Chapter 4: The d-and f-Block Elements Long Questions | chemistry Class 12 CBSE

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Chapter 4: The d-and f-Block Elements Long Questions | chemistry Class 12 CBSE | ByteNotes.in

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Question 1

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Compare the general electronic configurations of transition metals and inner transition metals. How do exceptions in the 3d series arise, and what is the basis for the IUPAC definition of transition metals?

Transition (d-block) elements have the general configuration (n–1)d1–10 ns1–2; inner transition (f-block) elements have (n–2)f1–14 (n–1)d0–1 ns2.
Exceptions in the 3d series occur for Cr (3d5 4s1) and Cu (3d10 4s1) because half-filled and completely filled d orbitals are more stable due to extra exchange energy.
The small energy gap between (n–1)d and ns orbitals allows one electron to shift from 4s to 3d to achieve these stable configurations.
IUPAC defines transition metals as those with an incomplete d subshell in the neutral atom or in their common ions; this excludes Zn, Cd, Hg (d10 in all states).
Inner transition metals are further divided into lanthanoids (4f filling, Ce–Lu) and actinoids (5f filling, Th–Lr), both placed at the bottom of the periodic table.

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Question 2

Long Form

Discuss the trends in ionisation enthalpies of first-row transition metals. Why do they not increase as steeply as in non-transition elements?

Ionisation enthalpies generally increase from Sc to Zn due to increasing nuclear charge, but the increase is less steep than in non-transition elements.
As electrons are added to 3d orbitals, these inner d electrons partially shield the 4s electrons from the increasing nuclear charge, slowing the increase in ionisation enthalpy.
Irregular trends arise due to varying stability of d configurations: extra-stable d5 makes Mn2+ hard to oxidise (high third ionisation enthalpy of Mn) while Fe3+ (d5) is relatively easy to form (low third ionisation enthalpy of Fe).
The second ionisation enthalpy shows unusually high values for Cr and Cu (whose M+ ions have stable d5 and d10 configurations) and a low value for Zn (removal gives stable d10).
Exchange energy in degenerate d orbitals contributes to stability; loss of exchange energy at d6 configuration (no loss compared to d5) means Fe2+ is easier to oxidise than Mn2+.

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Question 3

Long Form

Describe the preparation of potassium dichromate from chromite ore. Explain the effect of pH on dichromate solution and its role as an oxidising agent in acidic solution.

Chromite ore (FeCr2O4) is fused with Na2CO3 in air: 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2.
The yellow filtrate of sodium chromate is acidified with H2SO4 to crystallise orange sodium dichromate: 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O.
Sodium dichromate is treated with KCl to obtain K2Cr2O7: Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl.
On increasing pH, dichromate converts to chromate (Cr2O72– + 2OH– → 2CrO42– + H2O); decreasing pH reverses this.
In acidic solution, K2Cr2O7 is a powerful oxidising agent: Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O (Eo = 1.33 V), oxidising Fe2+ to Fe3+, I– to I2, and H2S to S.

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Question 4

Long Form

Describe the preparation of potassium permanganate and outline its oxidising action in acidic, neutral and alkaline media.

KMnO4 is prepared by fusing MnO2 with KOH and KNO3 to give dark green K2MnO4 (manganate), which is electrolytically oxidised in alkaline solution to KMnO4.
In the laboratory, Mn2+ is oxidised by peroxodisulphate: 2Mn2+ + 5S2O82– + 8H2O → 2MnO4– + 10SO42– + 16H+.
In acidic solution (Eo = +1.52 V): MnO4– + 8H+ + 5e– → Mn2+ + 4H2O; oxidises Fe2+→Fe3+, I–→I2, oxalates→CO2, SO32–→SO42–.
In neutral/faintly alkaline solution (Eo = +1.69 V): MnO4– + 4H+ + 3e– → MnO2 + 2H2O; oxidises I– to IO3– and thiosulphate to sulphate.
In strongly alkaline solution (Eo = +0.56 V): MnO4– + e– → MnO42–; the green manganate is the reduction product.

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Question 5

Long Form

Analyse the variation in oxidation states of the first-row transition metals. Why do elements at the extremes of the series exhibit fewer oxidation states than those in the middle?

Elements at the beginning of the series (Sc, Ti) have too few d electrons to exhibit many oxidation states; Sc shows only +3 (noble gas core), Ti is most stable in +4.
Elements near the middle (V, Cr, Mn) show the maximum range; Mn displays +2 to +7 as its 3d5 4s2 configuration provides seven electrons for bonding.
Elements at the end (Cu, Zn) have too many d electrons — fewer orbitals are available to share with others; Zn only shows +2 (no d electrons involved).
In d-block, higher oxidation states in heavier groups (e.g., Mo(VI), W(VI)) are more stable than in lighter ones (Cr(VI) is a strong oxidant), opposite to p-block inert pair effect.
Low or zero oxidation states occur in complexes with π-acceptor ligands: Ni(CO)4 (Ni = 0), Fe(CO)5 (Fe = 0), stabilised by back-bonding into CO antibonding orbitals.

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Question 6

Long Form

Examine the catalytic properties of transition metals with specific examples from industrial processes, explaining the mechanism of their catalytic action.

