:C O: σorbitals of CO. πorbitals of CO. Assumed knowledge. Chemistry 2. Learning outcomes. Lecture 2 Particle in a box approximation. C 2p.

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1 Chemistry 2 Lecture 2 Particle in a bo approimation Assumed knowledge Be able to predict the geometry of a hydrocarbon from its structure and account for each valence electron. Predict the hybridization of atomic orbitals on carbon atoms. Learning outcomes Be able to eplain why confining a particle to a bo leads to quantization of its energy levels Be able to eplain why the lowest energy of the particle in a bo is not zero Be able to apply the particle in a bo approimation as a model for the electronic structure of a conjugated molecule (given equation for E n ). πorbitals of CO More carbon character σorbitals of CO 2sσ 2sσ* 2pσ π* C 2p The unequal energies of C and O 2p orbitalsmeans that the miing is incomplete. This results in the p bonding orbital being polarized to the oygen, and the antibonding orbital polarized to the carbon. This is important for inorganic chemistry π O 2p More oygen character σ-bonding orbital Polarized to oygen lone pairs. ighest occupied σ orbital is localized on carbon lone pair this is important for inorganic chemistry :C O:

2 sp hybridization acetylene (ethyne) When one mies the 2s and the 2p to make an sp hybrid orbital, one must obtain, afterwards, two orbitals. These are the plus and the minus combinations. There are 1 valence electrons 2 go into C-C bonding σ orbital (sp + sp) 2 go into one of the C- bonding σ orbitals (sp + 1s) 2 go into the other C- bond We are left with pristine (unhybridized) 2p and 2p y orbitals. 2 electrons go into 2p π bonding orbital 2 electrons go into 2p y π bonding orbital CC bond is triple σ and 2 π minus combinations are anti-bonding and are not occupied. C C σ σ σ π π sp 2 hybridization In order to make two σ bonds, sp hybrids are constructed. In some molecules, such as ethylene (ethene), the carbon atom must make three sigma bonds. To achieve this, the 2s orbital is mied with two 2p orbitals to make three sp 2 hybrid orbitals. There are 12 valence electrons ethylene (ethene) 2 go into C-C bonding σ orbital (sp 2 + sp 2 ) 2 go into each the C- bonding σ orbitals (sp 2 + 1s) (8 electrons) We are left with pristine (unhybridized) 2p z orbitals. 2 electrons go into 2p π bonding orbital σ σ CC bond is double σ and π minus combinations are anti-bonding and are not occupied. π

3 sp 3 hybridization In order to make four bonds, sp 3 hybrids are constructed. All saturated carbon atoms are sp 3 hybridized. To achieve this, the 2s orbital is mied with two 3p orbitals to make four sp 3 hybrid orbitals. sp 3 hybridization The bonding orbitals in methane are sums of sp3 hybrids on the carbon, and 1s orbitals of the hydrogen. 1s There are four sp3 hybrids since they are constructed from four atomic orbitals, (2s + 3 2p). They point at the corners of a tetrahedron and are 19 apart. C sp 3 sp 3 hybridization All tetrahedral atoms are considered to be sp 3 hybridized. σ-framework The electronic structure of most molecules may be simplified by considering the σ framework to be constructed from σ-bonding orbitals, with anti-bonding combinations unoccupied. Where carbons are sp 2, or sp hybridized, there are unhybridized p orbitals which may be used to construct π-bonds. π-system O : N : : water ammonia σ-framework C C C

4 σ interactions: ethylene σ interactions no nodes ( ignoring inside) one node Same energy, interact and mi one node C-C σbond appears to be positive overlap of sp 2 hybrids two nodes two nodes σorbitals do not interact with πorbitals π orbital of ethylene ψ ˆ ψ d τ = 2 1 = cancels ψ σ ψ π There is no net interaction between these orbitals. The positive-positive term is cancelled by the positive-negative term C-C π bond appears to be positive overlap of unhybridized p orbitals

5 σ orbitals of cis-butadiene π orbitals of cis-butadiene three nodes two nodes The four p orbitals all have the same energy interact and mi one node no node (ignore plane) The σmolecular orbitals are mitures of the sp 2 hybrids π orbitals of heatriene The si p orbitals all have the same energy interact and mi Maybe we can model the π-electrons as confined to a one dimensional bo?

