Handout 8
Linear Combination of Atomic Orbitals (LCAO)
In this lecture you will learn:
H • An approach to energy states in molecules based on the linear combination of atomic orbitals
H C H H
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Energy Bands and Atomic Potentials in Crystals V r The potential energy of an electron due 0 to a single isolated atom looks like: Potential of an isolated Energy levels In a crystal, the potential energy due to atom all the atoms in the lattice looks like: x V r 0 Vacuum level 0
Energy Potential in bands a crystal
x 0 The lowest energy levels and wavefunctions of electrons remain unchanged when going from an isolated atom to a crystal
The higher energy levels (usually corresponding to the outermost atomic shell) get modified, and the corresponding wavefunctions are no longer localized at individual atoms but become spread over the entire crystal
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
1 Failure of the Nearly-Free-Electron Approach V r Vacuum level 0 3 2 Energy Potential in 1 bands a crystal
x 0 • For energy bands that are higher in energy (e.g. 2 & 3 in the figure above) the periodic potential of the atoms can be taken as a small perturbation
For higher energy bands, the nearly-free-electron approach works well and gives almost the correct results
• For energy bands that are lower in energy (e.g. 1 in the figure above) the periodic potential of the atoms is a strong perturbation
For lower energy bands, the nearly-free-electron approach does not usually work very well
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Nearly-Free-Electron Approach Vs LCAO for Germanium
Energy Energy Empirical Energy NFA (eV) LCAO (eV) Pseudopotential (eV)
1 2
2 1
1 1 3 4 2 2 2 2
1 1 1 1 1 1
1 1 1 1 1 1
FBZ (FCC lattice) • For most semiconductors, the nearly-free- electron approach does not work very well
• LCAO (or tight binding) works much better and provides additional insights
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2 LCAO: From Hydrogen Atom to Hydrogen Molecule Consider a Hydrogen atom with one electron in the 1s orbital: V r
One can solve the Schrodinger equation: 2 2 r V r r E r 2m Potential of 1s energy level a Hydrogen and find the energy of the 1s orbital and atom its wavefunction ˆ Ho 1s r E1s 1s r 0 r where: r 2 1s Hˆ 2 V r o 2m
Angular probability 0 r distribution for the 1 1s orbital r er ao 1s 3 ao Radial amplitude for the 1s orbital
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Linear Combination of Atomic Orbitals (LCAO) V r Now consider a Hydrogen molecule made up of two covalently bonded a atom b atom Hydrogen atoms sitting at a distance of 2d from each other, as shown:
Hamiltonian for an electron is: -d 0 d x 2 Hˆ 2 V r dxˆ V r dxˆ 2m The basic idea behind LCAO approach is to construct a trial variational solution in which the wavefunction is made up of a linear combination (or superposition) of orbitals of isolated atoms: r ca 1s r d xˆ cb 1s r dxˆ
And then plugging the trial solution into the Schrodinger equation to find the coefficients ca and cb and the new eigenenergies:
Hˆ r E r
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
3 LCAO: From Hydrogen Atom to Hydrogen Molecule Plug the LCAO solution: r ca 1s r d xˆ cb 1s r dxˆ 2 into: Hˆ r E r Hˆ 2 V r dxˆ V r dxˆ 2m STEP 1: take the bra of the equation first with 1 s r d x ˆ to get: ˆ 1s r d xˆ H ca 1s r d xˆ cb 1s r dxˆ E 1s r d xˆ ca 1s r d xˆ cb 1s r dxˆ
Note that: ˆ V r 1s r dxˆ H 1s r dxˆ E1s Let: a atom b atom
ˆ 1s r dxˆ H 1s r dxˆ Vss
-d 0 d x 1s r dxˆ 1s r dxˆ 0
Not exactly zero – but we will assume so for simplicity
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: From Hydrogen Atom to Hydrogen Molecule
So we get finally:
E1s ca Vss cb E ca STEP 2: take the bra of the equation now with 1 s r d x ˆ to get:
E1s cb Vss ca E cb Write the two equations obtained in matrix form:
E1s Vss ca ca E Vss E1s cb cb This is now an eigenvalue equation and the two solutions are:
E E1s Vss
ca 1 1 ca 1 1 cb 2 1 cb 2 1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
4 Bonding and Anti-Bonding Orbitals V r For the lower energy solution we have: EA E EB E1s Vss B 1 r r d xˆ r dxˆ B 2 1s 1s -d 0 d x B r This is called the “Bonding molecular orbital”
For the higher energy solution we have: -d 0 d x r EA E1s Vss A
1 r r d xˆ r dxˆ A 2 1s 1s -d 0 d x This is called the “Anti-bonding molecular orbital”
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: Energy Level Splitting and the Energy Matrix Element
V r V r Anti-bonding 2V 1s atomic ss Energy levels Bonding energy level of the molecule
0 x -d 0 d x
Energy level diagram going from two isolated atoms to the molecule:
The total energy is 1: EA 2 : E1s lowered when 2Vss Hydrogen atoms form 1: EB a Hydrogen molecule
The two 1s orbitals on each Hydrogen atom combine to generate two molecular orbitals – the bonding orbital and the anti-bonding orbital – with energy splitting related to the energy matrix element: ˆ 1s r dxˆ H 1s r dxˆ Vss
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
5 Atomic Orbitals
• Wavefunction amplitudes of the atomic s s and p orbitals in the angular directions are plotted
• The s-orbital is spherically symmetric
• The p-orbitals have +ve and –ve lobes and are oriented along x-axis, y-axis, and z-axis
pz px py
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Orbitals and Bonding There are two main types of co-valent bonds: sigma bonds (or -bonds) and pi-bonds (or -bonds) (1) Sigma bonds (or -bonds): s-s -bond ss (Example: Hydrogen molecule, semiconductors) ˆ s r r1 H s r r2 Vss r1 r2 p-p -bond pp (Example: Semiconductors) ˆ p r r1 H p r r2 Vpp r1 r2 s-p -bond p s (Example: Semiconductors) ˆ p r r1 H s r r2 Vsp r1 r2 p s s-p -bond (Example: Semiconductors) ˆ p r r1 H s r r2 Vsp r1 r2 ECE 407 – Spring 2009 – Farhan Rana – Cornell University
