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Chapter 24 Transition Metals and Coordination Compounds part 2 Bonding in Coordination Compounds Valence Bond Theory Coordinate covalent bond is between: completely filled atomic orbital from and an empty atomic orbital. in complex. The metal ion orbital becomes Coordination # Hybridization Geometry Example 2 sp linear [Ag(NH3)2]+ 4 sp3 tetrahedral [Zn(CN)4]2- 4 dsp2 of sp2d square planar [Ni(CN)4]2- 5 dsp3 trigonal bipyrimidal [CuCl5]3- 5 d2sp2 square pyramidal [Ni(CN)5]3- 6 d2sp3 or sp3d2 octahedral [Fe(CN)6]4- 25-13 Crystal Field Theory – color Focuses on interactions between ligands and unhybridized d orbitals as the ligands approach during bond formation. Originally there are __ degenerate (same energy level) d orbitals. 3 orbitals have configurations between the axes: 2 orbitals have configurations that put them along the axes;) If ligands approach the metal cation on a path that makes them overlap with the electrons in the d orbitals, this will cause ________________ This will cause those orbitals to be ________________ This will result in them having ______________ A. For an octahedral complex, there are 6 ligand attachment points (along the x, y, & z axes) Three of the orbitals are farther away from the ligands because they lie between the axes, not on them. These feel _____ destabilizing repulsive forces from the ligand electrons. Two of the orbitals are closer to the ligands because they lie on the axes. These feel _____ destabilizing repulsive forces from the ligand electrons. 25-14 For an octahedral complex, degenerate d orbitals _____ into __ levels. - All split levels are __________ in energy than the original orbitals due to ______________ from ligand electrons, some more than others. - The orbitals along the axes, with _____ destabilization due to greater repulsive forces end up ___________ in energy than those that are between the axes. - The difference between the two energy levels is referred to a ____, known as the ____________________________ . Crystal Field Theory – magnetic properties A substance that contains unpaired electrons and is attracted by a magnetic field is _____________. A substance that has electrons that are all paired and is weakly repelled by a magnetic field is _____________. Valence bond theory cannot always account for the observed magnetic behavior of complexes. Hund’s rule – electrons occupy degenerate orbitals singly as long as an empty orbital is available. When dealing with original metal atomic d orbitals this gives: ____ ____ ____ ____ ____ When dealing with non-degenerate d orbitals, the decision is more complex 25-15 ____ ____ ____ ____ ____ OR ___ ___ ____ ____ ____ B. Tetrahedral and Square Planar complexes 1. Tetrahedral Complexes: There are only ____ ligands instead of six. The ligands approach the metal cation ______ axis. energy splitting is just the opposite of that in octahedral complexes. (tetrahedral complexes) ½(octahedral complexes). < Pairing energy (P) These terms will be explained in a few minutes. all complexes are high-spin 2. Square Planar Complexes: two trans ligands along the z axis are missing (compared to octahedral). Large energy gap between the x2-y2 orbital and the four lower-energy orbitals. Most common for metal ions with ____ electronic configurations. Favors __________ complexes in which all four lower-energy orbitals are filled and the higher energy x2-y2 orbital is vacant 25-16 Problem: Consider [Ni(NH3)6]2+ a) What is the shape and configuration of the complex? b) What metal ion orbitals are used? a) To find the shape, start by looking at the coordination number b) To find the orbital used, start by finding the electron configuration. Ni atom (28 valence e-) = ________________ Ni2+ (26 valence e-) = ________________ Octahedral splits Filling in the 8 e- in the d orbitals: Fill lower level first? We’ll come back to this! All of the 3d orbitals are occupied. Is the complex diamagnetic or paramagnetic? Ligands must put their pair of electrons into _________ orbitals! Is the complex d2sp3 or sp3d2? If there are empty orbitals left in the current d level, it will be ______ . If there are NO empty orbitals left in the current d level, it will be ______. 