CBSE Important Questions Class 12 Chemistry Chapter 4 – The d and f Block Elements

Class 12 Chemistry Chapter 4 The d and f Block Elements is part of the CBSE Class 12 Chemistry syllabus. The d-block elements are the transition metals and the f-block elements are the lanthanoids and actinoids. Students learn their electronic configurations, atomic and ionic radii, variable oxidation states, why their compounds are coloured, magnetic properties, catalytic behaviour, lanthanoid contraction, and the preparation of potassium dichromate and potassium permanganate.

These CBSE important questions for Class 12 Chemistry Chapter 4 cover why the E° value of copper is positive, lanthanoid contraction, why atomic radii hardly change across a transition series, electronic configurations of elements and ions, identifying an oxidising agent from its reduction products, and comparing lanthanoids with actinoids. Every question has a clear step-by-step solution in simple English, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. The E°(M²⁺/M) value for copper is positive (+0.34 V). What is the possible reason for this?

Answer: The E°(M²⁺/M) value for any metal is the overall result of three energy steps:
1. Enthalpy of atomisation: M(s) → M(g)
2. Ionisation enthalpy: M(g) → M²⁺(g) + 2e⁻
3. Hydration enthalpy: M²⁺(g) + aq → M²⁺(aq)
For copper, the enthalpy of atomisation is high and the hydration enthalpy is low. These two are not compensated by the energy released on hydration.
So the overall process of converting solid copper into hydrated Cu²⁺ ions is not favourable, which makes E°(Cu²⁺/Cu) positive. This is also why copper does not displace hydrogen from acids.

Q2. What is lanthanoid contraction?

Answer: In the lanthanoid series, as the atomic number increases there is a steady decrease in the size of the atoms and of the trivalent ions. This is called lanthanoid contraction.
Reason: The electrons are being added to the inner 4f subshell. The 4f electrons shield the outer electrons from the nucleus very poorly. So as the nuclear charge increases along the series, the effective nuclear charge felt by the outer electrons also increases, pulling them in closer.
Consequence: Elements of the second and third transition series have almost the same size, for example Zr and Hf, which makes them very hard to separate.

Q3. With an increase in atomic number, the atomic radius does not change very much across a transition series. Why?

Answer: Two opposing effects balance each other out.
• As the atomic number increases, the nuclear charge increases, which tends to pull the electrons closer and reduce the size.
• At the same time, the added electron goes into the inner d subshell, which increases the screening (shielding) effect on the outermost electrons.
The increased screening counterbalances the increased nuclear charge, so the net pull on the outer electrons stays nearly the same and the atomic radius changes very little across the series.

Q4. Write the electronic configurations of the elements with atomic numbers 61, 91, 101 and 109.

Answer:

Atomic number Element Electronic configuration
61 Pm (Promethium) [Xe] 4f⁵ 5d⁰ 6s²
91 Pa (Protactinium) [Rn] 5f² 6d¹ 7s²
101 Md (Mendelevium) [Rn] 5f¹³ 6d⁰ 7s²
109 Mt (Meitnerium) [Rn] 5f¹⁴ 6d⁷ 7s²

Q5. Write down the electronic configurations of the following ions.
(i) Cr³⁺ (ii) Cu⁺ (iii) Co²⁺ (iv) Mn²⁺ (v) Pm³⁺ (vi) Ce⁴⁺ (vii) Lu²⁺ (viii) Th⁴⁺

Answer: (i) Cr³⁺ = [Ar] 3d³
(ii) Cu⁺ = [Ar] 3d¹⁰
(iii) Co²⁺ = [Ar] 3d⁷
(iv) Mn²⁺ = [Ar] 3d⁵
(v) Pm³⁺ = [Xe] 4f⁴
(vi) Ce⁴⁺ = [Xe]
(vii) Lu²⁺ = [Xe] 4f¹⁴ 5d¹
(viii) Th⁴⁺ = [Rn]
Note that in forming cations from transition metals, the 4s electrons are removed before the 3d electrons.

Q6. The electronic configuration of chromium is
(a) [Ar] 3d⁵ 4s¹
(b) [Ar] 3d⁴ 4s²
(c) [Ar] 3d⁶ 4s⁰
(d) [Ar] 3d³ 4s³

Answer: (a) [Ar] 3d⁵ 4s¹
The expected configuration would be 3d⁴ 4s², but chromium takes 3d⁵ 4s¹ instead.
Reason: A half-filled d subshell (d⁵) is especially stable because of its symmetrical distribution of electrons and the extra exchange energy it gains. So one 4s electron shifts into the 3d orbital.
Copper shows the same behaviour for the fully filled d¹⁰ configuration: [Ar] 3d¹⁰ 4s¹.

Q7. If one end of a piece of metal is heated, the other end becomes hot after some time. This is due to
(a) minor perturbation in the energy of atoms
(b) mobility of atoms in the metal
(c) resistance of the metal
(d) energised electrons moving to the other part of the metal

Answer: (d) Energised electrons moving to the other part of the metal
Metals contain a sea of free (delocalised) electrons. When one end is heated, those electrons gain kinetic energy and move rapidly to the cooler end, carrying the energy with them.
This same sea of mobile electrons is why metals are good conductors of both heat and electricity.

