Chapter 25 Transition Metals and Coordination Compounds Part 2

Transcription

1 Chapter 25 Transition Metals and Coordination Compounds Part 2 Bonding in Coordination Compounds Valence Bond Theory Coordinate covalent bond is between: completely filled atomic orbital and an empty atomic orbital. from ligand metal cation in complex. The metal ion orbital becomes hybridized. Coordination # Hybridization Geometry Example 2 sp linear [Ag(NH3)2] + 4 sp 3 tetrahedral [Zn(CN)4] 2-4 dsp 2 of sp 2 d square planar [Ni(CN)4] 2-5 dsp 3 trigonal bipyrimidal [CuCl5] 3-5 d 2 sp 2 square pyramidal [Ni(CN)5] 3-6 d 2 sp 3 or sp 3 d 2 octahedral [Fe(CN)6]

2 Crystal Field Theory color Focuses on interactions between ligands and unhybridized d orbitals as the ligands approach during bond formation. Originally there are _5_ 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 repulsions. This will cause those orbitals to be destabilized. This will result in them having higher energy. 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 less destabilizing repulsive forces from the ligand electrons. Two of the orbitals are closer to the ligands because they lie on the axes. These feel more destabilizing repulsive forces from the ligand electrons

3 For an octahedral complex, degenerate d orbitals split into _2_ levels. - All split levels are higher in energy than the original orbitals due to repulsions from ligand electrons, some more than others. - The orbitals along the axes, with more destabilization due to greater repulsive forces end up higher 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 splitting energy. Crystal Field Theory magnetic properties A substance that contains unpaired electrons and is attracted by a magnetic field is paramagnetic. A substance that has electrons that are all paired and is weakly repelled by a magnetic field is diamagnetic. 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: 25-15

4 When dealing with non-degenerate d orbitals, the decision is more complex OR High Spin & Low Spin Complexes In octahedral geometry complexes, with ions containing 4-7 d electrons: We have 2 possible choices for electron configurations. A. High Spin Complexes: 1. Makes early use of high 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 lower 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 paramagnetic. b) Those without unpaired electrons are diamagnetic C. How do molecules choose high or low spin? - Whichever takes less energy! 1. There is an energetic penalty for putting 2 negative particles (e - ) in the same space. A repulsive force must be overcome. force = P = pairing energy (energy to pair up) 2. There is an energetic penalty for placing an e - in a higher energy level. Energy = ( ) (energy to jump up) 3. Decision Process The complex that forms will be the one that takes less energy

5 If we have 6 e- to place, the first 3 e- will always go in the bottom level. If P < Then we have low spin: because of bigger. (This would be diamagnetic.) If P > Then we have high spin: because of smaller. (This would be paramagnetic.) P (pairing energy) does not change. It is the size of that changes. B. Tetrahedral and Square Planar complexes 1. Tetrahedral Complexes: There are only 4 ligands instead of six. The ligands approach the metal cation off axis. energy splitting is just the opposite of that in octahedral complexes. (tetrahedral complexes) ½ (octahedral complexes). < Pairing energy (P) all complexes are high-spin 25-17

6 2. Square Planar Complexes: two trans ligands along the z axis are missing (compared to octahedral). Large energy gap between the x 2 -y 2 orbital and the four lower-energy orbitals. Most common for metal ions with d 8 electronic configurations. Favors low-spin complexes in which all four lower-energy orbitals are filled and the higher energy x 2 -y 2 orbital is vacant 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 6 :NH 3 monodentate ligands = 6 total octahedral. b) To find the orbital used, start by finding the electron configuration. Ni atom (28 valence e-) = [Ar] 4s 2 3d 8 Ni 2+ (26 valence e-) = [Ar] 3d 8 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? paramagnetic 25-18

7 Ligands must put their pair of electrons into empty orbitals! Is the complex d 2 sp 3 or sp 3 d 2? If there are empty orbitals left in the current d level, it will be d 2 sp 3. If there are NO empty orbitals left in the current d level, it will be sp 3 d 2. The complex would look like: Color In Transition Metal Complexes A. The color of transition-metal complexes depends on the identity of the metal and the ligands. 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 E = E2 - E1=. 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/λ where is the wavelength of light absorbed. The color that we see is complementary to the color absorbed 25-19

8 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 λmax. (wavelength of max absorbance) Example: What wavelength (in nm) is absorbed if o = 2.75 x J? What color will the solution be? h = x J s c = x 10 8 m/s E = o = hc/λ λ = hc = (6.626 x J s)(2.998 x 10 8 m/s) = o 2.75 x J λ = 7.22 x 10-7 m x 10 9 nm = 722 nm (red absorbed, green soln) 1 m According to the color wheel, it should appear to be green. Note: is the energy to move 1 electron. If the question asks about the energy per mole, you need to multiply by Avogadro s Number! D. Spectrochemical Series of Ligands 1. Size of depends on the nature of the ligands. weak-field ligands - produce a relatively small value of resulting in high-spin complexes. strong-field ligands - produce a relatively large value of resulting in low-spin complexes

9 Spectrochemical Series of Ligands I - < Br - <Cl - <:SCN <F - OH - < C2O4 2- < H2O <:NCS < NH3 < en < phen < CN - CO weak field ligands medium field strong field 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? very little data on P-donors may be higher than N-donors P S Cl S-donors between Br and Cl Br I spectrochemical series follows arrows around starting at I and ending at C More complete list that is on the formula sheet Weak field ligands strong field ligands I - < Br - < S 2- < :SCN - < Cl - < F - < OH < C 2O 4 2 < :ONO < H 2O < :NCS < EDTA 4- < NH 3 < en < phen < :NO 2 < CN CO en = ethylenediamine phen = phenanthrolene C2O4 2- = oxalate ion 25-21

10 Small a. caused by weak 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 strong 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 [CoF 6 ] 3- vs [Co(CN) 6 ] 3- These are Co 3+ complexes. Co 3+ = [Ar] 3d 6 6 ligands = octahedral orientation and orbital splitting Weak field ligands will result in high spin complexes Strong filed ligands will result in low low spin complexes Starting orientation for original 6 electrons is: [CoF 6 ] 3- [Co(CN) 6 ] 3- (F - is weak field) (CN - is strong field) high spin low spin Complex is sp 3 d 2 d 2 sp

11 Practice How many unpaired electrons are there in the tetrahedral complex ion [CoCl 4 ] 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? Since there are 6 ligands it is octahedral