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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Previous HSE questions and Answers for the chapter ‘d and f block elements’, 1. Transition elements show variable oxidation states and many of the transition metal ions are attracted, by a magnetic field., a) Give reason for variability of oxidation state. (1), b) Name the two types of magnetic behaviour., (1), Ans: a) This is because in transition elements d and s electrons have comparable energies. So along with, s-electrons, d-electrons also participate in chemical reactions., b) Diamagnetism and paramagnetism, 2. The observed magnetic moment of Sc3+ was found to be zero. Calculate the magnetic moment of Sc3+, using the spin-only formula and compare the result of observed and calculated magnetic moment. (2), [March 2008], 3+, 0, Ans: The valence shell electronic configuration of Sc ion is 3d ., So the number of unpaired electrons (n) = 0, Magnetic mpment, µs = √n(n+2) = √0(0+2) = 0., i.e. the observed magnetic moment is in agreement with the calculated magnetic moment., 3. A list of Lanthanide ions are given:, La3+, Ce4+, Yb2+, Lu3+, Atomic numbers of La, Ce, Yb and Lu are 57, 58, 70 and 71 respectively., a), Give the number of unpaired electrons in each ion. (1), b), Identify the ions which are paramagnetic. Justify. (1½ ), c), Identify the ions which are colourless. Give reason. (1½) [SAY 2008], Ans:, a) The valence shell electronic configuration of La3+ ion is 4f 0. So the number of unpaired electrons = 0., The valence shell electronic configuration of Ce4+ ion is 4f 0. So the number of unpaired electrons = 0., The valence shell electronic configuration of Yb2+ ion is 4f 14. So the number of unpaired electrons = 0., The valence shell electronic configuration of Lu3+ ion is 4f 14. So the number of unpaired electrons = 0., b) All these ions are diamagnetic, since they contain only paired electrons., c) All these ions are colourless, since they do not contain any partially filled orbitals., 4. Potassium permanganate is a powerful oxidising agent in neutral, acidic and alkaline medium. In the lab,, students were asked to convert an iodide to iodate. One of the students obtained I 2 instead of iodate., a) i) What is the reaction to be carried out by the students who got iodate? Write the chemical, equation. (1½ ), ii) What may be the reaction carried out by the student who got I2 as one of the products? (1), b) i) Suppose you are going the same experiment with the iodide using Potassium dichromate (K 2Cr2O7)., What are the products going to be obtained? Write down the chemical equation., (1½ ), ii) What is Baeyer’s reagent? (1), [March 2009], Ans: a) i) In order to get iodate, the reaction should be done in alkaline or neutral medium. The, chemical equation for this reaction is: 2MnO4– + H2O + I –→ 2MnO2 + 2OH –+ IO3–, ii) The student who got I2 as one of the product conducted the reaction in acidic medium. The chemical, equation for this reaction is: 10 I – +2 MnO4– + 16H+ → 5 I2 + 2 Mn2+ + 8H2O, b) i) The products obtained are Iodine, Cr3+ and water as follows:, 6I – + Cr2O72– + 14H+ → 3I2 + 2Cr3+ + 7H2O, ii) Alkaline KMnO4 solution is called Baeyer’s reagent., 5. Potassium permanganate and Potassium dichromate are oxidising agents., a) Name the ores of the above compounds from which they are prepared., (½ ), b) Give one example each for the oxidising property of them. Write down the balanced chemical, equation. (2), [SAY 2009], d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 1

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , Ans: a) Potassium permanganate – Pyrolusite (MnO2) and Potassium dichromate – Chromite ore (FeCr2O4), b) Both KMnO4 and K2Cr2O7 oxidise iodides in acidic medium and liberate iodine vapours., , 10 I – +2 MnO4– + 16H+ → 5 I2 + 2 Mn2+ + 8H2O, 6I – + Cr2O72– + 14H+ → 3I2 + 2Cr3+ + 7H2O, 6. Potassium permanganate is a violet crystal. What are the products obtained on strong heating of KMnO 4, crystals? Write the balanced chemical equation., (2), [March 2010], Ans: The products obtained are Potassium manganate (K2MnO4), Oxygen and manganese dioxide (MnO2)., , 2KMnO4 → K2MnO4 + MnO2 + O2, 7. a) Transition elements are d-block elements. Write any 4 characteristic properties of transition, elements?, (2), b) Lanthanoids and actinoids are f-block elements., i), What is the common oxidation state of Lanthanoids?, (½), ii), Name the Lanthanoid with common oxidation state +4., (½), iii), It is difficult to separate Lanthanoids in the pure state. Explain., (1), [SAY 2010], Ans: a) Transition elements are all metals, they show variable oxidation states, they form coloured, compounds and most of them are paramagnetic., b) (i) +3, (ii) Cerium (Ce), (iii) Due to Lanthanoid contraction, lanthanoids have similar radii and hence similar physical, properties. So their isolation is difficult., 8. a) Atomic size increases as we come down a group, but in 