Transition metals' ability to adopt multiple oxidation states allows them to act as electron relays in redox reactions; e.g., Fe3+ catalyses 2I– + S2O82– → I2 + 2SO42– by cycling Fe3+/Fe2+.
At solid surfaces, transition metals form bonds with reactant molecules using 3d and 4s electrons, increasing reactant concentration and weakening bonds (lowering activation energy).
V2O5 catalyses oxidation of SO2 to SO3 in the Contact Process (manufacture of H2SO4) by cycling between V5+ and V4+.
Finely divided Fe catalyses N2 + 3H2 → 2NH3 in the Haber Process; Ni catalyses hydrogenation of vegetable oils.
PdCl2 catalyses oxidation of ethyne to ethanal (Wacker Process); TiCl4/Al(CH3)3 (Ziegler catalysts) polymerises ethylene.

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Question 7

Long Form

Compare the lanthanoids and actinoids with respect to electronic configurations, oxidation states, atomic sizes, and chemical reactivity.

Lanthanoids have configuration [Xe] 4fn 5d0–1 6s2; actinoids have [Rn] 5fn 6d0–1 7s2; 4f orbitals are deeply buried while 5f orbitals are more accessible for bonding.
Lanthanoids predominantly show +3 state (occasionally +2 or +4); actinoids show a wider range (+3 to +7) because 5f, 6d, 7s levels have comparable energies.
Both series undergo contraction (lanthanoid and actinoid contraction) due to poor f-electron shielding; actinoid contraction is greater per element due to poorer 5f shielding.
Lanthanoids are reactive metals similar to Ca (early members) to Al (later members); actinoids are highly reactive, attacked by HCl but passivated by HNO3.
Actinoids are all radioactive (unlike lanthanoids); their study is complicated by radioactivity and the uneven distribution of oxidation states across the series.

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Question 8

Long Form

Analyse the trends in standard electrode potentials (Eo M2+/M) of the first transition series metals. Why are the values for Mn, Ni and Zn more negative than expected?

Eo(M2+/M) becomes generally less negative from Ti to Cu as the sum of first and second ionisation enthalpies increases across the series.
Cu has a uniquely positive Eo (+0.34 V) because its high enthalpy of atomisation and large sum of ionisation enthalpies are not offset by its hydration enthalpy.
Mn has a more negative Eo (–1.18 V) than expected because Mn2+ has the stable half-filled d5 configuration, making it energetically demanding to reduce Mn2+ to Mn.
Zn has a more negative Eo (–0.76 V) than expected due to the extra stability of the completely filled d10 configuration in Zn2+.
Ni has a more negative Eo (–0.25 V) than expected because it has the highest negative hydration enthalpy (–2121 kJ mol–1) among the 3d metals, stabilising Ni2+(aq).

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Question 9

Long Form

Discuss the formation of complex compounds by transition metals. Why do transition metals form a large number of complexes compared to non-transition metals?

Complex compounds are species where a metal ion is bonded to a definite number of ligands (anions or neutral molecules), e.g., [Fe(CN)6]3–, [Cu(NH3)4]2+, [PtCl4]2–.
Transition metal ions have small size and high ionic charge, creating strong electrostatic attraction towards ligands and enabling coordination bond formation.
Available d orbitals allow acceptance of electron pairs from ligands to form coordinate bonds; the energy splitting of d orbitals in ligand fields also stabilises complexes.
Variable oxidation states enable formation of complexes in multiple states (e.g., [Fe(CN)6]3– and [Fe(CN)6]4–), and π-back bonding with certain ligands stabilises low or zero oxidation states.
Non-transition metals lack available d orbitals of suitable energy and have lower charge-to-radius ratios, limiting their ability to form stable complexes.

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Question 10

Long Form

Explain the magnetic properties of transition metal ions. How can the spin-only formula be used to determine the number of unpaired electrons? Give examples.

Transition metal ions are paramagnetic due to unpaired d electrons; each unpaired electron has a magnetic moment from spin angular momentum (orbital contribution is quenched in first row).
The spin-only formula is µ = √n(n+2) BM, where n is the number of unpaired electrons; a single unpaired electron gives µ = 1.73 BM.
For Ti3+ (d1, n=1): µ = 1.73 BM; for V2+ (d3, n=3): µ = 3.87 BM; for Mn2+ (d5, n=5): µ = 5.92 BM — maximum for the 3d series.
Sc3+ (d0) and Zn2+ (d10) have zero unpaired electrons and are diamagnetic (µ = 0).
Ferromagnetism (very strong paramagnetism) is exhibited by Fe, Co and Ni; the observed magnetic moment helps identify the number of unpaired electrons experimentally.

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Chapter 4: The d-and f-Block Elements | Long Questions

Compare the general electronic configurations of transition metals and inner transition metals. How do exceptions in the 3d series arise, and what is the basis for the IUPAC definition of transition metals?

Discuss the trends in ionisation enthalpies of first-row transition metals. Why do they not increase as steeply as in non-transition elements?

Describe the preparation of potassium dichromate from chromite ore. Explain the effect of pH on dichromate solution and its role as an oxidising agent in acidic solution.

Describe the preparation of potassium permanganate and outline its oxidising action in acidic, neutral and alkaline media.

Analyse the variation in oxidation states of the first-row transition metals. Why do elements at the extremes of the series exhibit fewer oxidation states than those in the middle?

Examine the catalytic properties of transition metals with specific examples from industrial processes, explaining the mechanism of their catalytic action.

Compare the lanthanoids and actinoids with respect to electronic configurations, oxidation states, atomic sizes, and chemical reactivity.

Analyse the trends in standard electrode potentials (Eo M2+/M) of the first transition series metals. Why are the values for Mn, Ni and Zn more negative than expected?

Discuss the formation of complex compounds by transition metals. Why do transition metals form a large number of complexes compared to non-transition metals?

Explain the magnetic properties of transition metal ions. How can the spin-only formula be used to determine the number of unpaired electrons? Give examples.

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