The d- and f- Block Elements Question Bank

The d- and f- Block Elements I. Answer the following questions. Each question carries one mark 1. Define transition elements. Ans. Transition element is defined as the one which has incompletely filled d orbitals in its ground state or in any one of its oxidation states. 2. What is the position of the d block elements in the periodic table? Ans. The d block elements are in the middle of s and p blocks, comprising the groups 3 to 12. They are the four rows of elements in the periods 4th (3d series), 5th (4d series), 6th ( 5d series) and 7th ( 6d series). 3. Zinc, cadmium and mercury of group 12 are not regarded as transition metals, Why ? Ans. Zinc, cadmium and mercury of group 12 have full d10 configuration ( d orbitals are completely filled ) in their ground state as well as in their common oxidation states and hence, are not regarded as transition metals 4. Why d- block elements are named as 8transition elements ‘ ? Ans. The d3block elements occupies the middle of the periodic table and their properties are transitional between s3 and p3 block elements. 5.Write the general electronic configuration of d block elements. Ans. [ Noble gas] (n-1)d1-10ns1-2 6. Write the general outer electronic configuration of d- block elements. . Ans. The general outer electronic configuration of d- block elements is (n-1)d1310 ns132 7. Write the general electronic configuration of f- block elements. Ans. The general electronic configuration of f- block elements (Lanthanoids) is [Xe] 4f1 3 14 5d 0- 1 6s2 8.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state. Ans. Cerium 9. The outer electronic configuration of Cr is 3d5 4s1 instead of 3d44s2, why? Ans. Half filled (3d5) orbitals are relatively more stable, hence one electron of 4s orbital jump to 3d orbital. 10. The outer electronic configuration of Cu is 3d10 4s1 instead of 3d94s2 , why? Ans. Completely filled (3d10 ) orbitals are relatively more stable, hence one electron of 4s orbital jumps to 3d orbital. 11. Account for high melting point and boiling points of transition metals. Ans. The melting and boiling points of transition metals are high because of the involvement of greater number of electrons from (n-1)d orbitals in addition to the ns electrons in the inter atomic metallic bonding. 12.What is the trend in melting points of transition metals in a series? Ans.The melting points of the transition metals in a series rise to a maximum at the middle of the series (i.e. Cr or Mo or W - element with d 5 configuration ) and fall regularly as the atomi number increases. 13.Why do transition metals have higher enthalpies of atomization? Ans. Involvement of a large number of unpaired electrons of d orbitals favour stronger inter atomic interactions resulting in stronger bonds between the atoms of a metal and higher enthalpies of atomization. 14.Name one 3d series elements, that do not show variable oxidation states. Ans. Sc (+3) 15.Transition metals exhibit variable oxidation states in its compounds, why? Ans. Transition metals exhibit variable oxidation states in its compounds due to the availability of both ns & (n 3 1 ) d electrons for bond formation. 16. Name 3d series metal which shows highest oxidation state. Ans. The highest oxidation state shown by 3d series transiNa tion metals is +7 by Mn 17. Name a metal in the 3d series of transition metals which exhibit +1 oxidation state most frequently. Ans. copper 18.What is the trend in oxidation state of transition metals ? Ans. The oxidation state increases with increase in atomic number & reaches a maximum in the middle and then decreases. 19. 3d series transition metals exhibit +2 as the most common oxidation state (except Sc) why? Ans. The +2 oxidation state, which commonly occurs for nearly all the transition metals is due to the loss of their outer 4s electrons 20. Why transition metals and their compounds shows paramagnetic behavior ? Ans. The transition metal ions are generally containing one or more unpaired electrons in them & hence their compounds are generally paramagnetic. 21. Name an of alloys of transition metals with non transition metals. Ans. Brass ( Cu & Zn) or Bronze ( Cu & Sn) 22.What is the action of neutral or faintly alkaline permanganate solution on iodide ? Ans. Alkaline permanganate solution oxidize iodide to iodate. 