25-17 The complex would look like: Color In Transition Metal Complexes A. The color of transition-metal complexes depends on the identity of _____________________________. B. Metal complex can absorb light by undergoing an electronic transition from its lowest energy state (E1) to a higher energy state (E2). Wavelength of absorbed light depends on the energy separation between the two states ______________. Visible light contains a continuum of wavelengths, one of which is the correct wavelength to accomplish the promotion of an electron from a lower energy d orbital to a higher energy d orbital. : =E= hc/ where is the wavelength of light absorbed. Complexes always absorb a range of wavelengths, not just a single one. The color we see approximates the compliment of the peak of the absorbance spectrum, known as ______. 25-18 Example: What wavelength (in nm) is absorbed if o = 2.75 x 10–19 J? What color will the solution be? h = 6.626 x 10–34 J•s c = 2.998 x 108 m/s According to the color wheel, it should appear to be ________. Note: is the energy to move 1 electron. If the question asks about the energy per mole, you need to multiply by _________________! High Spin & Low Spin Complexes In octahedral geometry complexes, with ions containing ____ d electrons: We have 2 possible choices for electron configurations. A. High Spin Complexes: 1. Makes early use of ________ energy orbitals. 2. Puts d electrons in as many different d orbitals as possible. 3. Creates as many unpaired e- as possible. B. Low Spin Complexes: 1. Completely fills __________ energy orbitals first. 2. Puts as many d electrons in the lowest energy d orbitals as possible. 3. Reduces the number of unpaired e-. a) Substances with unpaired electrons are ___________. b) Those without unpaired electrons are ___________ 25-19 C. How do molecules choose high or low spin? ________________________ 1. There is an energetic penalty for putting 2 negative particles (e-) in the same space. A repulsive force must be overcome. 2. There is an energetic penalty for placing an e- in a higher energy level. 3. Decision Process The complex that forms will be the one that takes _______ energy. If we have 6 e- to place, the first 3 will always go in the bottom level. If P < Then we have ______ spin: because of bigger . If P > Then we have ______ spin: because of smaller . P (pairing energy) does not change. It is the size of that changes. D. Spectrochemical Series of Ligands 1. Size of depends on the nature of the ligands. weak-field ligands - produce a relatively ________ value of resulting in __________________ complexes. strong-field ligands - produce a relatively ________ value of resulting in __________________ complexes. 25-20 Spectrochemical Series of Ligands I- < Br- <Cl- <:SCN <F- OH- < C2O42- < H2O <:NCS < NH3 < en < phen < CN- CO The place of a ligand in the spectrochemical series is determined largely by its donor atoms. Thus, all N-donor ligands are close to ammonia in the spectrochemical series, while all O-donor ligands are close to water. The spectrochemical series follows the positions of the donor atoms in the periodic table as: C N O F P S Cl ? very little data on P-donors – may be higher than N-donors S-donors ≈ between Br and Cl Br spectrochemical series follows arrows around starting at I and ending at C I More complete list that is on the formula sheet Weak field ligands strong field ligands I-< Br- < S2- < :SCN- < Cl- < F- < OH– < C2O42– < :ONO– < H2O < :NCS– < EDTA4- < NH3 < en < phen < :NO2– < CN– ≈ CO en = ethylenediamine phen = phenanthrolene C2O42- = oxalate ion 25-21 Small a. caused by _________ field ligands(i.e., halides , oxalate ion) b. lower energy wavelengths absorbed (longer wavelengths absorbed) (closer to the red end of the spectrum absorbed) (colors appear closer to the violet end of the spectrum) Large a. caused by ____________ field ligands (i.e., CN-, CO) b. higher energy wavelengths absorbed (shorter wavelengths absorbed) (closer to the violet end of the spectrum absorbed) (colors appear closer to the red end of the spectrum) Example: Compare [CoF6]3- vs [Co(CN)6]3- These are Co3+ complexes. Co3+ = [Ar] 3d6 6 ligands = octahedral orientation and orbital splitting Weak field ligands will result in ________ spin complexes Strong filed ligands will result in ________ low spin complexes Starting orientation for original 6 electrons is: [CoF6]3- [Co(CN)6]3- Complex is 25-22 Practice___________________________________________________________ How many unpaired electrons are there in the tetrahedral complex ion [CoCl4]2? Is it diamagnetic or paramagnetic? Practice___________________________________________________________ How many unpaired electrons are there in the complex ion [Cr(CN)6]4- ? Is it diamagnetic or paramagnetic? 25-23