CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Define alloy.
Answer:
Alloys are homogeneous solid solutions of two or more metals.

Teacher's Note:
1. Transition metals form alloys easily because their atomic radii are similar (within about 15%).

Question 2. Transition metals show zero oxidation state with ligands like CO. Explain.
Answer:
CO forms synergic bonding with metal atom.

Teacher's Note:
1. Synergic bonding = σ-donation from CO to the metal plus π-back donation from filled metal d-orbitals to CO.

Question 3. Why can’t HCl acid be used to acidify KMnO₄ solution?
Answer:
Because KMnO₄ oxidises HCl into Cl₂.

Teacher's Note:
1. Dilute H₂SO₄ is used instead because it is not oxidised by KMnO₄.

Question 4. Name one ore of Mn and Cr.
Answer:
Mn : MnO₂ (pyrolusite)
Cr : FeCr₂O₄ (chromite)

Teacher's Note:
1. Pyrolusite is also the starting material for KMnO₄ and chromite for K₂Cr₂O₇.

Question 5. Why are Mn²⁺ compounds more stable than Fe²⁺ compounds towards oxidation to their +3 state?
Answer:
Mn²⁺ has half-filled electronic configuration (3d⁵).

Teacher's Note:
1. Fe²⁺ (3d⁶) becomes the stable half-filled 3d⁵ on oxidation to Fe³⁺, so it is oxidised easily.

Question 6. Why do transition metals (elements) show variable oxidation states?
Answer:
Due to very small energy difference between (n − 1)d and ns orbitals, electrons of both can take part in bond formation.

Teacher's Note:
1. Successive oxidation states usually differ by one unit in transition metals, unlike p-block elements where they differ by two.

Question 7. Write any use of pyrophoric alloy.
Answer:
Making bullets, shells and lighter flints.

Teacher's Note:
1. The common pyrophoric alloy is misch metal, which sparks when struck.

Question 8. Which is more basic – La(OH)₃ or Lu(OH)₃? Why?
Answer:
La(OH)₃. Due to lanthanide contraction, Lu³⁺ has lower size, so Lu(OH)₃ has more covalent character and is least basic.

Teacher's Note:
1. Basicity of lanthanoid hydroxides decreases from La(OH)₃ to Lu(OH)₃ as the ionic size decreases.

Question 9. Find out number of Cr–O–Cr bond/bonds in Cr₂O₇²⁻ ion.
Answer:
1

Teacher's Note:
1. The two CrO₄ tetrahedra share one corner oxygen; the other six Cr–O bonds are terminal.

Question 10. What is effect of increasing pH on K₂Cr₂O₇ solution?
Answer:
It changes into K₂CrO₄ solution/orange colour changes into yellow colour.

Teacher's Note:
1. Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O; the oxidation state of Cr stays +6.

Short Answer Questions (2 Marks)

Question 11. Complete and balance the following chemical equations :
(a) Fe²⁺ + MnO₄⁻ + H⁺ →
(b) MnO₄⁻ + H₂O + I⁻ → [CBSE Delhi/OD 2018]
Answer:
(a) 5Fe²⁺ + MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(b) 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
(Half mark to be deducted in each equation for not balancing)

Teacher's Note:
1. The product depends on the medium: in acid MnO₄⁻ gives Mn²⁺ and iodide gives I₂, but in neutral/alkaline medium it gives MnO₂ and iodide is oxidised to iodate.

Question 12. Explain why :
(i) E° for Mn³⁺/Mn²⁺ couple is more positive than that for Fe³⁺/Fe²⁺. (At. Nos. Mn = 25, Fe = 26).
(ii) Ce³⁺ can be easily oxidised to Ce⁴⁺. (At. No. Ce = 58). [CBSE OD 2012]
Answer:
(i) Stable half-filled 3d⁵ configuration of Mn²⁺ results in high 3^rd ionisation enthalpy of Mn. While in case of Fe²⁺, configuration is 3d⁶. Hence, it can easily loose one electron to give stable configuration 3d⁵.
(ii) Ce³⁺, ions having the configuration 4f¹5d⁰6s⁰ can easily lose electron to acquire the configuration 4f⁰5d⁰6s⁰ and form Ce⁴⁺ ion.

