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physical - yr2, s1 > 1c - w (rotational spec) > Flashcards

1c - w (rotational spec) Flashcards

1
Q

energy of a photon can be given by

A

E = hcv-
E = hc/ wavelength
E = hv

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2
Q

radio frequency range

A

30Hz –> 300kHz

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3
Q

what are microwaves all about

A

finding bond lengths etc

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4
Q

describe electromagnetic radiation

A

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

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5
Q

rotation of molecules is high or low energy

A

rotational spec uses microwaves which are low energy and high wavelength

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6
Q

gross selection rule of rotational // microwave spec: aka what is needed // think of grocery store

A

u need a permanent dipole!! to produce rotational absorptions

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7
Q

what does the dipole moment in rotational spec dictate

A

it dictates the intensity of the absorption

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8
Q

what does the geometry and mass of the molecule dictate in rotational // microwave spec

A

it dictates the energy levels

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9
Q

nmr uses what field oscillation

A

nmr uses msgnetic field oscillation

aka nuclear magnetic resonance

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10
Q

uv-vis uses what field oscillation

A

it uses the electric field oscillation

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11
Q

energy diff between levels can be given by what

A

E = hv

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12
Q

describe the point charge thing diagram

A

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..

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13
Q

okay so what energy does rotational spec use and what phase must the sample be in

A

microwaves

must be in gas phase

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14
Q

if the gross selection rule for rotational spec is it needs a permanent electric dipole,, which molecules will be rotational spec active

A

heteronuclear diatomic!!!

bc they will have 2 different atoms with 2 different electronegativities giving u a permanent dipole

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15
Q

what is the rotational spec selection rule: aka what transition is allowed

A

J = +- 1

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16
Q

what is J

A

J is the rotational sspec quantum number

think: jesus turned life around

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17
Q

what do we assume when we start rotational spec

A

we assume the molecule is a rigid motor: aka the bond lengths do not change!! fixed bond lengths

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18
Q

how can a rigid motor molecule rotate

A

it has 3 cartesian axis,, the molecule can rotate on all 3 of these

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19
Q

describe the cartesian axis

A

z is up
y is right
x is towards u

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20
Q

if the moelcule can rotate along all 3 cartesian axis,, what dimension can it rotate in

A

in 3D

bc it can rotate in all directions along each axis

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21
Q

what is meant by moment of inertia

A

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

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22
Q

smt has how many moments of inertia

A

3!! Ia Ib Ic ,,, one for each axis

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23
Q

linear motors have what moment of inertia

A

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!!

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24
Q

Ia rotation for inertia is what rotation:

A

rotation about the bond!!!

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25
Q

spherical motor has what rotation

A

Ia = Ib = Ic

all rotations change it.

octahedral // tetrahedral

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26
Q

symmetrical rotors have waht rotation

A

Ia doesnt equal but Ib = Ic

C3v

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27
Q

asymmetrical rotor has what rotation

A

none of them are equal

ia no Ib no Ic are equal! theyre all different

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28
Q

when atoms move I =

A

1

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29
Q

when atoms dont move I =

A

0

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30
Q

describe a heteronuclear bond and what we need to do to find I - moment of inertia

A

atom,, with mass1 connected by a bond of length r ,, to another atom with mass2

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31
Q

what is c in the heteronuclear bond drawing for inertia + explain where this is and what it is

A

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.

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32
Q

from C –> the atoms on the heteronuclear bond drawing for inertia.

A

r1

is atom 1 —> C

r2 : atom 2 –> c

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33
Q

r,, in moment of inertia isss

A

r1 + r2

both distances from separate atoms to centre of gravity , c

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34
Q

what does I equalssss,, the equation

A

I = reduced mass r ^2

reduced mass = m1 x m2 // m1 + m2

in kg and divided by avo to get mass of one atom

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35
Q

what does I relate to in terms of a rigid motor

A

it related mass of atoms and the bond lengths!! we say that bond length doesnt change in a rigid motor

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36
Q

to characterise rotational energy,, what must we solve

A

solve the schrodinger equation for that excited state // rotation

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37
Q

what do we need to solve in the schrodinger equation to find rotational energy

A

we need to find Ej

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38
Q

what does Ej equal to and what units dows this have

A

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

39
Q

J can be what values

A

0, 1, 2, 3, 4, etccc

40
Q

wavenumber =

A

E / hc

41
Q

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

A

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.

42
Q

HOW ELSE CAN WE FIND ENERGY OF MOELCULE ROTATING IN CM-1 UNITS

A

Ej = BJ(J+1)

where B is the rotational constant ,, it simplifies the planks 8 pi equation

43
Q

does B change eith diff molecules and why

A

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.

44
Q

energy of molecule at J = 0

A

use Ej = BJ(J+1)

so u have

EJ = B x 0 ( 0+1) = 0B aka 0

45
Q

energy of molecule at J = 1

A

EJ = BJ(J+1)
= B x 1 ( 1+1)
= 2B in cm-1

46
Q

energy of molecule at J = 2

A

EJ = BJ(J+1)
= B x 2 ( 2+1 )
= 6B cm-1

we’re characterising rotational energy levels

47
Q

how do we characterise rotational energy levels =

A

E J = BJ(J+1)

energy of a molecule at THAT rotationl energy = XXX in cm-1

48
Q

as J increases,, what else increases< using EJ = BJ(J+1)

A

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.

