Molecular Orbitals in Inorganic Chemistry

Dr. P. Huntp.hunt@imperial.ac.uk

Rm 167 (Chemistry)http://www.ch.ic.ac.uk/hunt/

Outline

LCAO theory

orbital size matters

where to put the FO energy levels

the splitting energy

rules for combining FOs

origin of the splitting “rules”: making connections with QM

Simple MOs

forming simple MOs ◆ combine the orbitals of each fragment ◆ add once “as is”, add once with phase inverted ◆ draw a “cartoon” and write an equation

+

+

ψ 1saψ 1sb

(−ψ 1sb)ψ 1sa

ψσ =12(ψ 1sa

+ψ 1sb)

ψσ * =12(ψ 1sa

−ψ 1sb)

Real orbitals:

Fig. 1

See slide in your notes from your

Chemical Bonding course

Fig.2

Generalise: linear combination of atomic orbitals ◆ Γ = symmetry label of MO ◆ N = normalisation coefficient ◆ Ci tells us how large the contribution from each

atomic orbital Ψi will be

LCAOs

ψΓ = N(c1ψ 1 + c2ψ 2 +!cnψ n ) = N ciψ i∑

to determine N and Ci

Solve Schrödinger Equation!

solved in your Chemical Bonding

course!

LCAOs

theory: linear combination of atomic orbitals

so don’t combine AOs on different centers!

often we work backwards! ◆ from computed MO to understand the bonding

real MOLCAO don't combinecomponents

Important!

Fig.3

LCAOsC’s are represented in the size of the AO contributions

A

B

ψσ∗

ψσ

s

s

A B

Bonding MO: lower E FO

makes larger contribution

Antibonding MO: higher E FO makes larger contribution

Important!

Fig.4

from your 1st year Molecular

Structure course!

LCAOs

C’s are represented in the size of the AO contributions

A

B

ψσ∗

ψσ

s

s

A B

larger energy gab between the FOs = more ionic the bond

= greater the disparity in contributions

more ionic

more covalent

Fig.4

FO Positioning

MO diagrams are built by combining fragment orbitals ◆ fragment orbitals can be simple atomic orbitals ◆ or they can be more complex fragments (covered later)

where to put atomic fragment orbital energy levels? ◆ relative electronegativity ordering of first row (expected to know) ◆ transition metals and main group elements tend to be electropositive ◆ reference is H 1sAO

H Li Be B C N O F

2.20 0.98 1.57 2.04 2.55 3.04 3.44 3.98

Na Mg Al Si P S Cl

0.93 1.31 1.61 1.91 2.19 2.58 2.19

Pauling electronegativity

H 1s AO

BCNOF

E 2p AOs

Important!valence pAO

relative to valence H1sAO

Fig.5

FO Positioning

atomic orbitals s-p gap increases along the periodic table

2s AOs only interact for adjacent atoms

2p AO always interact Important!

Problematic

last year: some problems this year: a few examples!

Fig.6

Orbital Interactions: CN¯

C NC N

2s

2p

2s

2p

H 1sAO reference level

See the on-line tutorial!

Important!

Fig.7

Orbital Interactions: CN¯

C NC N

2s

2p

2s

2p

Nitrogen is more electronegative than

Carbon: N2p lie below C2p

Fig.7

Orbital Interactions: CN¯

C NC N

N is further along PT than C N has larger sp gap C has smaller sp gap2s

2p

2s

2p

Fig.7

pAO interactions not drawn in because we are

focusing on sAOs

Orbital Interactions: CN¯

C NC N

C is next to N on the PT so the 2s AOs interact2s

2p

2s

2p

Fig.7

additional example for CO in

your notes

Self-study document on web-site • links to revision from last year • includes more complex details from this year • emphasis is on interpretation as well as

construction

In-Class Activity

Where would you put the FOs for the MO diagram of MgCl2?

◆ once you have completed the quiz then:

Draw the fragments for the MO diagram

socrative quiz!

WHZ9KBWC3

In-Class Activity

Where would you put the FOs for the MO diagram of MgCl2?

What are the valence orbitals of Mg? ◆ 3s and 3p

What relative energy will the valence orbitals of Mg have and why? ◆ Mg is a main group metal and electropositive the orbitals will be high in energy ◆ Mg lies to the left of the periodic table and will have a small sp gap

more information in the explanation

in socrative

socrative quiz!

WHZ9KBWC3

In-Class Activity

Where would you put the FOs for the MO diagram of MgCl2?

When we form the Cl—•—Cl fragment how large will the interaction energy of the MOs be? ◆ the fragment orbitals are far apart and overlap will be small therefor the energy of

stabilisation and destabilisation will be small ◆ the fragment orbital diagram will look like that of O2 because Cl is electronegative

and has a large sp gap

more information in the explanation

in socrative

socrative quiz!