6 The Schrödinger equation The total energy is etracted by the amiltonian operator. These are the observable energy levels of a quantum particle The Schrödinger equation The amiltonian has parts corresponding to Kinetic Energy and Potential Energy. In one dimension, : Energy eigenfunction Potential Energy amiltonian operator Energy eigenvalue amiltonian operator Kinetic Energy The particle in a bo The bo is a 1d well, with sides of infinite potential, where the electron cannot be The particle in a bo The bo is a 1d well, with sides of infinite potential, where the electron cannot be E E L L

7 The particle in a bo The particle cannot eist outside the bo Ψ = {<;>L (boundary conditions) The particle in a bo Let s try some test solutions Ψ = sin(π/l) {>;<L E E?? L L The particle in a bo The particle in a bo Zero potential inside bo Lowest energy possible is ε = ħ 2 π 2 /2mL 2 This is called zero point energy (ZPE) V =εψ!!! L

8 The particle in a bo Other solutions? Ψ = sin(2π/l) {>;<L The particle in a bo Other solutions? Ψ = sin(3π/l) {>;<L V V L L The particle in a bo Other solutions? Ψ = sin(4π/l) {>;<L V The particle in a bo Other solutions? Ψ = sin(nπ/l) {>;<L V ε n = ħ 2 n 2 π 2 /2mL 2? L L

9 The particle in a bo Ψ = sin(nπ/l) {>;<L;n> ε n = ħ 2 n 2 π 2 /2mL 2 ε n The particle in a bo Ψ = sin(nπ/l) {>;<L;n> ε n = ħ 2 n 2 π 2 /2mL 2 Philosophical question: why is n = not an appropriate solution? int: what s the probability of observing the particle? Consequences of confinement For a particle in a bo, there eist solutions to the Schrödinger equation (eigenfunctions) of the form Ψ = sin(nπ/l) with energies (eigenvalues) ε n = ħ 2 n 2 π 2 /2mL 2 These are quantized energy levels! Increasing L decreases spacing Increasing m decreases spacing (correspondence principle) Boundary conditions We impose the boundary conditions on imaginary atoms one bond length beyond the edge of the π-system.

10 A word about nitrogen Application: Polymethine dyes When a nitrogen is adjacent to a π-system, it may possess sp 2 hybridized resonance forms which allow it to participate in that π-system. e.g. amides If the resonance forms are not favourable, they may only contribute in a minor way and the nitrogen will still have a pyramidal structure to some etent. In amides, the oygen stabilizes the resonance structure and the nitrogen is planar. Adapted from Brooker, JACS 62, 1116 (194). Application: Polymethine dyes Polymethine dyes Question: how many π-electrons in the a- Brooker dye? Question: what is the length over which the π- electrons are delocalized, if the average bond length is 1.4 Å? Answer: firstly, we should recognize that the π -system etends over both nitrogens. There are two resonance forms. By symmetry these will involve double bonds with either nitrogen. Looking carefully at the structure, there are 2a+3 carbon atoms which each contribute one electron each. There are two nitrogens, one which contributes two electrons, and one which contributes one electron, since it bears a positive charge. Therefore, there are 2a+6 electrons. A = ε.c.l (Beer s Law) Answer: Looking carefully at the structure there are 2a+4 bonds in the π -system. If we add a bond at each end of the system, we get 2a+6 bonds in length, which equates to L = (2a+6) 1.4 Å. Question: if the energy levels of the electrons are given by ε n = ħ 2 n 2 π 2 /2mL 2, what is the energy of the OMO in ev if a=1? Answer: since there are 2a+ 6 π-electrons, there must be 8 electrons, and therefore the OMO must have n=4. We know that L = Å = 11.2 Å. From these numbers, we get ε n = n 2 in Joules. The energy of the OMO is thus ε 4 = J = 4.79 ev.

11 Polymethine dyes Question: what is the energy of the LUMO, and thus the OMO-LUMO transition? ow does this compare to eperiment (see figure)? Effect of chromophore size Answer: ε n = n 2 in Joules. The energy of the LUMO is thus ε 5 = J = 7.49 ev. The energy of the OMO-LUMO transition is thus 2.69 ev. This corresponds to photons of wavelength λ = hc/( ) ~ 46nm. This is not so far from the eperimental value (about 55nm), which is about as good as we can epect. Indeed, the R groups in the picture are, in fact, themselves conjugated. Inclusion of these as part of the electron track will improve the model considerably by further delocalizing the electrons and thus bringing the energies down closer to the eperimental values. Chromophore Wavelength (nm) Something to think about Particle on a ring Must fit even wavelengths into whole cycle Summary The particle in a bo problem can be solved eactly and is a good first approimation for the electrons in a delocalized π- system. Confining a particle in a bo leads to quantization of its energy levels due to the condition that its wavefunctionis zero at the edges of the bo The lowest energy (ZPE) of a particle in a bo is not zero Be able to apply the particle in a bo approimation as a Be able to apply the particle in a bo approimation as a model for the electronic structure of a conjugated molecule (given equation for E n ).

12 Net lecture Particle-on-a-ring model Practice Questions 1. The energy levels of the particle in a bo are given by ε n = ħ 2 n 2 p 2 /2mL 2. (a) Why does the lowest energy correspond to n = 1 rather than n =? Week 1 tutorials Particle in a bo approimation you solve the Schrödinger equation. (b) (c) (d) What is the separation between two adjacent levels? (int: ε = ε n+1 - ε n ) The π chain in a heatriene derivative has L = 973 pm and has 6 π electrons. What is energy of the OMO LUMO gap? What does the particle in a bo model predicts happens to the OMO LUMO gap of polyenes as the chain length increases?