23. What happens when potassium permanganate is heated to 513 K ? Ans. Potassium permanganate decomposes at 513K to potassium manganate, manganese dioxide and oxygen. 24. What is the principal oxidation state exhibited by the lanthanoids? Ans. The principal oxidation state of lanthanoids is +3. 25.Write the spin-only formula used to calculate the magnetic moment of metal ions. Ans. The magnetic moment is determined by using the spin only formula, where n is the number of unpaired electrons and μ is the magnetic moment in units of Bohr magneton (BM). 26. Why is Sc3+ (or Zn2+ ) diamagnetic? Ans. Sc3+(Z=21) 3d0 (or Zn2+(Z=30) 3d10 no unpaired electron, n=0, μ=0. no unpaired electron, n=0, μ=0) 27. What is the most common oxidation state of lanthanoids and actionoids? Ans. The most common oxidation state of lanthanoids and actionoids is +3. 28.What is Actinoid contraction? Ans. There is a gradual decrease in the size of atoms or M3+ ions across the series. This is known as the actinoid contraction. 29.Actinoid contraction is more than lanthanoid contraction. Give reason. Ans. The actinoid contraction is, more than lanthanoid contraction due to poor shielding by 5f electrons from nuclear charge. 30. Actionoids show larger number of oxidation states than lanthanoids. Why? Ans. In actinoids 5f, 6d and 7s levels are of comparable energies ,hence electrons from these orbitals are available to lose or share. 31. Give one use of Mischmetall . Ans. Mischmetall is used in Mg-based alloy to produce bullets, shell and lighter flint. 32. Why transition metals forms alloys readily? Ans. Transition metals readily form alloys with other transition metals because of their similar radii 33. Give one use of transition metal alloy. Ans.Ferrous alloys containing chromium, vanadium, tungsten, molybdenum and manganese are used for the production of a variety of steels. II. Answer the following questions. Each question carries TWO marks. 34. Name two characteristic properties exhibited by d – block elements due to their partly filled d orbitals. Ans. The characteristic properties exhibited by d 3 block elements due to their partly filled d orbitals are variable; (i) Oxidation states (ii) Formation of coloured ions. 35. Name two typical metallic properties displayed by transition elements. Ans.High tensile strength, ductility malleability, high thermal and electrical conductivity an metallic luster etc. 36. What are interstitial compounds? Give example. Ans. Interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside the crystal lattices of transition metals. Example; TiC, Mn4N, Fe3H, VH0.56 and TiH1.7, etc. 37. Give any two physical characteristics of interstitial compounds. Ans.Two physical characteristics of interstitial compounds are: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard and they retain metallic conductivity. 38.Calculate the 8spin only9 magnetic moment of M2+ (aq) ion (Z = 27). Ans. M (z= 27 , 3d7 4s2) M+2 (3d7 4s0) hence it has 3 unpaired electrons n= 3 = 3(3+2) = 3.87 BM 39. The second ionisation enthalpy is high for Cr and Cu , why? Ans: The second ionisation enthalpy is unusually high values for Cr and Cu because when M+ ion ionize to M+2 ion , the d5 and d10 configurations of the M+ ions (i.e Cr+ or Cu+) ar disrupted, with considerable loss of exchange energy . 40. Why first ionisation enthalpy of Cr is lower than that of Zn ? IE1 of Cr is lower, because removal of an electron from Cr does not change the d (3d5 4s1 to 3d5 4s0 ) configuration . Cr (z= 24 , 3d5 4s1) Cr+ (3d5 4s0) ------ IE1 IE1 value for Zn is higher, because removal of electron from 4s level needs more energy. Zn (z= 30 , 3d10 4s2) Zn+ (3d10 4s1) ------ IE1  IE1 (Zn) > IE1 (Cr) 41.Give two characteristics of transition metal alloys. . The alloys are hard and have high melting points. 42.What is the action of heat on potassium permanganate ? Give equation. Ans. It decomposes at 513K to potassium manganate, manganese dioxide and oxygen. 