Teacher's Note:
1. A more positive E°(M³⁺/M²⁺) means M³⁺ is more easily reduced, i.e., M²⁺ is the more stable state – true for Mn²⁺ (3d⁵).

Question 13. When chromite ore FeCr₂O₄ is fused with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCl forms a orange coloured crystalline compound (C).
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of compound (C). [CBSE Delhi 2016]
Answer:
(i) 4FeCr₂O₄ + 16NaOH + 7O₂ → 8Na₂CrO₄ [A] + 2Fe₂O₃ + 8H₂O
2Na₂CrO₄ [A] + H₂SO₄ → Na₂Cr₂O₇ [B] + Na₂SO₄ + H₂O
Na₂Cr₂O₇ [B] + 2KCl → K₂Cr₂O₇ [C] + 2NaCl
A : Na₂CrO₄   B : Na₂Cr₂O₇   C : K₂Cr₂O₇
(ii) Use of K₂Cr₂O₇ (C) : It is used as a strong oxidizing agent in industries.

Teacher's Note:
1. Colour clues: yellow = chromate (A), orange = dichromate (B, C). K₂Cr₂O₇ crystallises out because it is less soluble than the sodium salt.

Question 14. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid ? Write the ionic equations for the reactions. [CBSE Comptt. OD 2015]
Answer:
Potassium permanganate is prepared by fusion of MnO₂ with an alkali metal hydroxide and an oxidising agent like KNO₃. This produces the dark green K₂MnO₄ which disproportionates in a neutral or acidic solution to give permanganate.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Oxalate ion or oxalic acid is oxidised at 333 K :
5C₂O₄²⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 10CO₂

Teacher's Note:
1. The oxalic acid titration is done warm (about 333 K) because the reaction is slow at room temperature until Mn²⁺ builds up and autocatalyses it.

Question 15. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with (i) an iodide (ii) H₂S. [CBSE Comptt. OD 2015]
Answer:
In acidic medium dichromate is a strong oxidising agent: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
(i) Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O
(ii) Cr₂O₇²⁻ + 3H₂S + 8H⁺ → 2Cr³⁺ + 3S + 7H₂O

Teacher's Note:
1. Each dichromate ion takes 6 electrons, so it oxidises 6 I⁻ or 3 H₂S; the orange solution turns green (Cr³⁺).

Question 16. Silver atom has completely filled d orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element? [NCERT]
Answer:
Those elements which have partially filled d or f subshell in any oxidation state are called transition elements. Silver (Ag) has a completely filled 4d orbital (4d¹⁰ 5s¹) in its ground state. Now, silver displays two oxidation states (+1 and +2). In the +1 oxidation state, an electron is removed from the s-orbital. However, in the +2 oxidation state, an electron is removed from the d-orbital. Thus, the d-orbital now becomes incomplete (4d⁹). Hence, it is a transition element.

Teacher's Note:
1. The same argument works for copper: Cu²⁺ is 3d⁹, which is why Cu is a transition element even though the atom is 3d¹⁰4s¹.

Question 17. Chromium is typical hard metal while mercury is a liquid. Explain why?
Answer:
Cr has five unpaired d-electrons. Hence metallic bonds are strong. In Hg, there is absence of unpaired electrons and size is larger.

Teacher's Note:
1. Hardness and melting point of transition metals rise with the number of unpaired d electrons available for metallic bonding.

Question 18. Why is KMnO₄ used in cleaning surgical instruments in hospitals?
Answer:
This is because KMnO₄ has a germicidal action.

Teacher's Note:
1. The germicidal action comes from its strong oxidising power.

