Please refer to The d – and f – Block Elements Class 12 Chemistry Important Questions with answers below. These solved questions for Chapter 8 The d – and f – Block Elements in NCERT Book for Class 12 Chemistry have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these solved problems properly as these will help them to get better marks in your class tests and examinations. You will also be able to understand how to write answers properly. Revise these questions and answers regularly. We have provided Notes for Class 12 Chemistry for all chapters in your textbooks.

Important Questions Class 12 Chemistry Chapter 8 The d – and f – Block Elements

All The d – and f – Block Elements Class 12 Chemistry Important Questions provided below have been prepared by expert teachers of Standard 12 Chemistry. Please learn them and let us know if you have any questions.

Very Short Answer Questions :

Question. Write complete chemical equations for 
(i) Oxidation of Fe²⁺ by Cr₂O^2–₇ in acidic medium.
(ii) Oxidation of S₂O₃^2– by MnO^–₄ in neutral aqueous medium. 
Answer : (i) Cr₂O^2–_7(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) → 2Cr³⁺_(aq)+ 6Fe³⁺_(aq) + 7H₂O_(l)
(ii) 8MnO^–_4(aq) + 3S₂O^2–_3(aq) + H₂O_(l) → 8MnO_2(aq) + 6 SO^2–_4(aq) + 2OH^–_(aq)

Question. What are different oxidation states exhibited by lanthanoids? 
Answer : Lanthanum and all the lanthanoids predominantly show +3 oxidation state. However, some of the lanthanoids also show +2 and +4 oxidation states in solution or in solid compounds. This irregularity arises mainly due to attainment of stable empty (4f⁰), half-filled (4f⁷) and fully filled (4f¹⁴) sub shell.
e.g. Ce⁴⁺ : 4f⁰ ,  Eu²⁺ : 4f⁷
      Tb⁴⁺ : 4f⁷ ,  Yb²⁺ : 4f¹⁴

Question. Compare the chemistry of the actinoids with that of lanthanoids with reference to
(i) electronic configuration
(ii) oxidation states
(iii) chemical reactivity.
Answer : (i) Electronic configuration : The general electronic configuration of lanthanoids is [Xe] 4f^1–14 5d^0–1 6s² where as that of actinoids is [Rn] 5f^1–14 6d^0–1 7s². Thus, lanthanoids involve the filling of 4f-orbitals whereas actinoids involve the filling of 5f-orbitals.
(ii) Oxidation states : Lanthanoids have principal oxidation state of +3. In addition, the lanthanoids show limited oxidation states such as +2, +3 and +4 because of large energy gap between 4f and 5d subshells. On the other hand, actinoids show a large number of oxidation states because of small energy gap between 5f and 6d subshells.
(iii) Chemical reactivity : (a) First few members of lanthanoids are quite reactive almost like calcium, where as actinoids are highly reactive metals especialy in the finely divided state.
(b) Lanthanoids react with dilute acids to liberate H₂ gas whereas actinoids react with boiling water to give a mixture of oxide and hydride.

Question. Complete the following chemical equations:
(i) Cr₂O^2–₇ + 6Fe2⁺ + 14H⁺ →
(ii) 2CrO₄^2– + 2H⁺ →
(iii) 3MnO₄^– + 5C₂O₄^2– + 16H⁺ → 
Answer : (i) Cr₂O^2–_7(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) → 2Cr³⁺_(aq)+ 6Fe³⁺_(aq) + 7H₂O_(l)
(ii) 2CrO₄^2– + 2H⁺ → Cr₂O^2–₇ + H₂O
(iii) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂

Question. Describe the reactions involved in the preparation of K₂Cr₂O₇ from chromite ore.
Answer :

Question. Assign reasons for the following :
From element to element actinoid contraction is greater than the lanthanoid contraction.
Answer : The actinoid contraction is more than lanthanoid contraction because of poor shielding by 5f-electrons.

Question. With reference to structural variability and chemical reactivity, write the differences between lanthanoids and actinoids.
Answer : Structure : All the lanthanoids are silvery white soft metals. Hardness of Lanthanoids increases with increasing atomic number.
The actinoid metals are all silvery in appearance but display a variety of structures. The structural variability is due to irregularities in metallic radii which are greater than that of lanthanoids.
Chemical reactivity : Earlier members of lanthanoid series are quite reactive similar to calcium but with increasing atomic number they behave more like aluminium.
The actinoids are highly reactive in finely divided state.

