An Introduction to Molecular Orbital Theory

Objectives of the course

• Wave mechanics / Atomic orbitals (AOs) – The basis for rejjgecting classical mechanics ( the Bohr Model) in treating An introduction to electrons – Wave mechanics and the Schrödinger equation Molecular Orbital Theory – Representatifion of atomi ibilc orbitals as wave f fiunctions – Electron densities and radial distribution functions – Understanding the effects of shielding and penetration on AO energies

6 Lecture Course • Bonding – Review VSEPR and Hybridisation Prof G. W. Watson – Linear combination of molecular orbitals (LCAO), bonding / antibonding Lloyd d Ins titut e 2 .36 – Lab elli ng of mol ecul ar orbi tal s (MO s) ( σ, π and)d g, u) [email protected] – Homonuclear diatomic MO diagrams – mixing of different AO’s

– More complex molecules (CO , H 2O)O ….) – MO diagrams for Inorganic complexes

Lecture schedule Literature

Lecture 1 Revision of Bohr model of atoms • Book Sources: all titles listed here are available in the Hamilton Library

Lecture 2 Schrödinger equation, atomic wavefunctions and radial – 1. Chemical Bonding, M. J. Winter (Oxford Chemistry primer 15) distribution functions of s orbitals Oxford Science Publications ISBN 0 198556942 – condensed text, excellent diagrams Lecture 3 More complex wavefunctions and radial distribution functions and electron shielding – 2. Basic Inorggy(y),,anic Chemistry (Wiley) F.A.Cotton, G. Wilkinson, P. L. Gaus – comprehensive text, very detailed on aufbau principle Lecture 4 Lewis bonding, Hybridisation, and molecular orbitals – 3. I norgani c Ch emi stry (P renti ce H all) C . H ousecrof t, A . G . Sh arpe – Lecture 5 Labelling MO’s. 1st row homonuclear diatomics comprehensive text with very accessible language. CD contains interactive energy diagrams

Lecture 6 MO approach to more complex molecules and CO bonding in transition metals complexes – Additional sources: h//ihttp://winter.group.sh hfk/bi/ef.ac.uk/orbitron/ - gallery o f AOs an d MOs

3 4 Tutorials

• Expectation – Tutorials are to gggpo through problems that students are havin g with the course An introduction to

– Tutorilials are NOT f or th e l ecturer to gi ive you th e answers to th e Molecular Orbital Theory questions – or to give you another lecture.

– All student must BEFORE the tutorial • Look at the notes for the course and try to understand them Lecture 1 The Bohr Model • Attempt the questions set – and hence find out what you can not do! • Bring a list of questions relating to aspects of the course which you could not understand (either from looking at the notes or attempting the Prof G. W. Watson questions) Lloyd Institute 2.05 [email protected] • It is a waste of both the lecturers and students time if the tutorial to ends up being a lecture covering questions.

Bohr model of the atom (1913) Adsorption / Emission spectra for Hydrogen http://www.youtube.com/watch?v=R7OKPaKr5QM

Johann Balmer (1885) measured line spectra for hydrogen Assumptions 364.6 nm (uv), 410.2 nm (uv), 434.1 nm (violet), 486.1 nm (blue), and 1) Rutherford (1912) model of the atom (Planetary model with central 656.3 nm (red). nucleus + electrons in orbit)

2) Planck (1901) , Einstein (1905) – the energy electromagnetic waves is quantised into packets called photons (particle like property).

E = hν Balmer discovered these lines occur in a series - both absorption and emission - where ℜ is the Rydberg constant (3.29 ×1015 Hz) Velocity, c Wavelength, λ ic rr ⎛ 1 1 ⎞ v = ℜ⎜ − ⎟ ⎜ 2 2 ⎟ etic field n n ing elect 1 2 tt

⎝ ⎠ nn

Balmer series n1=2 and n2=n1+1, n1+2, n1+3 ….. /mag Fluctua OtOtherher seseriesries forfor n1=11 ((LymanLyman – UV), n1=3 ((PaschenPaschen – IRIR)) etc. An stationary observer counts } i.e. frequency = v Hz , cycles/sec, ν waves passing per second Electrons must have specific energies – no model of the atom could explain this sec-1 7 8 Bohr model of the atom Bohr model of the atom

Speed of electromagnetic waves (c) is constant (ν and λ vary) 3) Electron assumed to travel in circular orbits.

c = ν λ, ν = c / λ, E = h ν, E = h c / λ 4) Apply quantisation to orbits - only orbits allowed have quantised angular momentum (comes from observation of spectra)

As frequency increases, wavelength decreases. Given λ Æ ν ⎛ h ⎞ mvr = n⎜ ⎟ ⎝ 2π ⎠ ege.g. radiowaves: λ = 01m0.1 m X-rays: λ = 1x101 x 10-12 m ν = 3 x 109 Hz ν = 3 x 1020 Hz 5) Classical electrodynamical theory rejected (charged particles E = 2 x 10-24 J E = 2 x 10-13 J undergoing acceleration must emit radiation)

6) Radiation adsorbed or emitted only when electrons jump from one orbit to another E – energy (J), h – Plancks constant (J s), ν – frequency (Hz), ΔE = E − E c – speed of light (m-1), λ – wavelength (m) a b

where a and b represent the energy of the start and finish orbits

9 10

Bohr model – calculating the energy and radius Bohr model NOT EXAMANABLE Electron travelling around nucleus in circular 2 • Energy − Ze 1 2 orbits – must be a balance between attraction = − mv 8πε0r 2 to nucleus and flying off (like a planets orbit) ⎛ h ⎞ mvr = n⎜ ⎟ Electron feels two forces – must be balanced • Quantised angular momentum πε ⎝ 2π ⎠ πε − Ze2 − Ze2 1) Centrapedal (electrostatic) F = PE = − Ze2 1 − ()mvr 2 − n2h2 2 = − mv2 = = 4 0r 4πε0r 2 2 2 • Combining the two 8 0r 2 2mr 8π mr 2) Centrafugal mv2 1 F = KE = mv2 r 2 2 2 2 2 2 r − n h 8πε0 n h ε0 • Rearranging to give r = 2 2 r = 2 Equalize magnitude of forces Resulting energy r 8π m (− Ze ) πmZe πε 2 2 2 2 2 πε 2 2 4 mv Ze 2 Ze 1 2 −πεZe − Ze 1 2 − Ze − mZ e = 2 ⇒ mv = E = mv + = = − mv • Substitute r into energy gives = r 4 r 4πε r 2 4 0r 8πε0r 2 2 2 2 0 0 8 0r 8n h ε0

