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

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

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Question 1

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Why are Zn, Cd and Hg not regarded as transition metals despite being members of the d-block?

Their d orbitals are completely filled (d10) in both the ground state and in their common oxidation states.
According to the IUPAC definition, transition metals must have an incomplete d subshell in the neutral atom or in their ions.
However, since they are end members of the 3d, 4d and 5d series, their chemistry is studied along with transition metals.

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Question 2

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Explain why chromium has the electronic configuration 3d5 4s1 instead of the expected 3d4 4s2.

The energy difference between the 3d and 4s orbitals is very small in Cr.
Half-filled d orbitals (3d5) are relatively more stable due to extra exchange energy and symmetrical charge distribution.
Therefore, one electron from 4s shifts to 3d to attain the extra stability of the half-filled configuration 3d5 4s1.

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Question 3

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What is lanthanoid contraction? State one important consequence.

Lanthanoid contraction is the steady decrease in atomic and ionic radii from La to Lu due to the poor shielding of one 4f electron by another as nuclear charge increases.
As a consequence, the radii of 5d transition elements are nearly equal to those of the corresponding 4d elements (e.g., Zr 160 pm ≈ Hf 159 pm).
This makes the separation of Zr and Hf very difficult as they co-occur in nature and have almost identical properties.

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Question 4

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Why do transition metals exhibit variable oxidation states?

Both (n–1)d and ns electrons have similar energies and can participate in bond formation or be lost.
The small energy difference between successive d-configurations makes loss of variable numbers of d electrons energetically feasible.
Unlike non-transition elements where oxidation states differ by 2, in transition metals they differ by 1 (e.g., VII, VIII, VIV, VV).

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Question 5

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Why does Mn exhibit the largest number of oxidation states among 3d series transition metals?

Manganese (Z = 25) has the electronic configuration 3d5 4s2, giving it the maximum number of unpaired electrons (7 electrons available for bonding).
It can use all its seven valence electrons (3d5 + 4s2) in bond formation, leading to oxidation states from +2 to +7.
It lies near the middle of the 3d series where the balance between available d electrons and ionisation enthalpy is most favourable.

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Question 6

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Calculate the spin-only magnetic moment of Mn2+ ion.

Mn2+ has the configuration [Ar] 3d5, which contains 5 unpaired electrons.
Applying the spin-only formula: µ = √n(n+2) = √5(5+2) = √35 = 5.92 BM.
This high value reflects the maximum number of unpaired electrons in the 3d series for a divalent ion.

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Question 7

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Why do transition metal ions form coloured complexes in aqueous solution?

Transition metal ions have partially filled d orbitals which split into two energy levels in the presence of ligands.
An electron from the lower energy d level is excited to the higher level, absorbing visible light whose energy corresponds to the energy difference.
The colour observed is the complementary colour of the absorbed light; ions with d0 (Sc3+) and d10 (Zn2+) configurations are colourless.

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Question 8

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Why do transition metals have high enthalpies of atomisation?

Transition metals have a large number of unpaired d electrons that participate in interatomic metallic bonding alongside ns electrons.
Greater the number of unpaired electrons available for bonding, stronger the interatomic interactions and thus higher the enthalpy of atomisation.
The maxima in enthalpy of atomisation near the middle of each series (about d5) reflect the configuration with one unpaired electron per d orbital, which is particularly favourable.

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Question 9

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Why is Cu2+(aq) more stable than Cu+(aq) in aqueous solution?

Cu2+ has a much more negative hydration enthalpy than Cu+ because the higher charge density of Cu2+ leads to stronger ion-dipole interactions with water.
The large negative hydration enthalpy of Cu2+ more than compensates for the higher second ionisation enthalpy required to form Cu2+.
Cu+ in aqueous solution undergoes disproportionation: 2Cu+(aq) → Cu2+(aq) + Cu(s), confirming the relative instability of Cu+.

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Question 10

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Why is Cr2+ a stronger reducing agent than Fe2+?

When Cr2+ (d4) is oxidised to Cr3+ (d3), it achieves a half-filled t2g configuration which is particularly stable.
When Fe2+ (d6) is oxidised to Fe3+ (d5), it reaches a half-filled d5 configuration; however, the driving force for Cr2+ → Cr3+ is greater.
In an aqueous medium, d3 (Cr3+) is more stable than d5 (Fe3+) in terms of crystal field stabilisation, making Cr2+ a stronger reducing agent.

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Question 11

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What are interstitial compounds? Give two characteristics.

Interstitial compounds are formed when small atoms like H, C or N are trapped inside the crystal lattices of transition metals; they are generally non-stoichiometric, e.g., TiC, Mn4N, VH0.56.
They have very high melting points, higher than those of the pure metals, and are very hard (some borides approach diamond in hardness).
They retain metallic conductivity and are chemically inert, unlike typical ionic or covalent compounds.

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Question 12

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Explain why the ionic radii of 4d and 5d series elements in the same group are nearly equal.

Before the 5d series begins, the 14 lanthanoid elements (4f series) are filled, causing a regular decrease in atomic and ionic radii known as lanthanoid contraction.
This contraction essentially compensates for the expected increase in atomic size with increasing atomic number going from 4d to 5d.
As a result, the 5d elements have nearly the same ionic radii as their 4d counterparts (e.g., Zr 160 pm ≈ Hf 159 pm).

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Question 13

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Why do transition metals act as good catalysts?

Transition metals can adopt multiple oxidation states, allowing them to facilitate electron transfer steps in reactions by alternating between oxidation states.
They can form complexes with reactant molecules, increasing their local concentration at the catalyst surface and weakening bonds in the reacting molecules.
First-row transition metals use 3d and 4s electrons to form bonds with reactants, effectively lowering the activation energy of the reaction.

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

Why are Zn, Cd and Hg not regarded as transition metals despite being members of the d-block?

Explain why chromium has the electronic configuration 3d5 4s1 instead of the expected 3d4 4s2.

What is lanthanoid contraction? State one important consequence.

Why do transition metals exhibit variable oxidation states?

Why does Mn exhibit the largest number of oxidation states among 3d series transition metals?

Calculate the spin-only magnetic moment of Mn2+ ion.

Why do transition metal ions form coloured complexes in aqueous solution?

Why do transition metals have high enthalpies of atomisation?

Why is Cu2+(aq) more stable than Cu+(aq) in aqueous solution?

Why is Cr2+ a stronger reducing agent than Fe2+?

What are interstitial compounds? Give two characteristics.

Explain why the ionic radii of 4d and 5d series elements in the same group are nearly equal.

Why do transition metals act as good catalysts?

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