Question. Lokesh is a social worker. A milkman in the village has been complaining that a factory in his nearby area dumps chemical waste in his field which has become a major cause of decreasing productivity. Lokesh visited that place and found after analyis that the major waste was potassium permanganate which is being absorbed by the soil. He advised the factory people that they should treat potassium permanganate solution before dumping it into the drain.
Comment in brief
(i) About the value/s displayed by Lokesh.
(ii) Write balanced chemical equations for the two reactions showing oxidizing nature of potassium permanganate.
Answer : (i) The values, associated with Lokesh are alertness, care, responsibility and scientific knowledge.

Question. Complete the following chemical equations :
(i) MnO^–_4(aq) + S₂O^2–_3(aq) + H₂O_(l) →
(ii) Cr₂O^2–_7(aq) + Fe²⁺_(aq) + H⁺_(aq) →
Answer : (i) 8MnO^–_4(aq) + 3S₂O^2–_3(aq) + H₂O_(l) → 8MnO_2(aq) + 6 SO^2–_4(aq) + 2OH^–_(aq)
(ii) Cr₂O^2–_7(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) → 2Cr³⁺_(aq)+ 6Fe³⁺_(aq) + 7H₂O_(l)

Question. Describe the preparation of potassium dichromate from chromite ore. What is the effect of change of pH on dichromate ion?
Answer : (i) The yellow solution of sodium chromate is acidified with sulphuric acid to give a orange solution of sodium dichromate Na₂Cr₂O₇ which is crystallised.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O
 SodiumChromate          Sodium dichromate 
The solution of sodium dichromate is treated with potassium chloride to obtain potassium dichromate.
₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
                       Potassium dichromate
(ii) When the pH of the solution of potassium dichromate is decreased, the colour of the solution changes from yellow to orange due to the conversion of CrO^2–₄ ions into Cr₂O^2–₇ ions.
2CrO^2–₄ + 2H⁺ ⇌ Cr₂O^2–₇ + H₂O
  Yellow                 Orange

Question. What are the different oxidation states exhibited by the lanthanoids? 
Answer : Lanthanum and all the lanthanoids predominantly show +3 oxidation state. However, some of the lanthanoids also show +2 and +4 oxidation states in solution or in solid compounds. This irregularity arises mainly due to attainment of stable empty (4f⁰), half-filled (4f⁷) and fully filled (4f¹⁴) sub shell.
e.g. Ce⁴⁺ : 4f⁰ ,  Eu²⁺ : 4f⁷
      Tb⁴⁺ : 4f⁷ ,  Yb²⁺ : 4f¹⁴

Question. Why is europium (II) more stable than cerium (II)?
Answer : Europium (II) has electronic configuration [Xe]4f⁷5d⁰ while cerium (II) has electronic configuration [Xe] 4f ¹ 5d¹. In Eu²⁺, 4f subshell is half filled and 5d-subshell is empty. Since half filled and completely filled electronic configurations are more stable, Eu²⁺ ions is more stable than Ce²⁺ in which neither 4f subshell nor 5d subshell is half filled or completely filled.

Question. What is lanthanoid contraction? What is its eeffect on the chemistry of the elements which follow the lanthanoids? 
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. How would you account for the following :
Zr (Z = 40) and Hf (Z = 72) have almost identical radii. 
Answer : Due to lanthanoid contraction the elements of 4d and 5d-series have similar atomic radii e.g.,
Zr = 145 pm and Hf = 144 pm.

Question. Why do actinoids show a wide range of oxidation states? Write one similarity between the chemistry of lanthanoids and actinoids.
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.
Only trends of magnetic properties and colour of lanthanoids are similar to actinoids.