Question 19. Most of the transition metals do not displace hydrogen from dilute acids, why?
Answer:
Due to their positive or less −ve reduction potential.

Teacher's Note:
1. Only metals with a sufficiently negative E° (more negative than 0.00 V of the SHE) can liberate H₂ from dilute acids.

Question 20. Explain why Cu⁺ is not stable in aqueous solution?
Answer:
Due to less −ve Δ_hydH^⊖ of Cu⁺/it cannot compensate 2nd ionisation enthalpy of Cu, whereas the much more −ve Δ_hydH^⊖ of Cu²⁺ compensates it. Hence Cu⁺ disproportionates: 2Cu⁺ → Cu²⁺ + Cu.

Teacher's Note:
1. Stability in water depends on hydration enthalpy, not just on configuration; Cu⁺ (3d¹⁰) is stable in the solid state but not in solution.

Question 21. Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?
Answer:
Oxygen and fluorine have small size and high electronegativity. They can oxidise the metal.

Teacher's Note:
1. Oxygen beats fluorine here because it can form multiple bonds, e.g., Mn₂O₇ (+7) but only MnF₄.

Short Answer Questions (3 Marks)

Question 22. The elements of 3d transition series are given as: Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following:
(i) Copper has exceptionally positive E°_M²⁺/M value. Why?
(ii) Which element is a strong reducing agent in +2 oxidation state and why?
(iii) Zn²⁺ salts are colourless. Why? [CBSE East 2016]
Answer:
(i) Because the sum of sublimation enthalpy and hydration enthalpy to convert Cu(s) to Cu²⁺(aq) is so high that it is not balanced by its hydration enthalpy.
(ii) Cr is strongest reducing agent in +2 oxidation state. Cr²⁺ has configuration 3d⁴. After losing one electron it forms Cr³⁺ which has stable half filled t_2g level.
(iii) Zn²⁺(3d¹⁰) has completely filled d-orbitals. As a result of this, d-d transition cannot occur and hence, Zn²⁺ salts are colourless.

Teacher's Note:
1. Cu is the only 3d metal with a positive E°_M²⁺/M; that is why it does not liberate H₂ from acids.

Teacher's Note:
1. Compare the measured moment with the spin only values (1.73 BM for one, 5.92 BM for five unpaired electrons) to decide high spin or low spin.

Question 24. Complete the following reactions:
(i) MnO₂ + KOH + O₂ →
(ii) I⁻ + MnO₄⁻ + H⁺ →
(iii) Cr₂O₇²⁻ + Sn²⁺ + H⁺ → [CBSE 2019 (56/4/1)]
Answer:
(i) 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(ii) 10I⁻ + 2MnO₄⁻ + 16H⁺ → 2Mn²⁺ + 8H₂O + 5I₂
(iii) Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O

Teacher's Note:
1. A ‘complete the reaction’ question expects a balanced equation; check that the electrons lost equal the electrons gained.

Question 25. When a brown compound of manganese (A) is treated with HCl it gives a gas (B). The gas taken in excess, reacts with NH₃ to give an explosive compound (C). Identify compounds A and B. [NCERT Exemplar] [HOTS]
Answer:
A = MnO₂ [Manganese (IV) oxide]
B = Cl₂ (Chlorine)
C = NCl₃ (Nitrogen trichloride)
MnO₂ (A) + 4HCl → MnCl₂ + Cl₂↑ (B) + 2H₂O
NH₃ + 3Cl₂ (excess) → NCl₃ (C) + 3HCl

Teacher's Note:
1. With excess NH₃, chlorine gives N₂ and NH₄Cl instead; only excess Cl₂ gives explosive NCl₃.

