Coordination Compounds Grade 12 Chemistry PDF

Coordination Compounds Grade 12 Chemistry Session 4 Recap. Recap. Postulates of Werner's theory 1. Metals show two types of linkages (valency)-primary and secondary. 2. The primary valency is normally ionisable and is satisfied by negative ions. 3. The secondary valency is non io...

Coordination Compounds Grade 12 Chemistry Session 4 Recap. Recap. Postulates of Werner's theory 1. Metals show two types of linkages (valency)-primary and secondary. 2. The primary valency is normally ionisable and is satisfied by negative ions. 3. The secondary valency is non ionisable. These are satisfied by neutral molecules or negative ions. The secondary valence is equal to the CN. 4. The ions/groups bound by the secondary linkages to the metal have characteristic spatial arrangements. EAN Rule E.A.N. = Atomic number – oxidation number + no. of e– from ligands Complex having EAN of the central atom equal to the atomic number of the next noble gas, are found to be extra stable. Learning Outcomes 1 Werner's Theory 2 Sidgwick's EAN Rule 3 VB Theory: Postulates and analysis of 6-coordination complexes Valence Bond Theory (V.B.T.) Postulates of V.B.T. The central metal ion has empty orbitals for accommodating electrons donated by the ligands. The formation of a complex involves reaction between a lewis base (ligand) and a lewis acid (metal or metal ion) with the formation of a coordinate- covalent (or dative) bond between them. The model utilizes hybridisation of (n-1)d, ns, np or ns, np, nd orbitals of metal atom or ion to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar etc. The number of unpaired electrons, measured by the magnetic moment of the compounds determines which d-orbitals are used. Where n = no. of unpaired electron Valence Bond Theory (V.B.T.) Postulates of V.B.T. Hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. Coordination Type of Distribution of hybrid orbitals in space number hybridisation sp3 Tetrahedral 4 dsp2 Square planar 5 sp3d Trigonal bipyramidal sp3d2 Octahedral 6 d2sp3 Octahedral Valence Bond Theory (V.B.T.) Postulates of V.B.T. If the complex contains unpaired electrons, it is paramagnetic in nature. If the complex does not contain unpaired electrons, it is diamagnetic in nature Under the influence of a strong ligand, the electrons can be forced to pair up against the Hund’s rule of maximum multiplicity Hybridisation Coordination number = 6 Octahedral sp3d2 d2sp3 nd orbital used (n-1)d orbital used Outer orbital complex Inner orbital complex Hybridisation Coordination number = 6 Example - [CoF6] 3– ( Paramagnetic) Oxidation state of Co = +3 Orbitals of Co 3d 4 4p s Orbitals of Co+3 3d 4 4p As this complex is paramagnetic , there should be sunpaired electron 3d 4 4p s Hybridisation Coordination number = 6 Example - [CoF6] 3– ( Paramagnetic) Orbitals of Co+3 ion 3d 4 4p 4d s sp3d2 hybridised orbitals 3d 4d 3d 4d Six pairs of electrons from six F- Hybridisation Coordination number = 6 Example - [CoF6] 3– ( Paramagnetic) sp3d2 hybridised orbitals 3d Six pairs of electrons from six F- 4d Unpaired electrons ⇒ (Paramagnetic) The outer d orbital (4d) is used in hybridisation, the complex is called an outer orbital. It is a high spin or spin free complex. Hybridisation Coordination number = 6 Example - [Co(NH3)6] 3+ (Diamagnetic) Oxidation state of Co = +3 Orbitals of Co 3d 4 4p s Orbitals of Co+3 3d 4 4p s As this