Question. Describe the preparation of potassium permanganate from pyrolusite ore. Write balanced chemical equation for one reaction to show the oxidizing nature of potassium permanganate.
Answer : Preparation of potassium permanganate : Potassium permanganate is prepared by the fusion of MnO₂ (pyrolusite) with potassium hydroxide and an oxidising agent like KNO₃ to form potassium manganate which disproportionates in a neutral or acidic solution to form permanganate.
2MnO₄^– + 5C₂O₄^2– + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O

Question. How would you account for the following :
Among lanthanoids, Ln(III) compounds are predominant. However, occasionally in solutions or in solid compounds , +2 and +4 ions are also obtained.
Answer : Lanthanum and all the lanthanoids predominantly show +3 oxidation state. However, some of the lanthanoids also show +2 and +4 oxidation states in solution or in solid compounds. This irregularity arises mainly due to attainment of stable empty (4f⁰), half-filled (4f⁷) and fully filled (4f¹⁴) sub shell.
e.g. Ce⁴⁺ : 4f⁰ ,  Eu²⁺ : 4f⁷
      Tb⁴⁺ : 4f⁷ ,  Yb²⁺ : 4f¹⁴

Question. Write the electronic configuration of Ce³⁺ ion, and calculate the magnetic moment on the basis of ‘spin-only’ formula. [Atomic no. of Ce = 58]
Answer :

Question. Describe the preparation of potassium permanganate from pyrolusite ore. What happens when acidified potassium permanganate solution reacts with ferrous sulphate solution? Write balanced chemical equations. 
Answer : Preparation of potassium permanganate : Potassium permanganate is prepared by the fusion of MnO₂ (pyrolusite) with potassium hydroxide and an oxidising agent like KNO₃ to form potassium manganate which disproportionates in a neutral or acidic solution to form permanganate.
5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ 4H₂O + 5Fe³⁺

Question. How does the acidified potassium permanganate solution react with (i) iron (II) ions and (ii) oxalic acid?
Write the ionic equations for the reactions.
Answer : (i) MnO^–₄ + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ +4H₂O
(ii) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂

Short Answer Questions :

Question. 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? 
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. Give reasons :
Actinoids show irregularities in their electronic configurations. 
Answer : The irregularities in the electronic configurations of actinoids are due to extra stabilities of the f⁰ f⁷ and f¹⁴ orbitals.

Question. Name an important alloy which contains some of the lanthanoid metals. Mention its two uses.
Answer : Mischmetal is well known alloy which consists of a lanthanoid metal (about 95%), iron (about 5%) and traces of S, C, Ca, Al etc.
Mischmetal is used in Mg based alloy to produce bullets shells and lighter flint.

Question. (i) Describe how potassium dichromate is prepared from sodium chromate.
(ii) The colour of potassium dichromate solution changes with the change of pH of the solution. Explain how.
Answer : (i) The yellow solution of sodium chromate is acidified with sulphuric acid to give a orange solution of sodium dichromate Na₂Cr₂O₇ which is crystallised.
2Na₂CrO₄ + H₂SO₄ → Na₂Cr₂O₇ + Na₂SO₄ + H₂O
 SodiumChromate          Sodium dichromate 
The solution of sodium dichromate is treated with potassium chloride to obtain potassium dichromate.
₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
                       Potassium dichromate
(ii) When the pH of the solution of potassium dichromate is decreased, the colour of the solution changes from yellow to orange due to the conversion of CrO^2–₄ ions into Cr₂O^2–₇ ions.
2CrO^2–₄ + 2H⁺ ⇌ Cr₂O^2–₇ + H₂O
  Yellow                 Orange

Question. Explain the following observations :
La³⁺ (Z = 57) and Lu³⁺ (Z = 71) do not show any colour in solutions. 
Answer : Because they have empty 4f subshell.

Question. How would you account for the following :
There is a greater range of oxidation states among the actinoids than among the lanthanoids.
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. Account for the following :
Zr and Hf have almost similar atomic radii.
Answer : Due to lanthanoid contraction the elements of 4d and 5d-series have similar atomic radii e.g.,
Zr = 145 pm and Hf = 144 pm.

Question. How would you account for the following :
Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. Complete the following chemical reaction equations :
(i) MnO^–_4(aq) + C₂O^2–_4(aq)  + H⁺_(aq) → 
(ii) Cr₂O^2–_7(aq) + Fe2+(aq) + H⁺_(aq) → 
Answer : (i) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂
(ii) Cr₂O^2–_7(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) → 2Cr³⁺_(aq)+ 6Fe³⁺_(aq) + 7H₂O_(l)

Question. Complete the following chemical equations :
(i) MnO₄^– + NO₂^– + H⁺ 
Answer : 2MnO₄^– + 6H⁺ + 5NO₂^– → 2Mn²⁺ + 5NO₃^– +3H₂O

Question. Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state.
Answer : Europium (Eu) is well known to exhibit +2 oxidation state due to its half-filled f orbital in +2 oxidation state.

