1c - w (rotational spec) Flashcards
energy of a photon can be given by
E = hcv-
E = hc/ wavelength
E = hv
radio frequency range
30Hz –> 300kHz
what are microwaves all about
finding bond lengths etc
describe electromagnetic radiation
2 waves
one flowing towards us
one flowing parallel to us (like a normal wave)
the one wavig towards us // parallel to us is the magnetic frequency oscillation
the one waving parallel to us is the electric field oscillation
rotation of molecules is high or low energy
rotational spec uses microwaves which are low energy and high wavelength
gross selection rule of rotational // microwave spec: aka what is needed // think of grocery store
u need a permanent dipole!! to produce rotational absorptions
what does the dipole moment in rotational spec dictate
it dictates the intensity of the absorption
what does the geometry and mass of the molecule dictate in rotational // microwave spec
it dictates the energy levels
nmr uses what field oscillation
nmr uses msgnetic field oscillation
aka nuclear magnetic resonance
uv-vis uses what field oscillation
it uses the electric field oscillation
energy diff between levels can be given by what
E = hv
describe the point charge thing diagram
as the molecule has a permanant dipole,, as it spins,, the dipole goes with it,,
think of the arrow with the positive end,, as this spins in a circle,, the positive spins with it..
okay so what energy does rotational spec use and what phase must the sample be in
microwaves
must be in gas phase
if the gross selection rule for rotational spec is it needs a permanent electric dipole,, which molecules will be rotational spec active
heteronuclear diatomic!!!
bc they will have 2 different atoms with 2 different electronegativities giving u a permanent dipole
what is the rotational spec selection rule: aka what transition is allowed
J = +- 1
what is J
J is the rotational sspec quantum number
think: jesus turned life around
what do we assume when we start rotational spec
we assume the molecule is a rigid motor: aka the bond lengths do not change!! fixed bond lengths
how can a rigid motor molecule rotate
it has 3 cartesian axis,, the molecule can rotate on all 3 of these
describe the cartesian axis
z is up
y is right
x is towards u
if the moelcule can rotate along all 3 cartesian axis,, what dimension can it rotate in
in 3D
bc it can rotate in all directions along each axis
what is meant by moment of inertia
the torque required for a given angullar acceleration about a rotational axis
aka moment of inertia = what is required to get a moelcule rotating!!1
smt has how many moments of inertia
3!! Ia Ib Ic ,,, one for each axis
linear motors have what moment of inertia
they have 2
they can rotate around the Ib,, and Ic but not the Ia (the y axis) bc that would just remain the same,, the bonds would move. think of the chickens outside the galeria in Poland.
I = 1 = when atoms move
I = 0 = when atoms stay i nthe same place // no moment of inertia.
Ib = Ic
Ia= 0 bc theyre both curvy!!
Ia rotation for inertia is what rotation:
rotation about the bond!!!
spherical motor has what rotation
Ia = Ib = Ic
all rotations change it.
octahedral // tetrahedral
symmetrical rotors have waht rotation
Ia doesnt equal but Ib = Ic
C3v
asymmetrical rotor has what rotation
none of them are equal
ia no Ib no Ic are equal! theyre all different
when atoms move I =
1
when atoms dont move I =
0
describe a heteronuclear bond and what we need to do to find I - moment of inertia
atom,, with mass1 connected by a bond of length r ,, to another atom with mass2
what is c in the heteronuclear bond drawing for inertia + explain where this is and what it is
c = centre of gravity!!
the atoms will have a different mass and this point is where u could hold it for them to be equal like a seesaw. this obvs depends on the mass and isnt in the middle of the bond.
from C –> the atoms on the heteronuclear bond drawing for inertia.
