CBSE Class 12 Chemistry The d and f Block Elements Important Questions Set 09

Very Short Answer Questions (1 Mark)

Question 1. Complete the following :
2 MnO₄⁻ (aq) + 5 H₂O₂ (aq) → 2 Mn²⁺ + 8 H₂O + .... (Manipur Board 2011)
Answer:
2 MnO₄⁻ + 5 H₂O₂ (aq) + 6 H⁺ (aq) → 2 Mn²⁺ + 8 H₂O + 5 O₂

Teacher's Note:
1. Here H₂O₂ acts as a reducing agent and all the O₂ evolved comes from H₂O₂, one O₂ per H₂O₂.

Question 2. What is the basic difference between the electronic configurations of transition and inner transition elements ?
Answer:
General electronic configuration of transition elements = [Noble gas] (n − 1) d¹⁻¹⁰ ns¹⁻² and for inner transition elements = (n − 2) f¹⁻¹⁴ (n − 1) d⁰⁻¹ ns². Thus, in transition elements, last electron enters d-orbital of penultimate shell while in inner transition elements, it enters f-orbital of ante-penultimate shell.

Teacher's Note:
1. Penultimate means (n − 1), the second last shell; ante-penultimate means (n − 2), the third last shell.

Question 3. Write any two uses of pyrophoric alloys.
Answer:
Pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters and in making of bullets and shells.

Teacher's Note:
1. The best known pyrophoric alloy is misch metal (about 95% lanthanoid metals with iron and traces of S, C, Ca, Al).

Question 4. Give the general electronic configuration of Actinides.
Answer:
5f¹⁻¹⁴ 6d⁰⁻¹ 7s²

Teacher's Note:
1. Write it with the noble gas core for full marks: [Rn] 5f¹⁻¹⁴ 6d⁰⁻¹ 7s².

Question 5. Write the electronic configuration of the element with atomic number 102.
Answer:
[Rn]⁸⁶ 5f¹⁴ 6d⁰ 7s².

Teacher's Note:
1. Count from radon (Z = 86): 102 − 86 = 16 electrons, which go as 5f¹⁴ 7s². The element is nobelium (No).

Question 6. What is the maximum oxidation state shown by actinides ?
Answer:
+ 7.

Teacher's Note:
1. The +7 state is shown by Np and Pu; the most common state of actinoids is still +3.

Question 7. Write the electronic configuration of Gadolinium (Z = 64) and its most stable oxidation state. (AIPMT Mains 2008)
Answer:
Gd (Z = 64) = [Xe] 4f⁷ 5d¹ 6s²
Most stable oxidation state = +3.

Teacher's Note:
1. Gd³⁺ is 4f⁷, a half-filled f-subshell, which is why +3 is so stable for Gd.

Question 8. Which is the most common oxidation state of lanthanides ?
Answer:
+3.

Teacher's Note:
1. Ce (+4) and Eu, Yb (+2) also show other states, but +3 is the characteristic state of all lanthanoids.

Question 9. What is meant by lanthanoid contraction ? (CBSE 2011, Assam Board 2013)
Answer:
The regular decrease in the atomic and ionic radii of lanthanoids with increasing atomic number is called lanthanoid contraction.

Teacher's Note:
1. It is caused by the poor shielding of one 4f electron by another, so the effective nuclear charge rises steadily.

Question 10. Why is the separation of lanthanide elements difficult ?
Answer:
Due to lanthanide contraction, the change in the atomic or ionic radii of these elements is very small. Hence, their chemical properties are similar. This makes their separation difficult.

Teacher's Note:
1. Lanthanoids are separated by ion-exchange methods that use their small differences in size and complexing ability.

Question 11. Explain why lanthanoids are paramagnetic in nature ?
Answer:
All lanthanoids except La³⁺ and Lu³⁺ contain unpaired electrons and hence are paramagnetic.

Teacher's Note:
1. La³⁺ is 4f⁰ and Lu³⁺ is 4f¹⁴; both have no unpaired electrons and are diamagnetic.

