An introduction to. Objectives of the course. Literature. Lecture schedule. 6 Lecture Course

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1 Objectives of the course An introduction to Molecular Orbital Theory 6 Lecture Course Prof G. W. Watson Lloyd Institute t.36 watsong@tcd.ie Wave mechanics / Atomic orbitals (AOs) The basis for rejecting classical mechanics (the Bohr Model) in treating electrons Wave mechanics and the Schrödinger equation Representation of atomic orbitals as wave functions Electron densities and radial distribution functions Understanding the effects of shielding and penetration on AO energies Bonding Review VSEPR and ybridisation Linear combination of molecular orbitals (LCAO), bonding / antibonding Labelling of molecular orbitals (MOs) (σ, π and g, u) omonuclear diatomic MO diagrams mixing of different AO s More complex molecules (CO, O.) MO diagrams for Inorganic complexes Lecture Lecture schedule Revision of Bohr model of atoms Literature Book Sources: all titles listed here are available in the amilton Library Lecture Lecture 3 Lecture 4 Lecture 5 Lecture 6 Schrödinger equation, atomic wavefunctions and radial distribution functions of s orbitals More complex wavefunctions and radial distribution functions and electron shielding Lewis bonding, ybridisation, and molecular orbitals Labelling MO s. st row homonuclear diatomics MO approach to more complex molecules and CO bonding in transition metals complexes. Chemical Bonding, M. J. Winter (Oxford Chemistry primer 5) Oxford Science Publications ISBN condensed text, excellent diagrams. Basic Inorganic Chemistry y( (Wiley) F.A.Cotton, G. Wilkinson, P. L. Gaus comprehensive text, very detailed on aufbau principle 3. Inorganic Chemistry (Prentice all) C. ousecroft, A. G. Sharpe comprehensive text with very accessible language. CD contains interactive energy diagrams Additional sources: - gallery of AOs and MOs 3 4

2 Tutorials Expectation Tutorials are to go through problems that students are having with the course Tutorials are NOT for the lecturer to give you the answers to the questions or to give you another lecture. An introduction to Molecular Orbital Theory All student must BEFORE the tutorial Look at the notes for the course and try to understand them 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 questions) It is a waste of both the lecturers and students time if the tutorial to ends up being a lecture covering questions. Lecture The Bohr Model Prof G. W. Watson Lloyd Institute.5 watsong@tcd.ie 5 Adsorption / Emission spectra for ydrogen Johann Balmer (885) measured line spectra for hydrogen nm (uv), 4. nm (uv), 434. nm (violet), 486. nm (blue), and nm (red). Bohr model of the atom (93) Assumptions ) Rutherford (9) model of the atom (Planetary model with central nucleus + electrons in orbit) ) Planck (9), Einstein (95) the energy electromagnetic waves is quantised into packets called photons (particle like property). Balmer discovered these lines occur in a series - both absorption and emission - where is the Rydberg constant (3.9 5 z) R ric Velocity, c E = hν Wavelength, λ v R n n = Balmer series n = and n =n +, n +, n +3.. Other series for n = (Lyman UV), n =3 (Paschen IR) etc. Electrons must have specific energies no model of the atom could explain this 7 Fluctuat ting electr /magn netic field An stationary observer counts ν waves passing per second } i.e. frequency = v z, cycles/sec, sec - 8

3 Bohr model of the atom Speed of electromagnetic waves (c) is constant (ν and λ vary) c = ν λ, ν = c / λ, E = h ν, E = h c / λ As frequency increases, wavelength decreases. Given λ ν eg e.g. radiowaves: λ = m. X-rays: λ = x - m ν = 3 x 9 z ν = 3 x z E = x -4 J E = x -3 J E energy (J), h Plancks constant (J s), ν frequency (z), c speed of light (m - ), λ wavelength (m) Bohr model of the atom 3) Electron assumed to travel in circular orbits. 4) Apply quantisation to orbits - only orbits allowed have quantised angular momentum (comes from observation of spectra) h mvr = n π 5) Classical electrodynamical theory rejected (charged particles undergoing acceleration must emit radiation) 6) Radiation adsorbed or emitted only when electrons jump from one orbit to another ΔE = E a E b where a and b represent the energy of the start and finish orbits 9 Bohr model Electron travelling around nucleus in circular orbits must be a balance between attraction to nucleus and flying off (like a planets orbit) Electron feels two forces must be balanced Bohr model calculating the energy and radius NOT EXAMANABLE Energy Ze = mv 8πε r h mvr = n Quantised angular momentum π Ze Ze ) Centrapedal (electrostatic) F = PE = 4πε r 4πε r ) Centrafugal mv KE mv F = = r Equalize magnitude of forces Resulting energy Combining the two Rearranging to give r ( mvr) Ze n h = mv = = 8πε r mr 8π mr r n h 8 = r 8π m Ze πε n h ε r = ( ) πmze mv r Ze Ze = mv = 4πε r πε r 4 Ze Ze Ze E = mv + = = mv 4πε r 8πε r Substitute r into energy gives Ze mz e = 8πε r n h ε 8 4 -Ze nuclear charge, e electron charge, ε - permittivity of free space, r - radius of the orbit, m mass of electron, v velocity of the electron Energy is dependent on n and Z (s and p the same only true for electron systems

