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CHM3122 (7)


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Kathy Focsaneanu

• slide 2: shape and sometimes intensity of spectrum will depend on what kind of species that's absorbing. • slide 3: atomic spectra - don't have mole orbitals, just have individual atoms, orbitals are atomic orbitals. atomic orbital are quantified 1s, 2s, etc. when absorb photon, going from a specific atomic orbital to another specific atomic orbital. results in corresponding delta E results in a narrow spike/line in absorption spectrum. energy corresponds to that exact wavelength. this is how Bohr constructed his model of the atom. each elements provides it's own atomic spectrum and can use these spectra to find out what elements are in a sample. • slide 5: 2nd atom and make a bond -diatomic molecule. have a bond that can vibrate, molecular orbitals and vibrational suborbitals. now have a ground state excited state and including vibrational sub levels in excited states. various vibrational modes. if starting with a molecule in its ground state (energy min) if looking at gaussian distribution of molecules, vast majority of them will be around the energy min meaning most polluted state will be v=0, lowest of low. a fraction will be in the v1 or v2 but vast bulk will be on the bottom of the potential energy well. have population of molecule at min energy and they absorb photons, now they can land in various vibration sub levels in the excited state and they are well separated. can go from zero to zero, zero to 1, ego to 2, 0 to 3 depending of energy of othon absorbed. • slide 6: series of spikes, represent different vibrational dub levels in the excited state. can use the spectrum the fine structure giving info about the vibrational sub levels in the excited state. reflects the speearatin. look at delta E/space in between the spikes, space is proportional to the energy levels between the vibrational sub levels of Sn. can use absorption spectrum to find out info about the excited state itself. notice don't have the same intensity for the molecules, why? has to todo with probably if the transition. most probable will give yo biggest intensity. geo of the excited state is the same as geo of ground star - optimal geo of ground state is wrought the same internuclear distances and bond angles. if it looks the same in ground and excited, lowest energy is most probable. easiest to do is 0-0 therefore most probably. as you go higher in energy, slightly less probable, thats why relative intensitydecreases the spikes go from right to left because it's the lowest energy, meaning highest wavelength. • perfect alignment of the morse curves. most likely transition is the one that's easiest to do. the easiest transition is the one from the most populated state in s0 which is the lowest vibrational level. least is the one going straight up to sN. smallest energy jump - most likely transition in the molecule. when you look at the intensities of the peak, going left, increasing in energy - increasingly high energies form the sublelvel excited states. because the 0-0 band is the most intense, it's the most favourable transition in the molecule i.e. they're both equivalent. • slide 7: plot of potential energy and internuclear distance. most diatomic molecules, 2 morse curves for the ground and edited state. morse curve for excited state is right on top of the ground state curve. i.e. preferred geometry in excited state is the same for both. if you go straight up, favourable band is the 0-0 band (0-0 absorption) . if starting off with ground state, excite and going to have some vibrational relaxing until you get to preferred planar geometry. morse curve will not but in the same position. • slide 8: no curve directly over other curvee. optima geo in ground state is not the same as optimal geo as other. internculear distance has changed. must look at wave functions. for each state, has corresponding wave functions to describe electrons in the orbital. as a result, in order to have absorption, need to have construvtice overlap between wave functions between ground and excited state. two waves being added together. why is it important to go straight up? diagonal line implies absorption of a photon but also implies that over the course of the absotion, atoms must have moves. went up in energy and at the same time moved the atoms. the diagonal line is not permitted. implies that we abrobed a photon and moved hr atoms which is not possible. may not have 0-0a as most favourable, because you must have constrictive overlap. land in v=4. bcaus exit's the most probably transition, gives you the most intense band. some other band than the 0-0 band is the most intense. when you see a pattern like that, geo of ground state is not the same as the excited state. must get downy o the owes energy possible. vabrationally relaxed until it reaches the optimal geometry relaxes from v4 to v0. now in the lowest viratianln sub level. singlet excited state is ready to make the choice to emit or not emit a photon or use something els.e • s;ode 9: absorption of a photon is about 1 femtosecond. if you look at the atoms themselves, nucleus is about 2000x the mass of an electron. on the time scale, not enough time to move the atoms themselves/nuclei because they're too heavy. if you look at vibrational time scales, can be a lot longer than a femtosecond. on the time scale of absorption, can imagine the nuveli are frozen; do not move. when you absorb a pton, you freeze a molecule (the franck-condon prinicple_ this implies on the diagrams hat only vertical;l transitions are allowed, must go straight up. must go vertically up whenever we absorb a photon. band intensity absorbed in the spectrum is strictly related to probability of the transition which