Molecular Orbitals in Inorganic Chemistry. Dr. P. Hunt Rm 167 (Chemistry)

Transcription

1 Molecular Orbitals in Inorganic Chemistry Dr. P. Hunt Rm 167 (Chemistry

2 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

3 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 1sa + ψ 1sb ψ σ = 1 2 (ψ 1s a + ψ 1sb ψ 1sa + ( ψ 1sb ψ σ * = 1 2 (ψ 1s a ψ 1sb Fig. 1 See slide in your notes from your Chemical Bonding course Fig.2 Real orbitals:

4 LCAOs Generalise: linear combination of atomic orbitals ψ Γ = N(c 1 ψ 1 + c 2 ψ 2 +!c n ψ n = N Γ = symmetry label of MO N = normalisation coefficient Ci tells us how large the contribution from each atomic orbital Ψi will be c i ψ i Solve Schrödinger Equation! to determine N and C i solved in your Chemical Bonding course!

5 LCAOs theory: linear combination of atomic orbitals so don t combine AOs on different centers! Important! LCAO don't combine components real MO Fig.3 often we work backwards! from computed MO to understand the bonding

6 LCAOs C s are represented in the size of the AO contributions Antibonding MO: higher E FO makes larger contribution ψ σ from your 1st year Molecular Structure course! A s Important! s B ψ σ Fig.4 A B Bonding MO: lower E FO makes larger contribution

7 LCAOs C s are represented in the size of the AO contributions ψ σ more ionic more covalent A s larger energy gab between the FOs = more ionic the bond = greater the disparity in contributions ψ σ s B A B Fig.4

8 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 Na Mg Al Si P S Cl Pauling electronegativity Important! Fig.5 H 1s AO B C N O F E 2p AOs valence pao relative to valence H1sAO

9 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

10 Orbital Interactions: CN H 1sAO reference level 2p 2s 2p See the on-line tutorial! Important! 2s C C N N Fig.7

11 Orbital Interactions: CN 2p Nitrogen is more electronegative than Carbon: N2p lie below C2p 2p 2s 2s C C N N Fig.7

12 Orbital Interactions: CN 2p 2s 2p N is further along PT than C N has larger sp gap C has smaller sp gap pao interactions not drawn in because we are focusing on saos 2s C C N N Fig.7

13 Orbital Interactions: CN 2p 2p 2s C is next to N on the PT so the 2s AOs interact 2s additional example for CO in your notes Fig.7 C C N N 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

14 In-Class Activity Where would you put the FOs for the MO diagram of MgCl2? socrative quiz! WHZ9KBWC3 once you have completed the quiz then: Draw the fragments for the MO diagram

15 In-Class Activity Where would you put the FOs for the MO diagram of MgCl2? socrative quiz! WHZ9KBWC3 more information in the explanation in socrative 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

16 In-Class Activity Where would you put the FOs for the MO diagram of MgCl2? socrative quiz! WHZ9KBWC3 more information in the explanation in socrative 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

17 In-Class Activity Where would you put the FOs for a MO diagram of MgCl2? combine Cl 2 fragment with Mg σ u + 3p π u FO reference Mg Mg is a main group metal: very electropositive Cl is very electronegative σ g + 3s FO reference Cl + σ u π g π u + σ g y Cl Mg Cl z x σ u + σ g + Cl Mg Cl Mg 2e 16e Cl 14e Cl

18 In-Class Activity Where would you put the FOs for a MO diagram of MgCl2? combine Cl 2 fragment with Mg σ u + 3p π u Mg is a main group metal: very electropositive Cl is very electronegative σ g + 3s small splitting energy + σ u π g π u + σ g 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 Cl y Mg Cl z x σ u + σ g + Cl Mg Cl Mg 2e 16e Cl 14e Cl

19 In-Class Activity Where would you put the FOs for a MO diagram of MgCl2? combine Cl 2 fragment with Mg Mg is a main group metal: very electropositive Cl is very electronegative σ u + 3p π u + σ g 3s small sp gap + σ u π g π u + σ g 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 Cl y Mg Cl z large sp gap x σ u + Mg is left of PT: small sp gap σ g + Cl is right of PT: large sp gap Mg 2e Cl Mg 16e Cl Cl 14e Cl

20 Splitting Energy How far apart are the MOs placed?? Splitting energy destabilisation is larger than stabilisation energy E=ΔEs+ΔEd ψ - ΔE d E- ε b destabilisation energy φ b Δε φ a ε a stabilisation energy ψ + ΔE s E+ energy difference between FOs splitting energy E=ΔE s +ΔE d Fig.8

21 Splitting Energy Depends on 3 quantities Important! energy difference between FO: Δε = ε i ε j orbital coupling: H ij = ψ i H ψ j (molecule mediated orbital overlap extent of orbital overlap: S ij = ψ i ψ j (direct orbital overlap

22 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 large splitting smaller splitting ionic system: electron transfer Fig.10 different electronegativity energy levels are too far appart to interact

23 Splitting Energy Go back to the Schrödinger equation solving gives the energy of MOs ψ i H ψ j = ψ i Hψ j dτ "overlap" mediated by the molecule H ab interaction of orbital(i with orbital(j modified by the local environment (H S ab direct orbital overlap Fig.9