4 Energy levels of ydrogen Energy levels of ydrogen n= n=5 n=4 n=3 n 3 Substitute quantised momentum into energy expression and rearrange in terms of r (radius) (see previous slide) r= n h ε π mze = F hhydrogen For d (Z (Z=) ) n a Z energy r = n a a (Bohr) radius of the s electron on ydrogen 5.9 pm (n =, Z =) Radius ((r)) depends p on n and Z Substitute r back into energy expression gives En = 4 mz e Z (in ev) = 8n h ε n Energy of s electron in is ev =.5 artree (ev =.6-9 J) Energy (E) depends on n energy (ev) and Z n En = n n= n= nucleus Ionization energy = ev Note. The spacing reflects the energy not the radius of the orbit. 3 4 n=4 Emission spectra Radius of orbits ( p ( y ) hv Energy of emission is Einitial - Efinal= n=3 For hydrogen (Z=) (Z ) ΔE = n inital n final distance r = n a En = n n energy (ev) r (pm) n= Same form as fitted to emission specta n= Balmer series (Æ nfinal=) nucleus n=3 Æ n= Æ λ = 656 nm n=4 4 Æ n= Æ λ = 486 nm n=5 Æ n= Æ λ = 434 nm Note The spacing reflects the radius of the Note. Orbit not the energy. 5 ℜ = ev / c = z Note. The spacing reflects the energy not the radius of the orbit. 6

5 Problems with the Bohr Model Only works for electron systems E.g., e +, Li + Can not explain splitting of lines in a magnetic field Modified Bohr-Sommerfield (elliptical orbits - not satisfactory) Can not apply the model to interpret the emission spectra of complex atoms Electrons were found to exhibit wave-like properties p e.g. can be diffracted as they pass through a crystal (like x-rays) considered as classical particles in Bohr model Wave / particle duality _ oixdf_ 8 de Broglie (93) By this time it was accepted that EM radiation can have wave and particle properties (photons) de Broglie proposed that particles could have wave properties (wave / particle duality). Particles could have an associated wavelength (l) No experimental at time. hc E = mc, E = λ = λ 95 Davisson and Germer showed electrons could be diffracted according to Braggs Law (used for X-ray diffraction) h mc Numerically confirm de Broglie s equation 7 8 Wave Mechanics For waves: it is impossible to determine the position and momentum of the electron simultaneously eisenberg Uncertainty principle Use probability of finding an electron from ψ (actually ψ*ψ but functions we will deal with are real) Where ψ is is a wave function and a solution of the Schrödinger equation (97). The time-independent form of the Schrödinger equation for the hydrogen atom is: h e Ψ Ψ 8 π m 4 πε rr = E Ψ Kinetic Potential Total energy energy energy = + + x y z What does this mean for atoms Wave mechanics and atoms Electrons in orbits must have an integer number of wavelengths E.g. n=4 and n=5 are allowed These create continuous or standing waves (like on a guitar string) E.g. n=4.33 is not allowed The wavefunction is not continuous The wave nature of electrons brings in the quantized nature of the orbital energies. 9

6 Atomic solutions of the Schrödinger equation for Schrödinger equation can be solved exactly for electron systems Solved by trial and error manipulations for more electrons 3 quantum numbers describing a three dimensional space called an atomic orbital: n, l, m (and spin quantum number describing the electron s) n = principal quantum number, defines the orbital size with values to l = azimuthal or angular momentum quantum number, defines shape. For a given value of n, l has values to (n-). m l = magnetic quantum number, defines the orbital orientation. For a given value of fl, m l has values from +l through h to l. l Solution of the Schrödinger equation for l has values to (n-) m has values from +l through to l n l m l - Orbital s s p p p n l m l Oribtal 3s 3p 3p 3p 3d 3d 3d 3d 3d Last Lecture An introduction to Molecular Orbital Theory Lecture Representing atomic orbitals - The Schrödinger equation and wavefunctions. Prof fg. W. Watson Lloyd Institute.5 watsong@tcd.ie Recap of the Bohr model Electrons Assumptions Energies / emission spectra Radii Problems with Bohr model Only works for electron atoms Can not explain splitting by a magnetic field Wave-particle duality Wave mechanics Schrödinger Solutions give quantum number n, l, m l atomic orbitals 4

7 Representations of Orbitals: For an atomic system containing one electron (e.g., e + etc.) The wavefunction, Ψ, is a solution of the Schrödinger equation It describes the behaviour of an electron in region of space called an atomic orbital (φ - phi) 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 changes as a function of angles Orbitals have φ xyz = φ radial (r) () φ angular (φ,θ) = R nl( (r) Y lm(φ, (φ,θ) Polar Coordinates To describe the wavefunction of atomic orbitals we must describe it in three dimensional space For an atom it is more appropriate to use spherical polar coordinates: Location of point tp Cartesian = x, y, z r, φ, θ SIZE determined by R nl (r) - radial part SAPE determined by Y lm(φ,θ) -angular part (spherical harmonics) ENERGY determined by the Schrödinger equation 5 java applet on polar coordinates at 6 Wavefunctions for the AO s of Radial Wavefunction General hydrogen like orbitals R nl (r) Y lm (φ,θ) 3 Z r s ( ρ ) e a 4π ρ = Z na R(r) of the s orbital of it decays exponentially with r R(r) = ( r) e it has a maximum at r = R(r) has no physical meaning 4 For hydrogen this simplifies as Z= and a o = (in atomic units) and thus ρ =. ence Normalisation Constants are such that R nl (r) Y lm (φ,θ) s ϕ τ = ( r ) e π that is the probability of the electron Angular component is a constant in an orbital must be when all space Spherical is considered Probability depends on R(r) 3 (s) ) in a.u. Misleading does not take into account the volume R(r) increases toward r = R(r s Volume very small so probability of being at small r is small Radius (a.u) 7 8