Question. What is lanthanoid contraction? List any two consequences of lanthanoid contraction.
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. Complete the following equations :
(i) 2CrO₄^2– + 2H⁺ →
(ii) KMnO₄ →^Heat 
Answer : (i) 2CrO₄^2– + 2H⁺ → Cr₂O^2–₇ + H₂O
(ii) 2KMnO⁴ → K₂MnO⁴ + MnO₂ + O₂

Question. What is lanthanoid contraction and what is it due to? Write two consequences of lanthanoid contraction. 
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. Give reason for the following :
Among the Lanthanoids, Ce (III) is easily oxidised to Ce (IV).
Answer : Ce(III) has outer configuration 4f¹5d⁰6s⁰. It easily loses an electron to acquire the configuration 4f⁰ and forms Ce(IV). In fact this is the only (+IV) lanthanoid which exists in solution.

Question. (i) Write ionic equation to represent the reaction of acidified KMnO₄ solution with oxalic acid. 
Answer : (i) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂

Question. Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
Answer : Lanthanoids showing +4 oxidation state are ₅₈Ce, ₅₉Pr, ₆₅Tb and ₆₆Dy.

Question. Complete the following chemical reaction equations:
(i) Fe²⁺_(aq) + MnO^–_4(aq) + H⁺_(aq) →
(ii) Cr₂O^2–_7(aq) + I^–_(aq) + H⁺_(aq) →
Answer : (i) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺
(ii) The transition metals and their compounds, are known for their catalytic activity. This activity is ascribed to their ability to adopt multiple oxidation states, ability to adsorb the reactant(s) and ability to form complexes. Vanadium (V) oxide (in Contact Process), finely divided iron (in Haber’s Process), and nickel (in catalytic hydrogenation) are some of the examples. Catalysis involves the formation of bonds between reactant molecules and atoms at the surface of the catalyst.

Question. Give reasons for the following :
Actinoids exhibit a greater range of oxidation states than lanthanoids.
Answer : Actinoids exhibit greater range of oxidation states than lanthanoids. This is because there is less energy difference between 5f and 6d orbitals in actinoids than the energy difference between 4f and 5d orbitals in case of lanthanoids.

Question. Complete the following chemical equations :
(i) MnO₄^– + C₂O₄^2– + H⁺ → 
(ii) KMnO₄ →^Heated
(iii) Cr₂O^2–₇ + H₂S + H⁺ → 
Answer : (i) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂
(ii) 2KMnO⁴ → K₂MnO⁴ + MnO₂ + O₂
(iii) Cr₂O^2–_7(aq) + 3H₂S_(g) + 8H⁺_(aq) → 2Cr³⁺_(aq)+ 7H₂O_(l) + 3S_(s)

Question. What is lanthanoid contraction? Mention its main consequences. 
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf. 

Question. Write the steps involved in the preparation of
(i) K₂Cr₂O₇ from Na₂CrO₄
(ii) KMnO₄ from K₂MnO₄. 
Answer : 

(ii) The potassium manganate is oxidised to potassium permanganate by oxidation with chlorine.
2K₂MnO_4(aq) + Cl_2(g) → 2KMnO_4(aq) + 2KCl_(aq)

Question. Explain the following observation :
Actinoids exhibit a much larger number of oxidation states than the lanthanoids.
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. What is Lanthanoid contraction? What are its two consequences?
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. Complete the following chemical reaction equations :
(i) MnO^–_4(aq)  + S₂O^2–_3(aq) + H₂O_(l) →
(ii) Cr₂O^2–_7(aq) + H₂S + H⁺_(aq) →
Answer : (i) 8MnO^–_4(aq) + 3S₂O^2–_3(aq) + H₂O_(l) → 8MnO_2(aq) + 6 SO^2–_4(aq) + 2OH^–_(aq)
(ii) Cr₂O^2–_7(aq) + 3H₂S_(g) + 8H⁺_(aq) → 2Cr³⁺_(aq)+ 7H₂O_(l) + 3S_(s)

Question. Explain giving reasons :
The chemistry of actinoids is not as smooth as that of lanthanoids.
Answer : The chemistry of actinoids is not as smooth as lanthanoid because they show greater number of oxidation states due to comparable energies of 5f, 6d and 7s orbitals.