Short Answer Questions (2 Marks)

Question 12. Give equation for chromyl chloride test.
Answer:
When a chloride is heated with solid K₂Cr₂O₇ and conc. H₂SO₄, red vapours of chromyl chloride are evolved :
K₂Cr₂O₇ + 4NaCl + 6H₂SO₄ → 2CrO₂Cl₂ + 2KHSO₄ + 4NaHSO₄ + 3H₂O
The vapours, when passed into NaOH solution, give a yellow solution of sodium chromate which gives a yellow precipitate of lead chromate with lead acetate :
CrO₂Cl₂ + 4NaOH → Na₂CrO₄ + 2NaCl + 2H₂O ; Na₂CrO₄ + (CH₃COO)₂Pb → PbCrO₄↓ + 2CH₃COONa

Teacher's Note:
1. The test is specific for ionic chlorides; covalent chlorides like HgCl₂ do not respond.

Question 13. Write an equation in ionic form to represent the oxidizing action of Cr₂O₇²⁻ in acidic medium. Also draw the structure of Cr₂O₇²⁻ ion.
Answer:
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Structure of Cr₂O₇²⁻ ion :

Teacher's Note:
1. Each Cr changes from +6 to +3, so one dichromate ion accepts 2 × 3 = 6 electrons.

Question 14. What chemical changes take place when
(i) MnO₂ is fused with KOH in air ?
(ii) pH of a chromate solution is progressively lowered ?
Answer:
(i) Green potassium manganate is formed : 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(ii) Yellow chromate changes to orange dichromate : 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

Teacher's Note:
1. Lowering pH means adding H⁺; it drives the chromate–dichromate equilibrium towards dichromate.

Question 15. Write complete chemical equations for :
(i) Oxidation of Fe²⁺ by Cr₂O₇²⁻ in acid medium.
(ii) Oxidation of S₂O₃²⁻ by MnO₄⁻ in neutral aqueous medium. (CBSE 2008)
Answer:
(i) Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O
(ii) 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻

Teacher's Note:
1. Check (ii) by electrons: 8 MnO₄⁻ gain 8 × 3 = 24 e⁻, and 3 S₂O₃²⁻ lose 3 × 2 × 4 = 24 e⁻ (S from +2 to +6).

Question 16. What is meant by disproportionation of oxidation state. Give one example. (Hr. Board 2011)
Answer:
When a particular oxidation state becomes less stable relative to other oxidation states, one lower and one higher, it is said to undergo disproportionation, i.e., the same species is simultaneously oxidised and reduced. For example, manganese (VI) becomes unstable relative to manganese (VII) and manganese (IV) in acidic solution :
3Mn^VIO₄²⁻ + 4H⁺ → 2Mn^VIIO₄⁻ + Mn^IVO₂ + 2H₂O
Similarly, 2Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s)

Teacher's Note:
1. To spot disproportionation, check that one element appears in the products with both a higher and a lower oxidation number than in the reactant.

Question 17. What happens when (i) H₂S reacts with acidified K₂Cr₂O₇ solution ? (ii) Ethanol is oxidized with acidified KMnO₄ solution ? (Bihar Board 2011)
Answer:
(i) H₂S is oxidised to sulphur (milky/yellow) and the solution turns green : Cr₂O₇²⁻ + 8H⁺ + 3H₂S → 2Cr³⁺ + 3S + 7H₂O
(ii) Ethanol is oxidised to ethanoic acid : CH₃CH₂OH \( \xrightarrow[\text{KMnO}_4/\text{H}^+]{[\text{O}]} \) CH₃CHO \( \xrightarrow{[\text{O}]} \) CH₃COOH

Teacher's Note:
1. With a strong oxidising agent like acidified KMnO₄, ethanol does not stop at the aldehyde stage but goes on to the carboxylic acid.

Question 18. What happens when
(i) (NH₄)₂Cr₂O₇ is heated ?
(ii) H₃PO₃ is heated ? (CBSE 2017)
Answer:
(i) (NH₄)₂Cr₂O₇ \( \xrightarrow{\text{Heat}} \) N₂ + 4H₂O + Cr₂O₃
(ii) 4H₃PO₃ \( \xrightarrow{\text{473 K}} \) 3H₃PO₄ + PH₃

Teacher's Note:
1. Both are internal redox reactions: in (ii) phosphorus(+3) disproportionates into +5 (H₃PO₄) and −3 (PH₃).