8 Radial distribution functions (RDF) Probability of an electron at a radius r (RDF) is given by probability of an electron at a point which has radius r multiplied by the volume at a radius of r Consider a sphere volume as we move at a small slice is 4πr δr By differentiation, 8 the volume of a sphere = so V = 4πr r RDF(r) = 4πr R(r) 4 r 3 π 3 6 Maximum for s at a (like Bohr!) Radius (a.u) ) 4πr R(r) 4 s 9 General Wave functions of s and 3s orbitals s(r) 3s(r) 3 Z ( ρ r ) 3 Z ( ρ r ) a ( ρr) e 9 3 a (6 6ρr + ( ρr) ) e Z ρ = Z ρ = na na For s Z=, n=, ρ = 3s Z=, n=3, ρ =/3 ( r ) ( r) e 9 6 4r + r 3 3 e ( r /3) The form of the wave functions is the important concept not the precise equation Note R(r) has functional form Normalisation constant * polynomial (increasing order with n) * exponential (-r/n) 3 Wave functions of ydrogen s and 3s orbitals ( r ) For s(r) = ( r) e For 3s(r) = 9 6 4r + r 3 3 e ( r /3) exponential decreases more slowly than s, ( / n) e r more diffuse orbital Wavefunctions changes sign R(r) = RADIAL NODE s at (-r) = (i.e. r= a.u.) 3s changes sign twice with two nodes (r =.9, 7. a.u.) Caused by the order of the polynomial! R(r) in a..u s s Radius (a.u) 3 What does a negative sign mean The absolute sign of a wave function is not important. The wave function has NO PYSICAL SIGNIFICANCE the electron density is related to the square of the wave function This is the same irrespective of the sign Wavefunction signs matter when two orbitals interact with each other (see later) Some books have the s as opposite sign you can see that the electron density R(r) is the same in a.u. R(r) i s s Radius (a.u) ) in a.u. R(r) 4.4. (s) Radius (a.u) 3

9 Radial Nodes Number of radial node = n l s = = s = = p = = 3s = 3 = 3p = 3 = 3d = 3 = Why are there radial nodes? Pauli exclusion principle no two electrons can have the same set of QN s RDF s of ns orbitals s peak. Maximum at r = a - Bohr Model radius of a s peaks Maximum at r 5 a - Bohr Model radius of 4 a 3s 3 peaks Maximum at r 3 a - Bohr Model radius of 9 a Shape important for orbital energies 8 6 s Actually no two electron can overlap p( (i.e. occupy same space) * Overlap integral = ϕ AϕB τ = (analogous to normalisation) AO s are said to be Orthogonal Satisfied for AO s with same l by having change(s) in the wave function sign Satisfied for AO s with different l angular component ensures no overlap 33 4πr R(r) 4 s 3s 5 5 Radius (a.u) 34 Representing atomic orbitals Boundary surface Represent orbitals, so far radial and angular terms In D we can use dot diagrams to look at the whole wave function s orbitals have no angular component spherical symmetry Dot diagrams show electron density within a plane no sign Can see where density goes to zero nodes Can see how greater volume as r increases makes most probable distance. s s 3s Represent the wave function in 3D Draw a 3D contour at a given value of φ Alternatively can define contour such that in enclosed a space which the electron spends most of its time Shows the shape and size of the orbital Can not see the inner structure of the wave function R(r) in a.u. s Boundary surface Radius (a.u) 35 36

10 p orbitals - wavefunctions There are three p orbitals for each value of n (p x, p y, p z ) The radial function is the same for all np orbitals The angular terms are different different shapes (orientations) Angular terms are the same for different n p x, 3p x, 4p x Wave function for p and 3p orbitals R(r) Y(θ φ) 3 3 Z r ) ( ) ( ( ρ Y p z = R p) = ρre 4π 6 a 3 ( ) 4π Y p x = 3 Z ( ρr ) R(3p) = (4 ρr) ρre a ( ) 4π Y p x = cos( θ ) Z ρ = na sin( θ )cos( φ) sin( θ )sin( φ) Note the functional form of R(r) Constant * polynomial * r * exponential 37 p orbitals radial functions Radial wave function for hydrogen p orbitals (Z=) for p n = ρ = for 3p n = 3 ρ = /3 ( r ) r r R( p) = re R(3p) = 4 e Polynomial nodes Equation for no. of radial nodes n l p =, 3p = Ensures p and 3p orthogonal All p orbitals are multiplied by r R(r) = at r = ) in a.u. R(r) Required to match the angular function angular node 3p p ( r 3) Radius (a.u) 38 p orbitals angular functions boundary surfaces All p orbitals have the same shape Angular function give rise to direction Can represent p orbital as dot diagrams or boundary surfaces Y( Y ( Y ( p z p x p x 3 ) 4π = 3 ) 4π = 3 ) 4π = angular nodal plane p x (yz plane), p y (xz plane) p z (xy plane) Ensures that p orbitals are orthogonal to s orbitals cos( θ ) sin( θ )cos( φ) sin( θ )sin( φ) p orbitals RDF s Radial distribution function show probability at a given radius p function no nodes, maximum at r = 4 a (same as n= for Bohr model) 3p function two peaks, maximum at r a (not the same as Bohr) (r) 4πr R( p 3p 5 5 Radius (a.u) 39 4