4th group of the periodic table, Zr and Hf have, almost similar atomic size. Why?, (1½), 0, b) E (std. electrode potential) values generally become less negative as we move across a transition, series, but E0 values of Ni/Ni2+ and Zn/Zn2+ values are exceptions. Justify. (2½) [March 2011], Ans: a) This is due to Lanthanoid contraction., b) This is due to the highest negative hydration enthalpy of Ni2+ ion and completely filled d10, configuration of Zn2+ ion., 9. Transition elements are d-block elements, with some exceptions. Usually they are paramagnetic. They, show variable oxidation states. They and their compounds show catalytic activity., a) Zn (atomic number = 30) is not a transition element, though it is a d block element. Why? (1), b) Which is more paramagnetic, Fe2+ or Fe3+? Why?, (1), c) Why do transition elements show variable oxidation states? (1), d) What is the reason for their catalytic property? (1), [SAY 2011], Ans: a) This is due to the absence of partially filled d orbitals in the ground state or in any of the oxidation, states of Zn., b) In Fe3+, there are 5 unpaired electrons. So it is more paramagnetic., c) Due to the participation of penultimate d electrons along with valence s electrons in chemical, reactions., d) Large surface area and ability to show variable oxidation states are the reason for catalytic property., 10. a) Potassium dichromate (K2Cr2O7) is an important compound of chromium. Describe the method of, preparation of potassium chromate from chromite ore., (3), b) The gradual decrease in the size of lanthanoid elements from lanthanum to lutetium is known as, lanthanoid contraction. Write any one consequence of lanthanoid contraction. (1), [March 2012], Ans: a) Potassium dichromate is generally prepared from chromite ore (FeCr2O4). The preparation, involves three steps., 1. Conversion of chromite ore to sodium chromate, Chromite ore is first fused with sodium carbonate in presence of air to form sodium chromate., 4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2, d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 2

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 2., , Acidification of sodium chromate to sodium dichromate, The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to orange, sodium dichromate., 2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O, 3. Conversion of sodium dichromate to potassium dichromate, The solution of sodium dichromate is treated with potassium chloride so that orange, crystals of potassium dichromate crystallise out., Na2Cr2O7 + 2 KCl → K2Cr2O7 + 2 NaCl, b) Due to Lanthanide Contraction the 2nd and 3rd row transition series elements have similar radii., 11. Assume that you are going to present a seminar on transition elements. Prepare a seminar paper by, stressing any four important properties of transition elements., (4), [SAY 2012], Ans: Some important properties of transition metals are:, 1. Variable oxidation states: Transition metals show variable oxidation states. This is because in these, , elements d and s electrons have comparable energies. So in chemical reaction along with s-electrons,, d-electrons also participate., 2. Magnetic properties: Transition metals show mainly two types of magnetic propertiesparamagnetism and diamagnetism. Some transition metals also show ferromagnetism., 3. Formation of coloured compounds or ions: Most of the Transition metals ions or compounds are, coloured in aqueous solution. This is because of the presence of partially filled d orbitals., 4. Catalytic properties: Transition metals act as catalysts in a large no. of chemical reactions. This is due, to their large surface area and their ability to show variable oxidation state., 12. Account for the following trends in atomic and ionic radii of transition elements., a) Ions of the same charge in a given series (3d, 4d or 5d) show progressive decrease in radii with, increasing atomic number., (1), b) The atomic radii of elements in 4d series are more than that of corresponding elements in 3d series., (1), c) The atomic radii of the corresponding elements in 4d series and 5d series are virtually the same., (2), [March 2013], Ans: a) This is due to the poor shielding effect and increase in nuclear charge., , b) This is due to increase in no. of shells and greater shielding effect., c) This is due to Lanthanoid contraction., 13. d block elements belong to groups 3 – 12 in the periodic table, in which the d orbitals are progressively, filled., a) What is their common oxidation state? (½ ), b) Name two important compounds of transition elements., (1), c) Transition elements form a large number of complex compounds. Why?, (1½ ), d) What is misch metal?, (1), [SAY 2013], Ans: a) The common oxidation state is +2., b) K2Cr2O7 and KMnO4, , c), , This is due to comparatively smaller size, high ionic charge, pesence of partially filled d orbitals and, ability to show variable oxidation state., d) It is an alloy of Lanthanoids which consists of a lanthanoid metal (~ 95%) and iron (~ 5%) and traces, of S, C, Ca and Al., 14. Potassium dichromate is an orange coloured crystal and is an important compound used as an oxidant in, many reactions., a) How do you prepare K2Cr2O7 from chromite ore? (3), b) How will you account for the colour of potassium dichromate crystals? (1), [March 2014], Ans: a) Refer the answer of Question no. 10 (a), d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 3