Question 19. Complete the following equations :
(i) 2MnO₄⁻ + 16H⁺ + 5S²⁻ →
(ii) KMnO₄ \( \xrightarrow{\text{heat}} \) (CBSE 2017)
Answer:
(i) 2MnO₄⁻ + 16H⁺ + 5S²⁻ → 2Mn²⁺ + 8H₂O + 5S
(ii) 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂

Teacher's Note:
1. In (ii) remember to double KMnO₄ to balance: two Mn atoms end up in two different products.

Question 20. Complete the following equations :
(a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ →
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → (CBSE 2018 (C))
Answer:
(a) 2MnO₄⁻ + 5SO₃²⁻ + 6H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻
(b) Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O

Teacher's Note:
1. In (a), sulphite (S = +4) is oxidised to sulphate (S = +6); 5 × 2 = 10 electrons match 2 × 5 = 10 gained by Mn.

Question 21. What is Misch metal ? Give its one use.
Answer:
Misch metal is an alloy which consists of a lanthanoid metal (about 95%) and iron (about 5%) along with traces of S, C, Ca and Al.
Use : It is used in Mg-based alloys to produce bullets, shells and lighter flints.

Teacher's Note:
1. Misch metal is pyrophoric: it gives sparks when struck, which is the property behind its use in flints.

Question 22. How would you account for the following observations ?
(i) The enthalpies of atomisation of the transition metals are high.
(ii) Of the lanthanoids, only cerium (Z = 58) is known to exhibit quite stable +4 state in solutions.
Answer:
(i) Transition metals have a large number of unpaired electrons in their (n − 1)d and ns orbitals. These take part in strong interatomic metallic bonding. Hence, their enthalpies of atomisation are high.
(ii) Ce⁴⁺ has the stable noble gas configuration [Xe] 4f⁰. Hence, Ce shows a stable +4 state (though Ce⁴⁺ is a good oxidising agent and changes slowly to the common +3 state).

Teacher's Note:
1. The metals in the middle of each series (about d⁵) have the highest enthalpies of atomisation because they have the most unpaired electrons.

Question 23. (a) Write two consequences of lanthanoid contraction.
(b) Name the element of 3d series which exhibits the largest number of oxidation states. Give reason. (CBSE 2022)
Answer:
(a) (i) The elements of the second and third transition series (e.g., Zr and Hf) have nearly the same atomic radii and similar properties. (ii) Basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃; lanthanoids are difficult to separate.
(b) Mn (+2 to +7). Mn has the configuration 3d⁵ 4s², so it has the maximum number of electrons (seven) available for bonding with nearly equal energies.

Teacher's Note:
1. The highest oxidation state of an early 3d element equals the total number of its 3d + 4s electrons, which is why Mn reaches +7.

Short Answer Questions (3 Marks)

Question 24. What happens when (i) KMnO₄ is heated (HP Board 2013)
(ii) K₂Cr₂O₇ is heated.
Answer:
(i) On heating to 513 K, KMnO₄ decomposes giving O₂ : 2KMnO₄ \( \xrightarrow{\text{513 K}} \) K₂MnO₄ + MnO₂ + O₂
(ii) On strong heating, K₂Cr₂O₇ decomposes giving O₂ : 4K₂Cr₂O₇ \( \xrightarrow{\text{Heat}} \) 4K₂CrO₄ + 2Cr₂O₃ + 3O₂

Teacher's Note:
1. Both decompositions release oxygen, which is why KMnO₄ heating is a laboratory source of O₂.

Question 25. What happens when acidified KMnO₄ reacts with (i) H₂S (ii) Oxalic acid (iii) KNO₂ Write ionic reactions. (Hr. Board 2012)
Answer:
(i) H₂S is oxidised to sulphur : 2MnO₄⁻ + 16H⁺ + 5S²⁻ → 2Mn²⁺ + 8H₂O + 5S
(ii) Oxalic acid is oxidised to CO₂ : 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ \( \xrightarrow{\text{333 K}} \) 2Mn²⁺ + 8H₂O + 10CO₂
(iii) Nitrite is oxidised to nitrate : 2MnO₄⁻ + 6H⁺ + 5NO₂⁻ → 2Mn²⁺ + 5NO₃⁻ + 3H₂O

Teacher's Note:
1. In every case the purple colour of MnO₄⁻ disappears because nearly colourless Mn²⁺ is formed, so KMnO₄ acts as its own indicator.