11 Last week An introduction to Molecular Orbital Theory Lecture 3 More complex wave functions, radial distribution functions and electron shielding Revision of Lewis bonding and hybridization Prof G. W. Watson Lloyd Institute.5 watsong@tcd.ie Solutions of the Schrödinger equation for atoms Atomic orbitals (φ) Defined three quantum number (n, l, m l ) Defined polar coordinates radial and angular terms Examined wavefuntions of the s orbitals Angular term constant for s orbitals Wavefunction as constant * polynomial * exponential Decays as ( rr / n ) e the larger n the more diffuse the orbital Defined radial nodes and examined there number (polynomial n l -) Discussed the requirement for radial nodes Pauli exclusion principle p orbitals Radial functions similar to s orbital (except additional r) R() = Angular terms define shapes p x, p y and p z same for different n Radial distribution function for p orbitals 4 d orbitals wave functions Five d orbitals for each value of n (n 3) l =, m l = -, -,,, Wave functions slightly more complicated (constant * polynomial * r * exp) Radial wave functions same for all 3d orbital Z R(3d ) = 9 3 a. Max probability at r = 9 a.8 AO s with nodes have max probability at same radius as Bohr model 4d orbital has node (r) in a.u. R( ( ρr) e 3d - R(r) ( ρr ) 3d - RDF 5 5 Radius (a.u) Note the functional form of R(r) Constant * polynomial * r * exponential.5.5 4πr R(r) 43 d orbitals angular functions Angular functions same for d xy, d yz, d xy, d z, d x y irrespective of n same shape for 3d,, 4d,, 5d orbitals using boundary surfaces Five different angular function e.g. 5 ( ) sin( θ )cos( θ )cos( φ) π Y d xz = 6π Two angular nodes planes orthogonal to s () and p () e.g. d xy Nodal planes in xy and xz d xy (yz) d z (xz) 44

12 f orbitals Almost no covalent bonding shape not really important l = 3 Seven different angular function for each n (n 4) f block is 4 element wide, same shape for 4f, 5f etc Radial functions same for all nf orbitals Three angular nodes (nodal planes) orthogonal to s, p and d orbitals Penetration The RDF s of AO s within a given principle QN (n) have different shapes Number of nodes n l - n = 3 3s nodes 3p node 3d nodes 3s has a peak very close to the nucleus 3p has a peak close to the nucleus.5 3p These close peaks have a very strong interaction with the nucleus 3s is said to be the most penetrating 4π πr R(r).5 3d 3s Note the functional form of R(r) Constant * polynomial * r 3 * exponential 45 Penetration 3s > 3p > 3d 5 5 Radius (a.u) 46 Multi electron atoms Can not analytically ll solve the Schrödinger equation for multi-electron l t atoms We assume hydrogen like wave functions for multi-electron atoms Nuclear charge increases with atomic No. electrons repel each other and shield the nucleus from other electrons Effective nuclear charge Z eff = Z S S = a screening or shielding constant E.g. Li atom why is the electronic configuration s s and not s p? s electrons shields the valence electrons from the nuclear charge s penetrates more effectively feels a greater nuclear charge p penetrates less effectively s is filled first E(s) < E(s) < E(p) E(ns) < E(np) < E(nd) 5 Radius (a.u) 4πr R(r) s p s 47 Periodic table Shielding and penetration E(ns) < E(np) < E(nd) < E(nf) This gives rise to electronic configuration of atoms and the order of elements in the periodic table Electrons are filled in increasing energy (Aufbau principle) and electrons fill degenerate (same energy) levels singularly first to give maximum spin (und s rule) E(4s) < E(3d) s s K, Ca s p 3s 3p 4s 3d E(6s) < E(5d) E(4f) La [Xe] 6s 5d 6s 5d Ce [Xe] 6s 4f 4f 48

13 More complex results of penetration and shielding Energy levels vs atomic number For (Z=) all orbitals within a principle QN have same energy For multi electron atoms penetration follows s > p > d > f 3d shielded very effectively by orbitals of n 3 and 3d almost does not change in energy with Z until Z = 9 The energy of the 4s and 3d orbitals For K and Ca the E(3d) > E(4s), Sc on the E(3d) < E(4s) (but close) 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 diagram Result is that TM s loose 4s electrons first when ionized Energy 4p { 3d 4s { 4p 4s filled before 3d owever n = 4 does not shield 3d effectively energy drops Similar pattern for 4d and 4f 49 K Ca Sc Ti 4s 3d Increasing Z 5 Drawing representations of AO s Need to be able to draw AO s when considering their interactions in MO s So far diagrams have been to help visualise the 3D nature of AO s Simple drawings are all you need!!!!! Making Bonds Localised Bond Pictures Revision i of JF of Lewis Bonding / VSEPR Localised view of fbonding 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 st and nd row atom attains a noble gas configuration (usually) Shape obtained by VSEPR (Valence Shell Electron Pair Repulsion) e.g. + Each has a share of electrons 5 5

