Atoms to Molecules

Atoms to Molecules Single electron atom or ion (arbitrary Z) orbitals known exactly n ml solutions characterized by QNs: n, l, ml l R(rn)  Consider 1s orbital: single electron (neutral) atom is hydrogen, Z = 1

Electrons in Molecules H2+ ionic molecule: 2 protons and 1 electron Take new molecular orbital to be a linear combination of atomic orbitals *symmetry about x = 0 requires coefficients to be equal in magnitude * Orbitals y b a b a position - ½R ½R Probability densities: y*y = |y|2 Antibonding |y|2 electron between nuclei Bonding electron on either side of nuclei b a b a position - ½R ½R “node” high energy intuitively: this orbital has lower coulombic energy

Types of MOs from LCAOs s s + pz pz + s bonds s pz + + px(py) p bond

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

The H2+ Ion Vary inter-nucleus distance, R, between the two protons R0 R anti-bonding E electron sees only one proton, no interaction -13.6 eV R   bonding To find equilibrium bond distance, R0

The H2 Molecule • Introduce 2nd electron • Solution has perturbations due to electron-electron interactions • Ignore these and place 2nd electron in ‘same’ bonding orbital but with opposite spin alternative depictions 1s 1s   R E From Molecules to Extended Solids  H H 2 anti-bonding states  -13.6 eV 2 protons 2 bonding states  two 1s states each   4 states total • Result: H2 covalent bond • Directional; typical of molecules

The N-atom Hydrogen Solid 1s1N-atom solid  N electrons Chemical Bonding  Continuous Bands R0 R E overlap of states discrete  continuous N statesunoccupied N anti-bonding states 1s 2N states N statesoccupied N bonding states H2 molecule: N = 2 1s orbital is close to nucleus, nuclear interaction prevents strong overlap

Lithium: A Simple Metal Li #3 1s22s1 N-atom solid 2s orbitals overlap without nuclear repulsion N 2s electrons, 2N states R0 R E overlap of states discrete  continuous N statesunoccupied anti-bonding 2s 2N states N statesoccupied bonding 1s 2N states All states occupied, independent

Silicon: A Semiconductor Si: #14 1s22s22p6 3s23p2 N-atom solid  4N relevant electrons hybrid orbital composed of 3s and all 3p orbitals: 3s: 2N states 3p: 6N states [3(sp3)4] [Ne] 8N states Hybridization: consider just 2 atoms R0 R anti-bonding E overlap of states discrete  continuous 4 states (+ 4) 4N statesunoccupied 3 1 3p 3p 6 states 4N anti-bonding states 3s 3s 2 states “sp3” 8N states Eg 1 3 4N statesoccupied bonding 4N bonding states 4 states (+ 4) N-atom Solid  Continuous Bands

Magnesium: A Metal? N atom solid, 2N electrons Mg: #12 1s22s22p6 3s2 [Ne] R0 R 3s and 3p overlap to create a band with 8N states; only 2N states occupied  yes, a metal E 3p 6N states 2N statesoccupied anti-bonding 3s 2N states bonding metal requires a partially occupied band??

LiF: An Ionic Solid  1s2  Li+1 + e- Li: #3 1s22s1 F: #9 1s22s22p5  2p6  F + e-  F-1 Energy of bonding for a hypothetical ion pair Zeff < Zact due to shielding DE = DEionization + DEcoulombic  5.4 eV > 0 requires energy  -3.7 eV < 0 releases energy +1 -1  -7.2 eV DEpair = 5.4 - 3.7 - 7.2 eV = 5.5 eV

N-atom Pair Solid of LiF  1s2 e: Li(2s)  F(2p) Li+1 + e- Li: #3 1s22s1 F: #9 1s22s22p5  2p6  E(F2p) < E(Li2s) 6N electrons R0 R LiF is a non-metal E Li 2s 2N states Eg 6N statesoccupied F 2p 6N states 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

Atoms to Molecules