2KMnO4  K2MnO4 + MnO2 + O2 43. What is the action of neutral or faintly alkaline permanganate solution on iodide ?Give equation. Ans. Alkaline permanganate solution oxidize iodide to iodate. I- + 2MnO4- + H2O  IO3- + 2MnO2 + 2OH44. Explain the oxidising action of acidified potassium dichromate on (iron(II) salts) Fe +2 ions and write the ionic equations for the reaction. Ans. Acidified potassium dichromate oxidises iron(II) salts to iron(III). Cr2O72− + 14H+ + 6Fe2+ → 2Cr3+ + 6Fe3+ + 7H2O 45. The transition metals generally form coloured compounds, why? Ans. The compounds of transition elements shows colour due to presence of unpaired electron & ability to undergo d-d transition. When an electron from a lower energy d orbital is excited to a higher energy d orbital, the energy of excitation corresponds to the frequency of light absorbed. This frequency generally lies in the visible region. 46. Give reason < transition metals and their many compounds acts as good catalysts=. Ans. Transition metals and their many compounds acts as good catalysts,it is due to (i) partially filled (n-1) d orbital (ii)variable oxidation state and provide a suitable surface for th reaction to take place. 47.Explain giving reason 2MnO2 + 2OH− + IO3− (b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO43 + 3S2O323 + H2O 44> 8MnO2 + 6SO423 + 2OH3 III. Answer the following questions. Each question carries THREE marks 61.Name the metal of the 1st row transition series that i) has highest value for magnetic moment ii) has zero spin only magnetic moment in its +2 oxidation state. iii) exhibit maximum number of oxidation states. Ans. i) Chromium ii) Zinc iii) Manganese 62.Transition metals form a large number of complex compounds.Give reason. Ans. Transition metals for complex compounds due to, i) small sizes of metal cations ii) their ionic charges and iii) availability of d orbitals for bond formation. 63.Explain the trend in atomic size of 3d series of transition elements with reason. Ans. With increase in atomic number in 3d series - atomic size decreases (Sc to Cr) , then remain almost constant (Cr to Cu) and increase slightly at the end (Cu to Zn). Reason: In the beginning of the series the screening (or shielding effect) effect of a d electron is not that effective, hence the net electrostatic attraction between the nuclear charge and the outermost electron increases , hence atomic size radius decreases. In the middle of the series, increase in nuclear charge and increase in screening effect balance each other. So atomic radii become almost constant. Increase in atomic radii towards the end is due to the electro 3 electron repulsions causes the expansion of electron cloud. 64. Explain trend in Ionisation Enthalpies of 3d series of transition elements . Ans.Ionisation enthalpy increase along each series of the transition elements from left to right. However many small variations, IE of Chromium is lower because removal of an electron from Chromium does not change in the d (3d5 4s1 to 3d5 4s0 ) configuration. I.E value for Zn (3d10 4s2) is higher because an electron is removed from 4s level which needs more energy. 65. How is potassium dichromate prepared from iron chromite ore? Ans.Potassium dichromate is manufactured from chromite ore (FeCr2O4). (i) Chromite ore is fused (FeCr2O4) with sodium or potassium carbonate in free access of air to get sodium chromate . 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 (ii) The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which orange sodium dichromate, Na2Cr2O7. 2H2O can be crystallised. 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O Potassium dichromate prepared by treating the solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl Orange crystals of potassium dichromate crystallise out. 66. Describe the preparation of potassium permanganate from manganous dioxide. 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 → 2MnO43 + MnO2 + 2H2O 3MnO423 + 4H+ 67.How does the acidified permanganate solution react with (a) iron(II) ions (b) oxalic acid and (c) hydrogen sulphide ? Write the ionic equations for the reactions In acid solutions: (a) Fe2+ ion (green) is converted to Fe3+ (yellow): 5Fe2+ + MnO43 + 8H+ 44> Mn2+ + 4H2O + 5Fe3+ (b) Oxalate ion or oxalic acid is oxidised at 333 K: 5C2O42− + 2MnO4- + 16H+ 44> 2Mn2+ + 8H2O + 10CO2 (c) Hydrogen sulphide is oxidised, sulphur being precipitated: H2S 4> 2H+ + S2− 5S2− + 2MnO4− + 16H+ 44> 2Mn2+ + 8H2O + 5S