

 Ans:

In the first series of transition metals, Cu exhibits +1 oxidation state very frequently. We know Cu has an electronic configuration of [Ar] 3d 10, that is, the completely filled d-orbital makes it highly stable.

Question 8.24:

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

Ans:

  Gaseous ions Number of unpaired electrons
(i) Mn2+, [Ar] 3d4 4
(ii) Cr3+,[Ar] 3d3 3
(iii) V3+,[Ar] 3d2 2
(iv) Ti3+, [Ar]3d1 1

Cr3+ is the most stable in aqueous solutions owing to a t2g3t^3_{2g} configuration.

Question 8.25:

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.

Ans:

(i) In the 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 atom 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 and so, they are unavailable. There is also a high effective nuclear charge. As a result, it can accept electrons and behave as an acid.

For example, MnIIOMn^{II}O  is basic and Mn2VIIO7Mn^{VII}_2O_7 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 higher oxidation states in oxides and fluorides.

For example, in OsF6 and V2O5, the oxidation states of Os and V are +6 and +5, respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in , the oxidation state of Mn is +7.

Q 8.26:

Indicate the steps in the preparation of:
(i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore.

 Ans:

(i) Potassium dichromate ( K2Cr2O7 ) is prepared from chromite ore (FeCr2O4) in the following steps.

Step (1): Preparation of sodium chromate

4FeCr2O4+16NaOH+7O28Na2CrO4+2Fe2O3+8H2O4FeCr_2O_4 + 16 NaOH + 7O_2 \rightarrow 8Na_2CrO_4 + 2 Fe_2O_3 + 8H_2O

Step (2): Conversion of sodium chromate into sodium dichromate

2Na2CrO4+Conc.H2SO4Na2Cr2O3+Na2SO4+H2O2Na_2CrO_4 + Conc. H_2SO_4 \rightarrow Na_2Cr_2O_3 + Na_2SO_4 + H_2O

Step(3): Conversion of sodium dichromate to potassium dichromate

Na2Cr2O7+2KClK2Cr2O7+2NaClNa_2Cr_2O_7 + 2KCl \rightarrow K_2Cr_2O_7 + 2NaCl

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

The dichromate ion Cr2O72Cr_2O^{2-}_7 exists in equilibrium with chromate ion CrO42CrO^{2-}_4 at pH 4.

However, by changing the pH, they can be interconverted.

NCERT Solutions for Class 12 Chemistry Chapter 8 D and F Block Elements Q 8.26(i)

(ii) Potassium permanganate ( KMnO4 ) can be prepared from pyrolusite (MnO2). The ore is fused with KOH in the presence of either atmospheric oxygen or an oxidising agent, such as KNO3 or KClO4, to give K2MnO4.

2MnO2+4KOH+O2heat2K2MnO4+2H2O2MnO_2 + 4KOH + O_2 \overset{heat}{\rightarrow} 2K_2MnO_4 + 2H_2O

The green mass can be extracted with water and then oxidized either electrolytically or by passing  chlorine/ozone into the solution. Electrolytic oxidation

K2MnO42K++MnO42K_2MnO_4\leftrightarrow 2K^+ + MnO^{2-}_4 H2OH++OHH_2O\leftrightarrow H^+ + OH^-

At anode, manganate ions are oxidized to permanganate ions.

MnO42MnO4+eMnO^{2-}_4\leftrightarrow MnO^-_4 + e^-

Oxidation by chlorine

2K2MnO4+Cl22KMnO4+2KCl2K_2MnO_4 + Cl_2 \rightarrow 2KMnO_4 + 2KCl 2MnO42+Cl22MnO4+2Cl2MnO^{2-}_4 + Cl_2 \rightarrow 2MnO^-_4 + 2Cl^-

Oxidation by ozone

2K2MnO4+O2+H2O2KMnO4+2KOH+O22K_2MnO_4 +O_2 + H_2O \rightarrow 2 KMnO_4 + 2KOH + O_2 2MnO42+O2+H2O2MnO42+2OH+O22MnO^{2-}_4 +O_2 + H_2O \rightarrow 2 MnO^{2-}_4 + 2OH^- + O_2

Question 8.27:

What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.