is depenedient of the constructive overlap i.e. can only go straight up. • slide 10: as soon as you add a third atom, have vibrational sub level, have rotational sub-sub levels. can land anywhere in the band, lose fine structure - well defined peaks. get a lagr broad band because any possible combo will land you in the continues band of energy in the excited state. can get broad bands. not in all cases though. • ex. pyrene (slide 11) some relatively sharp peaks separated by a constant value (delta E). because you have fine structure, this implies that you have well separated vibrational sub levels in the excited state. if you include the rotational sub levels, the definition disappeared which is why you get bands if you have nice peaks, the sub levels must be well defined meaning there's no rotation. it's a relatively rigid molecule. because this has a lack of rotational possibilities, you get a separation of excited states. the 0-0 band is the most intense, this implies that the excited state is the same as the ground state, essentially equal. therefore 0-0 band is the largest. geo of Sn is equal to the config of S0. • slide 12: relatively close. the energies between he bands are in the UV vis region. the possible transitions are the HUMO LUMO transition, sigma star orbital is accessible via UV vis photon, could put it in sigma * energetically speaking. another one: instead of HOMO electron, can take HOMO -1 is pie orbital, can take the lectern and put it in the next empty orbital. pie ti pie star. can take sigma electron and put it in sigma star orbital. has to do with some orbital difficulties - wave functions. can see the increase in energy in the transitions. easiest is the n to pi *.reason why is to realize that energy is a factor but orbital overlap is a factor as well. fi you take an electron from sigma to sigma stare, very little constrictive overlap. can rule out 2 because the orbital overlap is unfavourable although it's theoretically possible. n to pie* and pi to pi* stare are the 2 likely ones. for any carbonyl containing molecule. • slide 13: any molecule that has carbonyl molecule, two electrons in pie and two electrons in the n in ground state. just look at slide. for each of the states, two possibilities. could have signet n to pi* and triplet n to pi* same thing for pi to pi star. • slide 14: the behaviour of the excited state depends on the type of the excited state. behaves differently. in the n to pi*, single unpaired electron on the oxygen (n). this excited state behaves like an alkyl radical. has radical like properties because the oxygen does;t know where its electron went. all it knows is that it has a single unpaired electron therefore it will try to find another electron. pi pi* state will behave otherwise. put an electron in the non bonding orbital. bond order = # of bonding electrons - number of electrons in anti bonding orbitals divided by 2. 2-0/2=1 for pi bond. 1-1/2 = 0. effectively, destroyed the carbon oxygen double bond, weakened carbonyl which makes it susceptible to nucleophilic attack. • slide 16: look ay the delta Es. band gap (energy gap) between singlet and corresponding triplet state is bigger. as you get higher in E, gap between Sn and Tn is bigger and bigger. going from signet to triplet, is easiest between S1 and T1. no chem form Sn. • slide 17: HOMO is always the pi bond and the LUMO is always the Pi* because theres no oxygen. only transition you see is the pi to pi*. no oxygen means no n orbital which means no n to pi*. energies of pi to pi* shown. on righthand side, electromagnetic spectrum. at the top, short wavelengths, long wavelengths on the bottom, benzene's s1 lies at 110 k/cal. go across, you're in the UV region. if we add another aromatic ring i.e. increase conj. singlet state has dropped form 110 to 90 which absorbs in the UVA region. if you add another ring, singlet energy drops to 70 which is at 400nm, getting into the visible region. as you increase conjugations associated with pi pi*, decreases the s1 state, brining down the LUMO, HOMO and LUMO are getting closer in energy, delta E is getting smaller. this means that when we take the absorption spec, can work backwards. measure absorptions spectra. • slid 20: add negative sign in front of the equation. I0 is the amount of incident light on a substance (incoming beam of light). intensity of the transmitted light is It. if the solution has absorbed anything, It is lower then I0. take negative log, gives you a positive integer number. proportional to 3 factors, molar extinction coefficient, second factor is the concentration, 3rd factor is l, the path length. absorption has no units. conc has a unit of mol/L. mol extinct is molar-1cm-1 • has possibly value for absorbance is about 2 because eyou have 0% transmission. the consequence of this is that in a lab, generally speaking, when we get to an absortipion about 1, very little of the light is getting through, less that 10%. in the lab, it means the solution is absorbing most of the photons, can induce some errors. ie. when measuring absorption, stick to the middle of the graph. don't want values that are too low and don't want values that are too high. want to be in the middle. minimizes sources of error. • slide 21: value of 3 is unique for every molecule. it is unique at every wavelength. as concentration increases, aborption increases as well. pick the lamb max (max absorptions). 335 nm. make a beer lambert graph out of the data points. if you plot absorption at the wavelength vs conc and use beer lambert law, 3 is the slope. goes through 0. slope is delta Y divided by delta x which gives you 8500 M-1cm-1. x axis is in micro molar.use a subscript to indicate the wavelength 3 was calculated. sometimes it matters which
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