r1
is atom 1 —> C
r2 : atom 2 –> c
r,, in moment of inertia isss
r1 + r2
both distances from separate atoms to centre of gravity , c
what does I equalssss,, the equation
I = reduced mass r ^2
reduced mass = m1 x m2 // m1 + m2
in kg and divided by avo to get mass of one atom
what 2 things does I relate to in terms of a rigid motor
it related mass of atoms and the bond lengths!! we say that bond length doesnt change in a rigid motor
to characterise rotational energy,, what must we solve
solve the schrodinger equation for that excited state // rotation
what do we need to solve in the schrodinger equation to find rotational energy
we need to find Ej
what does Ej equal to and what units dows this have
Ej = (h^2 / 8 pi^2 I.c ) J(J+1)
where h = planks
I.c = moment of inertia about that cartesian axis
J = rotational quantum number
J units
energy of the moelcule rotating
J can be what values
0, 1, 2, 3, 4, etccc
wavenumber =
E / hc
so to find the energy of a molecule rotating,, the Ej equation in J,, what do we do to find the value in cm-1
we do the same thing but divide the first bit of the equation by c - speed of light. WHICH NEEDS TO BE IN CM UNITS!! 2.998X10^10
this gives cm-1 units!! in terms of the energy of the molecule rotating.
HOW ELSE CAN WE FIND ENERGY OF MOELCULE ROTATING IN CM-1 UNITS
Ej = BJ(J+1)
where B is the rotational constant ,, it simplifies the planks 8 pi equation
does B change eith diff molecules and why
yesss ,, B is molecule dependent.
bc B is used to simplify the planks, I 8 pi equation
I. = reduced mass x r^2
and reduced mass depends on the molecule.
energy of molecule at J = 0
use Ej = BJ(J+1)
so u have
EJ = B x 0 ( 0+1) = 0B aka 0
energy of molecule at J = 1
EJ = BJ(J+1)
= B x 1 ( 1+1)
= 2B in cm-1
energy of molecule at J = 2
EJ = BJ(J+1)
= B x 2 ( 2+1 )
= 6B cm-1
we’re characterising rotational energy levels
how do we characterise rotational energy levels =
E J = BJ(J+1)
energy of a molecule at THAT rotationl energy = XXX in cm-1
as J increases,, what else increases< using EJ = BJ(J+1)
as J increases,, rotational quantum number = enrgy increases
as energy increases,, the gaps get larger between energy levels!! if u clock it it makes sense.. bc the enrgy of each level gets larger so the spacing will be different.
selection rule for rotational spec =
change in J. = +-1 !!!! 1
we can only transition to neighbouring energy levels // neighbouring J values.
in rotational spec we like being in what energy state
we like being in the lower neergy state,, so most of the transitions we do using the J = +-1 will be excitations,, they will be +1. going to the next J value.
energy difference between 2 J energy levels,,
EJ+1 - EJ
aka 2B (J(small) + 1) cm-1
this is the wavenumber of the resonance peak in spectra,, aka the enrgy difference between 2 energy levels is what we see in the spectra!!!
what do we see in rotational spec as the peaks
the peaks represent the energy difference between 2 J energy levels.
2B ( J(small) + 1) in cm-1
energy differece from J0 to J1
u do 2B (J0 +1)
to give 2B(1) = 2B!!
so the first transition will be at 2B!!
energy difference between J 1 and J 2 ,, and what this gives us in spec
this will give us a peak at the energy we find!! in cm-1
so for J1 –> J2 the smallest J value is 1,, so this is the one we use in the equation :
E2-1 = 2B ( 1+ 1) = 4B in cm-1
so first transition was at 2B and this one will be at 4B!!
what is the separation between peaks at rotational spec
the difference between peaks in rotational spec = 2B
the energy for transitons are diff,, but the energy for transition is what is plotted.
and the difference between what is plotted = 2B!! u can do this girl.
periodic table values are given in what units
g/mol
spacing of lines in 27Al 1H is 12.60cm-1,, find I and bond length of the molecule
12.60cm-1 = 2B (bc spacing between peaks // lines in rot spec = 2B
12.60 / 2 = 6.30cm-1
so B = 6.30cm-1
6.30cm-1 = h / 8 pi^2 c I.c
rearrange to find I.c
I = h / 8 pi ^2 C (x10^10) x 6.30
I = 4.442 x 10^ -47 kg m^2
and I = red mass x r^2
find red mass then divide I by red mass to find r^2. then square root this to find ‘r’ aka bond length!!!
why does reduced mass have to be in kg
bc I = kg m^2
m^2 bc u do r^2
bond length values are usually to powers offf
10
I values are normally in powers offf
-47 kgm^2
intensity of lines depends on what
the boltzman distibution
aka the more populated an energy level is,, the more likely it is for a transition to occur
and the size of the dipole moment
equation for the population difference between 2 energy levels
Nj / No = exp(-Ej / KT)
= exp( -BJ (J+1) hc) // KT)
if B = 5cm-1,, J = 4 and T= 300k what is the population difference
0.61
slightly less than half the population of J=4
the J value u use is the one ur finding population of,, relative to J = 0
if J increases and B increases ,, what happens to NJ/N0
it decreases!!