 Ans :

An alloy is a solid solution of two or more elements in a metallic matrix. It can be a partial solid solution or a complete solid solution. Alloys generally are created as they have varied properties compared to their constituent elements. An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94−95%), iron (5%), and traces of Al, Ca, Si, C, S.

Uses

(a) It is used in tracer bullets and shells.

(b) Mischmetal is used in cigarettes and gas lighters.

(c) It is used in flame-throwing tanks.

Question 8.28:

What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

Ans :

Inner transition metals : elements in which the last electron enters the f-orbital.

The elements in which the 4f and the 5f orbitals are progressively filled are called f-block elements.

From the given atomic numbers, the inner transition elements are the ones with atomic numbers 59, 95, and 102.

Q 8.29:

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.

Ans :

Lanthanoids project 3 oxidation states (+2, +3, +4). From these, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6s orbital’s is quite large. While, the energy difference between 5f, 6d, and 7s orbital’s is very less. Hence, actinoids display a large number of oxidation states.

For example, neptunium displays +3, +4, +5, and +7 while uranium and plutonium display +3, +4, +5, and +6 oxidation states. The most common oxidation state in case of actinoids is also +3.

Q 8.30 :

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.

 Ans :

In the actinoid series, the last element is lawrencium, Lr. The atomic number of the element is 103 and its electronic configuration is [Rn] 5f14 6d1 7s2 . The most common oxidation state displayed by it is +3; because after losing 3 electrons it attains stable f14 configuration.

Q 8.31:

Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

 Ans :

Ce  :  1s2  2s2  2p6  3s2  3p6  3d10  4s2  4p6  4d10  5s2  5p6  4f1  5d1  6s2Ce \; : \; 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^{10} \; 4s^2 \; 4p^6 \; 4d^{10} \; 5s^2 \; 5p^6 \; 4f^1 \; 5d^1 \; 6s^2 Ce3+  :  1s2  2s2  2p6  3s2  3p6  3d10  4s2  4p6  4d10  5s2  5p6  4f1Ce^{3+} \; : \; 1s^2 \; 2s^2 \; 2p^6 \; 3s^2 \; 3p^6 \; 3d^{10} \; 4s^2 \; 4p^6 \; 4d^{10} \; 5s^2 \; 5p^6 \; 4f^1

Magnetic moment can be calculated as:

ν=n(n+2)\nu = \sqrt{n(n+2)}

Where,

n = number of unpaired electrons

In Ce3+, n = 1

ν=1(1+2)\nu = \sqrt{1(1+2)} =3)= \sqrt{3)}

= 1. 732 BM

Q 8.32:

Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Ans :

+2 +4
Nd(60) Ce (58)
Sm(62) Pr (59)
Eu (63) Nd (60)
Tm(69) Tb (65)
Yb (70) Dy (66)

In the parenthesis are the atomic numbers of the elements are given.

Tb after forming Tb4+ attains a stable electronic configuration of [Xe] 4f7.

Yb after forming Yb2+ attains a stable electronic configuration of [Xe] 4f14.

Eu after forming Eu2+ attains a stable electronic configuration of [Xe] 4f7.

Ce after forming Ce4+ attains a stable electronic configuration of [Xe].

Q 8.33 : 

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

Ans :

Atomic number Electronic configuration
61 [Xe]544f55d06s2
91 [Rn]865f26d17s2
101 [Rn]865f136d07s2
109 [Rn]865f146d77s2

Q 8.34:

Compare the general characteristics of the first series of the 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

 Ans :

(i) In the 1st, 2nd and 3rd transition series,

the 3d, 4d, and 5d orbitals are respectively filled. We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series,

two elements show unusual electronic configurations:

Cu (29) = 3d104s1

Cr (24) = 3d54s1

Similarly, there are exceptions in the second transition series. These are:

Ag(47) = 4d105s1

Pd(46) = 4d10 5s0

Rh(45) = 4d85s1

Ru (44) = 4d75s1

Tc (43) = 4d65s1

Mo (42) = 4d55s1

There are some exceptions in the third transition series as well. These are:

Au (79 ) = 5d10 6s1

Pt (78) = 5d9 6s1

W (74) = 5d4 6s2

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends. However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.