2 2 -Ze – nuclear charge, e – electron charge, ε0- permittivity of free space, • Energy is dependent on n and Z (2s and 2p the same – only true for 1 r - radius of the orbit, m – mass of electron, v – velocity of the electron electron systems 11 12 Energy levels of Hydrogen Energy levels of Hydrogen n=∞ Substitute quantised momentum into energy expression and rearrange n=5 in terms of r (radius) (see previous slide) n=4 nn3=3 n2h2ε n2a FhdFor hydrogen (Z1)(Z=1) r = 0 = 0 energy n=2 π mZe2 Z 2 −13.6056 n=1 a0 (Bohr) radius of the 1s electron on Hydrogen 52.9 pm (n =1, Z =1) r = n a0 En = n2 1 2 Radius (()r) de pends on n and Z n energy (eV) nucleus Substitute r back into energy expression gives 1 -13.6056 − mZ 2e4 2 2 -3.4014 E = 13.6056× Z n 2 2 2 = 2 (in eV) 3 -1.5117 8n h ε0 n 4 -0.8504 Energy of 1s electron in H is 13.6056 eV = 0.5 Hartree (1eV = 1.602 × 10-19 J) 5 -0.3779 1 ∞ 0.0000 Energy (E) depends on and Z2 Note. The spacing reflects the energy n2 Ionization energy = -13.6056 eV not the radius of the orbit. 13 14

n=4 Radius of orbits Emission spectra (httppy://www.youtube.com/watch?v=5z2ZfYVzefs) hv Energy of emission is Einitial -Efinal= For hydrogen (Z=1) n=3 ⎛ 1 1 ⎞ distance ΔE =13.6056 ⎜ − ⎟ ⎜ 2 2 ⎟ ⎝ ninital n final ⎠ 2 −13.6056 r = n a0 En = n=2 n2 Same form as fitted to emission specta n=1 Balmer series (Æ nfinal=2) 5 4 3 2 1 n energy (eV) r (pm) nucleus 1 -13.6056 52.9 2 -3.4014 211 3 -1.5117 476 n=3 Æ n=2 Æ λ = 656 nm 4 -0.8504 847 n=4 Æ n=2 Æ λ = 486 nm 5 -0.3779 1322 n=5 Æ n=2 Æ λ = 434 nm ∞ 0.0000 ∞ Note. The spacing reflects the radius of the Note. The spacing reflects the energy Orbit – not the energy. ℜ = 13.6056 eV / c = 3.29 ×1015 Hz not the radius of the orbit. 15 16 Problems with the Bohr Model Wave / particle duality httppy://www.youtube.com/watch?v=IsA __ oIXdF 8

• Only works for 1 electron systems de Broglie (1923) – E.gg,. H, He+, Li2+ Byyp this time it was accepted that EM radiation can have wave and p article properties (photons) • Can not explain splitting of lines in a magnetic field – Modified Bohr-Sommerfield (elliptical orbits - not satisfactory) dlidhilldhde Broglie proposed that particles could have wave properti i(/es (wave / particle duality). Particles could have an associated wavelength (l)

• Can not apply the model to interpret the emission spectra of complex atoms hc h E = mc2 , E = ⇒ λ = λ mc • Electrons were found to exhibit wave-like ppproperties – e.g. can be diffracted as they pass through a crystal (like x-rays) No experimental at time. – considered as classical particles in Bohr model 1925 Davisson and Germer showed electrons could be diffracted according to Braggs Law (used for X-ray diffraction)

Numerically confirm de Broglie’s equation

17 18

Wave Mechanics Wave mechanics and atoms

• For waves: it is impossible to determine the position and momentum of the • What does this mean for atoms electron simultaneously – Heisenberg ‘Uncertainty principle’ • Electrons in “orbits” must have an integer number of • Use probability of finding an electron from ψ2 (actually ψ*ψ – but functions wavelengths we will deal with are real) • E.g. n=4 and n=5 are allowed Where ψ is is a wave function and a solution of the Schrödinggqer equation – These create continuous or standing waves (like on a (1927). The time-independent form of the Schrödinger equation for the guitar string) hydrogen atom is: • E.g. n=4.33 is not allowed

2 2 2 2 2 − h 2 − e 2 ∂ ∂ ∂ – The wavefunction is not continuous ∇ Ψ Ψ = E Ψ ∇ = + + 2 ∂x 2 ∂y 2 ∂z 2 8π m 4πε 0 r • The wave nature of electrons brings in the quantized Kinetic Potential Total nature of the orbital energies. energy energy energy

19 20 Atomic solutions of the Schrödinger equation for H Solution of the Schrödinger equation for H

• Schrödinger equation can be solved exactly for 1 electron systems l has values 0 to (n-1) m has values from +l through 0 to –l – Solved by trial and error manipulations for more electrons

• 3 quantum numbers describing a three dimensional space called an atomic n 12222 orbital: n, l, m (and spin quantum number describing the electron s) l 00111

n = principal quantum number, defines the orbital size with values 1 to ∞ ml 00-101 Orbital 1s 2s 2p 2p 2p l = azimuthal or angular momentum quantum number, defines shape. For a gg,iven value of n, l has values 0 to (n-1). n 3 3 3 3 3 3 3 3 3 l 011122222 ml = magnetic quantum number, defines the orbital orientation. FiFor a given val lfue of l, ml hlf+has values from +l through h 0 t o –l. ml 0-101-2-1012 Oribtal 3s 3p 3p 3p 3d 3d 3d 3d 3d

21 22

Last Lecture

• Recap of the Bohr model – Electrons An introduction to – Assumptions – Energies / emission spectra Molecular Orbital Theory – Radii

• Problems with Bohr model – Only works for 1 electron atoms – Can not explain splitting by a magnetic field Lecture 2 – Representing atomic orbitals - The Schrödinger equation and wavefunctions. • Wave-particle duality

PfGWWtProf G. W. Watson • Wave mechanics Lloyd Institute 2.05 – Schrödinger [email protected] – Solutions give quantum number n, l, ml Æ atomic orbitals

24 Representations of Orbitals: Polar Coordinates

For an atomic system containing one electron (e.g. H, He+ etc.) • To describe the wavefunction of atomic orbitals we must describe it in The wavefunction, Ψ, is a solution of the Schrödinger equation three dimensional space It describes the behaviour of an electron in region of space called an atomic orbital (φ - phi) • For an atom it is more appropriate to use spherical polar coordinates:

Each orbital wavefunction (φ ) is most easily described in two parts radial term – which changes as a function of distance from the nucleus angular terms – which ch anges as a f uncti on of angl les LtifitPLocation of point P

Cartesian = x, y, z φ xyz= φradial(()r) φangular(φ,θ) = Rnl(()r) Ylm(φ,θ) Æ r, φ, θ Orbitals have

• SIZE determined by Rnl(r) - radial part • SHAPE determined by Ylm(φ,θ) -anggp(pular part (spherical harmonics) • ENERGY determined by the Schrödinger equation java applet on polar coordinates at http://qsad.bu.edu/applets/SPCExp/SPCExp.html 25 26