Question 26. Describe the oxidizing property of KMnO₄ in neutral or faintly alkaline medium for its reaction with iodide ions and thiosulphate ions.
Answer:
In neutral or faintly alkaline medium, MnO₄⁻ is reduced to MnO₂ :
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
(i) Iodide is oxidised to iodate : 2MnO₄⁻ + H₂O + I⁻ → 2MnO₂ + 2OH⁻ + IO₃⁻
(ii) Thiosulphate is oxidised to sulphate : 8MnO₄⁻ + 3S₂O₃²⁻ + H₂O → 8MnO₂ + 6SO₄²⁻ + 2OH⁻

Teacher's Note:
1. In neutral medium Mn goes from +7 to +4 (3 electrons), not to +2 as in acid; a brown precipitate of MnO₂ is seen.

Question 27. Write ionic equations for the conversion of
(a) manganate to permanganate
(b) permanganate to manganese (II)
(c) chromate to dichromate.
Answer:
(a) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
(b) MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
(c) 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O

Teacher's Note:
1. Reaction (a) is a disproportionation of Mn(+6) into Mn(+7) and Mn(+4).

Question 28. Write chemical equations for the following reactions :
(a) Oxidation of nitrite ion by MnO₄⁻ in acidic medium
(b) Acidification of potassium chromate solution
(c) Disproportionation of Mn (VI) in acidic solution.
Answer:
(a) 5NO₂⁻ + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 3H₂O + 5NO₃⁻
(b) 2K₂CrO₄ + 2H⁺ → K₂Cr₂O₇ + 2K⁺ + H₂O
(c) 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ (Permanganate ion) + MnO₂ + 2H₂O

Teacher's Note:
1. Charge check for (a): left −5 − 2 + 6 = −1, right +4 − 5 = −1.

Question 29. What are inner transition metals ? Why are they so called ? Write their general electronic configuration. Which of the following atomic numbers are the atomic numbers of inner transition elements ?
25, 59, 74, 95, 104. (Uttarakhand Board 2012)
Answer:
The f-block elements in which the last electron enters the (n − 2)f orbitals are called inner transition elements. They are so called because the differentiating electron enters the f-orbitals of the ante-penultimate shell, which lies deep inside (inner to) the penultimate d-subshell. They comprise lanthanoids (Ce to Lu) and actinoids (Th to Lr).
General electronic configuration : (n − 2)f¹⁻¹⁴ (n − 1)d⁰⁻¹ ns²
Inner transition elements are : 59 (Pr, lanthanoid) and 95 (Am, actinoid). (25 = Mn and 74 = W are d-block; 104 = Rf belongs to 6d series.)

Teacher's Note:
1. Lanthanoids have Z = 58 to 71 and actinoids Z = 90 to 103; any atomic number in these ranges is an inner transition element.

Question 30. What is lanthanoid contraction ? Give two examples of lanthanoid elements. What are oxidation states exhibited by lanthanoid elements ? (Jharkhand Board 2012)
Answer:
The steady decrease in the atomic and ionic (M³⁺) radii of lanthanoids with increase in atomic number from La to Lu is called lanthanoid contraction. It is due to the imperfect shielding of one 4f electron by another.
Examples of lanthanoids : Cerium (Ce) and Gadolinium (Gd).
Oxidation states : The common oxidation state is +3. Some also show +2 (Eu, Yb, Sm) and +4 (Ce, Tb, Pr).

Teacher's Note:
1. Eu²⁺ (4f⁷) and Ce⁴⁺ (4f⁰) are the standard examples of stable +2 and +4 states, because of their half-filled and empty f-subshells.