Some stable complexes are :

[FeII(Cn)6]4{\left [ Fe^{II}(Cn)_6 \right ]}^{4-},

[CoIII(NH3)6]3+{\left [ Co^{III}(NH_3)_6 \right ]}^{3+},

[Ti(H2O)6]3+{\left [ Ti(H_2O)_6 \right ]}^{3+}

The issue is that no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high.

For example: WCl6, ReF7, RuO4, etc.

(iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4f electrons in the third transition series. Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the 2nd transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the 1st transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first 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 contraction.

Q 8.35:

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).

Ans :

Metal ion Number of d-electrons Filling of d-orbitals

 

Cu2+ 9 t2g6eg3t^{6}_{2g}e^{3}_{g}
Ni2+ 8 t2g6eg2t^{6}_{2g}e^{2}_{g}
CO2+ 7 t2g5eg2t^{5}_{2g}e^{2}_{g}
Fe3+ 5 t2g3eg2t^{3}_{2g}e^{2}_{g}
Fe2+ 6 t2g4eg2t^{4}_{2g}e^{2}_{g}
Mn2+ 5 t2g3eg2t^{3}_{2g}e^{2}_{g}
Cr3+ 3 t2g3t^{3}_{2g}
V2+ 3 t2g3t^{3}_{2g}
Ti2+ 2 t2g2t^{2}_{2g}

Q 8.36:

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

Ans :

In many ways, the properties of elements of heavier transition elements differ from those of the first transition series.

(a) The elements of the first transition series form 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.

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

(c) The meltingand boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding (M−M bonding).

(d)  For heavier elements , the higher oxidation states are more common whereas for  first transition series elements the  +2 and +3 oxidation states are more common.

(e) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of 2nd and 3rd 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 contraction.

Q 8.37:

What can be inferred from the magnetic moment values of the following complex species?
Example            Magnetic Moment (BM)
K4[Mn(CN)6)    2.2
[Fe(H2O)6]2+   5.3
K2[MnCl4]         5.9

Ans :

Magnetic moment  . ν=n(n+2)\nu = \sqrt{n(n+2)} )

For value n = 1, ν=1(1+2)=3=1.732\nu = \sqrt{1(1+2)} = \sqrt{3} = 1.732

For value n = 2, ν=2(2+2)=8=2.83\nu = \sqrt{2(2+2)} = \sqrt{8} = 2.83

For value n = 3, ν=3(3+2)=15=3.87\nu = \sqrt{3(3+2)} = \sqrt{15} = 3.87

For value n = 4, ν=4(4+2)=24=4.899\nu = \sqrt{4(4+2)} = \sqrt{24} = 4.899

For value n = 5, ν=5(5+2)=35=5.92\nu = \sqrt{5(5+2)} = \sqrt{35} = 5.92

K4[Mn(CN)6]

n(n+2)=2.2\sqrt{n(n+2)} = 2.2

For in transition metals, the magnetic moment is calculated from the spin-only formula.

Therefore, we can see from the above calculation that the given value is closest to n = 1 . Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d-orbital.

Hence, we can say that CN is a strong field ligand that causes the pairing of electrons.

[Fe(H2O)6]2+

n(n+2)=5.3\sqrt{n(n+2)} = 5.3

Thus, the calculation given above proves that the closest value to n= 4  . In addition to this, the +2 oxidation state  in this complex is none other than Fe. The d-orbital for Fe would have 6 electrons.