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , b) The colour of K2Cr2O7 is due to charge transfer spectrum., , 15. Potassium permanganate and potassium dichromate are two transition metal compounds., a) Write any four characteristics of transition metals., (2), b) Write any two uses of potassium permanganate., (1), c) Draw the structure of dichromate ion. (1), [SAY 2014], Ans: a) Refer the answer of Question no. 7 (a)., , b) It is used as an oxidising agent in acidic, basic and neutral medium. It is used as a primary standard in, volumetric analysis., c), , 16. Fourteen elements following Lanthanum are called Lanthanoids:, a) What is Lanthanoid contraction? Give reason for it?, (2), b) KMnO4 is a purple coloured crystal and it acts as an oxidant. How will you prepare KMnO 4 from, MnO2?, (2), [March 2015], Ans: a) The regular decrease in the atomic and ionic radii along lanthanide series is known as lanthanide, contraction. It is due to the poor shielding effect of f – electrons and increase in nuclear charge., , b) The preparation of Potassium permanganate from Pyrolusite (MnO2) involves two steps. In the, first step MnO2 is fused with KOH to form potassium manganate (K2MnO4)., 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O, In the second step, K2MnO4 is electrolytically oxidised to potassium permanganate., MnO42– ⎯⎯⎯⎯⎯⎯ Electrolytic oxidation in alkaline solution ⎯⎯⎯⎯→MnO4 –, 17. Which of the following oxidation state is common for lanthanides?, i) +2, ii) +3, iii) +4, iv) +5, (1), b) Draw the structures of chromate and dichromate ions., (1), c) Zirconium (Zr) belongs to ‘4d’ and Hafnium (Hf) belongs to ‘5d’ transition series. It is difficult to, separate them. Explain., (2), [SAY 2015], Ans: a) +3, , b), , c) Due to lanthanoid contraction, Zr and Hf have similar radii and hence similar physical properties. So, their separation is difficult., 18. a) Which of the following oxidation state is not shown by Manganese?, (i), +1, (ii) +2, (iii) +4, (iv) +7, (1), b) Represent the structure of dichromate ion., (1), c) Potassium permanganate (KMnO4) is a strong oxidizing agent. Write oxidizing reactions of KMnO4. (2), [March 2016], Ans: a) +1, d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 4

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , b) Refer the answer of Question no. 15 (c), c) In acidic medium, KMnO4 oxidises ferrous salt to ferric salt., 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O, 19. Transition elements are d-block elements and inner transition elements are f-block elements., (i), Write any two properties of transition elements., (1), (ii), Name a transition metal compound and write one use of it. (1), (iii), What is Lanthanoid contraction? (1), (iv), Write any two consequences of Lanthanoid contraction., (1), [SAY 2016], Ans: (i) Refer the answer of Question no. 7 (a), (ii) KMnO4, used as an oxidising agent, (iii) It is the regular decrease in the atomic and ionic radii along lanthanide series., (iv) Due to Lanthanide Contraction the 2nd and 3rd row transition series elements have similar radii., Lanthanides have similar physical properties and they occur together in nature. So their isolation is, difficult., 20. a) Transition elements are‘d’ block elements., i) Write any four characteristic properties of transition elements. (2), ii) Cr2+ and Mn3+ have d4 configuration. But Cr2+ is reducing and Mn3+ is oxidising. Why? (1), b) Which of the following is not a Lanthanoid element?, i) Cerium, ii) Europium iii) Lutetium iv) Thorium (1), [March 2017], Ans: a) (i) Refer the answer of Question no. 7 (a), (ii) This is due to the extra stability of Cr3+ and Mn2+ ions., b) Thorium, 21. a) Zr and Hf are having similar chemical properties. This is due to …………….., (1), b) ‘Magnetic moments arise due to the presence of unpaired electrons’., Calculated magnetic moments of two transition metal ions are given below., Ion, Calculated magnetic moment, 3+, Sc, 0, 3+, Ti, 1.73, Justify these observations on the basis of spin only formula., (2), c) Transition metal ions are generally coloured. Why?, (1), [SAY 2017], Ans: a) Lanthanoid contraction, b) The spin only magnetic moment µs = √n(n+2), where n is the no. of unpaired electrons., For Sc 3+ the electronic configuration is 3d 0. So there is no unpaired electron and hence µs = 0., , Ti 3+ the electronic configuration is 3d 1. So there is one unpaired electron and hence µs = √1(1+2) = 1.73 BM, , c) This is due to the presence