Transition Elements Exercise # 5

TRANSITION ELEMENTS Exercise # 5 Part # I [Previous Year Questions 1. Cerium (Z = 58) is an important member of the lanthanoide. Which of the following statement about cerium is incorrect? (1) The common oxidation state of cerium are +3 and +4. (2) The +3 oxidation state of cerium is more stable than +4 oxidation state. (3) The +4 oxidation state of cerium is not known in solution. (4) Cerium (IV) acts as an oxidizing agent. 2. The lanthanide contraction is responsible for the fact that (1) Zr and Y have about the same radius (2) Zr and Nb have similar oxidation state (3) Zr and Hf have about the same radius (4) Zr and Zn have same oxidation state. Which of the following factors may be regarded as the main cause of lanthanide contraction ? (1) Greater shielding of 5d electrons by 4f electrons (2) Poorer shielding of 5d electron by 4f electrons (3) Effective shielding of one of 4f electrons by another in the sub-shell (4) Poor shielding of one of 4f electron by another in the sub-shell. 3. 4. The ‘‘spin-only’’ magnetic moment [in units of Bohr magneton, (B) of Ni2+ in aqueous solution would be (atomic number of Ni = 28) (1) 2.84 (2) 4.90 (3) 0 (4) 1.73 5. Lanthanoid contraction is caused due to : (1) the appreciable shielding on outer electrons by 4ƒ electrons from the nuclear charge (2) the appreciable shielding on outer electrons by 5ƒ electrons from the nuclear charge (3) the same effective nuclear charge from Ce to Lu (4) the imperfect shielding on outer electrons by 4f electrons from the nuclear charge 6. Identify the incorrect statement among the following. (1) The chemistry of various lanthanoids is very similar. (2) 4f and 5f orbitals are equally shielded. (3) d-block elements show irregular and erratic chemical properties among themselves. (4) La and Lu have partially filled d orbitals and no other partially filled orbitals. 7. The actinoids exhibit more number of oxidation states in general than the lanthanoids. This is because (1) The actinoids are more reactive than the lanthanoids. (2) The 5f orbitals extend farther from the nucleus than the 4f orbitals. (3) The 5f orbitals are more buried than the 4f orbitals (4) There is a similarity between 4f and 5f orbitals in their angular part of the wave function 8. Larger number of oxidation states are exhibited by the actinoids than those by the lanthanoids, the main reason being. (1) lesser energy difference between 5f and 6d than between 4f and 5d orbitals (2) more energy difference between 5f and 6d than between 4f and 5d orbitals (3) more reactive nature of the actinoids than the lanthanoids (4) 4f orbitals more diffused than the 5f orbitals 9. In context with the transition elements, which of the following statements is incorrect ? (1) In the highest oxidation states, the transition metal show basic character and form cationic complexes. (2) In the highest oxidation states of the first five transition elements (Sc to Mn), all the 4s and 3d electrons are used for bonding. (3) Once the d5 configuration is exceeded, the tendency to involve all the 3d electrons in bonding decreases. (4) In addition to the normal oxidation states, the zero oxidation state is also shown by these elements in complexes. 10. Knowing that the Chemistry of lanthanoids (Ln) is dominated by its +3 oxidation state, which of the following statement is incorrect ? (1) The ionic sizes of Ln (III) decrease in general with increasing atomic number. (2) Ln (III) compounds are generally colourless. (3) Ln (III) hydroxides are mainly basic in character (4) Because of the large size of the Ln (III) ions the bonding in its compounds is predominently ionic in character. 11.  2 The correct order of EM / M values with negative sign for the four successive elements Cr, Mn, Fe and Co is 12. 