Wavefunctions for the AO’s of H Radial Wavefunction

General hydrogen like orbitals • R(r) of the 1s orbital of H it decays exponentially with r (−r) Rnl(r) Ylm(φ,θ) R(r) = 2e it has a maximum at r = 0 3 1 ⎛ Z ⎞ 2 ⎛ 1 ⎞ 2 ⎛ 2Z ⎞ 1s 2⎜ ⎟ e(−ρr 2) ⎜ ⎟ ρ = ⎜ ⎟ • R(r) has no physical meaning ⎜ ⎟ ⎜ na ⎟ 4 ⎝ a0 ⎠ ⎝ 4π ⎠ ⎝ 0 ⎠ 2 • Probability depends on R(r) 2 3 (1s) For hydrogen this simplifies as Z=1 and ao=1 (in atomic units) and thus ρ = 2. Hence – Misleading – does not take Normalisation into account the volume 2 ChhConstants are such that a.u. in ) R (r) Y (φ,θ) R(r nl lm – R(r)2 increases toward r = 0 ϕ 2 1 1 ∫ ∂τ =1 1s 1s 2e(−r) 2 π – Volume very small so that is the probability of the electron probability of being at 0 Angular component is a constant in an orbital must be 1 when all space small r is small 0 1 2 3 4 5 Æ Spherical is considered Radius (a.u)

27 28 Radial distribution functions (RDF) Wave functions of 2s and 3s orbitals

• Probability of an electron at a radius r (RDF) is given by probability of an 2s(r) 3s(r) electron at a point which has radius r multiplied by the volume at a radius of r 3 3 2 2 General 1 ⎛ Z ⎞ ( r 2) 1 ⎛ Z ⎞ ρ ⎜ ⎟ (2 − ρr)e −ρ ⎜ ⎟ (6 − 6 r + (ρr)2 )e(−ρr 2) 2 ⎜ ⎟ ⎜ ⎟ • Consider a sphere – volume as we move at a small slice is 4πr δr 2 2 ⎝ a0 ⎠ 9 3 ⎝ a0 ⎠ ⎛ 2Z ⎞ ⎛ ⎞ ⎜ ⎟ 2Z – By differentiation, 8 ρ = ⎜ ⎟ ρ = ⎜ ⎟ ⎝ na0 ⎠ ⎝ na0 ⎠ 4 3 the volume of a sphere = π r 6 3 1s For H 2s Æ Z=1, n=2, ρ =1 3s Æ Z=1, n=3, ρ =2/3 2 V )) so ∂ 2 4 ⎡ 2 ⎤ = 4πr R(r 1 1 ⎛ 2 ⎞ (−r /3)

2 (−r 2) ∂r r (2 − r) e ⎢6 − 4r + ⎜ r⎟ ⎥e π 4 2 2 9 3 ⎣⎢ ⎝ 3 ⎠ ⎦⎥ • RDF(r ) = 4πr2 R(r) 2 2 The form of the wave functions is the important concept – not the precise equation 0 • Maximum for 1s at a 0 (like Bohr!) 0 1 2 3 4 5 Note R(r) has functional form Radius (a.u) Normalisation constant * polynomial (increasing order with n) * exponential (-r/n) 29 30

Wave functions of Hydrogen 2s and 3s orbitals What does a negative sign mean

⎡ 2 ⎤ • The absolute sign of a wave function is not important. 1 (−r 2) 1 ⎛ 2 ⎞ (−r /3) For H 2s(r) = ( 2 − r ) e For H 3s(r) = ⎢6 − 4r + ⎜ r⎟ ⎥e – The wave function has NO PHYSICAL SIGNIFICANCE 2 2 9 3 ⎣⎢ ⎝ 3 ⎠ ⎦⎥ – the electron density is related to the square of the wave function – This is the same irrespective of the sign exponential decreases more slowly than 1s, e ( − r / n ) Æ more diffuse orbital • Wavefunction signs matter when two orbitals interact with each other (see later) Wavefunctions changes sign • Some books have the 2s as opposite sign – you can see that the electron 050.5 density R(r)2 is the same R(r) = 0 Æ RADIAL NODE 0.4 2s 0.8

u. 0.6 .. 030.3 040.4 2s at (2-r) = 0 (i.e. r=2 a.u.) 0.4 0.2 0.2 2s 2 (r) in a (2s) n a.u. in a.u.

ii 0 )) 3s changes sign twice with two RR 010.1 3s 020.2 -0.2 0246810 nodes (r =1.9, 7.1 a.u.) R(r 0 R(r) -2s -0.4 0 2 4 6 8 10 12 14 16 ÆCaused by the order of -010.1 -0.6 0 the polynomial ! Radius (a.u) -0.8 0246810 Radius (a.u) Radius (a.u) 31 32 Radial Nodes RDF’s of ns orbitals

• Number of radial node = n – l –1 1s – 1 peak. Maximum at r = a0 - Bohr Model Æ radius of a0 2s – 2 peaks Maximum at r ≈ 5 a - Bohr Model Æ radius of 4 a 1s = 1 – 0 – 1 = 0 0 0 3s – 3k3 peaks MiMaximum at r ≈ 13 a 0 - BhMdlBohr Model Æ radius of 9 a 2s = 2 – 0 – 1 = 1 2p = 2 – 1 – 1 = 0 0 3s = 3 – 0 – 1 = 2 3p = 3 – 1 – 1 = 1 3d = 3 – 2 – 1 = 0 Shape important for orbital energies 8 • Why are there radial nodes ? 1s – Pauli exclusion principle – no two electrons can have the same set of QN’s 6 2

– Actually – no two electron can overlapp( (i.e. occu py same sp ace ) 4(r) RR

2 2s r π

ϕ 4 3s * ϕ 2 – Overlap integral = ∫ A B ∂ τ = 0 (analogous to normalisation) AO’s are said to be Orthogonal 0 0 5 10 15 20 – Satisfied for AO’ s with same l by having change(s) in the wave function sign Radius (a.u) – Satisfied for AO’s with different l angular component ensures no overlap 33 34

Representing atomic orbitals Boundary surface

• Represent orbitals, so far radial and angular terms • Represent the wave function in 3D – Draw a 3D contour at a given value of φ • In 2D we can use dot diagrams to look at the whole wave function – Alternatively can define contour such that in enclosed a space which the electron spends most of its time – s orbitals have no angular component – spherical symmetry – Shows the shape and size of the orbital – Dot diagrams show electron density within a plane – no sign – Can not see the inner structure of the wave function – Can see where density goes to zero – nodes – Can see how greater volume as r increases makes most probable 2 distance. 1s Boundary surface .u. 1s 2s 3s aa 1 R(r) in in R(r)

0 012345 Radius (a.u)

