NCERT Solutions for Class 12 Chemistry Chapter 8 – The d- and f-Block Elements (Exercises)

Question 8.1: Silver atom has a completely filled d-subshell (4d¹⁰) in its ground state. How is it considered a transition element?

Answer: Although silver (Ag) has a filled 4d¹⁰ configuration in its ground state, in its common oxidation state (+2), it has an incomplete d-subshell (4d⁹). Transition elements are those whose atoms or ions have an incomplete d-subshell; hence Ag is classified as a transition element.

Question 8.2: In the 3d transition series from Scandium (Z = 21) to Zinc (Z = 30), why does Zinc have the lowest enthalpy of atomisation?

Answer: Zinc (Zn) has a fully filled 3d¹⁰4s² configuration. The d-electrons do not contribute effectively to metallic bonding, resulting in weaker bonds and the lowest enthalpy of atomisation among 3d elements.

Question 8.3: Which element in the 3d series exhibits the highest number of oxidation states and why?

Answer: Manganese (Mn) exhibits the highest number of oxidation states (+2 to +7). Its [Ar] 3d⁵ 4s² configuration allows both 3d and 4s electrons to participate in bonding due to their similar energies.

Question 8.5: Why do the first and second ionisation enthalpies of 3d transition elements show an irregular trend?

Answer: The irregularity is due to the extra stability associated with half-filled (d⁵) and fully-filled (d¹⁰) electronic configurations. Also, reorganisation energy varies across the series, affecting ionisation enthalpies.

Question 8.6: Why is the highest oxidation state of transition metals observed only in their fluorides and oxides?

Answer: Fluorine and oxygen are highly electronegative and can oxidise metals to their highest oxidation states. Hence, the highest oxidation states are observed in oxides and fluorides.

Question 8.7: Between Cr²⁺ and Fe²⁺, which is a stronger reducing agent and why?

Answer: Cr²⁺ is a stronger reducing agent than Fe²⁺. Cr²⁺ readily oxidises to Cr³⁺ due to its negative electrode potential (E° = –0.41 V), making it a better electron donor.

Question 8.8: Calculate the spin-only magnetic moment of M²⁺ ion where atomic number Z = 27.

Answer: For Co²⁺ (Z = 27), configuration = [Ar] 3d⁷.
Unpaired electrons = 3
Magnetic moment (μ) = √[n(n + 2)] BM
μ = √[3(3+2)] = √15 ≈ 3.87 BM

Question 8.9: Why is the Cu⁺ ion not stable in aqueous solution?

Answer: Cu⁺ ion undergoes disproportionation:
2Cu⁺ → Cu²⁺ + Cu
This occurs because Cu²⁺ has a much higher hydration enthalpy, stabilising it in aqueous solution.

Question 8.10: Why is the actinoid contraction more significant than lanthanoid contraction?

Answer: Actinoid contraction is more significant because 5f-electrons shield poorly compared to 4f-electrons. The increasing nuclear charge thus pulls electrons inward more strongly across the actinide series.

NCERT Solutions for Class 12 Chemistry Chapter 8 – The d- and f-Block Elements (Exercises)

Question 8.1: Write the electronic configurations of the following ions:

Answer: (i) Cr³⁺ = [Ar] 3d³
(ii) Pm³⁺ = [Xe] 4f⁴
(iii) Cu⁺ = [Ar] 3d¹⁰
(iv) Ce⁴⁺ = [Xe]
(v) Co²⁺ = [Ar] 3d⁷
(vi) Lu²⁺ = [Xe] 4f¹⁴ 5d¹
(vii) Mn²⁺ = [Ar] 3d⁵
(viii) Th⁴⁺ = [Rn]

Question 8.2: Why are Mn²⁺ compounds more resistant to oxidation to the +3 state compared to Fe²⁺ compounds?

Answer: Mn²⁺ has a half-filled 3d⁵ configuration, which is very stable. Oxidizing Mn²⁺ to Mn³⁺ would break this stable arrangement, making oxidation unfavorable. Fe²⁺, however, achieves a stable 3d⁵ configuration when oxidized to Fe³⁺, thus oxidation is easier.

Question 8.3: Why does the +2 oxidation state become increasingly stable in the first half of the 3d series with rising atomic number?

Answer: As we move across the 3d series, after removing two 4s electrons, the increasing number of 3d electrons results in greater stability due to more unpaired electrons and the tendency to attain half-filled configurations.

Question 8.4: How do electronic configurations influence the stability of oxidation states in the 3d series? Give examples.

Answer: Electronic configurations with half-filled or fully-filled d-subshells are especially stable. For example, Mn²⁺ (3d⁵) and Fe³⁺ (3d⁵) are very stable due to half-filled d-orbitals.