bc uve got the exponential of a larger number!! so more negative
so larger J and B,, more population diff,, aka more moelcules in the excited state
what accccc // also effects peak intensity
degeneracy!!
the existance of 2 or more energy states with the same energy!! due to angular momentum!!
others: dipole + boltzman pop difference
symbol and equation for angular momentum
P = I.c w
w = angular // rotational frequency. aka the freq associated with rotation!!
energy of a rotator =
E = 1/2 I.c w^2
P =
p = I.c w
= root (2EI.c )
= root (J (J+1 ) h/2pi
= root ( J ( J+1 )
h/2 pi = unitssss for angular momentum.
units for angular momentum p
= h/2 pi
angular momentum , p, for J = 1
root ( 1 ( 1 + 1) ) = root2 = 1.4 units
and u go from -p –> p
so round down
-1 , 0 , +1 3 fold degeneracy
easier equation for angular momentum degeneracy
2J + 1
aka for J = 2
4 + 1 = 5
aka u get 5 degenerate levels.
and bc j = 2
-2, -1, 0, 1, 2 these are ur 5 degenerate levels for that energy // J value.
J = 2 = 5 degenerate levels
degeneracy and J value relationship
larger J value = more degenerate levels
bc 2J + 1 = levels of degeneracy for that level.
J relationship with degeneracy and population
as J increases: degeneracy goes up but population goes down
what is peak intensity proportionate to
intensity is prop to
( 2J +1 ) x exp( -Ej/KT)
degeneracy population
the total population in each energy level //J value =
(2J +1) exp( -Ej / KT )
max population = largest at the nearest integral J vaalue to : what equation is sued
max pop : J = root(KT/2hcB) - 1/2
from 1 resonance line// peak we can find the value of B!!
frequency or rotation and moment of inertia relationship
greater fundamental frequ = smaller bonds (think of iceskater spinning) = smaller I vale bc (I= red mas x r^2)
angular momentum,, freq of rotation , I value relationship
larger angular momentum = increase freq of rotation = smaller I bc r decreases
what force effects a bond when it spins
centrifugal force
how much does centrifugal force effect a bond
depends on k,, force constant // strength of bond.
non rigid motor:
bonds can vibrate // they can change lengths
energy levels of a diatomic is not equal to 2B ,, they get closer as J increases
larger J = greater centrifugal force = increase r,, extent of increase depends on k ,, force constant
k ,, force constant equation
k = 4 pi^2 w(cm-1)^2 c^2 red mass
force constant affects what value
effects B
what is B depeded on
bond distance
this changes with vibration
D =
centrifugal distortion
D=
4B^3 / w(cm-1) ^2
what does D effect
r, I, w (cm-1)
lowers energy
energy is lower for the same energy level but with a rigid motor.
aka energy levels have the same J value: but for non rigid motors,, bond length can change due to centrifugal distortion which lowers its energy bc it increases r (depending on k) which changes I and changes w(cm-1) aka rotational frequ!!
when is centrifugal distoriton, D, negligable
small J values
big J values are affected tho!!
okay so for a non rigid motor: what is Ej aka energy level of J
EJ = BJ ( J+1) - DJ^2 ( J + 1) ^2 cm-1
D= h^3 / 32pi^4 I.c^2 r^2 K c
D= 4B^3 / w^2 (cm-1)
cm-1
the centrifugal force correction value does what
it lowers the energy of the non rigid motor energy level!!
bc it increases its bond length etc
the larger the J value,, the xxxx effect of D
the larger j value = larger effect of D . aka that energy level will be decreased in energy by a greater amount
what does distortion of a non rigid motor change
the gaps between peaks // energy gaps between peaks are no longer 2B!!
for small J values its still approx 2B tho
finding anything about a molecule thats non rigid: what should we use
use the energy gaps between peaks of low J value: these are affacted less by distortion and enrgy gap will still be simiar to 2B.
the more isotopes an element has,,
the more resonance peaks there will be.
peak intensities differ with isotope abundance
more abundant isotope = larger peak = arger intensity etc
