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Published byConstance Wright Modified over 8 years ago
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Atoms to Molecules Single electron atom or ion (arbitrary Z ) orbitals known exactly Consider 1 s orbital: n mlml l solutions characterized by QNs: n, l, m l R(r n ) single electron (neutral) atom is hydrogen, Z = 1
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Electrons in Molecules H 2 + ionic molecule: 2 protons and 1 electron Take new molecular orbital to be a linear combination of atomic orbitals ab a b position - ½R½R½R *symmetry about x = 0 requires coefficients to be equal in magnitude Orbitals Probability densities: * = | | 2 ab a b 2 position - ½R½R½R Bonding Antibonding electron between nuclei electron on either side of nuclei “node” high energy * intuitively: this orbital has lower coulombic energy
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Types of MOs from LCAOs s s + pzpz pzpz + s pzpz + + px (py)px (py) bonds bond
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Types of MOs from LCAOs s s + pzpz pzpz + bonds Showed: symmetric = bonding Symmetric + + ++ + + + Antisymmetric Anti-bonding Bonding ++
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The H 2 + Ion Vary inter-nucleus distance, R, between the two protons E R -13.6 eV R electron sees only one proton, no interaction anti-bonding bonding R0R0 To find equilibrium bond distance, R 0
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The H 2 Molecule Introduce 2 nd electron Solution has perturbations due to electron-electron interactions Ignore these and place 2 nd electron in ‘same’ bonding orbital but with opposite spin E R -13.6 eV 2 anti-bonding states 2 bonding states 2 protons two 1 s states each 4 states total 1s1s 1s1s alternative depictions H H Result: H 2 covalent bond Directional; typical of molecules
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The N-atom Hydrogen Solid 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 1 s orbital is close to nucleus, nuclear interaction prevents strong overlap Chemical Bonding Continuous Bands
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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 2 s orbitals overlap without nuclear repulsion N 2 s electrons, 2N states
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Silicon: A Semiconductor 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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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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LiF: An Ionic Solid 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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