Question 8.5: What stable oxidation states might be expected for elements with the following ground state d-electron configurations?

Answer: 3d³ → +5
3d⁵ → +2, +7
3d⁸ → +2
3d⁴ → +2, +3, +6

Question 8.6: Name the oxometal anions of first-row transition metals where the oxidation state matches the group number.

Answer: Chromate (CrO₄²⁻), Dichromate (Cr₂O₇²⁻), and Permanganate (MnO₄⁻).

Question 8.7: What is lanthanoid contraction and what are its consequences?

Answer: Lanthanoid contraction is the gradual decrease in ionic radii from La³⁺ to Lu³⁺ across the lanthanoid series.
Consequences: Similar atomic radii of 2nd and 3rd transition series, difficulty in separation of lanthanoids, variation in basicity, and influence on chemical reactivity.

Question 8.8: What defines transition elements? Which d-block elements are not considered transition elements?

Answer: Transition elements have incomplete d-orbitals in atoms or ions. Zn, Cd, and Hg are not true transition elements because their d-orbitals are completely filled.

Question 8.9: How are the electronic configurations of transition elements different from those of non-transition elements?

Answer: Transition elements have partially filled d-orbitals (n-1)d¹–¹⁰ ns¹–², whereas non-transition elements lack partially filled d-orbitals.

Question 8.10: What are the common oxidation states of lanthanoids?

Answer: The most common oxidation state is +3. However, +2 and +4 states are also observed in some elements.

Question 8.11: Explain:

Answer: (i) Transition elements and their compounds are paramagnetic due to unpaired electrons.
(ii) They have high enthalpies of atomisation due to strong metallic bonds involving d-electrons.
(iii) Colored compounds arise from d-d electronic transitions.
(iv) Good catalysts due to multiple oxidation states and ability to adsorb reactants.

Question 8.12: What are interstitial compounds? Why do transition metals form them easily?

Answer: Interstitial compounds form when small atoms occupy spaces between metal atoms. Transition metals form them easily due to their close-packed structures and availability of interstitial voids.

Question 8.13: How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Answer: Transition metals exhibit multiple oxidation states because both (n-1)d and ns electrons participate in bonding. Non-transition elements typically show a fixed oxidation state (e.g., Ca²⁺). Examples: Mn (+2 to +7), Fe (+2, +3).

Question 8.14: Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Answer: Preparation involves fusing FeCr₂O₄ with Na₂CO₃ and O₂ to form Na₂CrO₄, acidifying to form Na₂Cr₂O₇, then treating with KCl to precipitate K₂Cr₂O₇. At higher pH, Cr₂O₇²⁻ converts to CrO₄²⁻, changing the color from orange to yellow.

Question 8.15: Describe the oxidising action of potassium dichromate and write ionic equations for its reaction with (i) iodide, (ii) iron(II) solution, and (iii) H₂S.

Answer: (i) Cr₂O₇²⁻ + 14H⁺ + 6I⁻ → 2Cr³⁺ + 3I₂ + 7H₂O
(ii) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(iii) Cr₂O₇²⁻ + 14H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O

Question 8.16: Describe the preparation of potassium permanganate and reactions with Fe²⁺, SO₂, and oxalic acid.

Answer: Preparation: MnO₂ fused with KOH and O₂ forms K₂MnO₄, which disproportionates to KMnO₄.
Reactions: (i) MnO₄⁻ + 8H⁺ + 5Fe²⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
(ii) 2MnO₄⁻ + 5SO₂ + 2H₂O → 2Mn²⁺ + 5SO₄²⁻ + 4H⁺
(iii) 2MnO₄⁻ + 5H₂C₂O₄ + 6H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Question 8.17: Which ion is easiest to reduce and which element oxidizes most easily?

Answer: Mn³⁺ is easiest to reduce (+1.5 V). Mn metal is easiest to oxidize (E° = –1.2 V).

Question 8.18: Which ions are coloured in aqueous solution?

Answer: Colored: Ti³⁺, V³⁺, Mn²⁺, Fe³⁺, Co²⁺
Colorless: Sc³⁺, Cu⁺

Question 8.19: Compare the stability of +2 oxidation state for the first transition series elements.

Answer: Stability increases from Sc to Mn due to half-filled configurations. Mn²⁺ and Zn²⁺ are exceptionally stable due to 3d⁵ and 3d¹⁰ configurations, respectively.

Question 8.20: Compare the chemistry of actinoids with lanthanoids.

Answer: Both show contraction, but actinoids have wider oxidation states (+3 to +7), poorer shielding, and higher reactivity compared to lanthanoids.