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

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

According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

NCERT Answers for Class 12 Chapter 8 d and f Block Elements Components are strong review materials that have replies to course reading practices and significant inquiries from earlier years and test papers. These NCERT Answers for Class 12 Chapter 8 d and f Block Elements are arranged in view of the most recent CBSE Prospectus for 2023-24 and are given in a straightforward manner for simplicity of understanding. Significant hints and deceives have additionally been spread out alongside these arrangements.

The NCERT Answer for Class Chapter 8 d and f Block Elements for the d and f block components gave here can assist the understudies with clearing every one of their questions and comprehend the essentials better. Further, these arrangements can be utilized for planning section notes and during modifications. Download the free PDF of the NCERT Answers for Class 12 Chapter 8 d and f Block Elements from the connection given underneath.


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: To what extent do the electronic configurations decide the stability of oxidation states in the first series of transition elements? Illustrate your answer with examples.

Ans : The oxidation states with the greatest number (+2 to +7) are displayed by the components of Mn oxidation expresses that are in the principal half of the change series. Ti With the expansion in nuclear number, the solidness of the +2 oxidation state increments. Yet again electrons are filled in the d-orbital, this happens generally. The +2 oxidation state isn’t shown by the Sc. Its electronic design is 4s2 3d1. It loses every one of the three electrons to shape Sc3+, +3 oxidation province of Sc is truly steady as by losing each of the three electrons, it achieves stable honorable gas setup, [Ar], Ti (+ 4) and V(+5) are entirely steady for a similar explanation. For Mn, the +2 oxidation state is truly steady as subsequent to losing two electrons, its d-orbital is precisely half-filled, [Ar] 3d5

Q 8.3: What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Ans: On moving along the lanthanoid series, the nuclear number slowly increments by one. With the expansion in nuclear number, the quantity of protons and electrons present in the particle likewise increments by one. The viable atomic charge increments as electrons are being added to a similar shell. Attributable to the proton expansion being articulated more over the interelectronic shocks coming about because of electron expansion, an atomic fascination increment occurs. Likewise, with the expansion in the nuclear number, the quantity of electrons in the 4f orbital additionally increments. The 4f electrons have an unfortunate safeguarding impact. Hence, the compelling atomic charge experienced by the external electrons increments. Thusly, the fascination of the core for the peripheral electrons increments.  Chapter 8 d and f Block Elements This outcomes in a consistent lessening in the size of lanthanoids with an expansion in the nuclear number. This is named lanthanoid constriction.

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 the second and third transition series.

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

Q 8.4: 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 transition elements?

Ans: Progress components are those components wherein the iotas or particles (in a steady oxidation state) contain to some extent filled d-orbital. These components are found in the d-block and show a change of properties between the s-block and the p-block. Consequently, these are called progress components.

Components, for example, Album, Hg and Zn can’t be called progress components since they have totally filled the d-sub shell.

Question 8.5: In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Ans : Change metals have a to some degree filled d−orbital. Thus, the electronic design of change components is (n − 1)d1-10 ns0-2.

The non-change components either don’t have a d−orbital or have a completely filled d−orbital.

Accordingly, the electronic arrangement of non-progress components is ns1-2 or ns2 np1-6.

Q 8.6: What are the different oxidation states exhibited by lanthanoids?

Ans : On account of lanthanoids, the +3 oxidation state is the most well-known.

That is, Ln(III) compounds are transcendent.

In any case, +2 and +4 oxidation states can likewise be tracked down in the arrangement or in strong mixtures.

Q 8.7 : 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 catalysts.

Ans : (I) Change metals show paramagnetic way of behaving. Paramagnetism emerges because of the presence of unpaired electrons, with every electron having an attractive second connected with its twist rakish energy and orbital precise force. Be that as it may, in the principal change series, the orbital rakish energy is extinguished. Accordingly, the subsequent paramagnetism is simply because of the unpaired electron.

(ii) Change components have an exceptionally compelling atomic charge and an enormous number of valence electrons. Consequently, they structure areas of strength for extremely bonds. Accordingly, the enthalpy of atomisation of progress metals is high.

(iii) The greater part of the edifices of change metals are hued. This is a direct result of the ingestion of radiation from the noticeable light locale to advance an electron starting with one of the d−orbitals then onto the next. Within the sight of ligands, the d-orbitals split up into two arrangements of orbitals having various energies. In this manner, the change of electrons can occur starting with one set then onto the next. The energy expected for these advances is tiny and falls in the noticeable locale of radiation. The particles of change metals ingest the radiation of a specific frequency, and the rest is reflected, bestowing variety to the arrangement.

(iv) The reactant action of the progress components can be made sense of by two essential realities.

(a) Progress metals give a reasonable surface to the responses to happen.

(b) Attributable to their capacity to show variable oxidation states and structure edifices, change metals structure unsteady middle mixtures. In this manner, they give another way lower actuation energy, Ea, for the response.

Q 8.8: What are interstitial compounds? Why are such compounds well-known for transition metals?
Ans : Progress metals are huge in size, and they likewise contain bunches of interstitial destinations.
These interstitial locales can be utilized to trap molecules of different components (that have little nuclear sizes), like H, C, N. The mixtures coming about are called interstitial mixtures.

