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

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

Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 1

Applied Concept

A chemist needs to separate Zr and Hf from a mixed ore. With reference to the properties of these elements and the concept of lanthanoid contraction, explain why this separation is extremely difficult.

Zr (Z = 40) and Hf (Z = 72) have nearly identical atomic radii (Zr 160 pm, Hf 159 pm) and ionic radii because lanthanoid contraction in the 4f series compensates for the expected increase in size from 4d to 5d elements.
Because their radii and electronic properties are so similar, Zr and Hf have almost identical crystal structures, ionisation enthalpies, electrode potentials and chemical reactivities — they co-occur naturally and form similar compounds.
Conventional separation techniques based on differences in solubility, volatility or complex formation are largely ineffective; expensive fractional distillation of volatile chlorides or liquid-liquid extraction with specific complexing agents must be used.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In the Contact Process for manufacturing sulphuric acid, V2O5 acts as a catalyst. Explain how the variable oxidation state of vanadium enables this catalytic action.

In the Contact Process, SO2 is oxidised to SO3: 2SO2 + O2 → 2SO3. V2O5 acts as the catalyst by cycling between V(V) and V(IV) oxidation states.
V2O5 (V in +5 state) oxidises SO2 to SO3 and is itself reduced to VO2 (V in +4 state): V2O5 + SO2 → 2VO2 + SO3. Then VO2 is reoxidised by O2 back to V2O5: 4VO2 + O2 → 2V2O5.
The ability of vanadium to readily interconvert between +4 and +5 states — facilitated by the small energy difference between these d configurations — makes it an effective catalyst that is regenerated in each catalytic cycle.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

A student adds dilute H2SO4 to copper metal but observes no reaction. However, when hot concentrated H2SO4 is used, the metal dissolves. With reference to the electrode potential of copper, explain this observation.

Copper has a positive standard electrode potential Eo(Cu2+/Cu) = +0.34 V, meaning Cu is a less reactive metal that cannot reduce H+ ions to H2; dilute H2SO4 provides only H+ ions as the oxidising agent, which Cu cannot reduce.
Hot concentrated H2SO4 is an oxidising acid that provides SO42– ions which act as a stronger oxidising agent (being reduced to SO2) than H+; copper reduces sulphate under these conditions.
The reaction is: Cu + 2H2SO4 (conc., hot) → CuSO4 + SO2 + 2H2O. The positive Eo of Cu explains why only oxidising acids (HNO3 and hot conc. H2SO4) can dissolve it.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A solution of potassium dichromate turns from orange to yellow on addition of NaOH. Explain the chemistry involved and predict what would happen on acidifying the yellow solution.

Potassium dichromate (Cr2O72–, orange) and potassium chromate (CrO42–, yellow) are interconvertible depending on the pH of the solution. On adding NaOH (increasing pH), the equilibrium shifts: Cr2O72– + 2OH– → 2CrO42– + H2O, converting dichromate to chromate and turning the solution yellow.
On acidifying the yellow chromate solution (decreasing pH), the reverse reaction occurs: 2CrO42– + 2H+ → Cr2O72– + H2O, and the solution turns back to orange as dichromate reforms.
Both ions contain Cr in the +6 oxidation state; the difference lies in structure — CrO42– is tetrahedral while Cr2O72– consists of two tetrahedra sharing one corner. The interconversion does not involve any change in oxidation state of chromium.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

A transition metal ion in aqueous solution shows a magnetic moment of 3.87 BM. Identify the number of unpaired electrons and suggest which 3d ion (from Mn2+ to Ni2+) could have this configuration.

Using the spin-only formula µ = √n(n+2): 3.87 = √n(n+2); solving, n(n+2) ≈ 15, which gives n = 3 unpaired electrons.
V2+ (d3) and Cr3+ (d3) both have 3 unpaired electrons and a calculated magnetic moment of 3.87 BM; Co2+ (d7) also has 3 unpaired electrons and approximately the same calculated value.
From the range Mn2+ to Ni2+: Co2+ (d7, 3 unpaired electrons) fits this value. The experimentally observed value for Co2+ is 4.4–5.2 BM due to some orbital angular momentum contribution, but the spin-only calculation gives 3.87 BM.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

In a leather tanning factory, potassium dichromate is used as an oxidising agent for the preparation of azo compounds. Write the balanced ionic equation for the oxidation of Fe2+ ions by acidified K2Cr2O7 and identify the role of H+ ions.