24 Orbital Overlap Sab overlap depends on relative location orientation diffusivity good overlap larger splitting (a ΔE decreases poor overlap smaller splitting A B A B r r < r' larger overlap smaller overlap r' Fig.11 (b sigma orbitals overlap better than pi orbitals overlap: s σ > p σ > sp σ > p π strong overlap weak overlap orbitals orientated toward each other overlap better reduced overlap S<<1 diffuse orbitals have poor overlap better overlap negative and positive overlap cancel out S=0

25 Making Connections with QM link qualitative arguments to mathematics solved in your Chemical Bonding course solve Schrödinger equation Eψ = Hψ ψ = c n φ n ψ = c a φ a ± c b φ b ψ = c(φ a ±φ b expand our MO in terms of the FOs H = T n +T e +V ee +V en +V nn See slides in your notes from your Chemical Bonding course Fig.12

26 Reminder from QM: determine the expression for c important for later require ψ = c φ a ± φ b ψ * ψ dτ = ψ ψ = 1 ( ( c( φ a ± φ b ψ ψ = c φ a ± φ b ψ ψ = c 2 φ a φ!"# + φ a b φ!"# ± 2 φ b a φ b!"# =1 =1 =S ab 1 = ψ ψ = c 2 2(1± S ab c 2 = ψ = 1 2(1± S ab 1 c = ( 2(1± S ab φ a ± φ b 1 2(1± S ab AOs are normalised as well!

27 Reminder from QM: rearrange the Schrödinger equation Hψ = Eψ ψ * Hψ = ψ * Eψ ψ * Hψdτ = E ψ * ψdτ because E is a number ψ * Hψdτ = ψ H ψ = E require E = ψ H ψ ψ * ψdτ = ψ ψ =1 cannot divide by wavefunction! in QM we premultiply and integrate

28 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ( two important pieces of information ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

29 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ( put in wavefunction ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

30 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 ( expand out brackets #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

31 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 ( #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] simplify the notation E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

32 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 ( #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] replace c E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

33 Making Connections with QM ready to solve! E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 ( #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] DONE! (not quite!! E + = H aa + 2H ab + H bb 2(1+ S ab E = H aa 2H ab + H bb 2(1 S ab

34 Making Connections with QM ready to solve! ( E = ψ H ψ and ψ = c φ a ±φ b ψ + = c + ( φ a +φ b ψ H ψ = c + 2 φ a +φ b +φ b ψ H ψ = c + 2 #$ φ a + φ a H φ b + φ b + φ b H φ b % & φ a = H aa assume H ab = H ba ψ H ψ = c + 2 E + = ψ H ψ [ H aa + 2H ab + H bb ] H aa = ε a and H bb = ε b ε a ε b E + = H aa + 2H ab + H bb 2(1+ S ab ε a ε b E = H aa 2H ab + H bb 2(1 S ab

35 Making Connections with QM set up model: diatomic degenerate FOs εa=εb=ε no overlap Sab =0 non-zero coupling Hab (Hab is negative! insert into the equations E + = ε a + ε b + 2H ab 2(1+ S ab and E = ε a + ε b 2H ab 2(1 S ab energy ψ = 1 ( 2 φ a φ b E = ε H ab H ab ε ε H ab E + = ε + H ab ψ + = 1 ( 2 φ +φ a b Fig.13

36 Making Connections with QM set up model: diatomic degenerate FOs εa=εb=ε no overlap Sab =0 non-zero coupling Hab energy ψ = 1 ( 2 φ a φ b E = ε H ab Important! ε H ab ε degenerate orbitals: splitting depends on H ab H ab splitting is large! E + = ε + H ab ψ + = 1 ( 2 φ +φ a b Fig.13

37 Making Connections with QM modify model degenerate FOs εa=εb=ε overlap allowed Sab 0 non-zero coupling Hab insert into the equations E + = ε a + ε b + 2H ab 2(1+ S ab and E = ε a + ε b 2H ab 2(1 S ab E = ε H ab 1 S ab H ab ε ε H ab E+ = ε + H ab 1+ S ab Fig.14 See slides in your notes from your Chemical Bonding course Fig.15

38 Making Connections with QM E + = ε + H ab 1+ S ab E = ε H ab 1 S ab 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 destabilisation energy ε stabilisation energy Important! ε Fig.14 general rule: antibonding MOs are destabilised by more than bonding MOs are stabilised Justified!

39 Making Connections with QM modify model εa εb no overlap Sab =0 See slides in your notes from your Chemical Bonding course Fig.17 E = ε b + H 2 ab Δε for non-degenerate orbitals the splitting energy depends on: c a = Δε c b H ab Δε ε b ε a Important! ( H 2 ab Δε 2 H E + = ε ab a Δε Fig.16 Course cross-over material: I expect you to be able to carry out the derivations!

40 Making Connections with QM general rule: splitting energy decreases as the energy between the FO increases Justified! degenerate orbitals: splitting energy depends on non-degenerate orbitals: splitting energy depends on ( H 2 ab Δε H ab = ψ a H ψ b degenerate large splitting smaller splitting ionic system: electron transfer Fig.10 different electronegativity energy levels are too far appart to interact

41 Key Points 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

42 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!!