35 36 p orbitals - wavefunctions p orbitals – radial functions

• There are three p orbitals for each value of n (px, py, pz) • Radial wave function for hydrogen p orbitals (Z=1) – The radial function is the same for all np orbitals for 2p n = 2 Æ ρ = 1 for 3p n = 3 Æ ρ = 2/3 – The angular terms are different Æ different shapes (orientations) 1 (−r 2) 1 ⎛ 2r ⎞ 2r (−2r 3) – Angular terms are the same for different n Æ 2px, 3px, 4px R(2 p) = re R(3p) = ⎜4 − ⎟ e 2 6 9 6 ⎝ 3 ⎠ 3 ⎛ 2Z ⎞ • Wave function for 2p and 3p orbitals ρ = ⎜ ⎟ ⎜ na ⎟ R(r) Y(θ φ) ⎝ 0 ⎠ • Polynomial Æ nodes 1 – Equation for no. of radial nodes 0.2 3 ⎛ 3 ⎞ 2 1 ⎛ Z ⎞ 2 Y( p ) = ⎜ ⎟ cos(θ ) – n – l –1 Æ 2p =0 , 3p =1 2p R(2 p) = ⎜ ⎟ ρre(−ρr 2) z 4π 0.15 ⎜ ⎟ ⎝ ⎠ – Ensures 2p and 3p orthogonal 2 6 ⎝ a0 ⎠ 1 ⎛ 3 ⎞ 2 θ Y ( p ) = sin( )cos(φ) 0.1 x ⎜ ⎟ • All p orbitals are multiplied by r 3 ⎝ 4π ⎠ a.u. in 2 )) 1 ⎛ Z ⎞ ρ (−ρr 2) π 1 Æ R(r) = 0 at r = 0 0.05

R(r 3p R(3p) = ⎜ ⎟ (4 − r)ρre ⎛ 3 ⎞ 2 θ 9 6 ⎝ a0 ⎠ Y ( px ) = ⎜ ⎟ sin( )sin(φ) ⎝ 4 ⎠ •Reqqguired to match the angular function 0 Æ angular node 0 5 10 15 20 Note the functional form of R(r) Æ Constant * polynomial * r * exponential -0.05 Radius (a.u) 37 38

p orbitals – angular functions boundary surfaces p orbitals – RDF’s

1 • All p orbitals have the same shape ⎛ 3 ⎞ 2 • Radial distribution function show probability at a given radius Y( pz ) = ⎜ ⎟ cos(θ ) ⎝ 4π ⎠ π 1 θ • Angular function give rise to direction ⎛ 3 ⎞ 2 • 2p function – no nodes, maximum at r = 4 a0 (same as n=2 for Bohr model) Y ( px ) = ⎜ ⎟ sin( )cos(φ) ⎝ 4 ⎠ • 3p function – two peaks, maximum at r ≈ 12 a0 (not the same as Bohr) • Can represent p orbital as dot diagrams or π 1 ⎛ 3 ⎞ 2 θ Y ( px ) = ⎜ ⎟ sin( )sin(φ) boundary surfaces ⎝ 4 ⎠ 2.5 2 2p • 1 angular nodal plane px (yz plane), py (xz plane) pz (xy plane) 1.5 2

– Ensures that p orbitals are orthogonal to s orbitals r) ((

R 3p 2 1 r π 4 0.5

0 0 5 10 15 20 Radius (()a.u)

39 40 Last week

• Solutions of the Schrödinger equation for atoms – Atomic orbitals (φ) An introduction to – Defined three quantum number (n, l, ml)

Molecular Orbital Theory • Defined polar coordinates Æ radial and angular terms

• Examined wavefuntions of the s orbitals Lecture 3 – More complex wave functions, radial distribution – Angular term constant for s orbitals functions and electron shielding – Wavefunction as constant * polynomial * exponential – Decays as e ( − r / n ) Æ the larger n the more diffuse the orbital Revision of Lewis bonding and hybridization – Defined radial nodes and examined there number (polynomial Æn – l -1) – Discussed the requirement for radial nodes Æ Pauli exclusion principle Prof G. W. Watson Lloyd Institute 2.05 • p orbitals wat@[email protected] – Radial functions similar to s orbital (except additional r) Æ R(0) =0 – Angular terms define shapes px, py and pz – same for different n – Radial distribution function for p orbitals 42

d orbitals – wave functions d orbitals – angular functions

• Five d orbitals for each value of n (n ≥ 3) Æ l = 2 , ml = -2, -1, 0, 1, 2 • Angular functions same for d xy , d yz , d xy , d z 2 , d x 2 − y 2 irrespective of n Æ same shape for 3d, 4d, 5d orbitals usinggy boundary surfaces 2 θ • Wave functions slightly more complicated (constant * polynomial * r * exp) 1 2 θ – Radial wave functions same for all 3d orbital Five different angular function e.g. ⎛ 15 ⎞ Y (dxz ) = ⎜ ⎟ sin( )cos( )cos(φ) 3 ⎝16π ⎠ 2 1 ⎛ Z ⎞ 2 (−ρr 2) R(3d) = ⎜ ⎟ (ρr) e Two angular nodes planes Æ 9 30 ⎝ a0 ⎠ orthogonal to s (0) and p (1) 0.1 1.5 • Max probability at r = 9 a0 0.08 3d - RDF 3d - R(r) e.g. dxy Nodal planes in 1 2 • AO’s with 0 nodes have 0.06 xy and xz R(r)

max probability at same 2 0.04 r (yz) d 2 π d r) in a.u. r) 0.5 xy z (( radius as Bohr model 44 R 0.02

0 0 (xz) • 4d orbital has 1 node 0 5 10 15 20 Radius (a.u)

2 Note the functional form of R(r) Æ Constant * polynomial * r * exponential 43 44 f orbitals Penetration

• Almost no covalent bonding Æ shape not really important • The RDF’s of AO’s within a given principle QN (n) have different shapes

• l = 3 Æ Seven different angular function for each n (n ≥ 4) • Number of nodes n – l -1 – f block is 14 element wide, same shape for 4f, 5f etc – n = 3 3s Æ 2 nodes 3p Æ 1 node 3d Æ 0 nodes – Radial functions same for all nf orbitals – Three angular nodes (nodal planes) Æorthogonal to s, p and d orbitals – 3s has a peak very close to the nucleus

– 3p has a peak close to the nucleus 1.5 3p

3d • These close ppygeaks have a very strong 3s 1

interaction with the nucleus 2 R(r) 2 r

ππ 0.5 • 3s is said to be the most penetrating 4

0 • Penetration 3s > 3p > 3d 0 5 10 15 20 Radius (a.u) 3 Note the functional form of R(r) Æ Constant * polynomial * r * exponential 45 46

Multi electron atoms Periodic table • CtltilllthCan not analytically solve the Sc hr ödinger equa tion f or multi -eltlectron at oms – We assume hydrogen like wave functions for multi-electron atoms • Shielding and penetration Æ E(ns) < E(np) < E(nd) < E(nf) • Nuclear charge increases with atomic No. • electrons repel each other and shield the nucleus from other electrons • This gives rise to electronic configuration of atoms and the order of elements in the periodic table • Effective nuclear charge Zeff = Z – S S = a screening or shielding constant • Electrons are filled in increasing energy (Aufbau principle) and electrons fill • E.g. Li atom – why is the electronic configuration 1s22s1 and not 1s2 2p1 ? degenerate (same energy) levels singularly first to give maximum spin 8 (Hund’s rule) – 1s electrons shields the valence electrons from the nuclear charge 6 1s –2s ppyenetrates more effectively • E(4s) < E(3d) 1s 1s