Question 31. What is lanthanide contraction ? What is its cause and what are its consequences ? (HP Board 2011, 2013, Hr. Board 2011, 2012, J & K Board 2012, Kerala Board 2012)
or
Write a note on lanthanoid contraction. (Maharashtra Board 2012, Uttarakhand Board 2012)
Answer:
Lanthanoid contraction : The regular decrease in the atomic and ionic radii of lanthanoids with increase in atomic number is called lanthanoid contraction.
Cause : As we move along the series, the nuclear charge increases by one unit at each step and one electron is added to the 4f subshell. The shielding of one 4f electron by another is very poor due to the diffused shape of f-orbitals. Hence, the effective nuclear charge increases and the size decreases.
Consequences : (i) The elements of the second and third transition series have nearly the same atomic radii (e.g., Zr 160 pm, Hf 159 pm) and similar properties. (ii) Lanthanoids have very similar properties, which makes their separation difficult. (iii) Basic strength of hydroxides decreases from La(OH)₃ to Lu(OH)₃. (iv) Ionisation enthalpy and electronegativity of 5d elements are higher than those of 4d elements.

Teacher's Note:
1. Lanthanoid contraction is a cause in many other answers; learn the Zr–Hf and La(OH)₃–Lu(OH)₃ examples with it.

Question 32. Write general electronic configuration of f-block elements. Write any two uses of lanthanides and any two uses of actinides. (MP Board 2011)
Answer:
General electronic configuration : (n − 2)f¹⁻¹⁴ (n − 1)d⁰⁻¹ ns²
Uses of lanthanoids : (i) Misch metal is used in making bullets, shells and lighter flints. (ii) Mixed oxides of lanthanoids are used as catalysts in petroleum cracking; some oxides are used as phosphors in television screens.
Uses of actinoids : (i) Uranium and plutonium are used as fuels in nuclear reactors and in atom bombs. (ii) Thorium oxide is used in incandescent gas mantles and thorium is used in atomic reactors.

Teacher's Note:
1. Keep lanthanoid uses (alloys, catalysts, phosphors) separate from actinoid uses (nuclear fuel), which come from radioactivity.

Question 33. How would you account for the following ?
(i) Many of the transition elements and their compounds can act as good catalysts
(ii) The metallic radii of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second series
or Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) There is a greater range of oxidation states among the actinoids than among the lanthanoids (CBSE 2009)
Answer:
(i) Transition elements show variable oxidation states and so form unstable intermediates with the reactants; they also provide a large surface area for adsorption of reactants. This provides a new path with lower activation energy.
(ii) This is due to lanthanoid contraction. The 14 lanthanoids come between the second and third series; the steady decrease in size caused by poorly shielding 4f electrons cancels the expected increase in size from the 4d to the 5d series. Hence, Zr and Hf have almost identical radii.
(iii) In actinoids, the 5f, 6d and 7s levels have comparable energies, so electrons from all of them can take part in bonding. In lanthanoids, the 4f electrons lie deep inside and are not available for bonding. Hence, actinoids show a greater range of oxidation states.

Teacher's Note:
1. 5f orbitals extend further out than 4f orbitals, which is the root cause of the wider range of oxidation states in actinoids.

Teacher's Note:
1. Write the comparison as a two-column table with matching points; examiners award a mark for each correctly paired difference.

Question 35. Account for the following :
(a) CuCl₂ is more stable than Cu₂Cl₂.
(b) Atomic radii of 4d and 5d series elements are nearly same.
(c) Hydrochloric acid is not used in permanganate titrations. (CBSE Foreign 2017)
or
Account for the following :
(a) Eu²⁺ is a strong reducing agent.
(b) Orange colour of dichromate ion changes to yellow in alkaline medium.
(c) E° (M²⁺/M) values for transition metals show irregular variation. (CBSE Foreign 2017)
Answer:
(a) Cu²⁺ (aq) is more stable than Cu⁺ (aq) because the much more negative hydration enthalpy of Cu²⁺ more than compensates for the second ionisation enthalpy of Cu. Hence, CuCl₂ is more stable than Cu₂Cl₂.
(b) This is due to lanthanoid contraction.
(c) HCl is oxidised by KMnO₄ to chlorine. Thus a part of KMnO₄ is used up in oxidising HCl, and the titration gives wrong results.
or (a) Eu²⁺ can lose an electron to form the more stable Eu³⁺. Hence, it is a strong reducing agent.
(b) In alkaline medium, dichromate changes into chromate : Cr₂O₇²⁻ (orange) + 2OH⁻ → 2CrO₄²⁻ (yellow) + H₂O
(c) E° (M²⁺/M) depends on enthalpy of atomisation, ionisation enthalpies and hydration enthalpy. Since these vary irregularly along the series, the E° values also show irregular variation.