Question 26. A solution of KMnO₄ on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out? [NCERT Exemplar]
Answer:
Oxidising behaviour of KMnO₄ depends on pH of the solution.
In acidic medium (pH < 7),
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ (Colourless) + 4H₂O
In alkaline medium (pH > 7),
MnO₄⁻ + e⁻ → MnO₄²⁻ (Green)
In neutral medium (pH = 7),
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ (Brown precipitate) + 4OH⁻

Teacher's Note:
1. Link each colour to the oxidation state of Mn: +2 colourless, +4 brown MnO₂, +6 green manganate.

Question 27. (i) How would you account for the following:
(a) Highest fluoride of Mn is MnF₄ whereas the highest oxide is Mn₂O₇.
(b) Transition metals and their compounds show catalytic properties. [CBSE 2020 (56/5/1)]
(ii) Complete the following equation: 3MnO₄²⁻ + 4H⁺ → [CBSE (F) 2015]
Answer:
(i) (a) As oxygen stabilises manganese more than fluorine by forming multiple bonds.
(b) The catalytic activity of transition metals and their compounds is attributed to the following reasons:
1. Due to their tendency to show variable oxidation states transition metals form unstable intermediate compounds and provide a new path for the reaction with lower activation energy.
2. In some cases, the transition metals provide a suitable large surface area with free valencies on which reactants are adsorbed.
(ii) 3MnO₄²⁻ (Manganate ion) + 4H⁺ → 2MnO₄⁻ (Permanganate ion) + MnO₂ + 2H₂O

Teacher's Note:
1. Equation (ii) is a disproportionation: Mn(+6) goes to Mn(+7) and Mn(+4).

Question 28. Explain the following observations:
(i) The enthalpies of atomisation of transition metals are quite high.
(ii) There is a close similarity in physical and chemical properties of the 4d and 5d series of the transition elements, much more than expected on the basis of usual family relationship.
(iii) The members in the actinoid series exhibit larger number of oxidation states than the corresponding members in the lanthanoid series. [NCERT Exemplar]
Answer:
(i) This is because transition metals have strong metallic bonds as they have large number of unpaired electrons.
(ii) This is because 5d and 4d-series elements have virtually the same atomic and ionic radii due to lanthanoid contraction. Due to equality in size of Zr and Hf, Nb and Ta, Mo and W, etc., the two elements of each pair have the same properties.
(iii) The members in the actinoid series exhibit larger number of oxidation states than the corresponding members in the lanthanoid series due to the fact that the 5f, 6d and 7s levels are of comparable energies.

Teacher's Note:
1. High enthalpy of atomisation is also why transition metals have high melting points and tend to be noble in their reactions.

Question 29. Account for the following:
(i) Mn₂O₇ is acidic whereas MnO is basic.
(ii) Though copper has completely filled d-orbital (d¹⁰) yet it is considered as a transition metal.
(iii) Actinoids show wide range of oxidation states. [CBSE (F) 2016]
Answer:
(i) Mn has +7 oxidation state in Mn₂O₇ and +2 in MnO. In low oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. On the other hand, in higher oxidation state of the metal, valence electrons are involved in bonding and are not available. Instead effective nuclear charge is high and hence it can accept electrons and behave as an acid.
(ii) Copper exhibits +2 oxidation state wherein it will have incompletely filled d-orbitals (3d⁹), hence a transition metal.
(iii) This is due to comparable energies of 5f, 6d and 7s orbitals.

Teacher's Note:
1. General rule: as the oxidation state of a metal rises, its oxide changes from basic to amphoteric to acidic.

Question 30. Complete the following reactions—
(i) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) 2CrO₄²⁻ + 2H⁺ →
(iii) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → [CBSE Delhi 2013]
Answer:
(i) Cr₂O₇²⁻(aq) + 6Fe²⁺(aq) + 14H⁺(aq) → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O
(iii) 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Teacher's Note:
1. Reaction (ii) is not a redox reaction: chromium stays +6 in both chromate and dichromate; only the pH changes.