Æ feels a greater nuclear charge 2 K, Ca 2s 2p – 2p penetrates less effectively 4 2p 3s 3p R(r) 2

r 2s 4s 3d

– 2s is filled first π 44 2 • E(6s) < E(5d) ≈ E(4f) 6s 5d • E(1s) < E(2s) < E(2p) La [Xe] 6s2 5d1 0 2 2 • E(ns) < E(np) < E(nd) 0510C[X]6Ce [Xe] 6s 4f 4f Radius (a.u)

47 48 More complex results of penetration and shielding The energy of the 4s and 3d orbitals Energy levels vs atomic number •For H (Z=1) all orbitals within a • For K and Ca the E(3d) > E(4s), Sc on the E(3d) < E(4s) (but close) principle QN have same energy – If 4s electron go into 3d orbital the extra e-e repulsion and shielding cause the 3d to rise above 4s again – hence the strange energy level • For multi electron atoms diagram penetration follows – Result is that TM ’s loose 4s electrons first when ionized s > p > d > f Energy 4p { • 3d shielded very effectively by orbitals of n ≤ 3 and 3d almost 4p does not change in energy with Z 3d { until Z = 19 4s

• 4s filled before 3d 4s • However n = 4 does not shield 3d effectively Æ energy drops 3d Increasing Z • Similar pattern for 4d and 4f K Ca Sc Ti 49 50

Making Bonds Drawing representations of AO’s Localised Bond Pictures Revi si on of JF of L ewi s B ondi ng / VSEPR • Need to be able to draw AO’s when considering their interactions in MO’s – So far diaggprams have been to help visualise the 3D nature of AO’s • LlidifbdiLocalised view of bonding – Simple drawings are all you need !!!!! – Views covalent bonds as occurring between two atoms – Each bond is independent of the others – Each single bond is made up of two shared electrons – One electron is usually provided by each atom – Each 1st and 2nd row atom attains a noble gas configuration (usually) – Shape obtained by VSEPR (Valence Shell Electron Pair Repulsion)

e.g. H2 H+HH + × H HHH × H

Each H has a share of 2 electrons Æ H ─ H

51 52 Lewis bonding Lewis bonding – polyatomics (H2O) • Oxygen atom has 6 valence electrons and each hydrogen has 1 electron • Octet rule for main group elements / 18 electron rule for transition metals – All atoms have (()()()or a share of) 8 electrons (main) or 18 electrons (TM) – Gives rise to noble has configuration ×× ×× – Stability since all valence levels filled 2 H + × O × H × O × H ×× ×× • Lewis bonding in H O • Diatomics ×× ×× 2

× – Oxygen has 8 electron, hy drog en has 2 electron Æ noble gggas config. – F × ×× 2 F F × F + F × × – Oxygen – hydrogen interactions share 2 electron Æ H ─ O ×× ×× – Oxygen also has two lone pairs 2 electron shared Æ bond order = 1 F ─ F • Shape – VSEPR – Electrons repel each other (lone pairs repulsion > than bonding pairs) – O2 ×× ×× × × × – Oxygen has 2 bond pairs and 2 lone pairs Æ 4 directions to consider O + O O × O ×× ×× – Accommodate 4 directions Æ Tetrahedral shape –HO is bent with H-O-H angle of 104.5o 4 electron shared Æ bond order = 2 2 – Compares with a perfect tetrahedral of 109.45o Æ lone pair repulsion O ═ O 53 54

Lewis bonding – polyatomics (ethene) Lewis structures – breaking the octet rule

• Used different symbols for electrons on adjacent atoms • Some structures to not obey the 8 electron rule.

– e.g PF5 H H o ×× F 90 F 4 H + 2 × C × C × C × × HH F o F × P × F FP 120 Carbon atoms share 4 electron Æ bond order = 2 Æ C ═ C × × F Carbon –hydrogen interactions share 2 electrons Æ C ─ H F F F

• Shape – VSEPR (only the electrons round the P are shown for clarity) – Electrons repel each other – Carbon atoms have 3 directions – bond to C and two bonds to H – F atoms have 3 lone pairs (6 electrons) + 2 in each bond Æ 8 – Accommodate 3 bond direction Æ 120o in a plane (molecule is flat) – P atom has 5 bond pairs with 2 electrons each Æ 10 electrons !

55 56 TUTORIAL 1

1. What is the relationship between the possible angular momentum quantum numbers to the principal quantum number? 2. How many atomic orbitals are there in a shell of principal quantum number An introduction to n ? 3. Draw sketches to represent the following for 3s, 3p and 3d orbitals. Molecular Orbital Theory a) the radial wave function b)) the radial distribution c) the angular wave function 4. Penetration and shielding are terms used when discussing atomic orbitals Lecture 4 Revision of hybridisation a) Explain what the terms penetration and shielding mean. Molecular orbital theory and diatomic molecules b) How do these concepts help to explain the structure of the periodic table Prof G. W. Watson 5. Sketch the d orbitals as enclosed surfaces, showing the signs of the Lloyd Institute 2.05 wavefunction. [email protected] 6. What does the sign of a wavefuntion mean ?

Last lecture Valence bond theory and hybridisation

• d orbitals • Valence bond theory (Linus Pauling) – Radial wavefunctions, nodes and angular wavefunctions (shapes) – Based on localised bonding – Hybridisation to give a geometry which is consistent with experiment. • f orbitals – Hybridisation constructs new hybrid atomic orbitals from the AO’s – Radial wavefunctions , nodes and angular wavefunctions (shapes)

• Multielectron atoms • Use Lewis model (number of electron pairs) Æhybridisation Æ shape. 2 2 – Penetration and shielding – E.g. BeH2, Be – 1s 2s – Atomic orbital energies, filling and the periodic table

HBeHH × Be ×× H • Valence bond theory (localised electron pairs forming bonds) – Lewis structures Æ number of electron pairs Æ bond order (electrons shares divided by 2) • Correctly predicted by VSEPR to be linear – can we explain it using AO’s

– VSEPR Æ repulsion of electron pairs (BP and LP) – Mix S with pz orbital Æ 2 sp hybridized orbitals Æ molecular shape

59 60 sp hybridisation Hybridisation – sp2 hybridisation

• sp hybridisation • Lewis structure Æ 3 directions

– Mix and a s and a p orbital – two combinations s + pz and s – pz – Two AO’s Æ two hybrid AO’s H H – Relative sign is important when mixing orbitals × – sp therefore means that the hybrid orbital is 50% s and 50% p CCC × C HH 1 ψ sp = (ϕ2s +ϕ2 p ) + • Molecular is planar 2 z

1 • Three directions each ψ sp = (ϕ2s −ϕ2 p ) + o 2 z at 120

Æ mix s with 2 p orbitals = - Æ sp2 hybridisation

61 62

Hybridisation – π bonds Hybridisation – sp3

• For ethene sp2 hybridisation Æ bonding in three directions • For tetrahedral molecules we have to mix the s with all the p orbitals (sp3) – This give rise 4 equally spaced orbitals e.g. methane H H

x C H H – Each local bond can hold 2 electrons z 4 electron pairs sp3 hybridisation tetrahedral – Have not accounted for the second pair of electron shared by the C atoms