WHZ9KBWC3

In-Class Activity

Where would you put the FOs for a MO diagram of MgCl2?

x

z

y

3s

3p

Mg

πu

πg

MgCl ClCl Cl

πu

2e 14e16e

Cl ClMg

σ g+

σu+

σ g+

σu+

σ g+

σu+

combine “Cl2” fragment with Mg

Mg is a main group metal: very electropositive

Cl is very electronegative

FO reference Mg

FO reference Cl

In-Class Activity

Where would you put the FOs for a MO diagram of MgCl2?

x

z

y

3s

3p

Mg

πu

πg

MgCl ClCl Cl

πu

2e 14e16e

Cl ClMg

σ g+

σu+

σ g+

σu+

σ g+

σu+

combine “Cl2” fragment with Mg

Mg is a main group metal: very electropositive

Cl is very electronegative

in the Cl-•-Cl fragments the Cl are far apart

overlap is very small thus energy separation is small know the MOs for a diatomic

small splitting energy

In-Class Activity

Where would you put the FOs for a MO diagram of MgCl2?

x

z

y

3s

3p

Mg

πu

πg

MgCl ClCl Cl

πu

2e 14e16e

Cl ClMg

σ g+

σu+

σ g+

σu+

σ g+

σu+

combine “Cl2” fragment with Mg

Mg is a main group metal: very electropositive

Cl is very electronegative

in the Cl-•-Cl fragments the Cl are far apart

overlap is very small thus energy separation is small know the MOs for a diatomic

large sp gap

small sp gap

Mg is left of PT: small sp gap

Cl is right of PT: large sp gap

Splitting Energy

How far apart are the MOs placed??

Splitting energy

stabilisation energy

destabilisation energy

E=ΔEs+ΔEd

εa

εb

φa

φb

E+

E-

splitting energy E=ΔEs+ΔEd

Δε

energy difference between FOs

ΔEs

ΔEd

ψ+

ψ-

Fig.8

◆ destabilisation is larger than stabilisation energy

Splitting Energy

energy difference between FO:

orbital coupling:

extent of orbital overlap:

Δε = εi − ε j

Hij = ψ i H ψ j

Sij = ψ i ψ j

Depends on 3 quantities

(direct orbital overlap)

(molecule mediated orbital overlap)

Important!

FO Energy Difference

degenerate orbitals have largest splitting

sliding scale

as FO shift apart splitting energy is reduced

point is reached at which there is no interaction ◆ form ionic compound

degenerate ionic system:electron transfer

largesplitting smaller

splitting

differentelectronegativity

energy levels are toofar appart to interactFig.10

Splitting Energy

Go back to the Schrödinger equation ◆ solving gives the energy of MOs

ψ i H ψ j = ψ iHψ j∫ dτ

interaction of orbital(i) with orbital(j) modified by the local

environment (H)

Sab

Hab

direct orbital overlap

"overlap" mediated by the molecule

Fig.9

Orbital Overlap

Sab overlap depends on ◆ relative location ◆ orientation ◆ diffusivity

better overlap

orbitals orientated toward each other overlap better

sigma orbitals overlap better than pi orbitals overlap: sσ > pσ > spσ > pπ

Fig.11

(b)

strong overlap weak overlap

negative and positive overlap cancel out

S=0

reduced overlapS<<1

larger overlap smaller overlap

A B A Br r'r < r'

ΔE decreases

(a)

good overlap larger splitting

poor overlap smaller splitting

diffuse orbitals have poor overlap

Making Connections with QM

link qualitative arguments to mathematics

solved in your Chemical Bonding course

solve Schrödinger equation

Eψ = Hψ

H = Tn +Te +Vee +Ven +Vnnψ = cn∑ φn

ψ = caφa ± cbφbψ = c(φa ±φb )

expand our MO in terms of the FOs

See slides in your notes from your

Chemical Bonding course

Fig.12

Reminder from QM:determine the expression for c

require ψ *ψ∫ dτ = ψ ψ = 1

ψ = c φa ±φb( )ψ ψ = c φa ±φb( ) c φa ±φb( )

ψ ψ = c2 φa φa=1

!"#+ φb φb

=1!"#

± 2 φa φb=Sab!"#

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥

1= ψ ψ = c2 2(1± Sab )

c2 = 12(1± Sab )

∴c = 12(1± Sab )

ψ = 12(1± Sab )

φa ±φb( )

AOs are normalised as well!

important for later

Reminder from QM:rearrange the Schrödinger equation

cannot divide by wavefunction! in QM we premultiply and

integrateHψ = Eψψ*Hψ =ψ*Eψ

ψ*Hψ∫ dτ = E ψ*ψ∫ dτ

because E is a number

ψ*Hψ∫ dτ = ψ H ψ = E

require ψ*ψ∫ dτ = ψ ψ =1

E = ψ H ψ

Making Connections with QM

ready to solve! two important pieces of information

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QM

ready to solve!

put in wavefunction

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QM

ready to solve!

expand out brackets

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QM

ready to solve!

simplify the notation

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QM

ready to solve!

replace c

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QMready to solve!