13. (1) Mn > Cr > Fe > Co (2) Cr > Fe > Mn > Co (3) Fe > Mn > Cr > Co (4) Cr > Mn > Fe > Co In context of the lanthanoids, which of the following statement is not correct ? (1) There is a gradual decrease in the radii of the members with increasing atomic number in the series. (2) All the member exhibit +3 oxidation state. (3) Because of similar properties the separation of lanthanoids is not easy. (4) Availability of 4f electrons results in the formation of compounds in +4 state for all the members of the series. The outer electron configuration of Lu (Atomic No : 71) is : (1) 4f3 5d5 6s2 (2) 4f8 5d0 6s2 (3) 4f4 5d4 6s2 (4) 4ƒ14 5d1 6s2 14. Iron exhibits +2 and + 3 oxidation states. Which of the following statements about iron is incorrect ? (1) Ferrous oxide is more basic in nature than the ferric oxide. (2) Ferrous compounds are relatively more ionic than the corresponding ferric compounds (3) Ferrous compounds are less volatile than the corresponding ferric compounds (4) Ferrous compounds are more easily hydrolysed than the corresponding ferric compounds. 15. Which of the following arrangements does not represent the correct order of the property stated against it ? (1) V2+ < Cr2+ < Mn2+ < Fe2+ : paramagnetic behaviour (2) Ni2+ < Co2+ < Fe2+ < Mn2+ : ionic size (3) Co3+ < Fe3+ < Cr3+ < Sc3+ : stability in aqueous solution (4) Sc < Ti < Cr < Mn : number of oxidation states 16. Four successive members of the first row transition elements are listed below with atomic numbers. Which one of them is expected to have the highest E (1) Cr(Z = 24) 0 value ? M3+/M2+ (2) Mn(Z = 25) (3) Fe(Z = 26) (4) Co(Z = 27) TRANSITION ELEMENTS [Previous Year Questions Part # II 1. 2. 3. 4. The product of oxidation of – with MnO4– in alkaline medium is : (A) O3– (B) 2 (C) O– The pair of compounds having metals in their highest oxidation state is : (A) MnO2, FeCl3 (B) [MnO4]– , CrO2Cl2 (D) O4– (C) [Fe(CN)6]2–, [Co(CN)6]3– (D) [NiCl4]2– , [Ni(CO)4]. Which of the following pair of compounds is expected to exhibit same colour in aqueous solution ? (A) FeCl3 , CuCl2 (B) VOCl2 , CuCl2 (C) VOCl2 , FeCl2 (D) FeCl2 , MnCl2 Give equations and describe the process for the developing of black and white photographic film. When sodium thiosulphate solution is treated with acidic solution turns milky white. Give the half reaction of the above described process. MCl4 (A) 5. Zn (B) purple colour compound Colourless liquid at room temperature, transition metal moist Air White fumes (C) Identify (A), (B) and (C). Also explain colour difference between MCl4 and (B). 6. 7. Match the reactions in Column I with nature of the reactions/type of the products in Column II. Column I Column II (A) O 2  O 2  O22 (p) Redox reaction (B) CrO24  H  (q) One of the products has trigonal planar structure (C) MnO4  NO2  H  (r) Dimeric bridged tetrahedral metal ion (D) NO3  H2SO 4  Fe 2  (s) Disproportionation Among the following, the coloured compound is : (A) CuCl (B) K3 [Cu(CN)4] (C) CuF2 (D) [Cu(CH3CN)4]BF4 8. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO2 is. 9. Reduction of the metal centre in aqueous permanganate ion involves : (A) 3 electrons in neutral medium (B) 5 electrons in neutral medium (C) 3 electrons in alkaline medium (D) 5 electrons in acidic medium 10. The colour of light absorbed by an aqueous solution of CuSO4 is : (A) organge-red 11. (B) blue-green (C) yellow (D) violet Which of the following halides react(s) with AgNO3(aq) to give a precipitate that dissolves in Na2S2O3(aq) ? (A) HCl (B) HF (C) HBr (D) HI EXERCISE - 5 : PART # I 1. 3 14. 4 2. 3 15. 1 3. 4 4. 1 5. 4 6. 2 7. 2 8. 1 9. 1 10. 2 11. 1 12. 4 13. 4 16. 4 PART # II 1. A 4. Developer is usually a weak reducing agent like potassium ferrous oxalate, an alkaline solution of pyrogallol or an alkaline solution of quinol. 2. B 3. B In the process of development of the photographic film, the exposed/activated AgBr grains are preferentially reduced by one of the reducing agent described above. HO O + 2e– + 2H+ HO  O Hydroquinone Quinone AgBr + e–  Ags + Br– The photographic film is then fixed by washing with hypo solution to remove the unreduced AgBr grains from the film. AgBr + 2Na2S2O3  Na3[Ag(S2O3)2] + NaBr.. S2O32– + H2O  2SO2 + 2H+ + 4e–. S2O32– + 6H+  2S  (white milky) + 3H2O. 5. (A) is TiCl4 as it has no unpaired electron and is liguid at room temperature on account of covalent character because of high polarising power of Ti+4. TiCl4 being covalent gets hydrolysed forming TiO2(H2O)2 and HCl (B) which fumes in air. In [Ti(H2O) 6]Cl3 complex Ti(III) has one unpaired electron(3d 1) which gives violet / purple colour due to d-d transition. TiCl4 (A) Colour less Zn TiCl3 Hydrolysis H2O [Ti (H2O)6]Cl3 purple or violet (B) moist air TiO2.(H2O)2 + HCl (white fumes) 6. A  p, s; B  r ; C  p, q ; D  p. 7. (C) 8. 6 9. ACD 10. A 11. ACD