49
Q

selection rule for rotational spec =

A

change in J. = +-1 !!!! 1

we can only transition to neighbouring energy levels // neighbouring J values.

50
Q

in rotational spec we like being in what energy state

A

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.

51
Q

energy difference between 2 J energy levels,,

A

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!!!

52
Q

what do we see in rotational spec as the peaks

A

the peaks represent the energy difference between 2 J energy levels.

2B ( J(small) + 1) in cm-1

53
Q

energy differece from J0 to J1

A

u do 2B (J0 +1)

to give 2B(1) = 2B!!

so the first transition will be at 2B!!

54
Q

energy difference between J 1 and J 2 ,, and what this gives us in spec

A

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!!

55
Q

what is the separation between peaks at rotational spec

A

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.

56
Q

periodic table values are given in what units

A

g/mol

57
Q

spacing of lines in 27Al 1H is 12.60cm-1,, find I and bond length of the molecule

A

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!!!

58
Q

why does reduced mass have to be in kg

A

bc I = kg m^2

m^2 bc u do r^2

59
Q

bond length values are usually to powers offf

A

10

60
Q

I values are normally in powers offf

A

-47 kgm^2

61
Q

intensity of lines depends on what

A

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

62
Q

equation for the population difference between 2 energy levels

A

Nj / No = exp(-Ej / KT)

= exp( -BJ (J+1) hc) // KT)

63
Q

if B = 5cm-1,, J = 4 and T= 300k what is the population difference

A

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

64
Q

if J increases and B increases ,, what happens to NJ/N0

A

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

65
Q

what accccc effects peak intensity

A

degeneracy!!
the existance of 2 or more energy states with the same energy!! due to angular momentum!!

66
Q

symbol and equation for angular momentum

A

P = I.c w

w = angular // rotational frequency. aka the freq associated with rotation!!

67
Q

energy of a rotator =

A

E = 1/2 I.c w^2

68
Q

P =

A

p = I.c w

= root (2EI.c )

= root (J (J+1 ) h/2pi
= root ( J ( J+1 )

h/2 pi = unitssss for angular momentum.

69
Q

units for angular momentum p

A

= h/2 pi

70
Q

angular momentum , p, for J = 1

A

root ( 1 ( 1 + 1) ) = root2 = 1.4 units

and u go from -p –> p

so round down

-1 , 0 , +1 3 fold degeneracy

71
Q

easier equation for angular momentum degeneracy

A

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

72
Q

degeneracy and J value relationship

A

larger J value = more degenerate levels

bc 2J + 1 = levels of degeneracy for that level.

73
Q

J relationship with degeneracy and population

A

as J increases: degeneracy goes up but population goes down

74
Q

what is peak intensity proportionate to

A

intensity is prop to

( 2J +1 ) x exp( -Ej/KT)
degeneracy population

75
Q

the total population in each energy level //J value =

A

(2J +1) exp( -Ej / KT )

76
Q

max population = largest at the nearest integral J vaalue to : what equation is sued

A

max pop : J = root(KT/2hcB) - 1/2

from 1 resonance line// peak we can find the value of B!!

77
Q

frequency or rotation and moment of inertia relationship

A

greater fundamental frequ = smaller bonds (think of iceskater spinning) = smaller I vale bc (I= red mas x r^2)

78
Q

angular momentum,, freq of rotation , I value relationship

A

larger angular momentum = increase freq of rotation = smaller I bc r decreases

79
Q

what force effects a bond when it spins

A

centrifugal force

80
Q

how much does centrifugal force effect a bond

A

depends on k,, force constant // strength of bond.

81
Q

non rigid motor:

A

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

82
Q

k ,, force constant equation

A

k = 4 pi^2 w(cm-1) c^2 red mass

83
Q

force constant affects what value

A

effects B

84
Q

what is B depeded on

A

bond distance

this changes with vibration

85
Q

D =

A

centrifugal distortion

86
Q

D=

A

4B^3 / w(cm-1) ^2

87
Q

what does D effect

A

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!!

88
Q

when is centrifugal distoriton, D, negligable

A

small J values

big J values are affected tho!!

89
Q

okay so for a non rigid motor: what is Ej aka energy level of J

A

EJ = BJ ( J+1) - DJ^2 ( J + 1) ^2 cm-1

D= h^3 / 32pi^4 I.c^2 r^2 K c
cm-1

90
Q

the centrifugal force correction value does what

A

it lowers the energy of the non rigid motor energy level!!

bc it increases its bond length etc

91
Q

the larger the J value,, the xxxx effect of D

A

the larger j value = larger effect of D . aka that energy level will be decreased in energy by a greater amount

92
Q

what does distortion of a non rigid motor change

A

the gaps between peaks // energy gaps between peaks are no longer 2B!!

for small J values its still approx 2B tho

93
Q

finding anything about a molecule thats non rigid: what should we use

A

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.

94
Q

the more isotopes an element has,,

A

the more resonance peaks there will be.

peak intensities differ with isotope abundance

physical - yr2, s1 (15 decks)

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