DONE! (not quite!!)

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Making Connections with QMready to solve!

E = ψ H ψ and ψ = c φa ±φb( )ψ+ = c+ φa +φb( )ψ H ψ = c+

2 φa +φb H φa +φb

ψ H ψ = c+2 φa H φa + φa H φb + φb H φa + φb H φb#$ %&

φa H φa = Haa assume Hab = Hba

ψ H ψ = c+2 Haa +2Hab +Hbb[ ]

E+ = ψ H ψ

E+ =Haa +2Hab +Hbb

2(1+Sab )E− =

Haa −2Hab +Hbb

2(1−Sab )

Haa = εa and Hbb = εb

εa εaεb εb

Making Connections with QMset up model: diatomic ◆ degenerate FOs εa=εb=ε ◆ no overlap Sab =0 ◆ non-zero coupling Hab (Hab is negative!)

ε ε

Hab

energy

E+ = ε +Hab

E− =ε −Hab

ψ+ =12φa +φb( )

ψ− =12φa −φb( )

Hab

Fig.13

insert into the equations

E+ =εa + εb + 2Hab

2(1+ Sab )and E− =

εa + εb − 2Hab

2(1− Sab )

Making Connections with QMset up model: diatomic ◆ degenerate FOs εa=εb=ε ◆ no overlap Sab =0 ◆ non-zero coupling Hab

degenerate orbitals: splitting depends on Hab

Important!

ε ε

Hab

energy

E+ = ε +Hab

E− =ε −Hab

ψ+ =12φa +φb( )

ψ− =12φa −φb( )

Hab

Fig.13

splitting is large!

Making Connections with QMmodify model ◆ degenerate FOs εa=εb=ε ◆ overlap allowed Sab ≠ 0 ◆ non-zero coupling Hab

Fig.14

ε ε

Hab

HabE+ =

ε +Hab

1+Sab

E− =ε −Hab

1−Sab

insert into the equations

E+ =εa + εb + 2Hab

2(1+ Sab )and E− =

εa + εb − 2Hab

2(1− Sab )

See slides in your notes from your

Chemical Bonding course

Fig.15

test! ◆ Sab=0.2 ◆ (ε-Hab)*(1/0.8)=(ε-Hab)*1.25 ◆ =increase in destabilisation ◆ (ε-Hab)*(1/1.2)=(ε-Hab)*0.83 ◆ =decrease in stabilsation

general rule: antibonding MOs are destabilised by more than bonding MOs are stabilised

Justified!

Making Connections with QM

E+ =ε + Hab

1+ SabE− =

ε − Hab

1− Sab

ε ε

stabilisation energy

destabilisation energy

Important!

Fig.14

Making Connections with QM

modify model ◆ εa ≠ εb ◆ no overlap Sab =0

for non-degenerate orbitals the splitting energy depends on:

Fig.16

εa

εb

E+ = εa −Hab

2

Δε

E− = εb +Hab

2

Δεcacb

= ΔεHab

Δε

See slides in your notes from your

Chemical Bonding course

Fig.17

Course cross-over material: I expect you to be able to carry out the derivations!

Hab( )2Δε

Important!

general rule: splitting energy decreases as the energy between the FO increases

non-degenerate orbitals:

splitting energy depends on

Hab( )2Δε

degenerate ionic system:electron transfer

largesplitting smaller

splitting

differentelectronegativity

energy levels are toofar appart to interact

Hab = ψ a H ψ b

degenerate orbitals:

splitting energy depends on

Fig.10

Justified!

Making Connections with QM

be able to explain LCAO, give the equation defining all the terms and illustrate LCAOs on a diagram

be able to explain and predict the relative size of FO contributions to a MO

be able to rationalise the position of FO energy levels

be able to define and illustrate the splitting energy

be able to discuss, employing equations, diagrams and examples Δε, Sab and Hab

be able to make a connection with QM including being able to derive the relevant equations

be able to compare and contrast the equations for E+ and E- for different levels of approximation

be able to employ this knowledge in forming MO diagrams

Key Points

Finally

See my web-site ✦notes AND slides ✦ link to panopto when it becomes available ✦optional background support for beginners ✦optional material to take you a little further ✦ links to interesting people and web-sites ✦ links to relevant research papers on MOs ✦model answers!!

http://www.huntresearchgroup.org.uk/