1. Chemistry

2. Session d and f Block Element - 1

3. Session Objectives 1.General properties of transition elements 2.Atomic radii 3.Ionisation energy 4.Oxidation states 5.Catalyst 6.Colour of complexes 7.Principle of extraction of Fe 8.Principle of extraction of Cu 9.Principle of extraction of Ag

4. d-block elements Not all d block elements have incomplete d sub-shells e.g. Zn has e.c. of [Ar]3d 10 4s 2, Zn 2+ ion [Ar] 3d 10 Similarly Sc forms Sc 3+ which has the stable e.c of Ar. Characteristic of d-block transition elements: incomplete filled d-subshell in common oxidation states.

5. Illustrative Example Why are Zn, Cd and Hg not considered transition metals? Solution : Because they do not have vacant d-orbitals neither in the atomic state nor in any stable oxidation state.

6. Illustrative Example What are coinage metals? Solution: Cu,Ag,Au are called coinage metals.

7. Atomic Radii

8. Illustrative Example In the transition series, with an increase in atomic number the atomic radius does not change very much. Why is it so? Solution : Since the electrons are added in inner d-orbitals which have poor screening effect,hence effective nuclear charge does not increase appreciably.So, atomic radius does not change much.

9. Electronic Arrangement ElementZ3d4s Sc21[Ar] Ti22[Ar] V23[Ar] Cr24[Ar] Mn25[Ar] Fe26[Ar] Co27[Ar] Ni28[Ar] Cu29[Ar] Zn30[Ar] Electronic Configuration

10. Melting Point The number of unpaired d-electrons increases up to the middle so metallic strength increases up to the middle. The dip in mp at Mn can be explained on the basis that it has stable half filled conf.so electrons are held tightly so delocalisation is less & metallic bond is weak

11. Ionisation Energy IE 2 :V Mn and Ni Zn IE 3 : Fe << Mn

12. Illustrative Example The sums of the first and second ionization energies and those of the third and fourth ionization energies of nickel and platinum are given below: IE 1 + IE 2 (KJ mol -1 )IE 3 + IE 4 (KJ mol -1 ) Ni 2.498.80 Pt 2.666.70 Taking these values into account write the following: (i)The most common oxidation state for Ni and Pt and its reasons. (ii)The name of the metal (Ni or Pt) which can form compounds in +4 oxidation state more easily and why?

13. Solution (i)Ni shows +2 oxidation state whereas Pt shows +4.(IE 1 +IE 2 ) of Ni is less than that for Pt whereas (IE 3 + IE 4 ) is less for Pt. (ii)From the given data it is clear that Pt (IV) is more easily attained while more energy would be required for obtaining Ni (IV) ion. Hence, Pt (IV) compounds are more stable than Ni (IV) compounds.

14. Oxidation States In solution,the stability of the compounds depends upon electrode potentials rather than ionisation energies.Electrode potential values depend upon factors such as energy of sublimation of the metal,ionisation energy and hydration energy.

15. Illustrative Example Out of Cr 2+ and Cr 3+,which one is stable in aqueous solution? Solution : Cr 3+ is more stable in aqueous solution due to higher hydration energy which is due to smaller size and higher charge

16. Illustrative Example Explain why E o for Mn +3 /Mn 2+ couple is more positive than that for Fe 3+ /Fe 2+. (Atomic numbers of Mn = 25, Fe = 26) Solution : Filled shell of Mn in +2 oxidation state (3d 5 ) makes it more stable.

17. Illustrative Example How do the oxides of the transition elements in lower oxidation states differ from those in higher oxidation states in the nature of metal-oxygen bonding and why? Oxides of transition metal in lower oxidation state are ionic and basic in nature whereas in higher oxidation state it forms covalent oxide which are acidic in nature. Solution

18. Catalysis Catalysis plays an essential aspect in about 90% of all chemical manufacturing. Ni and Pt are very heterogeneous catalysts. Pt, Rh, and Pd are used in catalytic converters. V 2 O 5 is used in conversion of SO 2 to SO 3. The catalytic properties of the transition elements are probably due to presence of unpaired electrons in their incomplete d- orbitals.In some cases the transition metals with their variable valency may form unstable intermediate compounds.In other cases the transition metal provides a suitable reaction surface.

19. Illustrative Example Why does V 2 O 5 acts as catalyst? Solution: V 2 O 5 acts as catalyst because it has large surface area.It can form unstable intermediates which readily change into products.

20. Magnetism Transition metal ions and their compounds show magnetic behaviour due to the presence of unpaired electrons in (n – 1) d-orbitals. n is the number of unpaired electrons.

