1. 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

2. 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

3. 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

4. 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

5. 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

6. 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

7. 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?

8. 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

9. 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 )

10. 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

11. 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