complex is diamagnetic so there should be no unpaired electron ⇒ pairing takes place against Hund’s rule. Orbitals of Co+3 3d 4 4p s Hybridisation Coordination number = 6 Example - [Co(NH3)6] 3+ (Diamagnetic) 3d 4 4p s d2sp3 hybrid orbitals 3d 3d Six pairs of electrons from six NH3 molecules Hybridisation Coordination number = 6 Example - [Co(NH3)6] 3+ (Diamagnetic) d2sp3 hybrid orbitals 3d Six pairs of electrons from six NH3 molecules No unpaired electron ⇒ Diamagnetic The inner d orbital (3d) is used in hybridisation, the complex is called an inner orbital. It is a low spin or spin paired complex. Hybridisation Coordination number = 6 In octahedral complex with configuration of CMA/CMI as d1, d2 and d3, always inner orbital complex is formed with d2sp3 hybridization regardless of the nature of ligand. i.e., [Ti(H2O)6]+2, [Ti(CO)6]+2, [Ti(ox)3]−4 [Sc(H2O)6]+2, [Sc(CO)6]+3, [Sc(NH3)6]+2 d2sp3 [V(CO)6]+2, [V(CN)6]−2, [V(H2O)6]+3 In octahedral complex with configuration of CMA/CMI as d8, d9 and d10, always outer orbital complex is formed with sp3d2 hybridization regardless of the nature of ligand. i.e., [Ni(H2O)6]+2, [Ni(CN)6]−4, [Ni(CO)6]+2 sp3d2 outer orbital complexes [Cu(NH3)6]+2, [Cu(H2O)6]+2, [Zn(NH3)6]+2 [Zn(H2O)6]+2, [Zn(CN)6]−4 Let’s Solve.. Let’s Solve.. Which of the following is a high spin complex? (CN– and NH3 are strong ligands and F– is a weak ligand) A [Co(NH3)6]3+ B [Fe(CN6)]4− C [Mn(NH3)6]2+ D [FeF6]3− Let’s Solve.. Which of the following is a high spin complex? (CN– and NH3 are strong ligands and F– is a weak ligand) A [Co(NH3)6]3+ B [Fe(CN6)]4− C [Mn(NH3)6]2+ D [FeF6]3− Solution 3d 4s 4p 4d sp3d2 in case of [FeF6]3– Outer orbital complex sp3d2 or High spin complex  ? Quiz.  ? Quiz. Which one of the following complexes is an outer orbital complex? A [Co(NH3)6]3+ B [Mn(CN)6]4- C [Fe(CN)6]4- D [Ni(NH3)6]2+  ? Quiz. Which one of the following complexes is an outer orbital complex? A [Co(NH3)6]3+ B [Mn(CN)6]4- C [Fe(CN)6]4- D [Ni(NH3)6]2+ Solution Since Ni2+ has [Ar] 3d8 configuration, it will always form outer orbital octahedral complexes. [Ni(NH3)6]2+ sp3d2 (outer)  ? Quiz. NEET 2017 Pick out the correct statement with respect to [Mn(CN)6]3−. A It is sp3d2 hybridised and tetrahedral. B It is d2sp3 hybridised and octahedral. C It is dsp2 hybridised and square planar. D It is sp3d2 hybridised and octahedral.  ? Quiz. NEET 2017 Pick out the correct statement with respect to [Mn(CN)6]3−. A It is sp3d2 hybridised and tetrahedral. B It is d2sp3 hybridised and octahedral. C It is dsp2 hybridised and square planar. D It is sp3d2 hybridised and octahedral.  ? Quiz. NEET 2017 Pick out the correct statement with respect to [Mn(CN)6]3−. Solution [Mn(CN)6]3− : Let oxidation state of Mn be x. x + 6 × (−1) = − 3 ⇒ x = +3 Electronic configuration of Mn : [Ar]4s2 3d5 Electronic configuration of Mn3+ : [Ar]3d4 3d 4s 4p [Mn(CN)6]3− :[Ar] xx xx xx xx xx xx CN- CN- CN- CN- CN- CN- d2sp3 hybridization Thus, [Mn(CN)6]3− has d2sp3 hybridization and has octahedral geometry.  ? Quiz. The number of unpaired electrons in d6, low spin, octahedral complex is: A 4 B 2 C 1 D 0  ? Quiz. The number of unpaired electrons in d6, low spin, octahedral complex is: A 4 B 2 C 1 D 0 Solution Low spin, d6, octahedral complex 3d 4s 4p d2sp3 No of unpaired electron (n) = 0 Learning Outcomes 1 VB Theory: Analysis of 4-coordination complexes 2 Magnetic Properties of Coordination Compounds 3 Crystal Field Theory Learning Outcomes 1 VB Theory: Analysis of 4-coordination complexes 2 Magnetic Properties of Coordination Compounds 3 Crystal Field Theory Hybridisation Coordination number = 4 C.N. =4 sp3 (tetrahedral) dsp2 (square planar) ns and np orbital used (n-1)d , ns and np orbital used Outer orbital complex Inner orbital complex Hybridisation Coordination number = 4 Example - [NiCl4] 2− ( Paramagnetic) Oxidation state of Ni = +2 Orbitals of Ni 3d 4s 4p Orbitals of Ni+2 3d 4s 4p As this complex is paramagnetic so there should be unpaired electron Hybridisation Coordination number = 4 Example - [NiCl4] 2– ( Paramagnetic) Orbitals of Ni+2 3d 4s 4p sp3 hybridised orbitals 3d 3d Four pairs of electrons from four Cl- Hybridisation Coordination number = 4 Example - [NiCl4]2− ( Paramagnetic) sp3 hybridised orbitals 3d Four pairs of electrons from four Cl− Unpaired electron ⇒ (Paramagnetic) Geometry - tetrahedral The outer ns and np orbital are used in hybridisation, the complex is called an outer orbital complex. It is a high spin or spin free complex Hybridisation Coordination number = 4 Example - [Ni(CN)4] 2− ( Diamagnetic) Oxidation state of Ni = +2 Orbitals of Ni 3d 4s 4p Orbitals of Ni+2 ion 3d 4s 4p As this complex is diamagnetic so there should be no unpaired electron 3d 4s 4p Hybridisation Coordination number = 4 Example - [Ni(CN)4] 2− ( Diamagnetic) Orbital of Ni2+ 3d 4s 4p dsp2 hybridised orbitals 3d 4p 3d Four pairs of electrons 4p from four CN− Hybridisation Coordination number = 4 Example - [Ni(CN)4] 2− ( Diamagnetic) dsp2 hybridised orbitals 3d 4p Four pairs of electrons from four CN− No unpaired electron ⇒ Diamagnetic Geometry - square planar The inner (n-1)d, ns and np orbital are used in hybridisation, the complex is called an inner orbital It is a low spin or spin paired complex. Hybridisation Coordination number = 4 Example - [Ni(CO)4] ( Diamagnetic) Oxidation state of Ni = 0 Orbitals of Ni 3d 4s 4p As this complex is diamagnetic so there should be no unpaired electron. Also, CO is exceptionally strong ligand. Orbitals of Ni 3d 4s 4p Orbitals of Ni 3d 4s 4p Hybridisation Coordination number = 4 Example - [Ni(CO)4] ( Diamagnetic) Oxidation state of Ni = 0 sp3 hybridised orbitals Orbitals of Ni 3d 4s 4p 3d 3d Four pairs of electrons from four CO Hybridisation Coordination number = 4 Example - [Ni(CO)4] ( Diamagnetic) sp3 hybridised orbitals 3d Four pairs of electrons from four CO No unpaired electron ⇒ Diamagnetic Geometry - tetrahedral The outer ns and np orbital are used in hybridisation, the complex is called an outer orbital complex it is a low spin or spin paired complex Hybridisation Special Note: With 4d and 5d series elements, all of the ligands acts as strong. Example - [PdCl4]2– Pd2+ = [Kr] 4d8 → Normally Cl– acts as weak ligand and thus the complex should have been a tetrahedral, paramagnetic, sp3 hybridised complex. → But in reality it is found to be square planar, diamagnetic, dsp 2 hybridised complex. Pd2+ = 4d 5s 5p dsp2 hybridised orbitals 4d 5p Limitations of VBT It involved number of assumptions. It does not give quantitative interpretation of magnetic data. It does not explain the colour exhibited by coordination compounds It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds. Example: Why [Cu(NH3)6]2+ is a stable compound whereas it forms an outer orbital complex? It does not make exact predictions regarding the tetrahedral and square- planar