6 Orbitals and Bonding What about this situation?
p r r Hˆ r r 0 s p 1 s 2 The Hamiltonian is up-down symmetric The s-orbital is up-down symmetric The p-orbital is up-down anti-symmetric The matrix element is zero! No bonding possible r1 r2 What about this situation? What should be the matrix element?
p p s p s = sin + cos s r1 r2 r r 1 2 r1 r2
ˆ p r r1 H s r r2 0 . sin Vsp . cos
Vsp cos
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Orbitals and Bonding (2) Pi bonds (or -bonds): p-p -bond (Example: graphene, carbon nanotubes, conjugated p p conducting molecules) ˆ p r r1 H p r r2 Vpp
r1 r2 What about this situation? What should be the matrix element?
p p p p p p sin + cos
r r r r r1 r2 1 2 1 2 ˆ p r r1 H p r r2 Vpp . sin 0 . cos
Vpp sin
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
7 LCAO: Methane Molecule
A methane molecule consists of one carbon atom covalently bonded to 4 hydrogen atoms in a tetrahedral configuration: H Carbon: Atomic number: 6 Electron Configuration: 1s2 2s2 2p2
Number of electrons in the outermost shell: 4 C H H Hydrogen: Atomic number: 1 H Electron Configuration: 1s1
Number of electrons in the outermost shell: 1
Methane: H The four electrons from the outermost shell of the carbon atom and the four electrons from the four H C H hydrogen atoms take part in covalent bonding H
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: Methane Molecule • The carbon atom is sitting at the origin r1 r2 • The position vectors of the four hydrogen atoms are:
d d d r1 1,1,1 r2 1,1,1 r3 1,1,1 r 3 3 3 r4 3 d r 1,1,1 Tetrahedral configuration 2 3
• The carbon atom has one 2s orbital and three 2p orbitals • Each hydrogen atom has one 1s orbital • One can write the solution for the methane molecule as a linear combination of all available orbitals 4 r c j 1s r r j c5 2s r c6 2px r c7 2py r c8 2pz r j 1
But we will pursue a different, and simpler, approach
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
8 LCAO: Methane Molecule – sp3 Hybridization For the carbon atom, do a change of basis and define 4 new sp3 atomic orbitals from the 4 existing (one 2s and three 2p) atomic orbitals r 1 1 r r r r r 1 2 2s 2px 2py 2pz 1 r r r r r 2 2 2s 2px 2py 2pz 1 r r r r r 3 2 2s 2px 2py 2pz 1 r r r r r 3 2 2s 2px 2py 2pz
r 2r 3r 4
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: Methane Molecule – sp3 Hybridization • Each carbon sp3 orbital forms a s-sp3 -bond with the 1s orbital of the hydrogen atoms towards which it is pointing: H s-sp3 Average energy of the sp3 orbital: -bond
E 3E r Hˆ r E 2s 2p j m sp3 j m 4 j m H C Important matrix element for the s-sp3 bond: H ˆ H 1s r r1 H 1r ˆ 1 1s r r1 H 2s r 2px r 2py r 2pz r 2 z 1 1V 1V 1V 1s sp sp sp 2pz Vss 2 2 3 2 3 2 3 x r1 Vss 3Vsp 2px 2 2py y
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
9 LCAO: Methane Molecule – sp3 Hybridization • Matrix elements for all s-sp3 -bonds are the same
H s-sp3 Important matrix elements: -bond
V 3V r r Hˆ r ss sp 1s 1 1 2 H C V 3V r r Hˆ r ss sp H 1s 2 2 2 H V 3V r r Hˆ r ss sp 1s 3 3 2 V 3V r r Hˆ r ss sp 1s 4 4 2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: Methane Molecule – sp3 Hybridization Write the solution for the methane molecule using the sp3 orbitals of the carbon atom: 4 r c j 1s r r j c5 1r c6 2r c7 3r c8 4 r j 1 And plug it into the Schrodinger equation: H ˆ r E r
To a first approximation, 1s orbital on each Hydrogen r r atom has a large matrix element only with the sp3 1 2 orbital pointing towards it, so instead of one giant 8x8 matrix equation one gets a set of four 2x2 matrix equations: r r4 3
E1s c1 c1 E1s c2 c2 E E Esp3 c5 c5 Esp3 c6 c6
E1s c3 c3 E1s c4 c4 E E Esp3 c7 c7 Esp3 c8 c8 V 3V ss sp r r Hˆ r Where: 1s 1 1 2 ECE 407 – Spring 2009 – Farhan Rana – Cornell University
10 LCAO: Methane Molecule – sp3 Hybridization
E1s c1 c1 E Esp3 c5 c5 r1 r2 The energy eigenvalues of any one of these four 2x2 equations are: r r4 3 2 E1s Esp3 E1s Esp3 2 EA 2 2 2 E1s Esp3 E1s Esp3 2 EB 2 2
• The higher energy corresponds to the anti-bonding state and the lower energy corresponds to the bonding state.