14 Lewis bonding Lewis bonding polyatomics ( O) Octet rule for main group elements / 8 electron rule for transition metals All atoms have (or a share of) 8 electrons (main) or 8 electrons (TM) Gives rise to noble has configuration Stability since all valence levels filled Diatomics F F + F F F electron shared bond order = F F O O + O O O 4 electron shared bond order = O O 53 Oxygen atom has 6 valence electrons and each hydrogen has electron + O O Lewis bonding in O Oxygen has 8 electron, hydrogen has electron noble gas config. Oxygen hydrogen interactions share electron O Oxygen also has two lone pairs Shape VSEPR Electrons repel each other (lone pairs repulsion > than bonding pairs) Oxygen has bond pairs and lone pairs 4 directions to consider Accommodate 4 directions Tetrahedral shape O is bent with -O- angle of 4.5 o Compares with a perfect tetrahedral of 9.45 o lone pair repulsion 54 Lewis bonding polyatomics (ethene) Lewis structures breaking the octet rule Used different symbols for electrons on adjacent atoms 4 + C C C Carbon atoms share 4 electron bond order = C C Carbon hydrogen interactions share electrons C Some structures to not obey the 8 electron rule. e.g PF 5 9 o F F P F F F F F P F o F F Shape VSEPR Electrons repel each other Carbon atoms have 3 directions bond to C and two bonds to Accommodate 3 bond direction o in a plane (molecule is flat) (only the electrons round the P are shown for clarity) F atoms have 3 lone pairs (6 electrons) + in each bond 8 P atom has 5 bond pairs with electrons each electrons! 55 56

15 TUTORIAL. What is the relationship between the possible angular momentum quantum numbers to the principal quantum number?. ow many atomic orbitals are there in a shell of principal quantum number n? 3. Draw sketches to represent the following for 3s, 3p and 3d orbitals. 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 a) Explain what the terms penetration and shielding mean. b) ow do these concepts help to explain the structure of the periodic table 5. Sketch the d orbitals as enclosed surfaces, showing the signs of the wavefunction. 6. What does the sign of a wavefuntion mean? An introduction to Molecular Orbital Theory Lecture 4 Revision of hybridisation Molecular orbital theory and diatomic molecules Prof G. W. Watson Lloyd Institute.5 watsong@tcd.ie 57 Last lecture d orbitals Radial wavefunctions, nodes and angular wavefunctions (shapes) f orbitals Radial wavefunctions, nodes and angular wavefunctions (shapes) Multielectron atoms Penetration and shielding Atomic orbital energies, filling and the periodic table Valence bond theory (localised electron pairs forming bonds) Lewis structures number of electron pairs bond order (electrons shares divided by ) VSEPR repulsion of electron pairs (BP and LP) molecular shape Valence bond theory and hybridisation Valence bond theory (Linus Pauling) Based on localised bonding ybridisation to give a geometry which is consistent with experiment. ybridisation constructs new hybrid atomic orbitals from the AO s Use Lewis model (number of electron pairs) hybridisation shape. E.g. Be, Be s s Be Correctly predicted by VSEPR to be linear can we explain it using AO s Mix S with p z orbital sp hybridized orbitals 59 6

16 sp hybridisation sp hybridisation Mix and a s and a p orbital two combinations s + p z and s p z Two AO s two hybrid AO s Relative sign is important when mixing orbitals sp therefore means that the hybrid orbital is 5% s and 5% p ψ sp = ( ϕs + ϕ p z ) + Lewis structure 3 directions C C Molecular is planar ybridisation sp hybridisation ψ = ( ϕ ) ϕ sp s p z = - + Three directions each at o mix s with p orbitals sp hybridisation 6 6 ybridisation π bonds For ethene sp hybridisation bonding in three directions Each local bond can hold electrons z ave not accounted for the second pair of electron shared by the C atoms Creates a π bond above and below the plane of the molecule Could think of the C as going from s p (sp ) 3 p x x ybridisation sp 3 For tetrahedral molecules we have to mix the s with all the p orbitals (sp 3 ) This give rise 4 equally spaced orbitals e.g. methane C 4 electron pairs sp 3 hybridisation tetrahedral O can also be thought ht of like this with two of the sp 3 orbitals occupied by lone pairs. O 63 4 electron pairs sp 3 hybridisation tetrahedral 64

17 Trigonal Bipyramidal sp 3 d 5 electron pairs (s + px + py + pz +dz ) ybridisation d orbitals Lone pairs equatorial ybridisation summary ybrid Atomic orbitals Geometry General Examples isation that are mixed formula octahedra sp 3 d 6 electron pairs (s+ px+ py+ pz+ dz + dx -y ) sp s + p linear AB Be sp s + p x + p y trigonal planar AB 3 BF 3, CO 3 - C 4 sp 3 s + p x + p + p y z tetrahedral AB 4 SO - 4, C 4, N 3, O, sp 3 d s+p +p +p +dz x y z dz Trigonal Bipyramidal AB 5 PCl 5, SF 4 s + p x + p y + p z + dx -y square pyramidal sp 3 d s + p x + p y + p z + dz + dx -y octahedral AB 6 SF 6 [Ni(CN) 4 ] - [PtCl 4 ] - Lone pairs trans Molecular orbital theory Molecule orbital theory (Robert Mullikan) Molecular orbital theory of - bonding molecule interaction of two hydrogen s orbitals ( ϕ a and ϕ b ) Electrons are delocalised Different to Lewis and hybridisation (these are not MO) Molecular orbitals are formed which involve all of the atoms of the molecule In phase interaction (same sign) ψ = ( ϕ a + ϕ b ) Constructive interference In Phase s + s Molecular orbital are formed by addition and subtraction of AO s Animation shows the in phase interaction of the s orbitals as they are brought together Linear Combination of Atomic Orbitals (LCAO) like hybrid AO s but the MO involves the whole molecule 67 68

