Chapter 15

2. Explain the similarities and differences between molecular fluorescence, phosphorescence and chemiluminescence

ALL of these are emission techniques of electrons. Fluorescence and phosphorescence both occur due to the excitation of an electron via absorption (absorption of photons), while chemiluminescence occurs via chemical reaction. Phosphorescence has a longer lifetime and lower energy than fluorescence (triplet state). Fluorescence has a higher energy and a shorter lifetime. The intensity of the excited state for both depends on a number of environmental factors. Chem: Background is low because there is no scattered low external source. Lowering the noise, you can detect the low signal.

Discuss the advantages and disadvantages of molecular luminescence over absorption techniques.

Absorption: high background since the electrons are hitting the detector directly but luminescence is at a 90 degree angle so the electrons are barely hitting the detector thus leading to a smaller background noise. An advantage is that only some species will flouresce but can also be an advnatage since you can find out which wavelength is but the way it fluoresce

Why is it not necessary to have a wavelength selector when performing a chemiluminescence experiment?

An external light source is not required, so there is no wavelengths involved. Generally, no wavelength selector is necessary for performing a chemiluminescence experiment since the only source of radiation is the chemical reaction between the analyte and reagent. Therefore, a wavelength selector would not be necessary because there is no external photons necessary. Also, since only one of the species in solution is emitting photons, an emission monochromator is not necessary, unless you want to measure to wavelengths emitted by the sample to get an emission spectra.

Discuss the effect of concentration on fluorescence intensity?

At low concentrations, fluorescence activity is linear, but as the concentration increases, linearity is lost. This is due to the phenomenon of self quenching and self absorption of molecules in the solvent. Self absorption is the occurrence of the wavelength emitted overlapping and interfering with absorption peaks. These activities decrease the chances of emitting a photon.

Compare resonance fluorescence to Stokes shift?

Atomic species undergo resonance fluorescence energy because you attract the atom at a certain wavelength and that same energy of the photon is being emitted. The reason for that is because they don't have any vibrational rotational states associated with them Molecular species undergo stokes shift. Once they are excited, they will lose energy before they are ever emit a photon· The reason for that is because they do have vibrational rotational states associated with them.

Discuss two fluorometric methods for the determination of inorganic species?

Direct method forms a type of "fluorescence complex" and measures its emission. This is similar to the absorption complex, in which two non fluorescing species are formed into a complex that fluoresces. (Molecule is not being absorbed but if attached to it then there is complex fluoresce) Indirect uses the quenching effect using heavy atoms. As the concentration of the heavy atom (Br) increases, the fluorescence will decrease. This will be seen as a negative calibration curve, which can be used to measure your analyte.

How are fluorescence and phosphorescence different in terms of intensity, lifetime, and energy? Test question!

Fl has a higher energy than ph because there is no intersystem crossing therefore there is no loss of energy and fl emits the same amount of energy that was being absorbed. The lifetime of ph is longer because the electrons stay in the excited state longer compared to Fl. The intensity depends on the experimental conditions like viscosity, rigidity, and etc.

Discuss the various factors that affect fluorescence and phosphorescence.

More aromatic rings= greater chance of fluorescence More resonance structures= greater chance of fluorescence Less substituents= greater chance of fluorescence Increase in structure rigidity= greater chance of fluorescence Decreases the opportunity of losing energy in vibrational/rotational states Environmental factors Decrease in temperature= greater chance of fluorescence Increase in solvent viscosity= less collisions; greater chance of fluorescence Lower concentration= greater chance to of fluorescence Increase in concentration increases the likelihood of self quenching and self absorption pH; Placing dissolved oxygen in a solvent promotes the conversion of excited molecules to a triplet state, encouraging phosphorescence · Heavy atoms; increase rate of triplet formation, therefore increasing phosphorescence and decreasing fluorescence.

Describe some of the structural features of a molecule that affect fluorescence. Will be on test!

Most unsubstituted aromatic compounds fluoresce in solution, with the quantum efficiency increasing with the number of rings and their condensation. Meaning that due to the increased amount of fused rings it will cause an increase in the probability that they will fluoresce. If in the case of heterocycles such as pyridine, furan, thiophene, and pyrrole which do not have fused rings will not exhibit fluorescence. Although if a hetero compound has an extra aromatic ring with it, it will fluoresce. Increased resonance structures lead to enhanced fluorescence.

Why are fluorescence methods more sensitive and selective than absorption techniques?

Sensitivity is due to the 90 degree angle of the fluorescence instrumentation, which allows for no scattering radiation.therefore there is very few photons that are going through the photons. The probability of measuring the photon through the external source, is low. Therefore is low sensitivity. Absorption through background with a bunch of photons. Through the sample through the detector but you are messing the decrease of it. So the background is high (Less photons hit detector) Selectivity is due to the limited amount of molecules that are able to fluoresce. Most molecules don't fluoresce, Most photons will absorb in some fashion (the viscosity, rigidity, resonance structures, etc. )

What effect does temperature, concentration, and solvent viscosity have on fluorescence and why?