Question 31. (i) Complete the following equations :
(a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ →
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ →
(ii) Based on the data, arrange Fe²⁺, Mn²⁺ and Cr²⁺ in the increasing order of stability of +2 oxidation state.
E°_{Cr³⁺/Cr²⁺} = −0.4 V
E°_{Mn³⁺/Mn²⁺} = +1.5 V
E°_{Fe³⁺/Fe²⁺} = +0.8 V [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) (a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) Cr²⁺ < Fe²⁺ < Mn²⁺

Teacher's Note:
1. The more positive E°(M³⁺/M²⁺), the harder it is to oxidise M²⁺, so the more stable the +2 state: −0.4 < +0.8 < +1.5 V.

Question 32. Write the preparation of following :
(i) KMnO₄ from K₂MnO₄
(ii) Na₂CrO₄ from FeCr₂O₄
(iii) Cr₂O₇²⁻ from CrO₄²⁻ [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (or any other correct equation)
(ii) 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
(iii) 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

Teacher's Note:
1. Balance the chromite equation by Fe and Cr first: 4 FeCr₂O₄ give 8 Na₂CrO₄ and 2 Fe₂O₃, then 8 CO₂ and 7 O₂.

Long Answer Questions (5 Marks)

Question 33. (i) Compare non transition and transition elements on the basis of their
(a) Variability of oxidation states
(b) Stability of oxidation states.
(ii) Give chemical reactions for the following observations:
(a) Potassium permanganate is a good oxidising agent in basic medium.
(b) Inter convertibility of chromate ion and dichromate ion in aqueous solution depends upon pH of the solution.
(c) Potassium permanganate is thermally unstable at 513K. [CBSE Sample Paper 2013]
Answer:
(i) (a) Oxidation states of transition elements differ from each other by unity. In non-transition elements oxidation states normally differ by a unit of two.
(b) In transition elements higher oxidation states are favoured by heavier elements whereas in non-transition elements lower oxidation state is favoured by heavier elements.
(ii) (a) [MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻] × 2
I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻
2MnO₄⁻ + I⁻ + H₂O → IO₃⁻ + 2MnO₂ + 2OH⁻
(b) 2CrO₄²⁻ + 2H⁺ \( \underset{\text{Alkali (pH more than 7)}}{\overset{\text{Acid (pH less than 7)}}{\rightleftharpoons}} \) Cr₂O₇²⁻ + H₂O
(c) 2KMnO₄ \( \xrightarrow{\text{513 K}} \) K₂MnO₄ + MnO₂ + O₂

Teacher's Note:
1. In alkaline medium each MnO₄⁻ takes 3 electrons (to MnO₂), compared with 5 electrons in acidic medium.

Question 34. (i) In the titration of FeSO₄ with KMnO₄ in the acidic medium, why is dil. H₂SO₄ used instead of dil. HCl?
(ii) Give reasons:
(a) Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal.
(b) Ce⁴⁺ is used as an oxidising agent in volumetric analysis.
(c) Zn²⁺ salts are white while Cu²⁺ salts are blue. [CBSE 2019 (56/2/3)]
OR
Why is Cu²⁺ ion coloured while Zn²⁺ ion is colourless in aqueous solution? [CBSE 2020 (56/3/1)]
Answer:
(i) Dil. H₂SO₄ is not oxidised by KMnO₄; it only provides the acidic medium in which KMnO₄ oxidises FeSO₄ to Fe₂(SO₄)₃. Dil. HCl is a reducing agent and liberates chlorine on reacting with KMnO₄ solution.
Hence, the part of the oxygen produced from KMnO₄ is used up by HCl.
(ii) (a) In these oxoanions the oxygen atoms are directly bonded to the transition metal. Since oxygen is highly electronegative, the oxoanions bring out the highest oxidation state of the metal.
(b) Ce⁴⁺ has the tendency to attain +3 oxidation state which is more stable and so it is used as an oxidising agent in volumetric analysis.
(c) Zn²⁺ ion has all its orbitals completely filled whereas in Cu²⁺ ion there is one half-filled 3d-orbital. Therefore, due to d-d transition Cu²⁺ has a tendency to form coloured salts whereas Zn²⁺ has no such tendency.

