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Published byRandall Watts Modified over 9 years ago
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Review: The H 2 Molecule E R -13.6 eV 2 anti-bonding states 2 bonding states 2 protons two 1 s states each 4 states total alternative representations H H Result: H 2 covalent bond Directional; typical of molecules R0R0
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The N-atom “Hydrogen Solid” Chemical Bonding Continuous Bands 1 s 1 N-atom solid N electrons E R N anti-bonding states N bonding states R0R0 1s1s 2N states H 2 molecule: N = 2 overlap of states discrete continuous N states occupied N states unoccupied
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Silicon – A ‘real’ N-atom Solid Si: #14 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 E R 4N anti-bonding states 4N bonding states R0R0 “sp 3 ” 8N states overlap of states discrete continuous 4N states occupied 4N states unoccupied N-atom solid 4N relevant electrons [Ne] N-atom Solid Continuous Bands [3( sp 3 ) 4 ] hybrid orbital composed of 3s and all 3p orbitals: 3s: 2N states 3p: 6N states 8N states 3p3p 3p3p 3s3s 3s3s Hybridization: consider just 2 atoms bonding anti-bonding 6 states 2 states 4 states (+ 4) 4 states (+ 4) 3 1 3 1 EgEg
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Lithium: A Simple Metal Li #31 s 2 2 s 1 N-atom solid E R anti-bonding bonding R0R0 2s2s 1s1s 2N states All states occupied, independent overlap of states discrete continuous N states occupied N states unoccupied
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Magnesium: A Metal? E R anti-bonding bonding 3s3s 2N states 2N states occupied R0R0 Mg: #12 1 s 2 2 s 2 2 p 6 3 s 2 [Ne] N atom solid, 2N electrons metal requires a partially occupied band?? 3p3p 6N states 3s and 3p overlap to create a band with 8N states; only 2N states occupied yes, a metal
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Ionic Solid Example: LiF Li: #3 1 s 2 2 s 1 F: #9 1 s 2 2 s 2 2 p 5 Energy of bonding for a hypothetical ion pair 5.4 eV Li +1 + e - 2 p 6 F + e - F -1 E = E ionization + E coulombic Z eff < Z act due to shielding -3.7 eV > 0 requires energy < 0 releases energy -7.2 eV +1 E pair = 5.4 - 3.7 - 7.2 eV = 5.5 eV 1 s 2
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N-atom Pair Solid of LiF E R F 2 p 6N states 6N states occupied R0R0 Li: #3 1 s 2 2 s 1 F: #9 1 s 2 2 s 2 2 p 5 Li 2 s 2N states 1 s 2 e: Li(2 s ) F(2 p ) Li +1 + e - 2 p 6 E(F2 p ) < E(Li2 s ) 6N electrons EgEg LiF is a non-metal Thoughts on how to transform it a metal?
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Summary: MO/LCAO Approach N atom solid –bonding and anti-bonding states –isolated states bands due to exclusion principle Metal –no energy gap between occupied and unoccupied states –many need to consider orbitals of slightly higher energy Semi-conductor –hybrid orbitals bands –‘small’ bandgap between occupied and unoccupied Ionic –electron transfer from electropositive to electronegative ion –orbitals bands –‘large’ bandgap between occupied and unoccupied states qualitative distinction
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Metals: Electron in a Box 0 V(x) L/2 -L/2 V(x) = 0 for | x | ≤ L/2 V(x) = for | x | > L/2 inside outside solve for (x) within the box solutions are of the form orwhere use boundary conditions to get A, B, k (or 0 )
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Metals: Electron in a Box A = B = 0 trivial solution subtract: add: there are no values of k that make these both true for arbitrary, non-zero A and B 2 solution sets set 1: A = 0 {0, , 2 , 3 , etc….} n = even set 2: B = 0 n = odd { 2, 3 2, 5 /2, etc….} n = principle quantum number k is quantized, not continuous
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Metals: Electron in a Box set 1: set 2: n = even n = odd to solve for A and B, normalize according to use n = even n = odd eigenfunctions general wave equation wave-length wave-vector
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