18 Molecular orbital theory of - antibonding Charge density associate with MO s in molecule interaction of two hydrogen s orbitas ( ϕ a and ϕ b ) Out of Phase Out of phase interaction (opposite sign) ( ϕ ) ψ = ϕ a ϕ b Destructive interference s - s In phase interaction - charge density given by ψ ( ) = ϕ a ϕ ψ = [ ϕ ] + [ ϕ ] [ ϕ ϕ ] b ψ + a b + This gives an enhanced density where the AO s overlap between the atoms referred to as positive overlap and pull the atoms together (σ bonding) ψ =ψσ a b Node between the atoms Animation shows the out of phase interaction (different colours) of the s orbitals as they are brought together Out of phase interaction ψ = ( ϕ a ϕ ) ψ = [ ϕ ] + [ ϕ ] [ ϕ ϕ ] b a b This leads to reduced density between the atoms ψ =ψ referred to as negative overlap and pushes the atoms apart (σ * anti-bonding) a b σ * Interaction of AO MO s A general rule is that n AO n MO s 69 Can not create electrons New wave functions must be normalised to ensure probability in! 7 Energy level diagram for What happens when the AO s have different energies? Interference between AO wave functions bonding Constructively bonding interaction Destructively anti-bonding interaction Energy level diagram represents this interaction Two s orbitals interaction to create a low energy bonding and high energy anti-bonding molecular orbital Electrons fill the lowest energy orbital (same rules as for filling AO s) Bonding energy = ΔE Energy ΔE 7 ypothetical molecule where the two s orbitals have different energies E( ϕ ) < E ( ϕ ) ( a b What would the MO s be like Bonding MO will be much more like the low energy orbital ϕ a Anti-bonding MO will be much more like high energy orbital We can say that the bonding MO is σ σ ( C ϕ + C ϕ ) ψ σ = + Energy ϕ b ψ (anti-bonding) = ϕ b a a b b Where the coefficients C, indicate the contribution tion of the AO to the MO So for ψ σ σ σ a C b C > C ϕ a ψ (bonding) A AB B 7

19 Linear Combination of Atomic Orbitals - LCAO LCAO We wrote an equation using coefficients for the contribution of AO s to the bonding MO, we can do the same for the anti-bonding MO ( ) * * σ σ = ( ) σ σ ψσ Ca ϕa + Cb ϕ ψ * = C b a ϕa Cb ϕ σ b where the coefficients are different are reflect the contribution to each MO Generally we can write ψ No AO' s n n = C x x= a... x=abc a,b,c. (alloftheao s in the molecule) n =,,3..(the resulting MO s) ϕ x σ σ a C b C > * σ σ * a C b C < The sign can be adsorbed into the coefficient and we can write all of the MO s in a general way n = ψ = ( C ϕ C ϕ ) C a.8, C =. n n n ( C ϕ C ϕ ) ψ = + a a b b a a + b b ( C ϕ C ϕ ) = a a + b b = b n = ψ C a =., Cb =. 8 So MO() = C x MO() = MO(3) = ψ = Caϕa + Cbϕb + Ccϕc + Cdϕd +... ψ = C a ϕ a + C b ϕ b + C c ϕ c + C d ϕ d +... ψ = Caϕa + Cbϕb + Ccϕc + Cdϕd + - coefficients for MO(), - coefficients for MO() etc. C x... The coefficients contains both phase (sign) of the AO s and how big their contribution (size) is to a particular MO 73 And an examination of the coefficients tells us the bonding characteristics of the MO s 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 reversing the sign negative overlap Positive Overlap Positive overlap (constructive interference) on one side in cancelled by negative overlap (destructive interference) on the other E.g. s + s and p x + p x +ve s s and p x p x -ve s + p x positive overlap above the axis is cancelled by negative overlap below Same is true for the other interactions below Must define orientation and stick to it for all orbitals. Thus p z + p z -ve p z p z +ve i.e. for sigma bond between P orbital need opposite sign coefficients! Negative Overlap Zero overlap 75 76