The relevant half-reactions are: Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O (reduction, Eo = 1.33 V) and Fe2+ → Fe3+ + e– (oxidation; multiply by 6: 6Fe2+ → 6Fe3+ + 6e–).
The balanced ionic equation is: Cr2O72– + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O.
H+ ions are essential because they participate directly in the reduction of dichromate; 14 H+ ions are needed per formula unit of Cr2O72–. Without acidic conditions, dichromate cannot oxidise Fe2+ effectively as the reaction is thermodynamically driven by the proton concentration.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 7

Applied Concept

With reference to the first ionisation enthalpies and electronic configurations, explain why the enthalpies of atomisation of Zn is the lowest (126 kJ mol–1) in the 3d transition series.

In zinc (3d10 4s2), all d orbitals are completely filled and no d electrons participate in metallic bonding; only the two 4s electrons contribute to interatomic bonding.
In contrast, all other metals in the 3d series (Sc to Cu) involve electrons from (n–1)d in addition to ns electrons in interatomic metallic bonding, creating stronger interactions.
With only 2 bonding electrons per atom versus up to 7 in elements like Cr or Mn, the metallic bonding in Zn is weaker, resulting in the lowest enthalpy of atomisation (126 kJ mol–1) and explaining why Zn is also the most volatile 3d metal.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 8

Applied Concept

A student wants to perform a permanganate titration but uses HCl instead of H2SO4 to acidify the solution. Explain why the results would be erroneous.

Permanganate (MnO4–) in acidic solution is a very strong oxidising agent with Eo = +1.52 V; it can oxidise Cl– ions from HCl to Cl2 (a side reaction).
This means some of the permanganate is consumed in oxidising Cl– rather than the analyte being titrated, leading to a higher consumption of KMnO4 than expected and an erroneously high result.
Therefore, H2SO4 is used to acidify permanganate titrations because sulphate ions are not oxidised by permanganate under normal conditions, ensuring that all the KMnO4 reacts only with the intended reducing agent.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

Cerium(IV) is used as an analytical reagent in volumetric analysis despite having a very positive reduction potential (Eo = +1.74 V for Ce4+/Ce3+). Explain its suitability.

Ce4+ is a strong oxidising agent (Eo = +1.74 V) because it strongly tends to revert to the stable +3 state by gaining one electron; Ce4+ has the configuration 4f0 (noble gas, [Xe]) and Ce3+ has 4f1.
Although the high Eo means Ce4+ could theoretically oxidise water, the reaction is extremely slow kinetically, making Ce(IV) a stable and reliable reagent in practice.
Ce(IV) reacts cleanly and quantitatively with reducing agents and the end point is easily detected; its stability in solution and predictable stoichiometry make it suitable as a primary standard in cerimetry (cerium-based titrations).

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 10

Applied Concept

Explain why titanium carbide (TiC) is used as a coating for cutting tools, with reference to the properties of interstitial compounds.

TiC is an interstitial compound formed when carbon atoms are trapped inside the crystal lattice of titanium metal; it is non-stoichiometric and its formula does not correspond to any normal oxidation state of Ti.
Interstitial compounds are very hard (some borides approach diamond in hardness), have high melting points higher than the pure metal, and are chemically inert — all ideal properties for cutting tool coatings.
TiC also retains metallic conductivity, ensuring heat generated during cutting is dissipated; its extreme hardness and chemical inertness prevent the tool surface from wearing away or reacting with the material being cut.

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

A chemist needs to separate Zr and Hf from a mixed ore. With reference to the properties of these elements and the concept of lanthanoid contraction, explain why this separation is extremely difficult.

In the Contact Process for manufacturing sulphuric acid, V2O5 acts as a catalyst. Explain how the variable oxidation state of vanadium enables this catalytic action.

A student adds dilute H2SO4 to copper metal but observes no reaction. However, when hot concentrated H2SO4 is used, the metal dissolves. With reference to the electrode potential of copper, explain this observation.

A solution of potassium dichromate turns from orange to yellow on addition of NaOH. Explain the chemistry involved and predict what would happen on acidifying the yellow solution.

A transition metal ion in aqueous solution shows a magnetic moment of 3.87 BM. Identify the number of unpaired electrons and suggest which 3d ion (from Mn2+ to Ni2+) could have this configuration.

In a leather tanning factory, potassium dichromate is used as an oxidising agent for the preparation of azo compounds. Write the balanced ionic equation for the oxidation of Fe2+ ions by acidified K2Cr2O7 and identify the role of H+ ions.

With reference to the first ionisation enthalpies and electronic configurations, explain why the enthalpies of atomisation of Zn is the lowest (126 kJ mol–1) in the 3d transition series.

A student wants to perform a permanganate titration but uses HCl instead of H2SO4 to acidify the solution. Explain why the results would be erroneous.

Cerium(IV) is used as an analytical reagent in volumetric analysis despite having a very positive reduction potential (Eo = +1.74 V for Ce4+/Ce3+). Explain its suitability.

Explain why titanium carbide (TiC) is used as a coating for cutting tools, with reference to the properties of interstitial compounds.

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