– H2O can a lso b e th ought ht of lik e thi s with t wo of th e • Creates a π bond above and below the plane of the molecule sp3 orbitals occupied by lone pairs. 2 2 2 3 1 – Could think of the C as going from s p Æ (sp ) px ××

H × O × H ××

4 electron pairs sp3 hybridisation tetrahedral 63 64 Hybridisation – d orbitals Hybridisation – summary

Trigonal Bipyramidal Lone pairs equatorial 3 sp d Æ 5 electron pairs Hybrid Atomic orbitals Geometry General Examples (+(s + px + py + pz + +d dz2) isation that are mixed formula

spp s + p linear AB2 BeH2 AB BF , CO 2- sp2 s + p + p trigonal planar 3 3 3 x y C2H4 SO 2-, CH , 3 4 4 sp s + p + p + p tetrahedral AB4 x y z NH3, H2O, octahedra 3 2 3 2 Trigonal Bipyramidal AB PCl , SF sp d Æ 6 electron pairs sp d s+ps + px +p+ py +p+ pz +dz+ dz 5 5 4 (s+ px+ py+ pz+ dz2+ dx2-y2) 2 2 square pyramidal s + px + py + pz + dx -y

SF6 3 2 2 2 2 2- sp d s + p + p + p + dz + dx -y octahedral AB6 [Ni(CN)4] x y z 2- [PtCl4]

Lone pairs trans 65 66

Molecular orbital theory Molecular orbital theory of H2 - bonding

ϕ ϕ • Molecule orbital theory (RobertRobert Mullikan) •H2 molecule – interaction of two hydrogen 1s orbitals ( a and b )

In phase interaction (same sign) In Phase • Electrons are delocalised 1s + 1s – Different to Lewis and hybridisation (these are not MO) ψ1 = (ϕa +ϕb ) Æ Constructive interference – Molecular orbitals are formed which involve all of the atoms of the molecule

Animation shows the in phase interaction of the s orbitals as they are brought – Molecular orbital are formed by addition and subtraction of AO ’ s together

Æ Linear Combination of Atomic Orbitals (()LCAO)

– like hybrid AO’s but the MO involves the whole molecule

67 68 Molecular orbital theory of H2 - antibonding Charge density associate with MO’s in H2 • In phase interaction - charge density given by ψ2 •H2 molecule – interaction of two hydrogen 1s orbitas ( ϕ a and ϕ b )

Out of Phase 2 2 2 2 2 ψ1 = (ϕa +ϕb ) ψ1 = [ϕa ] + [ϕb ] + 2[ϕaϕb ] Out of phase interaction (opposite sign) 1s - 1s Ö – This gives an enhanced density where the AO’s overlap between the atoms ψ 2 = (ϕa −ϕb ) ψ =ψ Æ Destructive interference – referred to as positive overlap and pull the atoms together (σ bonding) 1 σ

Node between the atoms • Out of phase interaction Animation shows the out of phase interaction (different colours) of the s orbitals as theyygg are brought together 2 2 2 2 2 ψ 2 = (ϕa −ϕb ) Ö ψ 2 = [ϕa ] + [ϕb ] − 2[ϕaϕb ] – This leads to reduced density between the atoms ψ 2 =ψσ * – referred to as negative overlap and pushes the atoms apart (σ* anti-bonding)

• CttltCan not create electrons Æ New wave func tions must b e normali sed t o ensure probability in 1 ! Interaction of 2 AO Æ 2 MO’s – A general rule is that n AO Æ n MO’s 69 70

Energy level diagram for H2 What happens when the AO’s have different energies?

• Interference between AO wave functions Æ bonding • Hypothetical molecule where the two s orbitals have different energies – Constructively Æ bonding interaction E(ϕa ) < E(ϕb ) – Destructively Æ anti-bonding interaction • What would the MO’s be like

– Bondi ng MO w ill b e much more lik e th e l ow energy orbit al ϕa

• Energy level diagram represents this interaction – Anti-bonding MO will be much more like high energy orbital ϕb – Two s orbitals interaction to create a low energy bonding and high energy ψ anti-bonding molecular orbital • We can say that the bonding MO is Energy (anti-bonding) – Electrons fill the lowest energy orbital (same rules as for filling AO’s) ψ = (Cσϕ + Cσϕ ) ϕb – Bonding energy = 2 ΔE σ a a b b • Where the coefficients C, indicate the contribu tion of the AO to the MO

ΔE ϕa ψ Energy σ σ (bonding) So for ψσ Ca > Cb A AB B H H2 H 71 72 Linear Combination of Atomic Orbitals - LCAO LCAO

• We wrote an equation using coefficients for the contribution of AO’s to the • Generally we can write bonding MO, we can do the same for the anti-bonding MO No AO's σ ϕ ψ = ∑ C nϕ ϕ * * n x x σ σ σ x=a... * ( ) ψσ = (Ca a + Cb ϕb ) ψσ = Ca a − Cb ϕb where the coefficients are different are reflect the contribution to each MO x=abcx = a,b,c …. (all (alloftheAO of the AO’ s in the molecule) n = 1, 2, 3….. (the resulting MO ’s) σ σ σ * σ * Ca > Cb Ca < Cb 1 1 1 1 • So MO(1) = ψ1 = Caϕa + Cbϕb + Ccϕc + Cdϕd +.... • The sign can be adsorbed into the coefficient and we can write all of the MO’s in a general way 2ϕ 2 2 2 ϕ MO(2) = ψ 2 = Caϕa + Cbϕb + Cc ϕc + Cdϕd +.... 1 1 1 1 ϕ n = 1 ψ1 = (Ca a + Cbϕb ) C = 0.8,C = 0.2 ϕ a b ϕ n n MO(3) = ψ = C3 + C3 + C3 + C3ϕ +.... ψ n = (Ca a + Cbϕb ) 3 a a b b c c d d ϕ 2 2 n = 2 2 2 C = 0.2,C = −0.8 2 ψ 2 = (Ca a + Cbϕb ) a b 1 - coefficients for MO(1), - coefficients for MO(2) etc. Cx Cx

• The coefficients contains both phase (sign) of the AO’s and how big their • And an examination of the coefficients tells us the bonding characteristics of contribution (size) is to a particular MO the MO’s 73 74

What interactions are possible What interactions are NOT possible

• We have seen how s orbitals interact – what about other orbitals • Some orbital can not interact – they give rise to zero overlap

• If you have positive overlap • Positive overlap (constructive interference) on one side in cancelled by Æ negative overlap (destructive interference) on the other reversing the sign negative overlap Positi ve Ove rla p

• s + p positive overlap above the axis is cancelled by negative overlap below E.g. s + s and px + px Æ +ve x – Same is true for the other interactions below s – s and px – px Æ -ve