20 An introduction to Molecular Orbital Theory Lecture 5 Labelling MO s. st row homonuclear diatomics Prof G. W. Watson Lloyd Institute.5 watsong@tcd.ie Last lecture ybridisation combining AO s on one atom to hybrid orbitals hybridisation made consistent with structure Molecular orbital theory (delocalised view of bonding) LCAO all AO s can contribute to a MO n AO s n MO s Filled in same way as AO s Example of Energy Molecular orbitals for AO s of different energy Linear Combination of Atomic Orbitals (LCAO) Use of coefficient to describe (i) phase of interaction and (ii) size of contribution of a given AO No AO s n ψ = ' ϕ n C x x= a... x ΔE 78 Labelling molecular orbitals ) Symmetry of bonding σ = spherical symmetry along the bond axis - same symmetry as s orbital no nodes pass through the bond axis (can be at right angles σ*) Labelling molecular orbitals ) bonding and anti-bonding (already met this label) Nothing if bonding g( (no nodes between bonded atoms) σ π π = one nodal plane which passes through the bond axis δ = two nodal plane which pass through the bond axis (end on d xy or d ) x y d zx + p x 79 Additional * if a nodal plane exits between the atoms, that is if the wavefunction changes sign as you go from one atom to the other. σ* π* 8

21 Labelling molecular orbitals 3) Is there a centre of inversion? i.e. is it Centrosymmetric? The final label indicated whether the MO has a centre of inversion symmetry y Li-Li Ne-Ne nd row homonuclear diatomics Possible interactions between s, s and p p x + p x As you go from one side of wave function through the centre of the bond the sign of the wavefunction reverses not centrosymmetric u = ungerade or odd p x - p x Wave function does not change sign centrosymmetric g = gerade or even σ bonding s s, p z p z π bonding px px, py - py MO s sometimes labelled with the type of AO forming them e.g. σ s or σ p 8 8 Energy level diagram for O Other possible interactions s and p energies sufficiently spaced to give little interaction Simple picture of the MO Unpaired electrons 6σ u 4σ u Paramagnetic Will σ interactions between s and p z be important Depends on energy difference bt between s and p z If large then no effect p π g π u p π g π u Label MO s starting from the bottom although often only valence orbitals ow does the energy of the s and p vary with Z (shielding / penetration) Energy s 5σ g 4σ u 3σ g σ u s s s σ O g O 3σ g σ u σ g Energy difference too big to interact with valence orbitals s AO s very small very small overlap in lower levels (small ΔE) 83 Energy p s sp mixing ii No sp mixing Li Be B C N O F Ne Gap increases p more effectively shielded - critical point between O and N 84

22 MO diagram for N s and p energies sufficiently close for interaction more complex σ and σ shift to lower energy 3σ shifted 4σ shifted to high energy 3σ now above π 4σ u π levels unaffected π g p p 3σ g π u Making sense of N Take basic model for oxygen no s p interaction Examine how the MO s can interact π and σ can not interact zero overlap π level remain the same Examine σ σ interactions p 4σ u π u π g 3σ u 3σ Bonding interactions can interact with each other σ u and 3σ u σ u σ s s σ u σ g s σ u Thus σ u goes down in energy and 3σ u goes up in energy Making sense of N Now examine the anti-bonding interaction (σ and 4σ) 4σ * MO diagrams for nd row diatomics The effect of the overlap between s and p is greatest for the Li. The MO diagram changes systematically as you go across the periodic table 4σ u p π u π g σ * Thus σ u goes down in energy and 4σ u goes up in energy 3σ u For bonding with anti-bonding (σ 4σ σ u or σ 3σ) the sign changes on one wave function zero overlap. s σ u s p mixing B paramagnetic and C diamagnetic 87 88

23 MO treatment of Be VSEPR linear molecule, Be Be s s p s Examine interaction of 6 AO with each other s, Be s and Be p x, Be p y Be p z 6MO s Z Energy level diagram for Be 4σ * u 3σ * g Interaction between s and Be s Interaction between s and Be p z p Non bonding p x and p y bonding anti-bonding bonding anti-bonding s σ u Each of these is delocalised over three atoms and can hold up to two electrons p x and p y have zero overlap non bonding 89 Be Be σ g Compare these two MO s with no sp mixing with the localised model of two Equivalent bonds formed via sp hybridization 9 Alternative approach Alternative energy level diagram for Be Step wise approach (ligands first) Mix hydrogen s orbitals first two Molecular orbitals Be Be Z 4σ * u 3σ * g Be Be p Non bonding p x and p y Then mix with s (zero overlap with pz) Then mix with p z (zero overlap with s) s σ u σ g Be Be 9 9

24 Last lecture An introduction to Molecular Orbital Theory Lecture 6 More complex molecules and CO bonding in transition metal complexes Prof G. W. Watson Lloyd Institute.5 watsong@tcd.ie LCAO Interaction of AO s with different energy lower AO has bigger contribution Representing contribution as coefficient AO interactions that were possible MO s positive, negative and zero overlap labelling of MO s (σ / π, *, g / u) nd row homo nuclear diatomics s p mixing occurs up to N energy different too big after this (O, F ) Difference in MO diagram for N and O Molecular orbital treatment of Be 94 MO treatment of O z O O is not linear but why? We will examine the MO s for an non linear tri-atomic and find out. What orbitals are involved s + O s O p x O p y and O p z x Three orbitals three MO s anti - bonding MO s of O z O x approx. -p z + Lets start by creating MO s from the hydrogen s orbitals. Taking the in phase pair first- it will interact with the O s and O p z (zero overlap with O p x and O p y ) p z s Approximately non-bonding approx. p z -s + + Bonding approx. s + p z (a little) + O Problem - This is mixing three orbitals must produce three orbital 95 96