Q8. An oxidising agent A is reduced under two different conditions at the same temperature. In acidic medium the deep violet solution becomes nearly colourless, with five electrons involved per mole. In neutral medium the solution turns brown with a precipitate, and only three electrons are involved. Identify A and give balanced half-reactions.

Answer: The deep violet colour and these two reduction patterns identify A as the permanganate ion, MnO₄⁻ (potassium permanganate, KMnO₄).
In acidic medium (5 electrons, violet to colourless):
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
Manganese goes from +7 to +2. The Mn²⁺ ion is very pale pink, so the solution looks almost colourless.
In neutral or faintly alkaline medium (3 electrons, brown precipitate):
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
Manganese goes from +7 to +4, forming the brown insoluble MnO₂.
Why the difference: In acidic medium plenty of H⁺ ions are available to combine with the oxygen, allowing the full reduction to Mn²⁺. In neutral medium there are not enough H⁺ ions, so the reduction stops at MnO₂.

Q9. A newly isolated metallic element X is a soft silvery-white metal that tarnishes rapidly in air, mainly forms X₂O₃ and XCl₃, has E°(X³⁺/X) of about −2.3 V, and most of its aqueous ions are coloured though one oxidation state gives a colourless ion. Is X a lanthanoid or a d-block transition element?

Answer: X is most likely a lanthanoid. The reasons:
1. Soft silvery-white metal that tarnishes rapidly — lanthanoids are soft and highly reactive, whereas most transition metals are hard and much less reactive in air.
2. Mainly +3 compounds (X₂O₃ and XCl₃) — the lanthanoids are strongly dominated by the +3 oxidation state. Transition metals typically show several oxidation states.
3. Very negative E° of −2.3 V — this shows a highly electropositive metal, comparable to the alkaline earth metals. Lanthanoid E° values lie around −2.2 to −2.4 V, while transition metal values are far less negative.
4. Coloured ions with one colourless state — lanthanoid ions are coloured because of f-f transitions, except those with f⁰ and f¹⁴ configurations, which are colourless.

Q10. Two f-block series P and Q are compared. In P almost all elements form +3 ions and the ionic radii decrease gradually. In Q the elements show a wider range of oxidation states and all are radioactive, with a more pronounced contraction. Identify P and Q and explain the difference.

Answer: P is the lanthanoid series and Q is the actinoid series.
Why lanthanoids mostly show +3: Removing the two 6s electrons and one 4f electron is energetically easy, and the resulting +3 state is stable. The 4f orbitals are buried deep inside and are not available for bonding, so other oxidation states are rare.
Why actinoids show many oxidation states: The 5f, 6d and 7s orbitals are very close in energy, so all three can take part in bonding. This is especially true for the early members such as U, Np and Pu, which show +3 to +6.
Why the contraction is greater in actinoids: The 5f electrons shield even more poorly than the 4f electrons, so the effective nuclear charge rises faster and the contraction is more pronounced.
Why all actinoids are radioactive: Their nuclei are too large to be stable, so they decay spontaneously.

Q11. Two lanthanoids X and Y show contrasting behaviour. X forms a stable X²⁺ ion with configuration 4f⁷, while Y forms a stable Y⁴⁺ ion with configuration 4f⁰. Identify X and Y and explain.

Answer: X is europium (Eu) and Y is cerium (Ce).
For X (europium): Eu has the configuration [Xe] 4f⁷ 6s². Losing the two 6s electrons gives Eu²⁺ with 4f⁷, which is a half-filled f subshell. Half-filled subshells have extra stability because of their symmetry and exchange energy, so Eu²⁺ is unusually stable in aqueous solution.
For Y (cerium): Ce has [Xe] 4f¹ 5d¹ 6s². Losing all four of these electrons gives Ce⁴⁺ with 4f⁰, which is the stable noble gas configuration of xenon. This makes Ce⁴⁺ stable, and it is widely used as an oxidising agent in volumetric analysis.
So both ions owe their stability to reaching a specially stable electronic arrangement.

Q12. Why do transition metals form coloured compounds, and why do they act as good catalysts?

Answer: Coloured compounds: Transition metal ions have partly filled d orbitals. In a compound, the ligands split the d orbitals into two sets of slightly different energy. An electron can absorb light from the visible range to jump from the lower set to the higher set — this is a d-d transition. The compound shows the complementary colour of the light absorbed.
Ions with a d⁰ or d¹⁰ configuration, such as Sc³⁺ and Zn²⁺, are colourless, because no d-d transition is possible.
Good catalysts: Two reasons:
1. They show variable oxidation states, so they can easily accept and donate electrons during a reaction, providing a lower-energy path.
2. They can form intermediate complexes with the reactants, bringing the reacting molecules together on the metal surface.
Examples are iron in the Haber process, vanadium pentoxide in the Contact process, and nickel in hydrogenation.

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