• Must define orientation and stick to it for all orbitals. Zero overlap Thus Neggpative Overlap pz + pz Æ -ve

pz – pz Æ +ve i.e . for sigma bond between P orbital need opposite sign coefficients ! 75 76 Last lecture

• Hybridisation Æ combining AO’s on one atom to Æ hybrid orbitals Æ hybridisation made consistent with structure An introduction to Molecular Orbital Theory • Molecular orbital theory (delocalised view of bonding) – LCAO – all AO’s can contribute to a MO – n AO’s Æ n MO’s – Filled in same way as AO’s ΔE – Example of H2 Energy Lecture 5 Labelling MO’s. 1st row homonuclear diatomics H H2 H • Molecular orbitals for AO’s of different energy Prof G. W. Watson Lloyd Institute 2.05 • Linear Combination of Atomic Orbitals (LCAO) [email protected] – Use of coefficient to describe (i) phase of interaction and (ii) size of No AO's contributi on of a gi iven AO n ψ n = ∑ Cxϕ x x=a... 78

Labelling molecular orbitals Labelling molecular orbitals 1) Symmetry of bonding 2) bonding and anti-bonding (already met this label) σ = spherical symmetry along the bond axis - same symmetry as s orbital no nodes pass through the bond axis (can be at right angles Æ σ*) – Nothinggg( if bonding (no nodes between bonded atoms ) σ π

π = one nodal plane which passes through the bond axis – Additional * if a nodal pp,lane exits between the atoms, that is if the wavefunction changes sign as you go from one atom to the other. dzx + px σ* δ = two nodal plane which pass π* through the bond axis

d 2 2 (end on dxy or x − y ) 79 80 Labelling molecular orbitals 2nd row homonuclear diatomics

3) Is there a centre of inversion ? i.e. is it Centrosymmetric ? •Li-Li Æ Ne-Ne – The final label indicated whether the MO has a centre of inversion syyymmetry – Possible interactions between 1s, 2s and 2p

px + px px - px As you go from one side of Wave function does σ bonding s – s, pz –pz π bonding px – px, py - py wave function through the not change sign centre of the bond the sign of Æ centrosymmetric the wavefunction reverses Æ g = gerade or Æ not centrosymmetric even Æ u = ungerade or odd

• MO’ s sometimes labelled with the type of AO forming them e .g . σs or σp

81 82

Energy level diagram for O2 Other possible interactions

• Will σ interactions between s and pz be important • 2s and 2p energies sufficiently spaced to give little interaction – Depends on energy difference – Simppple picture of the MO UidltUnpaired electrons btbetween s and p 6σ∗ 4σ∗ z u u ÆParamagnetic – If large then no effect ∗ 2π g 2π∗ 2p 2p g Label MO’s starting from the bottom although often only 1πu 1π • How does the energy of the 2s and 2p vary with Z (shielding / penetration) u valence orbitals 5σ 3σ g g 2p No sp mixing ∗ ∗ 4σ u 2σ u 2s 2s 2s

3σg 1σg Energy difference too big to interact with valence orbitals sp miiixing

Energy ∗ Energy 2σ u 1s AO’s very small Æ very Li Be B C N O F Ne 1s 1s small overl ap i n l ower l evel s 1σ (small ΔE) O g O • Gap increases – 2p more effectively shielded - critical point between O and N 83 84 MO diagram for N2 Making sense of N2 • Take basic model for oxygen – no s p interaction • 2s and 2p energies sufficiently close for interaction Æ more complex – Examine how the MO’s can interact – 1σ and 2σ shift to lower energy – π and σ can not interact – zero overlap Æ π level remain the same –3σ shifted 4σ shifted to high energy – Examine σ – σ interactions –3σ now above 1π ∗ ∗ 4σ u 4σ u Bonding interactions can interact – π levels unaffected ∗ 2π with each other 1σu and 3σu ∗ u 2π g 2p 3σ 2p 2p 3σg 1πg 1πu 3σu

2σ∗ u 1σ 2σ∗ 2s 2s u 2s Thus 1σu goes down in energy and 3σu goes up in energy 1σg 1σu 85 86

nd Making sense of N2 MO diagrams for 2 row diatomics • The effect of the overlap between 2s and 2p is greatest for the Li. The MO • Now examine the anti-bonding * diagram changes systematically as you go across the periodic table interaction (2σ and 4σ) 4σ

∗ 4σ u 2π∗ u 2σ* 2p ∗ ∗ Thus 2σ u goes down in energy and 4σ u goes up in energy 1πg 3σ u For bonding with anti-bonding (1σ –4σ or 2σ –3σ) the sign changes on one 2σ∗ u wave function Æ zero overlap.

• s – p mixing Æ B2 – paramagnetic and C2 diamagnetic 1σu 87 88 MO treatment of BeH2 Energy level diagram for BeH2 • VSEPR Æ linear molecule, H –Be –H Z 4σ* –Be –1s2 2s2 2p0 H – 1s1 u – Examine interaction of 6 AO with each other * –2 H 1s, Be 2s and Be 2px, Be 2py Be 2pz Æ 6MO’s 3σ g 2p Interaction between H 1s and Be 2s Interaction between H 1s and Be 2p z Non bonding px and py

2s 2σ bonding anti-bdibonding bdibonding anti-bonding u

1σg Be BeH2 2 H Each of these is delocalised over three atoms and can hold up to two electrons Compare these two MO’s with no sp mixing with the localised model of two Equivalent bonds formed via sp hybridization px and py have zero overlap Æ non bonding 89 90

Alternative approach Alternative energy level diagram for BeH2 • Step wise approach (ligands first) H –Be –H Z Mix hydrogen 1s orbitals first 4σ* Æ two Molecular orbitals u Be

* 3σ g Be Be 2p Non bonding px and py

Then mix with s (zero overlap with pz) Then mix with pz (zero overlap with s) 2s

2σu

1σg Be BeH 2 2H2 H

91 92 Last lecture

•LCAO – Interacti on of AO’ s wi th diff erent energy Æ lAOhbilower AO has bigger An introduction to contribution Molecular Orbital Theory – Reppgresenting contribution as coefficient

• AO interactions that were possible Æ MO’s – positive, negative and zero overlap – labelling of MO’s (σ / π , *, g / u) Lecture 6 More complex molecules and CO bonding in transition metal complexes •2nd row homo nuclear diatomics – 2s – 2p mixinggp occurs up to N Æ energy different too big after this ( O , F ) Prof G . W . Watson 2 2 – Difference in MO diagram for N and O Lloyd Institute 2.05 2 2 [email protected] • Molecular orbital treatment of BeH 2

MO treatment of H2O z O MO’s of H2O z O H H H H •H2O is not linear – but why ? x • Three orbitals Æ three MO’s x – We will examine the MO’s for an non linear tri-atomic and find out. anti - bonding approx. -2pz +H+ H – What orbitals are involved – 2 H 1s + O 2s O 2px O 2py and O 2pz

• Lets start by creating MO ’ s from the hydrogen 1s orbitals .

