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by: Ashley Kunze


Ashley Kunze
GPA 3.61

R. Little

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R. Little
Class Notes
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This 5 page Class Notes was uploaded by Ashley Kunze on Thursday October 22, 2015. The Class Notes belongs to CHEM 227 at University of California Santa Barbara taught by R. Little in Fall. Since its upload, it has received 21 views. For similar materials see /class/226943/chem-227-university-of-california-santa-barbara in Chemistry and Biochemistry at University of California Santa Barbara.

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Date Created: 10/22/15
39 11c h edu MATERIALS ZlSCHEMISTRY 227 COVALENT INTERACTIONS ENERGY BANDs 1 MATRL 218CHEM 227 Class Vll Covalent Intereractions energy bands Ram Seshadri seshadrimrlucsbedu Orbitals and crystal fields 0 The shapes of s and p orbitals 0 mg i y 5 py pz pX o The shapes of d orbitals Please see the handout Z X 0 Crystal field theory for d orbital complexes d orbitals on free atoms are fivefold degenerate This means that all d orbital in a free atom have the same energy When atoms with d valence orbitals transition metals are placed in a crystalline solid the energies of the different d levels are no longer degenerate In the simplest theory 7 electrostatic Crystal Field Theory the assumption is made that the ligand atoms the anions are point charges and when these approach a d orbital on the metal that d orbital is destabilized l stabilized ligand atom ligand atom l In the above figure ligand atoms such as 02 ions in an oxide approach orbitals along the z axis The dzz orbital is relatively destabilized as a result while the dxZ orbital is relatively stabilized The precise manner in which the degeneracy of the d orbitals are lifted therefore depends on the lo cal coordination of the transition metal cation 7 the precise geometrical arrangement of ligand atoms around the metal See the handout for the nature of crystal field splittings corresponding to different coordinations 39 in h edu MATERIALS ZlSCHEMISTRY 227 COVALENT INTERACTIONS ENERGY BANDS 1 o Familiarize yourself with the different ways in which these d levels can be filled by electrons in different transition metals 0 High spin and low spin crystal fields High spin 7L Low spin AAA AAA The figure shows how four d electrons such as in an Mn3 compound can be distributed over the octahedral d orbitals On the left the configuration has four unpaired electrons distributed as tgg 87 This is the highspin configuration On the right the configuration is low spin with only two unpaired 4 electrons distributed as tzg eg 0 Examples Octahedral compounds LaCrOs t3g lt 82 LaNiOs th lt 8 Tetrahedral compounds Mn in MnFezO4 82 lt t Square planar compounds Pd in PdS diZ d z lt d lt diy lt dgziyz M olecules 0 Molecular orbitals can be formed through linear combinations of atomic orbitals O antibonding molecular orbital atomic orbital Energy bonding molecular orbital The circles are the s orbitals on the H atoms We use two different colors black and white to represent the different possibilities of making linear combinations A addition to bond and subtraction to antibond Chains 0 Instead of a pair of H atoms we can consider chains with 3 4 5 etc 39 in h edu MATERIALS ZlSCHEMISTRY 227 COVALENT INTERACTIONS ENERGY BANDS 1 3 H atoms 5 H atoms MD 0OO gqp 00FOC W m M mquot we mo At the bottom are the most bonding molecular orbitals and at the top are the most antibonding The little yellow circles indicate nodes The more nodes in an orbital the higher the energy 1D Solids One can now extend this to an infinite chain 000000000000 most antibonding 000 O 000 less antibonding Energy O O 000 less bonding most bonding What one then has are not molecular orbitals but crystal orbitals The nodes in these form a pattern Use them to identify wavelengths There are zero nodes in the most bonding orbital so the repeat distance between nodes which is half the wavelength is 00 For the most antibonding orbital the repeat distance is u the interorbitalatornic distance so the wavelength is 2a The wavevector k appears for the first time as a solution to the free electron Schrodinger equation k becornes quantized as a result of applying boundary conditions see chapter 2 of Ashcroft and Merrnin for a discussion 39 in h edu MATERIALS ZlSCHEMISTRY 227 COVALENT INTERACTIONS ENERGY BANDS 1 We identify the wavelength described by the nodes as the deBroglie wavelength 27T A 7 k The most bonding crystal orbital formed from s orbitals then corresponds to k 27Too 0 and the most antibonding to k 27T 2H 7TH In 1 D these points are labeled T and X respectively 0 Dispersion relations or band structures or spaghetti diagrams are plots of the energy as a function of the wavevector For the 1D chain of s orbitals we know simply by counting nodes that the dispersion should look like Energy E ne rgy k0 kJa F k X The center of gravity of a band on the energy axis is determined by the energies of the participating orbitals The spread dispersion is determined by the strength of the interaction EMT gt DOSE F k X For example the interaction between p orbitals can be endon 7 of sideon 7T 7 are stronger so one obtains rnore disperse bands In the scheme above the p orbitals forming the 7139 bonds are at a higher energy to start with than those that form the 7 bonds l in h edu MATERIALS ZlSCHEMISTRY 227 COVALENT INTERACTIONS ENERGY BANDS 1 2D Solids o In 2D Similar logic applies For a square lattice of s orbitals yL bonding along x antibonding along x antibonding along y antibonding along y CCCCC COCOC OOOOO OCOCO CCCCC COCOC OOOOO OCOCO CCCCC COCOC CCCCC COCOC CCCCC COCOC CCCCC COCOC CCCCC COCOC CCCCC COCOC bonding along x bonding along y ing along y antibonding along x bond Let the lattice parameter be n Then we can consider the wavelengths in the four different cases Bonding along both at and y The wavelengths are 00 co and k 0 0 This is the T point Bonding along x and antibonding along y The wavelengths are 0025 and k 0 7TH This is the X point Antibonding along x and bonding along y The wavelengths are 25100 and k 7Tu 0 This is the Y point Antibonding along both at and y The wavelengths are 2H 2H and k 7Tu 7TH This is the M point The electronic band structure would look like F X M Y F For a square lattice X Y so there is no need to show the Y point However if the lattice is distorted to a rectangle the X and Y points are no longer degenerate


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