3c - w - electronic spec Flashcards
how can we respresent electronic transitions : what can we draw
we can draw a jablonski energy level diagram
s0 , s1, s2, s3
where s0 is the ground electronic state .
s0 –> s1 = less energy that s0 —> s4
between electronic transtions there areeee,, aka in the jabloski diagram,, between the different s levels there areeee
there are vibrational energy levels.
finding energy difference between peaks in abs spec// electronic spec
convert wavelength to wavenumber : 1/wavelength
then do the larger energy - the lower energy.
gross selection rule of electronic spec
there are no gross selection rules: u just move the e- to higher energy levels.
interaction with the electric field always induces a change in dipole moment!!!!
wait so can homonuclear diatomics give an electronic spec
yessss!! a homonuclear diatomic can give an electronic spec: bc interaction with the EM always induces a change in the dipole moment
what radiation is used in electronic spec
uv - vis region!!!!
okay so whats the acc total eergy of the system ,,, what do we acc need to take into consideration + what do they tell us not to take into consideration
the energy of the electric trnasition,,,, vibrational transition and rotational transition.
we dont need to take rotational spec into consideration bc its energy is much lower than vibrational and electronic energy.
we consider them independently and then add them
total energy of the system: Etotal =
E elec + we( v+1/2) - ( v+1/2)^2 wexe.
cm-1
the vib levels get closer or further
vib levels get closer = think of anharmonic oscillator
do rotational levels get closer or further
they get further
bc larger J = higher energy bc its squared
wait so for the spec with v=0 and J’’ values,, what can we now write them as
lower electronic state = s0 = E’‘=0
higher electronic state = s1 = E’=1
where do moelcules exist in
they exist in the lowest E’’ and V’’ levels
what transitions between vibrational and electronic levels are possible
all transitions are possible ,, there are no sleection rules in electronic spec
think about the lymann and all those series and how ionisation energy can literally occur and how emission spec can be from any energy level and we find this using the rydberg equation.
what is a progression
a set of transitions between electronic levelsss
(v’ ,, v’’) aka upper state and lower state. so the little piano type graph also have lines closer and closer together,, and theyre labelled at 0,0 1,0 2,0 3,0 4,0
where (upper state,, lower state)
so the overall change in energy of the system issss
change in total energy = change in electronic spec + change in vib specc
in anharmonic spec the vib lines are straight however this isnt correcttttt, what should they beeeeee
they should be curvesss
think like particle in a box but like in between the parabola lines of the full anharmonic curve.
okay so 1 electronic level has vibrational levels,, so one electronic level corresponds tooooooo
one anharmonic curveeeee
with the vib levels being the lines in the morse curve (we know this already)
but the lines are curvyyyyyfor the vibrational levels.
curve for v=0
a hill shape
curve for v=1
2 hillssss
the same size
curve for v=2
3 hills
big small big
curve for v=3
4 curves
big small small big
v=4 curves
5 hills
big small small small big
okay so describe the hill trend
so number of hills = V+1
u always have 2 big ones,, and the rest are small.
so for the electronic spec transition line and arrow thing and the diff vibrational levels
so u have the 1st electronic level made up of v’’ lines,,,, then a gap and then u have the 2nd electronic level,, with v’ vibrational levels. how do we draw the anharmonic curve for this whole thing
u have 2 electronic levels so u have 2 morse curves.
and they both have the curly lines for the vibrational levels.
one morse curve for v’’ (lower energy) and one morse curve for v’ (upper energy)
why do we use curves and hills for the vibrational energy levels + what does this represent
bc we use |Y|^2 ,,, aka the probability distribution.
which shows us that molecules normally spend more time at the 2 extremes of the stretching and vibrating energy levels.
which makes sense bc when we have a pendulum the ball doessss spend more time at the 2 extremes.
okay wait so higher and lower energies in the morse spec correspond to which mechanics
higher vib states = classical mechanics followed
lower vib levels = quantum mechanics followed.
do molecules move during a transition
nopeeeee
explain what happens when a molecule absorbs energy
the transition occurs
and then the molecule changes shape.
the molecule needs to make up for the change that occurs when it absorbs radiation.
so the moelcule isnt changing when it absorbs EMR.
excited and ground states sometimes haveeeeee (think of 2 different morse curves that correspond to 2 different electronic levels)
they sometimes have the correct geometries.