2pz approx. 2pz -2s + H Approximately

•Takinggpp the in phase pair first- it will interact with the O 2s and O 2pz ((pzero overlap non-bonding + with O 2px and O 2py) 2s

Bonding approx. 2s + 2pz (a little) + H

O 2H2 H • Problem - This is mixing three orbitals Æ must produce three orbital 95 96 MO’s of H2O z O Energy level diagram for H2O z O H H H H • Out of phase H 1s orbitals x x – Only interact with p Æ 2 MO’s x There are not two lone pairs ! – Zero overlap with py

H H anti - bonding 2p , n.b. 2p y Slightly bonding Pz σ n.b. Non bonding Py 2σ Very different to the VB non bdibonding py 2s concept of two identical sp3 filled orbital 2px 1σ H H H H Bonding O H2O2 H O 2 H MO theory correct. 97 98

Comparison of H2O and BeH2 π MO’s of Benzene • • Both cases of XA2 Æ same MO – different No of electrons π bonding is more important for reactivity –independent of σ (zero overlap) – Bonding MO’s as a function of A-X-A bond angle –six px orbitals Æ combine to form six MO’s

• Different ways of arranging six px orbitals on a ring – Lowest e ne rgy – aapasell in phase

BeH2 linear – Degenerate levels (1 nodal plane Æ 2 nodes) – Degenerate levels (2 nodal planes Æ 4 nodes) o BH2 131 – Highest energy – all out of phase (3 nodal planes Æ 6 nodes) o CH2 (t) 134 – Energy increases with number of nodes – as in AO’s CH (s) 102o 2 – Also the number of nodes on a ring must be even Æ continuous wavefunction

o NH2 103 • Lowest energy all in phase o OH2 105 – All coeffcients the same

90ºH2O CH2(T) 180º 99 100 π MO’s of Benzene MO diagram for CO • Next occupied degenerate pair Æ 1 nodal plane – Two ways of doing this – between atoms and through a pair of atoms • Same orbitals as homo nuclear diatomics – different energggies give rise to sig nificant 2s - 2p mixing – confusing set of orbitals 4σ C-O anti - bonding (more C)

2π∗ π∗ (uneven – more carbon) 2p 2p – As the wavefunction goes through 0 (at the node) the smooth wavefunction 3σ Primarily carbon (pz) has smaller coefficients next to the node zero at the node 1π π bond (uneven – more oxygen)

• 2 electron per MO spread over 6 atoms – Compare with Lewis structure with individual double bonds 2s 2σ Primarily oxygen (pz) – With local bonding have to resort to resonance structures to explain benzene 2s 1σ C-O bonding interaction (more O)

101 102

The HOMO and LUMO of CO Interaction of the CO 3σ with d orbitals

• For chemical reaction the HOMO (Highest Occupied Molecular Orbital) and • Three sets of interaction based on symmetry of ligand AO’s the LUMO (Lowest unoccupied Molecular Orbital) are the most important. – Generallyypp applicable to σ bondinggg TM ligands

HOMO – 3σ LUMO – 2π* •a1g all ligand AO’s in phase lbillow energy Oxygen orbitalsComes fddfrom standard π iiinteraction – Interaction with s orbital Æ 1 makes 2σ Æ mainly O pz however lower oxygen orbital Æ in 3 σ mainly C pz means π has has more oxygen and – t1u ligands in one axis contribute π* more carbon – With opposite phase – one nodal plane Some anti-bonding mixes in – Interaction with p orbitals Æ 3 due to sp mixing π π∗ 3σ C O

C O C O 103 104 Interaction of the CO 3σ with d orbitals MO diagram for Tm (σ-L)6

•eg ligand phases have two nodal planes * • Electrons from filled σ orbitals on 2 d 2 2 t 1u – Interact with d z x − y Æ 2 the ligands fill all the bonding

* orbitals a 1g 4p • d electrons fill t2g (non bonding) * e* and e g (antibonding) 4s g Δ oct • Example is d6 – e.g. Co3+ 3d t2g

• Three remaining d orbitals point between ligands • These are the orbitals considered in ligand filed theory. Note the e* – zero overlap (t ) 6 σ ligand g 2g eg is anti -bonding ! orbitals t1u

• What really decided Δoct is the π a1g interaction

105 106

π interactions with TMs Low ligand field situation • LiLigand orbi tal s are l ow energy and fill ed ( e.g. F) • Orbitals with π character can interact with the t2g d orbitals – Filled orbitals interact in a π fashion

– Must be correct symmetry (t2g) Æ 3 arrangements – Bondingggy(g) combinations are reduced in energy and filled (like ligand orbitals)

possible using dxy, dyx, dxz – Antibonding combination are raised in energy and filled (like d orbitals)

• Two extreme situations * * e g e g – Ligand orbitals are low energy and filled (e.g. F) Δoct – Ligand orbitals are high energy and empty (e .g . CO) Δoct t* Interaction with 2g Interaction with t high energy 2g e* low energy empty ligand g filled ligand t orbitals orbitals 2g Δoct without π with π ligand t2g ? ? – Strong interaction with filled orbitals with π interaction leads to reduction in Δoct (box shows the orbitals considered in ligand field theory) 107 108 High ligand field situation Tutorial 2 – part a • Ligand orbitals are high energy and empty (e.g. CO 2π*) 1. Explain the MO approach for the interaction of – Filled orbitals interact in a π fashion a) two s orbitals of identical energy – Bondi ng combi nati ons are red uced i n energy (lik e d orbit al s) – Antibonding combination are raised in energy and empty (like ligand b) two s orbitals of slightly different energy orbitals) c) two s orbital of very different energy.

e* e* g g 2. Consider the bonding in the molecule O2

a) Draw a Lewis structure for O 2 Δoct Δoct b) Determine the hybridization c) Perform an MO treatment of O t 2 2g (i) What orbitals are involved? t2g (ii) what interactions are possible? without π with π ligand (iii) what do the resulting MO’s look like? (iv) sketch an MO energy level diagram. – Strong i nt eracti on with empty t orbit al s with π itinteracti on l ead dts to i ncrease d) What difference are there in the details of the bonding diagram in Δoct (box shows the orbitals considered in ligand field theory) between the Lewis and MO treatments 109 110

Tutorial 2 – part b

3. Consider the molecule BeH2 a) Draw a Lewis structure b) Determine the hybridization

c) Perform an MO treatment of O2 (i) What orbitals are involved . (ii) Generate appropriate ‘ligand’ MO’s and interactions with the central atom (iii)what do th e resul ti ng MO’ s l ook lik e? THE END (iv)sketch an MO energy level diagram. d) What difference are there in the details of the bondinggg diagram between the Lewis and MO treatments

4. Perform the same analysis for BeH 2, HF, BH 3, and CH 4

5. Use molecular orbital theory to explan a) The splitting of the d orbitals by sigma interactions with ligands b) The effect of π interaction on the ligand field strength.

111 112