structures of 4-coordinate complexes. It does not distinguish between strong and weak ligands. Example: Why [Mn(CN)6]3- form inner orbital complex whereas [MnCl6]3- form outer orbital complex? Let’s Solve.. Let’s Solve.. NEET 2015 The hybridization involved in complex [Ni(CN)4]2− is (At. No. Ni = 28) A sp3 B d2sp2 C d2sp3 D dsp2 Let’s Solve.. NEET 2015 The hybridization involved in complex [Ni(CN)4]2− is (At. No. Ni = 28) A sp3 B d2sp2 C d2sp3 D dsp2 Let’s Solve.. NEET 2015 The hybridization involved in complex [Ni(CN)4]2− is (At. No. Ni = 28) Solution [Ni(CN)4]2− : Oxidation number of Ni = +2 Electronic configuration of Ni2+ = 3d8 4s0 [Ni(CN)4]2− = 3d 4 4p s Pairing occurs 3d 4 4p s dsp2 hybridised orbitals Let’s Solve.. Which one of the following has a square planar geometry? (Co = 27, Ni = 28, Fe = 26, Pt = 78) A [CoCl4]2- B [FeCl4]2- C [NiCl4]2- D [PtCl4]2- Let’s Solve.. Which one of the following has a square planar geometry? (Co = 27, Ni = 28, Fe = 26, Pt = 78) A [CoCl4]2- B [FeCl4]2- C [NiCl4]2- D [PtCl4]2- Solution Hybridisation of [CoCl4]2-, [FeCl4]2- and [NiCl4]2- is sp3 and for Pt+2 , Cl─ act as a strong field ligand and form inner orbital complex. So, hybridisation is dsp2 and geometry is square planar.  ? Quiz.  ? Quiz. Which of the following complexes has a geometry different from others? A [NiCl4]2- B Ni(CO)4 C [Ni(CN)4]2- D [Zn(NH3)4]2+  ? Quiz. Which of the following complexes has a geometry different from others? A [NiCl4]2- B Ni(CO)4 C [Ni(CN)4]2- D [Zn(NH3)4]2+ Solution [NiCl4]2- , sp3 hybridization, tetrahedral geometry Ni(CO)4, sp3 hybridization, tetrahedral geometry [Ni(CN)4]2-, dsp2 hybridization, square planar geometry [Zn(NH3)4]2+, sp3 hybridization, tetrahedral geometry  ? Quiz. Statement–1: [Ni(CN)4]2– is a diamagnetic complex Statement–2: Compound is low spin complex. A Both statements are true B Both statements are false C S-1 is true, but S-2 is false D S-1 is false, but S-2 is true  ? Quiz. Statement–1: [Ni(CN)4]2– is a diamagnetic complex Statement–2: Compound is low spin complex. A Both statements are true B Both statements are false C S-1 is true, but S-2 is false D S-1 is false, but S-2 is true  ? Quiz. Statement–1: [Ni(CN)4]2– is a diamagnetic complex Statement–2: Compound is low spin complex. Solution Oxidation number of Ni = +2 Electronic configuration of Ni2+ = 3d8 4s0 3d 4 4p s Pairing occurs in case of CN- ion : 3d 4 4p s dsp2 hybridised orbitals Thus, complex is both diamagnetic and low spin.  ? Quiz. Which of the following molecules is not tetrahedral? (NH3, en and CO are strong ligands) A [Ni(en)2]2+ B Ni(CO)4 C [Zn(NH3)4]2 + D [NiCl4]2–  ? Quiz. Which of the following molecules is not tetrahedral? (NH3, en and CO are strong ligands) A [Ni(en)2]2+ B Ni(CO)4 C [Zn(NH3)4]2 + D [NiCl4]2–  ? Quiz. Which of the following molecules is not tetrahedral? (NH3, en and CO are strong ligands) Solution E.C. of E.C. in presence of Complex Hybridisation Shape CA/CI. ligand [Ar] 3d8 with e– [Ni(en)2]2+ [Ar] 3d8 dsp2 Square planar paired [Ar] 3d8 [Ar] 3d10 with e– Ni(CO)4 sp3 Tetrahedral 4s2 paired [Zn(NH3)4]2+ [Ar] 3d10 [Ar] 3d10 sp3 Tetrahedral [NiCl4]2– [Ar] 3d8 [Ar] 3d8 - general sp3 Tetrahedral Learning Outcomes 1 VB Theory: Analysis of 4-coordination complexes 2 Magnetic Properties of Coordination Compounds 3 Crystal Field Theory Magnetic properties Coordination compounds generally have partially filled d orbitals and show characteristic