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A2 CHEMISTRY TRANSITION METALS TRANSITION METALS Atoms that can form stable ions which have incomplete 3d orbitals o Zn does not qualify as transition elements as Zn+2 = 1s2, 2s2, 2p6, 3s2, 3p6, 3d10, 4s0 o Sc does not qualify as transition elements as Sc+3 = 1s2, 2s2, 2p6, 3s2, 3p6, 3d0, 4s0 Chemical Properties o Variable Oxidation States o Form colored compounds o Form Metal Complexes o Catalysts Electronic Configuration Atomic Electronic Number Configuration Sc 21 [Ar], 4s2, 3d1 Ti 22 [Ar], 4s2, 3d2 V 23 [Ar], 4s2, 3d3 Cr 24 [Ar], 4s1, 3d5 Mn 25 [Ar], 4s2, 3d5 Fe 26 [Ar], 4s2, 3d6 Co 27 [Ar], 4s2, 3d7 Ni 28 [Ar], 4s2, 3d8 Cu 29 [Ar], 4s1, 3d10 Zn 30 [Ar], 4s2, 3d10 Cr and Cu promote electrons from lower energy orbital (4s) to higher energy orbital 3d to form more stable half-filled and completely filled orbitals. Sc +3 Element Electrons are lost from the outer most 4s electrons first when ions are formed. Fe+3 = [Ar], 3d5, 4s0 Transition elements can have variable oxidation states. 4s and 3d orbitals exist at approximately the same energy levels, hence there is very little difference in successive ionization energies. Ti +1 +2 +3 +4 V +1 +2 +3 +4 +5 Cr +1 +2 +3 +4 +5 +6 Mn Fe Co Ni +1 +1 +1 +1 +2 +2 +2 +2 +3 +3 +3 +3 +4 +4 +4 +4 +5 +5 +5 +6 +6 +7 Underlined Oxidation States are more stable than the rest Cu +1 +2 +3 From left-to-right period 2 (transition metals), proton number increases which would suggest IE should increase and Atomic Radii should decrease. But, each new electron gets added in the inner 3d orbital which increases shielding and decreases IE and increases Atomic Radii as well as the outer 4s electrons are now held less tightly Transition Metals vs Calcium Oxidation States Why can’t Na have variable oxidation states. o Na = 1s2, 2s2, 2p6, 3s1 Na+1 can be formed by removing 1 electron from Na, but removing the second electron becomes extremely difficult from the lower energy and less shielded 2p orbital. Invariant Atomic Radii and 1st Ionization Energies Transition Metal Ions FAHAD H.AHMAD Melting Point: Higher MP for Transition Metals due to Stronger Metallic Bonding as more Electrons get delocalized. Density: Atomic Radii of transition elements are smaller than Ca and masses of elements increase, hence transition elements have a much higher density. Atomic Radius: Atomic and Ionic Radius of Transition elements are smaller due to higher proton number, plus the increasing Zn +2 A2 CHEMISTRY TRANSITION METALS nuclear charge is poorly shielded by the inner incomplete 3d orbitals Conductivity: Conductivity increases for transition metals as more electrons get delocalized. LIGAND: Ion or Molecule that has a lone pair of electrons which can form a coordinate bond with a metal. List of Ligands Degenerate-Orbitals (D-Orbitals) FAHAD H.AHMAD dxy, dyz, dxz orbitals lie between axis dx2-y2 and dz2 lie along the axis Monodentate Ligands donate only one lone pair to form a single coordinate bond with central metal ion. Metal Complexes Fe+3 has the following electronic configuration 3d ↑ ↑ ↑ ↑ ↑ H2O: :NH3 :OH-1 hexa aqua metal complex hexa ammine complex hexa hydroxo complex 4s 4p 4d This ion has a high Charge Density due to its smaller size. The empty 4th shell orbitals can therefore attract ligands and datively bond to them. The six orbitals shaded above can be used to dative bonding. The number of dative bonds formed depends on the charge density and the size of the attached ligand. These six orbitals then hybridize, and in this case this is known as sp3d2 hybridization. (The above ligands are smaller in size, hence six ligands can fit around the central metal atom forming octahedral complexes) :Cl-1 tetra chloro metal complex :Br-1 tetra bromo metal complex :I-1 tetra iodo metal complex A2 CHEMISTRY TRANSITION METALS (The above ligands are larger in size, hence only four ligands can fit around the central metal atom forming tetrahedral complexes) FAHAD H.AHMAD Following shows an example of a hexadentate ligand forming six coordinate bonds with the central Cu+2 metal ion Bidentate ligands donate two lone pairs to form two coordinate bonds with the central metal ion Ethane dioate –OOCCOO1,2-diamino ethane NH2CH2CH2NH2 Transition Metals and