Molecular Orbitals in Inorganic Chemistry€¦ · Simple MOs forming simple MOs combine the...
Transcript of Molecular Orbitals in Inorganic Chemistry€¦ · Simple MOs forming simple MOs combine the...
Molecular Orbitals in Inorganic Chemistry
Dr. P. [email protected]
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/