meaning they are both alligned with the req 9aka the equilibrium bond length)
when we have 2 morse curves aligned perfectly with identical geometries,, what do we want and when do we get it
the v’‘0 –> v’0 gives good and intense overlap,, bc ur going from vo to vo and both of these just have a hill as their probaility distribution.
and bc ur going from the centre of one hill to the other,, u have a high probs density at both levels.
so u get the most intense peak in vib specccc.
bc u had a good overlap of both curves,, bc their probs distributed was high in both of them!! so it adds up to to give an intense peak in vib spec.
lower morse curve =
E’’ = E0 = S0
the higher morse curve =
E’ = E1 = S1
if the ground and excited states are aligned with identical geometries,, will each transition between them have high intensity
nopeeeee
some transitions are less likely to occur bc the overlap isnt that good between probaility distributions!!
aka if ur going from v’‘0 (the one hill) to v’1 (2 hills) ur going from the peak of one of the hills to the trough//valley of the 2 hills. so the probaility distribution is lower for the excited state.
so u have (upper, lower) and so u get the taller lines // more intense peaks whennnnn
when the geometries are identicalllll!!!!!!!!
when u get the best overlap between between the probaility distribution of the ground and excited states. (the vib waves)
what is the franck condon principle explaining
when theres no change in nuclear separation occuring during electronic transitions
bc they occur more rapidly compared to vibrational frequencies.
aka electronic transitions occur so rapidly in comparison to vibrational frequencies so that no change in nuclear separation occurs during the electronic transition. : the molecule changes after the transition to make up for changes made.
OHHHHHH wait so no change in nuclear sep =
no change in bond length
and aka the molecule desnt move when a transition occurs.
and molecules are more likey to have a transition when the excited and ground states have the same geometries : aka the same bond lengths.
aka the morse curves look exactly the same
classic theory dictates what vib level
classical dictates high vib levels
quantume theory dictatessss
low vib levelssss
wait so what do the graphs look like when they dont have identical geometries
one is more shifted to the right,, the higher energy one is normally shifted.
what transitions are more likey to occur when the geometries are NOT identical
u just draw a line up from the v’‘0 to upper energy levels.
and whichever one has the best overlap is the best,, most intense peak!!! sometimes its the v’3,, or v’4 etcccc. just depends how much the higher morse curve is shifted.
transitions occur predominantly from which ground states
they occur predominantly from the vibrational and electronic ground states.
probability distribution of the internuclear separation is dependednt on what
the vibrational level.
bc u get v+1 hills.
and the intensity of the hills correspond to goof or bad overlap
and this gives us an intense transition if theyre both tall hills.
what dictates vertical transitions
the frank condon principle dictates vertical transitions!!!!
Deq
dissasociation equilibrium energy
Do
this is the enerrgy gap between the lowest vibrational level and the plateau of the morse curve.
what is ZPE
zero point energy
all molecules in ground state have vibrational energy!!
this is bc v=0 there is still vibrational energy associated with it.
Deq =
Do + ZPE
Do =
Deq - (1/2 we - 1/4 we xe)
what is the continuum limit
the energy of the transition is more than the molecule dissociating in the excited state,, so u get a broad peak.
u get lines and then u get a line that kinda gives a broad peak type thing where the continuum limit isssss.
AKAAAAA ur going from V’‘0 —–»> draw ur line up but u dont hit a a curve for probaility distribution of vibrational levels. u just hit the edge of the morse curve and the place u hit is above the plateau.
Do’’
ground state
v’‘0 —> plateau’’
Do’
excited state
v’0 —> plateau’
what happens if a molecule is super unstable
u dont even get a morse curve
u just get a slope type beat
whats Eex
from plateau’’ –> plateau’
Vcl =
Eex + Do’’
Do’ =
Vcl - V(0,0)
where v(0,0) is the vo’’ –> vo’ line
when the molecule is so unstable it doesnt have a curve,, where is Eex
from plateau’’ to where the line ends (aka the bottom of the line when the line has a negative gradient)
okay so if Ev = we(v+1/2) - we(v+1/2)^2 xe what is change in energy at dissociation level
change in E = Ev+1 - Ev = 0
change in E =
we(1-2 xe ( vmax +1)) = 0
bc the lines are so close together
vmax equation
= (1/2xe) - 1
what is vmax
the max vibrational level aka where the morse curve plateausssssssss!!!
finding v max tells us what
the energy at dissociation
u can find thid by finding xe. and putting it into the change E = we(1-2 xe ( vmax +1)) = 0
which finds the energy at dissociation level.
