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CBSE Class 12 Chapter Test: THE d- & f-BLOCK ELEMENTS
General Instructions:
- This question paper consists of 15 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 5 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
Directions for Q6 & Q7: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): Zr and Hf lie in the same group of the periodic table.
Reason (R): The much more negative hydration enthalpy of Cu2+(aq) more than compensates for the second ionization enthalpy of Cu.
(i) Transition metals and their many compounds act as good catalysts.
(ii) The enthalpies of atomization of transition elements are high.
(iii) Transition metals generally form coloured compounds.
(i) Iron(II) ions (Fe2+)
(ii) Oxalate ions (C2O42-)
(iii) Iodide ions (I-)
(b) Give an example of a disproportionation reaction in an aqueous solution of copper ion (Cu+).
The f-block elements consist of the two series, lanthanoids (the fourteen elements following lanthanum) and actinoids (the fourteen elements following actinium). A regular decrease in the atomic and ionic radii is observed among these elements as the atomic number increases. This regular decrease in size among the lanthanoids is a key feature of this series. Furthermore, the chemistry of the actinoids is relatively more complicated compared to the lanthanoids because actinoids exhibit a wider range of oxidation states and are radioactive in nature.
(a) What is the general electronic configuration of the f-block elements? [1 Mark]
(b) What is the common oxidation state shown by all lanthanoids? [1 Mark]
(c) Name the phenomenon responsible for the regular decrease in the size of lanthanoids. State one critical consequence of this phenomenon on the d-block elements. [2 Marks]
(a) The E°(M2+/M) value for copper is positive (+0.34 V).
(b) Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state.
(c) Transition metals have a strong tendency to form complex compounds.
(d) Interstitial compounds are well known for transition metals.
(e) The lowest oxide of a transition metal is usually basic, whereas the highest oxide is usually amphoteric or acidic.
(b) What happens when KMnO4 crystals are heated strongly? Write the balanced chemical equation. [2 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (d) Copper (Cu)
Reasoning: Copper has the configuration [Ar] 3d10 4s1. Even though its d-orbital is completely filled in the ground state, it exhibits a +2 oxidation state where it has an incomplete d-orbital (3d9), thus qualifying as a transition element.
2. Answer: (b) Ligand to metal charge transfer (LMCT)
Reasoning: In the MnO4- ion, Manganese is in its highest oxidation state (+7) and has no d-electrons (d0 configuration). Therefore, d-d transition is impossible. The deep purple colour is due to the transfer of an electron from the oxygen ligand to the empty d-orbital of the metal (LMCT).
3. Answer: (c) Zr and Hf
Reasoning: Zirconium (Zr, period 5) and Hafnium (Hf, period 6) belong to Group 4. Due to lanthanoid contraction (poor shielding of 4f electrons), the expected increase in size down the group is canceled out, giving them nearly identical atomic and ionic radii.
4. Answer: (b) Cerium (Ce)
Reasoning: Cerium (Z=58) has the configuration [Xe] 4f1 5d1 6s2. By losing 4 electrons (to form Ce4+), it achieves the highly stable noble gas configuration of Xenon [Xe].
5. Answer: (c) 3.87 BM
Reasoning: The spin-only magnetic moment μ is given by μ = √[n(n+2)] BM, where n is the number of unpaired electrons. For n=3, μ = √[3(3+2)] = √15 ≈ 3.87 BM.
6. Answer: (b) Both A and R are true but R is not the correct explanation of A.
Reasoning: It is true that Zr and Hf are difficult to separate and that they lie in the same group. However, the exact reason they are difficult to separate is because they have virtually identical atomic/ionic sizes and chemical properties due to Lanthanoid Contraction, not merely because they are in the same group.
7. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: Cu2+ is much smaller and carries a higher charge than Cu+, giving it a highly negative hydration enthalpy when dissolved in water. This massive release of energy more than compensates for the energy required to remove the second electron (IE2), making Cu2+(aq) more stable.
8. Solution:
A transition element is defined as one which has incompletely filled d-orbitals in its ground state or in any of its oxidation states. Zinc has the configuration [Ar] 3d10 4s2 and forms only Zn2+ ions with a [Ar] 3d10 configuration. Since its d-orbitals are completely filled in both states, it is not considered a transition element.
The other two elements are Cadmium (Cd) and Mercury (Hg).
9. Solution:
Atomic number of Metal (Z) = 27 (Cobalt).
Electronic configuration of Co = [Ar] 3d7 4s2.