Temperature: in decreasing the temperature we slow down the molecule and decrease the probability for collisions which create a loss of energy in a non radiative form which decreases the probability for fluorescence. Viscosity: by increasing the viscosity we are slowing down the molecule in another way which decreases the probability for collisions and a loss of energy in a non radiative form and thus decreases the probability for fluorescence. Heavy atom: this decreases the probability for fluorescence ( increasing the probability for phosphorescence) by adding heavy atoms. Increase the intersystem crossing so by adding the heavy atoms we get the formation of triple state electrons causing a higher probability of phosphorescence and decrease in fluorescence. Concentration: if there is an increase in concentration we see the decrease in the probability for fluorescence. This is due to 2 factors: Self-quenching: when the electrons is ready to emit a photon and collides with another causing the energy to be passed to on reducing the probability for fluorescence. Self-absorption: when the electrons is ready to emit a photon and as it emits it is absorbed by another electron before it is able to exit the cell also reducing the probability of fluorescence. These are also causes for an increase in the rate at which we lose linearity from beer's law.

Draw a Jablonski diagram (Figure 15-1 page 357). Label and discuss each of the diagram. Be as thorough as possible.

The Jablonski diagram represents what occurs during fluorescence and phosphoresce in terms of absorption and relaxation. The absorption process is a two-step process, the first step is the absorption of a photon, and the second step is relaxation. The solid lines going up are the absorption of a photon, and the lines going down are the loss of energy due to some type of radiative or non-radiative relaxation. The wiggly lines coming down are nonradiative relaxation, in other words the loss of energy through some other means than the emission of a photon. The solid lines coming down are the loss of energy due to the emission of a photon (radiative). Excited ground state = if an electron is in an excited orbital, each of those orbitals are going to have vibrational and rotational associated with them. If the electron loses all the vibrational and rotational energy in that particular orbital, then its going to be in the excited ground states (as depicted by the bold lines in the diagram) Intersystem conversion = once a particular orbital is in its excited ground state, if one of the vibrational or rotational states associated with another molecular orbital matches, then the energy can be transferred to another part of the molecule. This is a nonradiative form of relaxation is known as internal conversion, in other words the energy is being transferred to another part of a molecule (remember this is quantized). S1 and T1 = S1 and S2 represent the singlet states. This is when the photons absorb and excites a molecule up to a higher molecular orbital, but the spin states for the particular electrons are matching (in other words, opposite spins). If one of the electrons flip spins, and the spins are the same, then the molecule cannot lose its energy until one of the electrons flips spins, this is known as the triplet state. Electrons stay in the triplet state much longer than in singlet state. Intersystem crossing = this is where the electron is going from the excited ground state of S1 to one of the vibrational rotational states of T1. Electrons cannot transfer their energy from the excited ground state of S1 to the excited vibrational states of T1 unless the energy associated with the excited ground state for S1 matches the energy associated with the vibrational/rotational states of T1. External conversion = external conversion is once a molecule becomes excited it can lose its energy to another molecule through collision.

Why is fluorescence seldom observed from sigma* to sigma transitions?

The single bond is difficult to detect in a required UV vacuum, and these bonds are more easily ruptured than by double or triple bonds. Single bonds in the vacuum UV range (sigma to sigma*) have the greatest energy difference, so bond rupture is easily achieved. Molar absorptivity of single bonds is very low. Single bonds are more difficult to detect because the atmospheric components can be easily absorbed by the vacuum UV.

Explain the relationship between vibrational relaxation and fluorescence.

Vibrational energy is going to be lost before fluorescence, with fluorescence being emitted at the excited ground state. Fluorescence lifetime is much longer than vibrational states, so fluorescence always occurs from the excited ground state. The vibrational energy is relaxed between energy states because it is lost too quickly to fluorescence.

List and explain the various ways an excited molecule can lose its energy (relaxation/ deactivation processes).

Vibrational relaxation: Occurs quickly involves the transition from the lowest vibrational level of an excited electrons state ( one of the reasons for stokes shift). Happens from some excited ground state Internal conversion: Transfer of energy from one part of the molecule to another. The excited ground state matches the vib/ roto state of another molecular orbital. Can cause predissociation where an electron in the molecule is at a high energy state and moves to a lower energy level and causes enough vib/ roto in that state to cause bond rupture. Disassociation: when a photon is absorbed it causes a electron to jump up to a higher energy orbital with associated vib/roto state that directly causes bond rupture. External conversion: Collisional quenching A molecule gets excited and collides with another an causes that molecule to transfer its energy to another molecule. Done by increasing the viscosity or lowering the temperature to slow down the number of collisions Intersystem crossing: Going from a singlet state to a triplet state Flipping of electrons Can be enhanced by the heavy atom effect Chlorine, borine, fluorine, iodine

Why is chemiluminescence such a sensitive technique?

You are decreasing the background noise because you don't have any external light coming through. Therefore, you don't have those carrying the photons that hit that detector.

Discuss how the Pauli Exclusion Principle is related to fluorescence and phosphorescence.

he Pauli Exclusion Principle says that an electron orbital must contain opposing spins of two electrons. In phosphorescence, an electron is kicked from a singlet to a triplet state, where the electrons now have the same directed spins. The electrons will maintain this position for a long time, and as per the Pauli Exclusion Principle, the electrons cannot come back down (be emitted) unless the electrons flip their spins back to opposing directions. In fluorescence, the electrons are not flipped, therefor the Pauli Exclusion Principle is always obeyed, and the electrons are not emitted