explain why peaks in uvvis are broad
peaks = electronic transition,, between ground and excited state.
ut remember that each electric level has manyyyy vibrsational levels. think of the 2 anharminic graphs stacked,, one for excited electronic state and one for ground electronic state.
now,, when a transition occurs,, the e- can go to any vib level in the excited anharmonic graph if there is overlap between the vibrational level curves of the ground and excited state.
meaning,, each one will have a different energy depending on the vib level chosen. and the amount of overlap dictates the probs of that transition occuring, more overlap = more probable so a larger extinction coefficient ,, so a larger abs intensity!!!
if the most intense peak isnt 0,0 ,, what does that mean
the excited and ground states have different geometries.
bc the greatest overlap was between ground 0 and excited 4!!
describe a birge sponer extrapolation
a graph of change in V (freq) aka change in energy
against vibrational levels
whic hgives a stright line graph
where the x axis intercept is the vibrational energy level that dissocition occurs at.
bc in an anharmonic oscillator,, energy levels get closer and closer,, and at the plateau,, change in energy is 0.
area under the graph for a birge sponer extrapolation issss
dissociation energy
not the vibrational level the dissociation occurs atbut the actual energy.
types of emission
flourescence
phosphorescence
what can a system do once its in its excited state
form a photochemical product
decay back into its ground electronic state
ground state =
s0
excited state
s1
from s0 to s1 name and rate
absorption of energy
10^15 // 10^16 s-1
from s1 to s0 : 2 diff things can occur
internal conversion
flourescence
internal conversion rate and what it is
going from S1 ground vibration level. S1 V0’ to S0 V1’
lowering electronic energy level but increasing vibrational energy level
10^1 to 10^7 s-1
flourescence rate
10^6 to 10^12 s-1
other types of emission
flourescence
internal conversion
external conversion - same electronic enrgy level but lower vibrational energy level
intersystem crossing - going from S1 to T1 : 10^4 to 10^12
phosphorescence : going from T1 after intersystem crossing to S0 10^1 to 10^6 s-1
is flourescence slower or faster than internal conversion
flourescence is faster than internal conversion!!
how do u go from S1 to T1
T1 = triplet excited state,, where one e- is in T0 and one in T1 and both point and have the same spin
giving 2(1) + 1 = 3 ,, triplet state.
IS T1 OR S1 lower in energy
T1 is lower in energy
does phsophorescence T1–>S0 or flourescnece S1–>S0 have a longer lifetime
phsophorescence has a longer lifetime bc its spin forbidden : going from T to S
what do we want to molecule to undergo
fluorescence
what is primary quantum yield
the number of fluorescence events vs number of absorptions
shows how good the system is at emitting
larger primary quantum yield =
system is better at emitting
to find primary quantum yield what must we think
the tate at which S1 is removed (we take into account intersystem crossing bc its still removing S1 character)
intersystem crossing
internal conversion
external conversion
fluorescence
u add their rates up
To =
observed fluorescence lifetime
1/ rate of Kf + Kic + KISC
theta o =
fluorescence quantum yield
Kf / kf + kic + kisc
we want to know thisss
what should we thinl about when we see quenching
a transfer of enrgy
what is a quencher
a molecule that absorbs energy to help smt dissociate
S1 = S0 + hv
energy transfer between S1 S0 and a quencher
S1 + Q –> S0 + Q*
goes to ground state but quencher now has additional energy
whats T
lifetime of smt with a quencher
1/kf + kic + kisc + kq[Q]
rate of quenching x conc of quencher
larger conc = larger rate as there are more things to abs energy to help S1 get to S0
whats T0
lifetime of smt without a quwncher
[Q] = 0
when conc of quencher is 0
T = T0
aka observed fluorescence lifetime with quench = that without quench
[Q]>0
with a quencher
T (observed fluorescence lifetime with quencher) < To (observed fluroescence lifetime without a quencher)
bc a quencher increases the rate that S1 goes to S0
so less time that fluorescence can occur for.
do we like quenchers
no
they limit fluorescnece time
okay so what do we do with the fluorescence quantum yield for both with snd without quencher
u do without quencher (longer) / with quencher
to give u 1+Tokq[Q]
so u can plot I0/I = 1+Tokq[Q]
y axis = Io/I
a axis = [Q]
y intercept = 1
gradient = Tokq