Teacher's Note:
1. Stability in water is decided by hydration enthalpy as well as ionisation enthalpy; the small, highly charged Cu²⁺ ion is hydrated much more strongly.

Long Answer Questions (5 Marks)

Question 36. (a) What is meant by the term 'lanthanoid contraction' ? What is it due to and what consequences does it have on the chemistry of elements following lanthanoids in the periodic table ?
(b) Explain the following observations :
(i) Cu⁺ ion is unstable in aqueous solution
(ii) Although Co²⁺ ion appears to be stable, it is easily oxidised to Co³⁺ ion in the presence of a strong ligand
(iii) The E°_Mn²⁺/Mn value for manganese is much more than expected from the trend for other elements in the series (CBSE 2009)
Answer:
(a) The steady decrease in the atomic and ionic radii of the lanthanoids with increase in atomic number is called lanthanoid contraction. It is due to the imperfect shielding of one 4f electron by another, so that the effective nuclear charge increases with atomic number.
Consequences : (i) The elements of the third transition series (5d) following lanthanoids have almost the same atomic radii as the corresponding elements of the second series (4d), e.g., Zr (160 pm) and Hf (159 pm); hence they have very similar properties and occur together in nature. (ii) The 5d elements have higher ionisation enthalpies and densities than the 4d elements.
(b) (i) Cu⁺ (aq) disproportionates to Cu²⁺ (aq) and Cu (s) : 2Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s). This is because the hydration enthalpy of Cu²⁺ is much more negative than that of Cu⁺, which more than compensates for the second ionisation enthalpy of Cu.
(ii) Strong ligands cause a large crystal field splitting. Co³⁺ (3d⁶) in a strong field gets the very stable t_2g⁶ configuration, and the large crystal field stabilisation energy more than compensates for the third ionisation enthalpy. Hence, Co²⁺ is easily oxidised to Co³⁺ in the presence of strong ligands.
(iii) Mn²⁺ has the stable half-filled 3d⁵ configuration, so the second ionisation enthalpy is relatively low but the third is very high; moreover, Mn has a low enthalpy of atomisation. The overall effect makes E° (Mn²⁺/Mn) (−1.18 V) more negative than expected from the trend.

Teacher's Note:
1. Hydration enthalpy and crystal field stabilisation can outweigh ionisation enthalpy, so do not judge ionic stability from configurations alone.

Question 37. (a) Complete the following chemical equations :
(i) Cr₂O₇²⁻ (aq) + H₂S (g) + H⁺ (aq) →
(ii) Cu²⁺ (aq) + I⁻ (aq) →
(b) How would you account for the following ?
(i) The oxidizing power of oxoanions is in the order : VO₂⁺ < Cr₂O₇²⁻ < MnO₄⁻.
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
(iii) Cr²⁺ is stronger reducing agent than Fe²⁺. (CBSE 2010)
Answer:
(a) (i) Cr₂O₇²⁻ (aq) + 3H₂S (g) + 8H⁺ (aq) → 2Cr³⁺ (aq) + 7H₂O (l) + 3S (s).
(ii) 2Cu²⁺ (aq) + 4I⁻ (aq) → Cu₂I₂ (s) + I₂ (aq).
(b) (i) This is due to increasing stability of the lower species to which they are reduced.
(ii) Because 3rd electron has to be removed from stable half-filled 3d-orbitals (₂₅Mn = 3d⁵ 4s²).
(iii) E° (Cr³⁺/Cr²⁺) = −ve (− 0.41 V), E° (Fe³⁺/Fe²⁺) = +ve (+ 0.77 V).
Hence, Cr²⁺ is easily oxidized to Cr³⁺ but Fe²⁺ cannot be easily oxidized to Fe³⁺.

Teacher's Note:
1. A more negative reduction potential for M³⁺/M²⁺ means M²⁺ is a stronger reducing agent; Cr³⁺ (t_2g³) is especially stable.