25 Out of phase s orbitals Only interact with p x MO s Zero overlap with p y MO s of O z O x Energy level diagram for O z O x There are not two lone pairs! p x anti - bonding p s p y, n.b. σ n.b. σ non bonding p y Slightly bonding Pz Non bonding Py Very different to the VB concept of two identical sp 3 filled orbital σ O Bonding O O MO theory correct Comparison of O and Be Both cases of XA same MO different No of electrons Bonding MO s as a function of A-X-A bond angle π MO s of Benzene π bonding is more important for reactivity independent of σ (zero overlap) six p x orbitals combine to form six MO s Be linear B 3 o C (t) 34 o C (s) o N o 3 O 5 o Different ways of arranging six px orbitals on a ring Lowest energy e all in phase Degenerate levels ( nodal plane nodes) Degenerate levels ( nodal planes 4 nodes) ighest energy all out of phase (3 nodal planes 6 nodes) Energy increases with number of nodes as in AO s Also the number of nodes on a ring must be even continuous wavefunction Lowest energy all in phase All coeffcients the same 9º O C (T) 8º 99

26 π MO s of Benzene Next occupied degenerate pair nodal plane Two ways of doing this between atoms and through a pair of atoms MO diagram for CO Same orbitals as homo nuclear diatomics different energies give rise to significant s - p mixing confusing set of orbitals 4σ C-O anti - bonding (more C) p π π (uneven more carbon) As the wavefunction goes through (at the node) the smooth wavefunction has smaller coefficients next to the node zero at the node p 3σ Primarily carbon (p z ) π π bond (uneven more oxygen) electron per MO spread over 6 atoms Compare with Lewis structure with individual double bonds With local bonding have to resort to resonance structures to explain benzene s s σ Primarily oxygen (p z ) σ C-O bonding interaction (more O) The OMO and LUMO of CO For chemical reaction the OMO (ighest Occupied Molecular Orbital) and the LUMO (Lowest unoccupied Molecular Orbital) are the most important. Interaction of the CO 3σ with d orbitals Three sets of interaction based on symmetry of ligand AO s Generally applicable to σ bonding TM ligands OMO 3σ LUMO π* low energy Oxygen orbitals Comes from standard π interaction makes σ mainly O pz however lower oxygen orbital in 3σ mainly C pz means π has has more oxygen and π* more carbon Some anti-bonding mixes in due to sp mixing π π a g all ligand AO s in phase Interaction with s orbital t u ligands in one axis contribute With opposite phase one nodal plane Interaction with p orbitals 3 3σ C O C O C O 3 4

27 Interaction of the CO 3σ with d orbitals MO diagram for Tm (σ-l) 6 e g ligand phases have two nodal planes Interact with d d z x y t * u a * g Electrons from filled σ orbitals on the ligands fill all the bonding orbitals 4p 4s 3d Δ oct e * g t g d electrons fill t g (non bonding) and e * g (antibonding) Example is d 6 e.g. Co 3+ Three remaining d orbitals point between ligands zero overlap (t g ) These are the orbitals considered in ligand filed theory. Note the e * e 6 σ ligand g g is anti-bonding orbitals! t u a g What really decided Δ oct is the π interaction 5 6 π interactions with TMs Orbitals with π character can interact with the t g d orbitals Must be correct symmetry (t g ) 3 arrangements g possible using d xy, d yx, d xz Low ligand field situation Ligand orbitals are low energy and filled (e.g. F) Filled orbitals interact in a π fashion Bonding combinations are reduced in energy and filled (like ligand orbitals) Antibonding combination are raised in energy and filled (like d orbitals) Two extreme situations Ligand orbitals are low energy and filled (e.g. F) Ligand orbitals are high energy and empty (e.g. CO) e * g e * g Δ Interaction with high energy empty ligand orbitals e * g Δ oct Interaction with low energy filled ligand orbitals Δ oct t * Δ oct t g t g g?? t g 7 without π with π ligand Strong interaction with filled orbitals with π interaction leads to reduction in Δ oct (box shows the orbitals considered in ligand field theory) 8

28 igh ligand field situation Ligand orbitals are high energy and empty (e.g. CO π*) Filled orbitals interact in a π fashion Bonding combinations are reduced d in energy (like d orbitals) Antibonding combination are raised in energy and empty (like ligand orbitals) Tutorial part a. Explain the MO approach for the interaction of a) two s orbitals of identical energy b) two s orbitals of slightly different energy c) two s orbital of very different energy. e * g e * g Δ oct Δ oct t g tg without π with π ligand Strong interaction ti with empty orbitals with π interaction ti leads to increase in Δ oct (box shows the orbitals considered in ligand field theory). Consider the bonding in the molecule O a) Draw a Lewis structure for O b) Determine the hybridization c) Perform an MO treatment of O (i) What orbitals are involved? (ii) what interactions are possible? (iii) what do the resulting MO s look like? (iv) sketch an MO energy level diagram. d) What difference are there in the details of the bonding diagram between the Lewis and MO treatments 9 Tutorial part b 3. Consider the molecule Be a) Draw a Lewis structure b) Determine the hybridization c) Perform an MO treatment of O (i) What orbitals are involved. (ii) Generate appropriate ligand MO s and interactions with the central atom (iii)what do the resulting MO s look like? (iv)sketch an MO energy level diagram. d) What difference are there in the details of the bonding diagram between the Lewis and MO treatments TE END 4. Perform the same analysis for Be, F, B 3, and C 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.