21. Colour of Complexes due to the presence of unpaired d- electrons. Charge transfer spectra (V) +5 +4 +3 +2

22. Illustrative Example Of the ions Co 2+, Sc 3+, and Cr 3+, which one will give coloured aqueous solution and respond to a magnetic field? Co 2+ and Cr 3+ are coloured and attracted in magnetic field because they have unpaired electrons whereas Sc 3+ does not have any unpaired electron hence it will be repelled by magnetic field. Solution :

23. Illustrative Example Give reasons for the following features of transition metal chemistry. (a) Most of the transition metal ions are coloured in solution (b) Transition metals are well known to form complex compounds (c) The second and third members in each group of the transition elements have very similar atomic radii Solution : (a)This is attributed due to presence of unpaired electrons,they undergo d-d transitions by absorbing light from visible region and radiate complementary colour. (b) Small size and high charge of cation and presence of vacant d- orbitals. (c)Due to lanthanide contraction.

24. Illustrative Example Account for the following: Scandium forms no coloured ions and yet it is regarded as a transition element. Solution : Sc has incompletely filled d-orbital hence it is regarded as transition metal.It forms no coloured ion due to absence of unpaired electron in Sc 3+ ion.

25. Alloy formation d-block elements have similar atomoc sizes due to which atoms of one metal can easily take up positions in the crystal lattice of other. Interstitial compounds The transition elements form a large number of interstitial compounds in which small atoms such as hydrogen,carbon,boron and nitrogen occupy the empty spaces in their lattices.

26. Metallurgy of Iron Ore: The most commonly used iron ores are haematite, Fe 2 O 3, and magnetite, Fe 3 O 4 Concentration of ore: by magnetic separation. Roasting: Moisture, sands are removed as oxide. Smelting: Done in blast furnace. Roasted ore, limestone and coke are added in blast furnace for smelting.

27. Metallurgy of Iron

28. Different type of iron PropertiesCast iron SteelWrought iron % carbon2.2-4.50.15-1.50.12-0.25 Melting point 12001200-14001500 HardnessHardHard and softSoft BrittlenessBrittleMalleable WeldabilityCan’t be welded Can be welded

29. Steel Mild steel: 0.1-0.5% C Hard steel: 0.6-1.5% C NameCompositionSpecial propertiesUses Manganese steel 12-13% MnExtremely hardGrinding machinary Nickel steel2-4% NiResists corrosionDrive shafts, gears Invar36% NiLow coefficient of expansion Pendulum rods,watches Stainless steel 46 %NiResists corrosionutensils Chrome- nickel 0.5%-25 Cr, 1-4% Ni High tensile strengthAxles,ball,bearing,cutting tools Chrome- vanadium 1-10% Cr, 0.14- 0.15%V High tensile strengthAutomobile axles Duriron15-18% SiHigh resistance to the attack of acids Acid pumps, pipe lines etc

30. Methods of preparation Three fundamental changes from pig iron Removal of Si, Mn, P and other minor impurities. through slag formation Reduction of the C content. 3-4% in pig iron, 0-1.5% in steel. Addition of alloying elements Cr, Ni, Mn, V, Mo, and W. Give the steel its desired properties. The oxygen top blowing process The electric arc process The high frequency induction process Old processes Bessemer process Open hearth process Modern processes

31. Reactions Occurring in Steelmaking Processes

32. Heat treatment of steel Quenching or hardening Process of heating steel to red hot followed by sudden cooling by plunging to oil or water. Makes steel hard and brittle due to formation of iron carbide (Fe 3 C). Annealing Process of heating steel to red hot then cooling slowly. Annealing makes steel soft and ductile. Tempering Process of heating quenched steel to temperature much below redness followed by slow cooling. Tempering retains the hardness but brittleness disappears. Nitriding Process of producing hard coating or iron nitride on surface of steel.

33. Metallurgy of copper Ore: Copper pyrite CuFeS 2 Concentration: done by froth floatation. Slag is removed from upper layer. Molten mass containing Cu 2 S and Fe is called copper mate. Smelting: Smelting at 800 o C converts CuS to CuO. Since iron has more affinity for oxygen, therefore FeO is formed first. FeO(s) + SiO 2 (s) → FeSiO 3 (fusible slag) Roasting:

34. Metallurgy of copper Molten matte is mixed with sand (SiO 2 ) and heated in Bessemer converter. Bessemerisation Refining of copper By poling or electrolytic refining. Blister copper (impure metal).

35. Metallurgy of Silver Extraction: By Mac Arthur Forrest Process. Refining: By electrolytic method. Ore: Argentite AgS Concentration: By froth flotation process. The sodium argentocyanide solution is filtered and the silver is precipitated from the solution by the addition of zinc.

36. Thank you