Question 38. (a) Complete the following chemical equations :
(i) MnO₄⁻ (aq) + S₂O₃²⁻ (aq) + H₂O (l) →
(ii) Cr₂O₇²⁻ (aq) + Fe²⁺ (aq) + H⁺ (aq) →
(b) Explain the following observations :
(i) La³⁺ (Z = 57) and Lu³⁺ (Z = 71) do not show any colour in the solutions.
(ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
(iii) Cu⁺ ion is not known in aqueous solution. (CBSE 2010)
Answer:
(a) (i) 8MnO₄⁻ (aq) + 3S₂O₃²⁻ (aq) + H₂O → 8MnO₂ (s) + 6SO₄²⁻ (aq) + 2OH⁻ (aq)
(ii) Cr₂O₇²⁻ (aq) + 6Fe²⁺ (aq) + 14H⁺ (aq) → 2Cr³⁺ (aq) + 6Fe³⁺ (aq) + 7H₂O (l).
(b) (i) Because La³⁺ has an empty 4f subshell (4f⁰) and Lu³⁺ has a completely filled 4f subshell (4f¹⁴), so no f-f transition is possible.
(ii) Mn²⁺ has maximum number of unpaired electrons (3d⁵, five unpaired).
(iii) Because Cu⁺ in aqueous solution undergoes disproportionation :
2 Cu⁺ (aq) → Cu²⁺ (aq) + Cu (s)
(as E° value is favourable for this reaction)

Teacher's Note:
1. For colour of lanthanoid ions the rule is the same as for d-block ions: f⁰ and f¹⁴ ions are colourless.

Question 39. (a) How do you prepare :
(i) K₂MnO₄ from MnO₂
(ii) Na₂Cr₂O₇ from Na₂CrO₄ ?
(b) Account for the following :
(i) Mn²⁺ is more stable than Fe²⁺ towards oxidation to +3 state.
(ii) The enthalpy of atomisation is lowest for Zn in 3d series of transition elements.
(iii) Actinoid elements show wide range of oxidation states.
Answer:
(a) (i) By fusing MnO₂ with KOH in the presence of air : 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
(ii) By acidifying sodium chromate solution with H₂SO₄ : 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
(b) (i) Mn²⁺ = d⁵, Fe³⁺ = d⁵. Oxidation of Mn²⁺ to Mn³⁺ would destroy the stable half-filled configuration, whereas oxidation of Fe²⁺ (d⁶) to Fe³⁺ gives the stable d⁵ configuration.
(ii) In Zn (3d¹⁰ 4s²), all the electrons are paired and the 3d electrons do not take part in metallic bonding. Hence, the metallic bonding is weakest and the enthalpy of atomisation is lowest.
(iii) In actinoids, the 5f, 6d and 7s subshells have comparable energies, so electrons from all of them can take part in bonding. Hence, they show a wide range of oxidation states.

Teacher's Note:
1. Enthalpy of atomisation follows the number of unpaired electrons available for metallic bonding; Zn with none is the lowest.

Question 40. (a) When a chromite ore (A) is fused with an aqueous solution of sodium carbonate in free excess of air, a yellow solution of compound (B) is obtained. The solution is filtered and acidified with sulphuric acid to form compound (C). Compound (C) on treatment with solution of KCl gives orange crystals of compound (D). Write the chemical formulae of compounds A to D.
(b) Describe the cause of the following variations with respect to lanthanoids and actinoids.
(i) Greater range of oxidation states of actinoids as compared to lanthanoids
(ii) Greater actinoid contraction as compared to lanthanoid contraction
(iii) Lower ionization enthalpy of early actinoids as compared to the early lanthanoids. (CBSE Sample Paper 2019)
Answer:
(a) A = FeCr₂O₄, B = Na₂CrO₄, C = Na₂Cr₂O₇, D = K₂Cr₂O₇
(b) (i) 5f, 6d and 7s subshells of actinoids have comparable energies.
(ii) Due to poorer shielding by 5f electrons in actinoids as compared to shielding by 4f electrons in lanthanoids.
(iii) In actinoids, 5f electrons are more effectively shielded from nuclear charge than the 4f electrons of the corresponding lanthanoids. As the outer electrons are less firmly held, they are available for bonding in the actinoids.

Teacher's Note:
1. Learn the colour sequence of the steps: chromite → yellow chromate (B) → orange dichromate (C, D).