of partially filled d-orbitals or due to d-d transition., 22. What is the structure of chromate ion ((CrO42-)?, (1), Ans: a) Tetrahedral, , 23. Give reasons for the following :, (a) Transition metals and many of their compounds act as catalyst., (1), (b) Scandium (Z = 21) does not exhibit variable oxidation state and yet it is regarded as a transition, element. (1), (c) Write the steps involved in the preparation of Na2CrO4 from chromite ore. (1), [March 2018], Ans: a) (a) Due to their large surface area and their ability to show variable oxidation state., (b) Due to the presence of partially filled d-orbitals in Scandium., (c) Chromite ore is fused with sodium carbonate in presence of air to form sodium chromate., Or, the equation: 4 FeCr2O4 + 8 Na2CO3 + 7 O2 → 8 Na2CrO4 + 2 Fe2O3 + 8 CO2, 24. What is the magnetic moment of an atom having d10 configuration?, (1), Ans: zero, d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 5

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 25. Describe lanthanoid contraction. Write any two consequences of it., (3), [SAY 2018], Ans: Refer the answers of Question no. 16 (a) and 19 (iv)., 26. MnO4- and ……………. are formed by the disproportionation of MnO42- in acidic medium. (1), Ans: MnO2 (Manganese dioxide), 27. Write any three applications of d- and f- block elements. (3), [March 2019], Ans: d and f block elements and their compounds are used as catalysts in many chemical reactions. Iron, and steels are the most important construction materials. Alloys of d and f block elements are used in, various fields. Cu, Ag, Au and some alloys are used for making coins. TiO is used in pigment industry., Zn, Ni, Cd, MnO2 etc are used in making batteries. Compounds of Ag are used in photography., [Only 3 required], 28. Which element of the 3d series exhibits the largest number of oxidation states? Why? (2), Ans: Manganese (Mn), 29. What is Lanthanoid contraction? Give reason for it. (2), [SAY 2019], Ans: Refer the answers of Question no. 16 (a), 30. (a) In d-block elements the radii of elements of third transition series are similar to those of the elements, of second transition series. Give reason., (1), 2+, (b) Outer electronic configuration of Cu ion is 3d9. Calculate its spin only magnetic moment. (1), Ans: (a) Due to Lanthanoid contraction., (b) For 3d 9 configuration, there is only one unpaired electron and hence µs = √1(1+2) = 1.73 BM, 31. Give the steps involved in the preparation potassium dichromate (K2Cr2O7) from chromite ore. (3), [March 2020], Ans: Ans: Refer the answers of Question no. 10 (a), 32. (a) What is the common oxidation state of Lanthanoids? (1), (b) Atomic sizes increases as we move down a group, but in 4 th group of the periodic table Zr and Hf have, almost the same atomic sizes. Why?, (1), Ans: (a) +3, (b) Due to Lanthanoid contraction., 33. Transition elements show various oxidation states and many of the transition metal ions are attracted by, a magnetic field., (a) Give reason for variability of oxidation state., (1), (b) Name the two types of magnetic behaviour., (1), 2+, (c) Calculate the ‘Spin only’ magnetic moment of M (aq) ion (Z = 27)., (1), [SAY 2020], Ans: (a) This is because in transition elements d and s electrons have comparable energies. So along with, s-electrons, d-electrons also participate in chemical reactions., (b) Diamagnetism and Paramagnetism., (c) For M2+ ion with atomic number 27, the electronic configuration is 3d 7. So there are 3 unpaired, electrons and hence µs = √3(3+2) = 3.87 BM, 34. Potassium dichromate is a very useful oxidizing agent., (i), Name the ore of Potassium dichromate. (1), (ii), Explain the preparation of Potassium dichromate from Sodium chromate., , (2), , Ans: (i) Chromite ore (Fe2CrO4), , (ii) First sodium chromate is acidified with sulphuric acid to produce sodium dichromate., d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 6

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Join Telegram Channel: https://t.me/hsslive, , Downloaded from www.Hsslive.in ®, , 2Na2CrO4 + 2 H+ → Na2Cr2O7 + 2 Na+ + H2O, Then the solution of sodium dichromate is treated with potassium chloride so that orange crystals of, potassium dichromate crystallise out., Na2Cr2O7 + 2 KCl → K2Cr2O7 + 2 NaCl, 35. (i) Account for the following :, A. Zr and Hf have identical radii., (1), B. Transition metals are very good catalysts., (1), (ii) Calculate the spin only magnetic moment of M2+(aq) ion (Z = 27). (1), [March 2021], Ans: (i) A. Due to lanthanoid contraction/lanthanide contraction., B. This is due to their large surface area and their ability to show variable oxidation state., (ii) For M2+ ion with atomic number 27, the electronic configuration is 3d 7. So there are 3 unpaired, electrons and hence µs = √3(3+2) = 3.87 BM, *******************************************************************************, , d and f Block Elements – Prepared by Anil Kumar K L, GHSS Ashtamudi, Kollam, , Page 7