For M2+ (Co2+), the configuration is [Ar] 3d7.
In a 3d subshell with 7 electrons, the arrangement is: ↑↓ | ↑↓ | ↑ | ↑ | ↑ .
Number of unpaired electrons (n) = 3.
Magnetic moment μ = √[n(n+2)] = √[3(3+2)] = √15 = 3.87 BM.
10. Solution:
(i) They act as good catalysts because they have the ability to adopt multiple (variable) oxidation states and can form unstable intermediate complexes. They also provide a large surface area for adsorption of reactants.
(ii) They have high enthalpies of atomization because they have a large number of unpaired electrons in their d-orbitals, which leads to very strong interatomic metallic bonding.
(iii) Coloured compounds are formed due to the presence of unpaired d-electrons. When visible light falls on the compound, an electron absorbs energy and gets promoted to a higher energy d-orbital (d-d transition). The transmitted/reflected light imparts colour.
11. Solution:
Acidified KMnO4 acts as a strong oxidizing agent. The MnO4- ion is reduced to Mn2+.
(i) Fe2+ to Fe3+:
MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O
(ii) C2O42- (Oxalate) to CO2:
2MnO4- + 5C2O42- + 16H+ → 2Mn2+ + 10CO2 + 8H2O
(iii) I- to I2:
2MnO4- + 10I- + 16H+ → 2Mn2+ + 5I2 + 8H2O
12. Solution:
(a) Disproportionation is a specific type of redox reaction in which an element in a particular oxidation state is simultaneously oxidized and reduced, yielding two different products containing the same element in two different oxidation states.
(b) Aqueous Cu+ is unstable and disproportionates into Cu2+ and solid Cu.
2Cu+(aq) → Cu2+(aq) + Cu(s)
13. Solution (Case-Based):
(a) The general electronic configuration is: (n-2)f1-14 (n-1)d0-1 ns2.
(b) The most common and stable oxidation state for all lanthanoids is +3.
(c) The phenomenon is called Lanthanoid Contraction. It occurs due to the imperfect/poor shielding effect of the 4f electrons.
Consequence: The atomic and ionic radii of the second and third transition series elements (e.g., Zr and Hf, Nb and Ta) become almost identical, making their chemical properties remarkably similar and their separation extremely difficult.
14. Solution:
(a) The E° value is determined by the sum of sublimation, ionization, and hydration enthalpies. For Copper, the energy required to sublimate and ionize it (high atomization and high IE) is not balanced by its hydration enthalpy. The net energy is positive, meaning it requires energy rather than releasing it, giving a positive E°.
(b) Mn2+ has a [Ar] 3d5 configuration, which is a highly stable exactly half-filled d-subshell. Fe2+ is 3d6 and readily loses one electron to become the more stable Fe3+ (3d5). Hence, Mn2+ resists oxidation.
(c) Transition metals form complexes readily because of their (i) small ionic sizes, (ii) high ionic charges, and (iii) the availability of empty d-orbitals to accept lone pairs of electrons from ligands.
(d) The crystal lattices of transition metals have small empty spaces (interstices). Small non-metal atoms like H, C, or N can easily trap themselves in these spaces, forming interstitial compounds without altering the chemical nature of the metal.
(e) In lower oxidation states, the metal has low charge and its bonds with oxygen are primarily ionic, acting as a base (donating electrons/reacting with acids). In higher oxidation states, the high effective nuclear charge strongly polarizes the electron cloud, making the M-O bonds largely covalent and highly acidic in nature.
15. Solution:
(a) Preparation of KMnO4: It is prepared on a large scale from pyrolusite ore (MnO2) in two steps:
Step 1: Fusion of MnO2 with an alkali metal hydroxide (KOH) and an oxidizing agent like KNO3 or O2. This produces the dark green potassium manganate (K2MnO4).
Equation: 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
Step 2: The green potassium manganate is extracted with water and then oxidized to potassium permanganate (KMnO4) either electrolytically or by passing chlorine or ozone gas (disproportionation in acidic/neutral medium).
Equation (in acidic medium): 3MnO42- + 4H+ → 2MnO4- + MnO2 + 2H2O
(b) Action of Heat on KMnO4:
When potassium permanganate crystals are heated strongly (at ~513 K), they decompose to give potassium manganate, manganese dioxide, and oxygen gas.
Equation: 2KMnO4 →(Δ) K2MnO4 + MnO2 + O2↑
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