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Bonding in Metal complexes Valence Bond Theory

Bonding in Metal complexes Valence Bond Theory. Postulates. 1.The central metal ion provides adequate number of vacant orbitals for the formation of bond with ligands . 2. The suitable ( s,p & d) pure atomic orbitals hybridise to give an equal number

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Bonding in Metal complexes Valence Bond Theory

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  1. Bonding in Metal complexes Valence Bond Theory www.smitaasthana.com

  2. Postulates 1.The central metal ion provides adequate number of vacant orbitals for the formation of bond with ligands. 2. The suitable ( s,p & d) pure atomic orbitalshybridise to give an equal number of hybridisedorbitalsof equivalent energy . 3. The d- orbitals involved may be inner, (n-1)d or outer, ndorbitals. The complexes thus formed are Inner , low spin complexes or outer, high spin complexes. The hybridisation may be d2sp3 or sp3 d2. 4. Each ligand provides at least one orbital containing lone pair of electrons. 5. The empty hybrid orbital overlaps with filled ligand orbital forming a sigma coordinate bond. 6. The compleses will be paramagnetic if it contains unpaired electron and diamagnetic if there are no unpaired electrons. 7. Sometimes under the influence of strong ligands electrons in metal orbitals are forced to pair up against Hund’s rule of maximum multiplicity. www.smitaasthana.com

  3. Examples Coordination no. - 4 • 1. Tetrahedral Complexes • a) [Ni(CN)4]2- Tetracyanonickelate (II) ion Ni 0 –[Ar] 3d8 4s2 Ni 2+ ion – [Ar] 3d8 4s0 • Ni(II) ion 3d8 4s0 4p0 sp3 hybridisation 2. [Ni(Cl)4]2- ion X X X X 4 ligands give 4 lone pairs of electrons sp3 hybridisation Paramagnetic, 2 unpaired electrons, Tetrahedral www.smitaasthana.com

  4. Examples Coordination no. - 4 • Tetrahedral Complexes • b) [Zn(NH3)4] 2+Tetraamminezinc (II) ion Zn 0 –[Ar] 3d10 4s2 Zn 2+ ion – [Ar] 3d10 4s0 Zn 2+ ion 3d10 4s0 4p0 sp3 hybridisation [Zn(NH3)4] 2+ X X X X 4 ligands give 4 lone pairs of electrons sp3 hybridisation Diamagnetic, Tetrahedral www.smitaasthana.com

  5. Examples Coordination no. - 4 2. Square Planar Complexes a) [Ni(CN) 4 ] 2-Tetracyanonickelate (II) ion Ni 0 –[Ar] 3d8 4s2 Ni 2+ ion – [Ar] 3d8 4s0 1. Ni(II) ion - 3d8 4s0 4p0 2. Pairing against Hund’s rule1 dsp2 hybridisation X X X X 3. [Ni(CN) 4 ] 2- 4 ligands give 4 lone pairs of electrons dsp2 hybridisation Diamagnetic, Square planar www.smitaasthana.com

  6. Examples Coordination no. - 4 Square Planar Complexes b) [Cu(NH3)4] 2+Tetraamminecopper (II) ion Cu 0 –[Ar] 3d10 4s1 Cu 2+ ion – [Ar] 3d9 4s0 1. Cu(II) ion - 3d94s0 4p0 Option 1 X X X X sp3 hybridisation X X X X Option 2 4 ligands give 4 lone pairs of electrons dsp2 hybridisation X- Ray studies confirm Square planar shape Paramagnetic www.smitaasthana.com

  7. Examples Coordination no. - 6 1. Octahedral – Low spin - Inner orbital complex [Fe(CN)6]4- Hexacyanoferrate(II) ion Fe 0 –[Ar] 3d6 4s2 Fe 3+ ion – [Ar] 3d6s0 • Fe(II) ion 2. Electronic rearrangement against Hund’s rule 3d6 4s0 4p0 X X X X X X 3. [Fe(CN)6]4- ion 6 ligands give 6 lone pairs of electrons d2sp3 hybridisation Octahedral Diamagnetic, Inner, Low spin complex www.smitaasthana.com

  8. Examples Coordination no. - 6 Octahedral – Low spin - Inner orbital complex [Cr(NH3)6]3+ Hexaamminechromium(III) ion Cr 0 –[Ar] 3d5 4s1 Cr 3+ ion – [Ar] 3d3s0 1. Cr 3+ ion 3d3 4s0 4p0 2. [Cr(NH3)6]3+ X X X X X X 6 ligands give 6 lone pairs of electrons d2sp3 hybridisation Octahedral Paramagnetic, 3 unpaired electrons, Inner orbital www.smitaasthana.com

  9. Examples Coordination no. - 6 2. Octahedral – High spin – Outer orbital complex [Fe(H2O)6]3+ Hexaaquoiron(III) ion Fe 0 –[Ar] 3d6 4s2 Fe 3+ ion – [Ar] 3d5s0 • Fe(III) ion 3d5 4s0 4p0 4d0 sp3d2 hybridisation 2. [Fe(H2O)6]3+ X X X X X X 6 ligands give 6 lone pairs of electrons sp3d2 hybridisation Octahedral Paramagnetic, 5 unpaired electrons, High spin, Outer complex www.smitaasthana.com

  10. Examples Coordination no. - 6 Octahedral – High spin – Outer complex [Co(F)6]3+ Hexafluoridecobolt(III) ion Co 0 –[Ar] 3d7 4s2 Co 3+ ion – [Ar] 3d6 4s0 Co 3+ ion 3d5 4s0 4p0 4d0 sp3d2 hybridisation [Co(F)6]3+ X X X X X X 6 ligands give 6 lone pairs of electrons sp3d2 hybridisation Octahedral Paramagnetic, 4 unpaired electrons, High spin, Outer complex www.smitaasthana.com

  11. Practice problem The magnetic moment of [Mn(Cl)6]3-is 2.8 B.M. while that of [Mn(Br)4]2-is 5.9 B.M. Predict the geometries of the complex ions. Solution – [Mn(Cl)6]3- U = [n(n+2)], 2.8 = [n(n+2)], so, n = 2 Complex should have only 2 unpaired electrons. The possible electronic arrangement would be - Mn 0 –[Ar] 3d5 4s2 Mn 3+ ion – [Ar] 3d4 4s0 Mn 3+ ion 3d4 4s0 4p0 d2sp3 hybridisation [Mn(Cl)6]3- X X X X X X 6 ligands give 6 lone pairs of electrons d2sp3 hybridisation Octahedral Paramagnetic, 2 unpaired electrons, Inner complex www.smitaasthana.com

  12. Practice problem [Mn(Br)4]2- Mn 0 –[Ar] 3d5 4s2 Mn 2+ ion – [Ar] 3d5 4s0 Solution – U = [n(n+2)], 5.9 = [n(n+2)], so, n = 5 Complex has 5 unpaired electrons. The possible electronic arrangement would be - Mn 2+ ion 3d5 4s0 4p0 sp3 hybridisation [Mn(Br)4]2- X X X X 4 ligands give 4 lone pairs of electrons sp3 hybridisation, Tetrahedra Paramagnetic, 5 unpaired electrons, Tetrahedra www.smitaasthana.com

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