• In this case, the bonding and anti-bonding states are made up of a linear combination of the hydrogen 1s state and one of the carbon sp3 state.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: Methane Molecule – sp3 Hybridization
Energy level diagram going from isolated hydrogen and carbon atoms to the orbitals of the methane molecule is:
3 : E 2p 4 : Esp3
1: E2s sp3 orbitals 4 : EA Atomic of C orbitals of C
4 : EB
4 : E1s H Atomic orbitals of H C H H H
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
11 LCAO: Boron Trifluoride Molecule
A BF3 molecule consists of one boron atom covalently bonded to 3 fluorine atoms and all atoms lie in the same plane y
Boron: x 120o F Atomic number: 5 F B Electron Configuration: 1s2 2s2 2p1
Number of electrons in the outermost shell: 3 F Fluorine: Atomic number: 9 Electron Configuration: 1s2 2s2 2p5
Number of electrons in the outermost shell: 7
BF3: 120o F All three electrons from the outermost shell of the boron F B atom and only one of the 7 electrons from each fluorine atom take part in covalent bonding F
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: BF3 Molecule – sp2 Hybridization For the boron atom, do a change of basis and define 3 new sp2 atomic orbitals from the 3 existing (one 2s and two 2p) atomic orbitals
1 1 1 1r 2s r 2px r 2py r 3 6 2 y r1 1 1 1 120o 2r 2s r 2px r 2py r 3 6 2 r3 120o x 1 2 3r 2s r 2px r 120o 3 3 r2
1r 2r 3r
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
12 LCAO: BF3 Molecule – sp2 Hybridization • In the fluorine atom, the 2s orbital and the two 2p orbitals that are perpendicular to the line joining the fluorine atom to the boron atom are all filled with 2 electrons each and do not participate in bonding
• The remaining p-orbital in fluorine that is pointing towards the boron atom, and contains one electron, forms a -bond with the sp2 orbital pointing towards it
Average energy of the sp2 orbital:
E 2E p-sp2 r Hˆ r E 2s 2p F j m sp2 j m 3 j m -bond 120o
Important matrix elements: V 2V B r r Hˆ r sp pp 2p 1 1 3 F V 2V r r Hˆ r sp pp 2p 2 2 F 3 V 2V r r Hˆ r sp pp 2p 3 3 3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: BF3 Molecule – sp2 Hybridization
Write the solution for the BF3 molecule using the sp2 orbitals of the boron atom: 3 r c j 2p r r j c4 1r c5 2r c6 3 r j 1 ˆ y And plug it into the Schrodinger equation: H r E r r o 1 120 To a first approximation, the 2p orbital on each r3 120o x fluorine atom has a large matrix element only with the 120o sp2 orbital pointing towards it, so instead of one giant r2 6x6 matrix equation one gets a set of three 2x2 matrix equations:
E2p c1 c1 E2p c2 c2 E E Esp2 c4 c4 Esp2 c5 c5
E2p c c Where: 3 E 3 E c c sp2 6 6 V 2V sp pp r r Hˆ r 2p 1 1 3
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
13 LCAO: BF3 Molecule – sp2 Hybridization
E2p c1 c1 E Esp2 c4 c4 y r The energy eigenvalues of any one of these three 2x2 o 1 equations are: 120 r3 120o x 2 E E E E 2p sp2 2p sp2 2 120o EA r2 2 2
2 E2p Esp2 E2p Esp2 2 EB 2 2
• The higher energy corresponds to the anti-bonding state and the lower energy corresponds to the bonding state.
• In this case, the bonding and anti-bonding states are made up of a linear combination of the fluorine 2p state and one of the carbon sp2 state.
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
LCAO: BF3 Molecule – sp2 Hybridization
Energy level diagram going from isolated fluorine and carbon atoms to the orbitals of the BF3 molecule is:
2 : E 2p 3 : Esp2
1: E2s sp2 orbitals 3 : EA Atomic of B orbitals of B
3 : EB
3 : E2p o F Atomic 120 orbitals F B of F
F
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
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