Teacher's Note:
1. If HCl is used, part of the KMnO₄ is consumed in oxidising Cl⁻, so the titre value comes out too high.

Question 35. A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H₂SO₄ and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved. [NCERT Exemplar] [HOTS]
Answer:
A = KMnO₄ (Potassium permanganate), B = K₂MnO₄ (Potassium manganate), C = MnO₂ (Manganese (IV) oxide), D = MnCl₂ (Manganese (II) chloride)
2KMnO₄ (A) \( \xrightarrow{\Delta} \) K₂MnO₄ (B) + MnO₂ (C) + O₂
MnO₂ (C) + 2KOH + KNO₃ → K₂MnO₄ (B) + H₂O + KNO₂
MnO₂ + 4NaCl + 4H₂SO₄ → MnCl₂ (D) + 4NaHSO₄ + 2H₂O + Cl₂

Teacher's Note:
1. ‘Violet compound of manganese’ almost always points to KMnO₄; build the rest of the chain from it.

Question 36. When an oxide of manganese (A) is fused with KOH in the presence of an oxidising agent and dissolved in water, it gives a dark solution of compound (B). Compound (B) disproportionates in neutral or acidic solution to give purple compound (C). An alkaline solution of compound (C) oxidises potassium iodide solution to a compound (D) and compound (A) is also formed. Identify compounds A to D and also explain the reactions involved. [NCERT Exemplar] [HOTS]
Answer:
A = MnO₂ (Manganese (IV) oxide), B = K₂MnO₄ (Potassium manganate), C = KMnO₄ (Potassium permanganate), D = KIO₃ (Potassium iodate)
2MnO₂ (A) + 4KOH + O₂ → 2K₂MnO₄ (B) + 2H₂O
3K₂MnO₄ + 4H⁺ → 2KMnO₄ (C) + MnO₂ + 2H₂O + 4K⁺
2MnO₄⁻ + H₂O + KI → 2MnO₂ (A) + 2OH⁻ + KIO₃ (D)

Teacher's Note:
1. Remember that in alkaline or neutral medium iodide is oxidised to iodate, not to iodine.

Question 37. (i) Write balanced equations to represent what happens when
(a) Cu²⁺ is treated with KI.
(b) Acidified potassium dichromate solution is reacted with iron (II) solution. (ionic equation)
(ii) (a) The figure given illustrates the first ionization enthalpies of first, second and third series of transition elements. Answer the question that follows:
Which series amongst the first, second and third series of transition elements have the highest first ionization enthalpy and why?
(b) Separation of lanthanoid elements is difficult. Explain. [CBSE 2020 (56/5/1)]
(c) Sm²⁺, Eu²⁺ and Yb²⁺ ions in solutions are good reducing agents but an aqueous solution of Ce⁴⁺ is a good oxidising agent. Why? [CBSE Sample Paper 2015]

Answer:
(i) (a) 2Cu²⁺ + 4I⁻ → Cu₂I₂ + I₂
(b) Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
[Fe²⁺ → Fe³⁺ + e⁻] × 6
Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) (a) Third series has the highest first ionization energy due to poorest shielding effect of the fully filled 4f subshell.
(b) Due to lanthanoid contraction, some pair of elements have almost similar size and as a result, such pairs have very similar properties which makes their separation difficult.
(c) Sm²⁺, Eu²⁺ and Yb²⁺ ions are good reducing agents as they tend to acquire common oxidation state of +3 shown by lanthanoids by the loss of one electron while Ce⁴⁺ gains one electron to attain +3. Hence, Ce⁴⁺ is an oxidising agent.

Teacher's Note:
1. Read the graph first: the 5d (third series) curve lies above the other two for almost every group.