magnetic properties. Magnetic properties depend upon the oxidation state, electronic configuration, coordination number of the central metal and the nature of the ligand field. The number of unpaired electrons in any complex can be easily calculated from the configuration of the metal ion, its coordination number and the nature of the ligands involved. Magnetic moment of the complexes can then be calculated using: n = number of unpaired electrons Magnetic properties For octahedral complexes For d1 complexes 3d 4s 4p Orbitals of Ti3+ d2sp3 n=1 3d 4s 4p For d2 complexes d2sp3 n=2 Magnetic properties For octahedral complexes For d3 complexes 3d 4s 4p Orbitals of Cr3+ d2sp3 n=3 Thus for d1, d2, d3 electronic configuration, inner orbital complexes are formed Magnetic properties For octahedral complexes For d4 complexes 3d 4s 4p 4d sp3d2 in case of [MnCl6]3– Outer orbital complex sp3d2 n=4 3d 4s 4p d2sp3 in case of [Mn(CN)6]3– Inner orbital complex d2sp3 n=2 Magnetic properties For octahedral complexes For d5 complexes 3d 4s 4p 4d sp3d2 in case of [FeF6]3– Outer orbital complex sp3d2 n=5 3d 4s 4p d2sp3 in case of [Fe(CN)6]3– Inner orbital complex d2sp3 n=1 Magnetic properties For octahedral complexes For d6 complexes 3d 4s 4p 4d sp3d2 in case of [CoF6]3− Outer orbital complex sp3d2 n=4 3d 4s 4p d2sp3 in case of [Co(C2O4)3]3− Inner orbital complex d2sp3 n=0 Magnetic properties For octahedral complexes In case of d7, the magnetic moment may depend on the strength of ligand. In case of d8, d9 and d10 configuration, magnetic moment is independent of strength of ligand. No. of unpaired electrons No. of unpaired electrons in presence of weak ligand in presence of strong ligands d7 3 1 d8 2 2 d9 1 1 d10 0 0 For tetrahedral and square planar complexes Similar analysis can be done for tetrahedral and square planar complexes. Let’s Solve.. Let’s Solve.. The magnetic moment (spin only) of [NiCl4]2-is A 1.82 BM B 5.46 BM C 2.82 BM D 1.41 BM Let’s Solve.. The magnetic moment (spin only) of [NiCl4]2-is A 1.82 BM B 5.46 BM C 2.82 BM D 1.41 BM Solution [NiCl4]2-: Ni2+ = 3d8 4s0 4s 4p [NiCl4]2- ⥮ : ⥮ ⥮ ↿ ↿ ‥ ‥ ‥ ‥ No. of unpaired e− = 2 Cl- Cl- Cl- Cl- 𝜇S = n (n + 2) 𝜇S = 2 (2 + 2) = 8 = 2.82 BM  ? Quiz.  ? Quiz. NEET 2013 A magnetic moment at 1.73 BM will be shown by one among of the following A TiCl4 B [CoCl6]4– C [Cu(NH3)4]2+ D [Ni(CN)4]2–  ? Quiz. NEET 2013 A magnetic moment at 1.73 BM will be shown by one among of the following A TiCl4 B [CoCl6]4– C [Cu(NH3)4]2+ D [Ni(CN)4]2– Solution Oxidation state of Cu in [Cu(NH3)4]2+ is + 2 Cu2+ = 3d9 ⥮ ⥮ ⥮ ⥮ ↿ It has one unpaired electron (n = 1). 𝜇 = n (n + 2) = 1(1 + 2) = 3 = 1.73 BM  ? Quiz. The species that has a spin-only magnetic moment of 5.9 BM, is: (Td = tetrahedral) A [Ni(CN)4]2- (square planar) B [NiCl4]2- (Td) C Ni(CO)4 (Td) D [MnBr4]2- (Td)  ? Quiz. The species that has a spin-only magnetic moment of 5.9 BM, is: (Td = tetrahedral) A [Ni(CN)4]2- (square planar) B [NiCl4]2- (Td) C Ni(CO)4 (Td) D [MnBr4]2- (Td)  ? Quiz. The species that has a spin-only magnetic moment of 5.9 BM, is: (Td = tetrahedral) Solution [MnBr4]2- ⇒ Mn2+ ⇒ d5 (Td) [∵ Br– is weak field ligand] d5(Td) is high spin complex. So, μ = 5(5 + 2) = 5.91 B.M.  ? Quiz. The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) [PdCl2(PPh3)2] A (A) ≈ (C) < (B) ≈ (D) B (C) < (D) < (B) < (A) C (C) ≈ (D) < (B) < (A) D (A) ≈ (C) ≈ (D) < (B)  ? Quiz. The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) [PdCl2(PPh3)2] A (A) ≈ (C) < (B) ≈ (D) B (C) < (D) < (B) < (A) C (C) ≈ (D) < (B) < (A) D (A) ≈ (C) ≈ (D) < (B)  ? Quiz. The correct order of the calculated spin-only magnetic moments of complexes (A) to (D) is: (A) Ni(CO)4 (B) [Ni(H2O)6]Cl2 (C) Na2[Ni(CN)4] (D) [PdCl2(PPh3)2] Solution Complex No. of unpaired e- 𝛍 (A) Ni(CO)4 Ni = 3d84s2 (SFL) 0 0 (B) [Ni(H2O)6]Cl2 Ni2+ = 3d8 (WFL) 2 8 BM (C) Na2[Ni(CN)4] Ni2+ = 3d8 (SFL) 0 0 (D) [PdCl2(PPh3)2] Pd2+ = 4d8 (SFL) 0 0 Correct order of the calculated spin only magnetic moments of complexes A to D is (A) ≈ (C) ≈ (D) < B Learning Outcomes 1 VB Theory: Analysis of 4-coordination complexes 2 Magnetic Properties of Coordination Compounds 3 Crystal Field Theory Crystal Field Theory (CFT) Postulates of CFT It is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interaction between the metal ion and the ligand. e_ Metal Ligand Crystal Field Theory (CFT) Postulates of CFT Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. 𝛅– L- 𝛅+ 𝛅+ Point charge Dipole Crystal Field Theory (CFT) Postulates of CFT The metal-ligand bond is considered to be ionic arising purely from electrostatic interaction e_ Metal Ligand Electrostatic attraction due to Electrostatic repulsion due to e– positive and negative charge present in metal and ligand Due to ligand field, energy of d orbitals is increased. Crystal Field Theory (CFT) Postulates of CFT The five d orbitals in an isolated gaseous metal atom/ion have same energy. (they are degenerate) This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal. M M L L or + M + Free metal atom/ion L L Symmetrical field of ligands Degenerate Energy of all the d orbitals is raised All d orbitals are of same energy Crystal Field Theory (CFT) Postulates of CFT Because of unsymmetrical field of ligands the degeneracy of the d orbitals is lost and cause splitting of the d orbitals. The pattern of splitting depends upon the nature of the crystal field. L M M L L or + M + Free metal atom/ion L L L Degenerate unsymmetrical field of ligands All d orbitals are of same energy Splitting of d- orbital CFT: Octahedral complexes Shape of d orbitals Non-axial d-orbitals CFT: Octahedral complexes Shape of d orbitals Axial d-orbitals dz2 dx2 -y2 CFT: Octahedral complexes In an octahedral coordination entity electrostatic repulsion is due to e– present in metal’s d-orbitals and electrons of ligand. In octahedral complexes ligand approaches the metal through the 3 axes. z y x Axial orbitals (dx2-y2, dz2) which point towards the axis along the direction of the ligand will experience more repulsion and will be raised in energy relative to the average energy in the spherical crystal field. Non axial orbitals (dxy, dyz and dzx) orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. CFT: Octahedral complexes Energy L L L M M M L L L dz2 dx2- y 2 eg 3/5Δo Δo Barycenter dxy dyz dzx dz2 dx2- y 2 2/5Δo t2 g dxy dyz dzx dxy dyz dzx d2 z dx2- y 2 Metal atom surrounded Splitting of d-orbital in Free metal atom by Octahedral crystal field Spherical crystal field Δo is energy separation due to Crystal Field Splitting  CFT: Octahedral complexes Degeneracy of the d orbitals has been removed due to ligand electron-metal electron repulsions. Octahedral complex yield- three orbitals of lower energy, t2g set two orbitals of higher energy, eg set. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. Energy separation is denoted by Δo (the subscript o is for octahedral). It is also called crystal field splitting energy (CFSE). The energy of the two eg orbitals will increase by (3/5)Δo and that of the three t2g will decrease by (2/5)Δo. NOTE: CFSE also has the meaning crystal field stabilization energy, which is defined differently. Unless it is mentioned, we will consider CFSE to be crystal field splitting energy.  CFT: Octahedral complexes The crystal field splitting, Δo depends upon - - The charge on the metal ion: increases with increasing charge on the metal. - Whether the metal is in the first, second, or third row of transition elements. (or has high Zeff ): It increases with increase in Zeff. - The nature of the ligands: M L Weak field Ligand Strong field Ligand Less attraction and less More attraction and more repulsion repulsion Less rise in energy of eg and More rise in energy of eg and decrease in energy of t2g, i.e., decrease in energy of t2g, i.e., less crystal field splitting more crystal field splitting CFT: Octahedral complexes Spectrochemical series Ligands can be arranged in a series in the orders of increasing field strength. I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH3 < en < CN– < CO Such a series of ligands is termed as spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands. Strong field ligand leads to absorption of light of more energy (smaller wavelength) Weak field ligand leads to absorption of light of less energy (greater wavelength) Let’s Solve.. Let’s Solve.. Among the following pairs of complexes, in which case the 𝚫0 value is higher for the first one? A [Co(NH3)6]3+ and [Co(CN)6]3- B [CoF6]3- and [Co(NH3)6]3+ C [Co(H2O)6]2+ and [Co(H2O)6]3+ D [Rh(H2O)6]3+ and [Co(H2O)6]3+ Let’s Solve.. Among the following pairs of complexes, in which case the 𝚫0 value is higher for the first one? A [Co(NH3)6]3+ and [Co(CN)6]3- B [CoF6]3- and [Co(NH3)6]3+ C [Co(H2O)6]2+ and [Co(H2O)6]3+ D [Rh(H2O)6]3+ and [Co(H2O)6]3+ Let’s Solve.. Among the following pairs of complexes, in which case the 𝚫0 value is higher for the first one? Solution In complexes [Rh(H2O)6]3+ and [Co(H2O)6]3+, central metal cations have same oxidation state as well as same ligands and they fall in same group, but 𝚫0 of [Rh(H2O)6]3+ > 𝚫0 of [Co(H2O)6]3+ because Rh3+ has high Zeff value than Co3+.  ? Quiz.  ? Quiz. NEET 2020 Which of the following is the correct order of increasing field strength of ligands to form coordination compounds? A SCN– < F– < C2O42– < CN– B SCN– < F– < CN– < C2O42– C F– < SCN– < C2O42– < CN– D CN– < C2O42– < SCN– < F–  ? Quiz. Which of the following is the correct order of increasing field strength of ligands to form coordination compounds? A SCN– < F– < C2O42– < CN– B SCN– < F– < CN– < C2O42– C F– < SCN– < C2O42– < CN– D CN– < C2O42– < SCN– < F– Solution According to spectrochemical series, order of increasing field strength is : SCN– < F– < C2O42– < CN–  ? Quiz. Among the following, which one has higher CFSE? [HINT: CFSE ∝ strength of ligand] A [Co(CN)6]3– B [Co(C2O4)3]3– C [Co(H2O)6]3+ D [Co(NH3)6]3+  ? Quiz. Among the following, which one has higher CFSE? [HINT: CFSE ∝ strength of ligand] A [Co(CN)6]3– B [Co(C2O4)3]3– C [Co(H2O)6]3+ D [Co(NH3)6]3+ Solution Among the given complexes, CN– is the strongest ligand and will cause maximum splitting.  ? Quiz. Three complexes, [Co(NH3)5Cl]2– (I), [Co(NH3)5(H2O)]3+ (II) and [Co(NH3)6]3+ (III) absorb light in the visible region. The correct order of the wavelength of light absorbed by them is: [HINT: CFSE ∝ strength of ligand ∝ 1/λ] A (III) > (I) > (II) B (III) > (II) > (I) C (II) > (I) > (III) D (I) > (II) > (III)  ? Quiz. Three complexes, [Co(NH3)5Cl]2– (I), [Co(NH3)5(H2O)]3+ (II) and [Co(NH3)6]3+ (III) absorb light in the visible region. The correct order of the wavelength of light absorbed by them is: [HINT: CFSE ∝ strength of ligand ∝ 1/λ] A (III) > (I) > (II) B (III) > (II) > (I) C (II) > (I) > (III) D (I) > (II) > (III) Solution Complex (III) has all ammine ligands which is the strongest. Thus, it will have maximum CFSE and will absorb light of minimum wavelength. Among Cl– and H2O, H2O is stronger. Thus, complex (II) will absorb light of smaller wavelength than (I). Summary. Summary. Valence bond theory (V.B.T.) Octahedra Square planar Tetrahedral l Inner orbital complex Inner orbital Outer orbital complex (d2sp3) complex (sp3) E.g. [Fe(CN)6]4- (dsp2) E.g. [NiCl4]2-- Outer orbital complex E.g. [PtCl4]2- (sp3d2) E.g. [Fe(H2O)6]2+ Summary. Magnetic properties of coordination compounds Magnetic properties depend upon the oxidation state, electronic configuration, coordination number of the central metal and the nature of the ligand field. Magnetic moment of the complexes can be easily calculated using 𝛍S = n (n + 2) For d1, d2, d3 electronic configuration, inner orbital complexes are formed Since there is unpaired e_ are present so they are paramagnetic. For d4, d5, d6 electronic configuration, inner or outer orbital complexes are formed and they can be paramagnetic or diamagnetic. Summary. Crystal Field Theory Splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For octahedral complexes, splitting occurs as follows: dz2 dx2- y 2 eg 3/5Δo Δo Barycenter dxy dyz dzx dz2 dx2- y 2 2/5Δo t2 g dxy dyz dzx dxy dyz dzx d2z dx2- y 2 Metal atom surrounded Splitting of d-orbital in Free metal atom by Octahedral crystal field Spherical crystal field Summary. Spectrochemical series Ligands can be arranged in a series in the orders of increasing field strength. I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH3 < en < CN– < CO Vedantu Daily Practice Problems VDPP Number VDPP Name Wavebook subtopics covered 1 Coordination Compounds - VDPP-1 Introduction and components of coordination compounds 2 Coordination Compounds - VDPP-2 Classification of Ligands 3 Coordination Compounds - VDPP-3 Nomenclature of Coordination Compounds 4 Coordination Compounds - VDPP-4 Werner's Theory 5 Coordination Compounds - VDPP-5 Sidgwick's EAN Rule VBT: Analysis of 6-coordination complexes 6 Coordination Compounds - VDPP-6 VBT: Postulates 7 Coordination Compounds - VDPP-7 Analysis of 4-coordination complexes Vedantu Daily Practice Problems VDPP Number VDPP Name Wavebook subtopics covered 8 Coordination Compounds - VDPP-8 Magnetic Properties of Coordination Compounds Crystal Field Theory Electronic Configuration and CFSE in octahedral Complexes 9 Coordination Compounds - VDPP-9 Electronic Configuration and CFSE in tetrahedral Complexes Colour in complexes 10 Coordination Compounds - VDPP-10 Organometallic Compounds 11 Coordination Compounds - VDPP-11 Isomerism in Coordination Compounds 12 Coordination Compounds - VDPP-12 Geometrical Isomerism in Coordination Compounds 13 Coordination Compounds - VDPP-13 Optical Isomerism in Coordination Compounds 14 Coordination Compounds - VDPP-14 Importance of Coordination Compounds Feedback Thanks.