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

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

Types of MOs from LCAOs s s + pzpz pzpz + s pzpz + + px (py)px (py)  bonds  bond

Types of MOs from LCAOs s s + pzpz pzpz +  bonds Showed: symmetric = bonding Symmetric +  +  ++  +  + +  Antisymmetric Anti-bonding Bonding ++ 

The H 2 + Ion Vary inter-nucleus distance, R, between the two protons E R eV R   electron sees only one proton, no interaction anti-bonding bonding R0R0 To find equilibrium bond distance, R 0

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 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

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

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

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) EgEg

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

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 = eV = 5.5 eV  1 s 2

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?

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