Thus, we can assume that H2O is a weak field ligand and does not induce electron pairing.

K2[MnCl4

n(n+2)=5.9\sqrt{n(n+2)} = 5.9

From the equation, we can see that the given value is nearest to n = 5. Mn is in the +2 oxidation state in this complex. This implies that in the d-orbital, Mn has 5 electrons.

We may also assume that Cl – is a weak field ligand and does not induce electron pairing.

The elements lying in the middle of the periodic table between the s-block and p-block elements are known as d-block or transition elements. The f-block elements are called inner-transition series. The d and f block elements introduce various important topics to the students like general properties of the transition elements (d block), physical properties, variation in atomic and ionic sizes of transition metals, ionization enthalpies, oxidation states, magnetic properties, the formation of coloured ions and complex compounds. To explore more about these concepts in detail access the NCERT Solutions to this chapter.

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

Chapter 8 d- and f-Block Elements of Class 12 Chemistry is categorized under the term – II CBSE Syllabus for 2021-22. This chapter of NCERT Solutions for Class 12 deals with elements of the d and f block of the modern periodic table. In this chapter, students shall first deal with the electronic configuration, occurrence and general characteristics of transition elements with special emphasis on the trends in the properties of the first row (3d) transition metals along with the preparation and properties of some important compounds. This will be followed by consideration of certain general aspects such as electronic configurations, oxidation states and chemical reactivity of the inner transition metals.

Subtopics of Chemistry Chapter 8 – The d and f Block Elements

  1. Position in the Periodic Table
  2. Electronic Configurations of the d-Block Elements
  3. General Properties of the Transition Elements (d-Block)
  4. Some Important Compounds of Transition Elements
  5. The Lanthanoids
  6. The Actinoids
  7. Some Applications of d- and f-Block Elements

Why Opt to BYJU’S?

While solving the Class 12 Chemistry NCERT textbook questions of Chapter 8 with the help of NCERT Solutions for Class 12 Chemistry Chapter 8, students should also solve lots of sample papers and previous year questions. Solving previous year questions and sample papers will help you to get acquainted with the latest term – II exam pattern as well as the marking scheme.

Along with NCERT Solutions, students should also follow some good study materials, NCERT textbooks are one of the best study materials for the students who are studying in Class 12. Students are also advised to make study notes so that they can go through the notes for quick revision.

Frequently Asked Questions on NCERT Solutions for Class 12 Chemistry Chapter 8

Will I be able to get my doubts cleared using the NCERT Solutions for Class 12 Chemistry Chapter 8?

Students who are not able to get their doubts cleared during class hours can make use of the NCERT Solutions at BYJU’S. The experts frame the solutions with utmost care with the main objective of helping students with their second term exam preparation. To score good marks in Chemistry, learning the concepts on a daily basis is very much important. The concepts are explained in an interactive manner to help students grasp them and perform well in the term – II exams.

What kind of questions can I expect from Chapter 8 of NCERT Solutions for Class 12 Chemistry in the term – II exams?

For the first and second term exams, units 6, 7 and 8 come under one section which carries 19 marks. So, Chapter 8 carries 5 marks which consist of both short and very short answer questions. To score full marks in this chapter, students are advised to solve the textbook questions using the solutions available at BYJU’S. The solutions of this chapter are framed by expert faculty having vast experience in the relevant subject. They create answers which are concept focused rather than question focused to build a strong foundation of basic concepts which are important from the exam perspective.

Why should I refer to the NCERT Solutions for Class 12 Chemistry Chapter 8 from BYJU’S?

You should refer to the NCERT Solutions for Class 12 Chemistry Chapter 8 from BYJU’S to understand the concepts easily. The solutions are framed by the teachers based on the understanding abilities of students. Pictorial representation of certain topics improves visual learning skills among students which helps them to grasp the concepts. NCERT Solutions are created in a stepwise manner in order to meet the CBSE standards. Every year the solutions are updated based on the changes occurring in the CBSE Syllabus and its guidelines.

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