Colored Metal Complexes Following shows an example of a bidentate ligand forming coordinate bonds with a central metal atom Hexadentate Ligands donates six lone pairs to form six coordinate bonds with the central metal ion e.g. Increased repulsion on the inner 3d orbital due to coordinate bond formation of the outer 4s, 4p, 4d orbitals splits the 3d orbital into two sets of two and three orbitals. The type of splitting depends on the structure of the metal complex formed. For an octahedral complex (6 ligands), dx2 and dx2-y2 split to a A2 CHEMISTRY TRANSITION METALS FAHAD H.AHMAD higher energy levels. For tetrahedral complex, dxy, dxz, dyz split to a higher energy level. Violet-Blue-Green-Yellow-Orange-Red Color Blue Cu+2 ion has a 3d9 orbital which splits as shown below when it forms a hexa aqua complex with water ligands. Electrons are unevenly distributed in the two sets of 3d orbitals. Electrons can now be promoted from the lower energy 3d orbital set to the higher energy 3d orbital set. A small amount of energy is required for this promotion which can be obtained from the visible spectrum. green violet red orange yellow The above colors are arranged in a circle. If violet, blue, green, yellow, orange get absorbed, you will see red reflected and so on. LIGAND COLORS Hexa Aqua Complexes Other metal compounds do not have split 3d orbitals with a small energy gap. If a sodium ion promotes an electron from the outer 2p orbital to the 3s orbital then a huge amount of energy is required and a photon this energetic cannot be obtained from the visible spectrum, instead it gains that energy from a photon from a higher frequency photon. (E = h*frequency) [V (H2O)6]+2 violet [V (H2O)6]+3 green [Cr (H2O)6]+3 blue/violet [Mn (H2O)6]+2 light pink [Fe (H2O)6]+2 light green [Fe (H2O)6]+3 red/brown [Co (H2O)6]+2 pink [Ni (H2O)6]+2 green [Cu (H2O)6]+2 blue [Zn (H2O)6]+2 colorless The energy gap between the 3d orbital sets is determined by the strength of the ligand field (repulsion exerted by the ligand) and the geometry and charge density of the complex ion. Hence each complex ion absorbs a different photon and form different colored compounds. Tetra Ammine Complex Zn+2 has complete 3d orbital, hence even orbital splitting does not lead to colored compounds. [Cu (H2O)6]+2 blue solution [Co (NH3)6]+2 brown [Ni (NH3)6]+2 blue [Cu (NH3)4 (H2O)2]+2 deep blue Ammonia Ligand Exchange for Copper ions If small amount of ammonia is added COMPLEMENTARY COLORS The colors that we see are the complementary colors, A metal complex absorbs a certain wavelegnth and we see the reflected white light with a missing wavelength. Following is a list of complementary colors NH3 + H2O NH4+1 + OH-1 OH-1 ions produced replace the H2O ligands +2 -1 [Cu (H2O)6] + 2OH [Cu (OH)2 (H2O)4] + 2H2O Which gives a pale blue ppt of [Cu (OH)2 (H2O)4] A2 CHEMISTRY TRANSITION METALS If ammonia is added in excess then NH3 ligands replace the OH-1 and H2O ligands FAHAD H.AHMAD If small amount of ammonia is added NH3 + H2O NH4+1 + OH-1 +2 [Cu (OH)2 (H2O)4] + NH3 ↔ [Cu (NH3)4 (H2O)2] + 2OH-1 + 2H2O OH-1 ions produced replace the H2O ligands +2 [Cu (NH3)4 (H2O)2]+2 gives a deep blue solution Chloro Ligand Exchange for Cu+2 ions When concentrated HCl is added to blue solution of [Cu (H2O)6]+2 -1 [Ni (H2O)6] + 2OH [Ni (OH)2 (H2O)4] + 2H2O Which gives a grey-green ppt of [Cu (OH)2 (H2O)4] If ammonia is added in excess then NH3 ligands replace the OH-1 and H2O ligands +2 +2 -1 -2 [Cu (H2O)6] + Cl ↔ [CuCl4] + 6H2O 4H2O -2 Yellow/green ppt of [CuCl4] is obtained which changes to blue solution when excess water is added which favours the reverse reaction LIGAND EXCHANGE IN COBALT [Co (NH3)6]+2 + H2O ↔ [Co (H2O)6]+2 + NH3 brown pink Above reversible reaction shifts to the right or to the left depending on the concentration of ammonia [Co (H2O)6]+2+4Cl-1 ↔ [CoCl4]-2+6H2O pink blue The above equilibrium shifts according to the concentration of Chloride ions Ligand Exchange for Ni+2 Green Solution of Ni+2 forms a grey-green precipitate when a small amount of ammonia is added, the precipitate dissolves in excess of ammonia to give a blue violet solution [Ni (H2O)6]+2 green solution -1 [Ni (OH)2 (H2O)4] + NH3 ↔ [Ni (NH3)6] + 2OH + [Ni (NH3)6]+2 gives a blue violet solution