Q 8.9: How is the variability in oxidation states of transition metals different from that of non-transition metals? Illustrate with examples.

Ans : On account of progress components, the oxidation state can shift from +1 to the most elevated oxidation state by eliminating all its valence electrons. Likewise, on the move components, the oxidation states contrast by 1 (Fe2+ and Fe3+; Cu+ and Cu2+). In non-progress components, the oxidation states vary by 2, for instance, +2 and +4 or +3 and +5, and so on.

8.10. In the series Sc(Z = 21) to (Z = 30), the enthalpy of atomisation of zinc is the lowest i.e., 126 kJ mol-1. Why?
Ans: The enthalpy of atomisation is straightforwardly connected with the steadiness of the gem cross section and furthermore the strength of the metallic bond. In the event of zinc (3d104s2 setup), no electrons from the 3d-orbitals are engaged with the arrangement of metallic bonds since all the orbitals are filled. Nonetheless, in any remaining components having a place with 3d series at least one d-electrons are engaged with the metallic bonds. This implies that the metallic bonds are very powerless in zinc and it has accordingly, least enthalpy of atomisation in the 3d series.

8.11. How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?
Ans: There is an abnormality in the IE’s of 3d-series because of rotation of energies of 4s and 3d orbitals when an e-1 is taken out. In this way, there is a revamping energy going with ionization. This outcomes into arrival of trade energy which increments as the quantity of e-1 s expansions in the dn arrangement. Cr has low first IE since deficiency of 1 e-gives stable EC (3d6). Zn has extremely high IE in light of the fact that e~ must be taken out from 4s orbital of the steady design (3d10 4s2) After the deficiency of one e-, expulsion of second e-, becomes troublesome. Thus, second IE’s are higher and by and large, increment from left to right. Nonetheless, Cr and Cu show a lot higher qualities in light of the fact that second e-must be taken out from stable setup of Cr+ (3d5) and Cu+ (3d10)

8.12. Why is the highest oxidation state of a metal exhibited by its fluoride and oxide only? (C.B.S.E. Delhi 2010)
Ans: Both fluorine and oxygen have exceptionally high electronegativity values. They can oxidize the metals to the most noteworthy oxidation state. Accordingly, the most elevated oxidation states are shown by the fluorides and oxides of the metals; progress metals specifically.

8.13. 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.
Sol: (I) Electronic design: The overall electronic arrangement of lanthanoids is [Xe]54 4f1-14 5d0-1 6s2 and that of actinoids is [Rn]86 5f0-14 6d0-1 7s2, lanthanoids . have a place with 4 f series while actinoids have a place with 5f-series.
(ii) Nuclear and ionic sizes: The two lanthanoids and actinoids show decline in size of their iotas or particles in + 3 oxidation state as we go from left to right. In lanthanoids, the diminishing is called lanthanoid compression while in actinoids, it is called actinoid withdrawal. The contratibn is more prominent from one component to another in actinodes because of less fortunate protecting by 5f electrons.
(iii)Oxidation state: Lanthanoids show restricted oxidation states (+ 2, + 3, + 4) out of which + 3 is generally normal while actinoids show +3, +4, +5, +6, +7 oxidation states.This is a direct result of enormous energy hole between 4f 5d and 6s orbitals. Notwithstanding, actinoids show an enormous number of oxidation states due to little energy ap-between 5f 6d and Is orbitals.
(iv) Compound reactivity: The previous individuals
of the lanthanoids series are very receptive like calcium however, with expansion in nuclear number, they act more like aluminum. The metals consolidate with hydrogen when . tenderly warmed in the gas. Carbides, Ln3C, Ln2C3 and LnC2 are framed when the metals are warmed with carbon. They free hydrogen from weaken corrosive and consume in incandescent light to frame halides. They structure oxides M203 and hydroxides M(OH)3.
Actinoids are exceptionally receptive metals, particularly when finely partitioned. The activity of bubbling water on them gives a combination of oxide and hydride and blend with most non-metals happen at moderate temperatures. HCl goes after all metals yet most are somewhat impacted by nitric corrosive attributable to the arrangement of defensive oxide layers, antacids have no activity. Actinoids are more receptive than lanthanoids because of greater nuclear size and lower ionization energy.

8.14 . How would you account for the following:
(i) Of the d4 species, Cr2+ is strongly reducing while manganese (III) is strongly oxidizing.
(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.
Sol: (I) E° an incentive for Cr3+/Cr2+ is negative (- 0-41 V) though E° values for Mn3+/Mn2+is positive (+1.57 V). Consequently, Cr2+ particle can without much of a stretch go through oxidation to give Cr3+ particle and, in this way, go about areas of strength for as specialist while Mn3+ can undoubtedly go through’ decrease to give Mn2+ and subsequently go about as an oxidizing specialist.
(ii) Co (III) tends .to frame coordination buildings than Co (II). Subsequently, within the sight of ligands, Co (II) changes to Co (III), i.e., is effectively oxidized.
(iii) The particles with dx setup tend to lose the main electron present in d-subshell to procure stable d° arrangement. Consequently, they are unsteady and go through oxidation or disproportionation.