Question. What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
Answer : Lanthanoid contraction : The steady decrease in the atomic and ionic radii of lanthanoid elements with increase in atomic number is called lanthanoid contraction. It is caused due to imperfect shielding of nuclear charge by 4f-electrons.
Consequences of lanthanoid contraction :
(i) The basic strength of oxides and hydroxides of lanthanoids decrease with increasing atomic number.
(ii) Atomic and ionic sizes of 4d transition series elements and 5d series elements are similar. e.g., atomic radii of zirconium(Zr) is same as that of hafnium Hf.

Question. How would you account for the following :
The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series.
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. Give reason for the following :
The second and third transition series elements have almost similar atomic radii.
Answer : Due to lanthanoid contraction the elements of 4d and 5d-series have similar atomic radii e.g.,
Zr = 145 pm and Hf = 144 pm.

Question. Give reason :
There is a gradual decrease in the size of atoms with increasing atomic number in the series of lanthanoids. 
Answer : As the atomic number increases, each succeeding element contains one more electron in the 4f orbital and one extra proton in the nucleus. The 4f electrons are rather ineffective in screening the outer electrons from the nucleus. As a result, there is gradual increase in the nuclear attraction for the outer electrons. Consequently, the atomic size gradually decreases. This is called lanthanoid contraction.

Question. How would you account for the following:
Actinoid contraction is greater than lanthanoid contraction?
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. Complete the following reactions in an aqueous medium :
(i) MnO₄^– + C₂O₄^2– + H⁺ →
(ii) Cr₂O₇^2– + H₂S + H⁺ → 
Answer : (i) 2MnO^–₄ + 5C₂O^2–₄ + 16H⁺ → 2Mn²⁺ + 8H₂O+ 10CO₂
(ii) Cr₂O^2–_7(aq) + 3H₂S_(g) + 8H⁺_(aq) → 2Cr³⁺_(aq)+ 7H₂O_(l) + 3S_(s)

Question. What is meant by ‘lanthanoid contraction’?
Answer : The steady decrease in the atomic and ionic radii (having the same charge) with increase in atomic number across the series from lanthanum to lutetium is known as lanthanoid contraction.

Question. Describe the oxidising actions of potassium dichromate and write the ionic equations for its reaction with (i) iodide (ii) iron (II) solution (iii) H₂S. 
Answer : (i) Cr₂O^2–₇ + 14H⁺ + 6I^– → 2Cr³⁺ + 3I₂ + 7H₂O
(ii) Cr₂O^2–_7(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) → 2Cr³⁺_(aq)+ 6Fe³⁺_(aq) + 7H₂O_(l)
(iii) Cr₂O^2–_7(aq) + 3H₂S_(g) + 8H⁺_(aq) → 2Cr³⁺_(aq)+ 7H₂O_(l) + 3S_(s)

Question. Explain the following observation :
The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. 
Answer : The actinoid contraction is more than lanthanoid contraction because 5f-electrons are more poorly shielding than 4f-electrons.

Question. Complete the following chemical equations :
(i) Fe³⁺ + I^– →
(ii) CrO₄^2–  + H⁺ → 
Answer : (i) 2Fe³⁺ + 2I^– → 2Fe²⁺ + I²
(ii) 2CrO₄^2– + 2H⁺ → Cr₂O^2–₇ + H₂O

Question. Describe the preparation of potassium permanganate from pyrolusite ore. Write the ionic equation for the reaction that takes place between acidified KMnO₄ solution and iron (II) ions.
Answer : Preparation of potassium permanganate : Potassium permanganate is prepared by the fusion of MnO₂ (pyrolusite) with potassium hydroxide and an oxidising agent like KNO₃ to form potassium manganate which disproportionates in a neutral or acidic solution to form permanganate.
5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ 4H₂O + 5Fe³⁺