sym 6 Flashcards
what do characters represent
what happens to vibrations // orbitals when an operation occurs
1 =
unchanged orbital
-1 =
reversed orbital
random rotation =
cos( rotation)
0 =
when the orbital has changed but not by being reversed.
aka from x –> -y THAT = 0
bc ur not getting -x -> x = -1
or x -> x = 1
ur going to something completely different so u give it a 0
when draeing matrices,, when trying to write down where each axis has gone,, what do u give each thing
u have x’ y’ z’ going vertical on the lhs. these are the new positions thats why they have a ‘.
then u have a matrix. and x y z going vertical on the rhs.
these are the original ones.
so on ur molecule u would label x, y, z, and as u do the operation u see where they end up. the x might have gone to -y,, so for x’ u put a -1 on the y part. remmeber ur going across the matrix,, and down the xyz one. so u would put 0 for x, -1 for y and 0 for z.
bc x’ didnt land on x or z (the old x r z) but landed on the old y,, but the negative side of it,, aka the extended bit, when u go past the central atom.
if theres a 1 that means the new veector is that vector,, the one with the 1.
basis vactor = enchanged
1
bases vector changesto another in the set
0
basis vector is reversed
-1
for orbtials to bond they must have what
they must have the same symmetry.
a1 and a1
b1 and b1
character
amount of the original orbital present after the operation.
complete change = 0 original present
no chnage = 1 //100% presetn
reversed. = -1, ,same orbital but different phases
classes is what
the operations ,, the coloumns
up down
if characters are the amount of the original obrital in the new one ,, after an operation,, in a matrix,, when do we use 1 and 0
0 = none of that orbital is in the new one
1 = all of that orbital is in the new one,, so this is where the orbital is moved to
u put a 1 in the new place and a 0 in the other places bc they have no contribution to the new position
thats why we give a 0 when things change,, bc their old thin g has no contribution to their new position.
D4h and representations and matrices
U HAVE X, Y AND Z
z pointing up,, and tbh no operation chnages z,, z remains in its z position
so the matrix with x, y and z = REDUCIBLE REPRESENTATION!! bc we can cancel out z bc its always the same
the new matrix with just the relationship between y and x = IRREDUCIBLE REPRESENTATION
in D4h the xy are seen as (x,y) what does this mean
theyre degenerate.
they have the same energy
they have the same frequency when changing their dipoles
looking at the point group character table for D4h,, (x,y) are in the Eu irreducible representation row,, what does that mean
ungerade.
when u do the inversion operator,, u reverse the positions.
-1 for each!! then add up -1 and -1 to get -2!!!!
bc x goes to -x and y goes to -y.
what does Eg mean
gerade
when u do the inversion operator ,, i doesnt change the vectors. it doesnrt reverse or invert them.
u get a positive number
Eu and i
inversion = always a negative number
Eg and i
inversion = positive number
x y and z tell u what
they tell u about IR // vibrational spec
products, x^2, y^2, z^2 and xy tell us what
about raman spec
so if u have a point of inversion,, what should u lok for
raman aka products of xyz,, will be Eg
and IR x, y , z will be Eu
the same irreducible reps cannot be both IR and raman active
when can the same irreducible reps be both raman and IR active
when u dont have a centre of inversion
gerade orbitals
opposite lobes have the same phase.
s and d orbitals
circle and closer
ungerade orbitals
opposite lobes have opposite phases
p orbitals
bumbbell shape
normally irriducible reps can be found using matrices and looking which orbital we can get rid of bc they dont change during an operation. how else can we do this tho
by doing the r thing.
we drae the molecule and number the atoms and draw an arrow coming out of each one,, extending the bond.
u then do all the operations and see how the molecule and arrows change with each operation.
1: all original = new
0: no original = new
do vthis for each arrow!!!! and like add their effects.
if 1/3 things change u do 0+1+1=2
gc =
stoichiometry of the operation at the top of that column.
xi =
the character we get from the point group table
xj =
the character we get from the point group table from a different irreducible representation than the one from xi.
how to find gamma ‘r’
find the point group using the flow chart.
look at all the symmetry operations.
draw an arrow on the atoms (not the central one) and see how each operation effects the arrows. add their effects and thats ur gamma, ‘r’.
how do we do the reducable table to find what irreducable things gamma, ‘r’ is made up of.
find gamma like we said.
do gc (stoic) x gamma x number from character table. for each one of the oeprations.
find ‘h’ = order = add all the stoich.
add all the characters in a row up!!.
divide the totals by ‘h’.
thats how many of each oepration are in gamma.
use their characters in the table given and add then to see if they make up gamma values we found.
r = x + x (include how many of each)
describe the symmetry seen in H2O
u have the C2 axis
the sigma v (following the C2 axis and splitting the molecule in half,, separating the H’s.
then u have sigma v’
dash bc this plane of symmetry is in the plane of the molecule.!!!!
it still has a z component
when to think whenever youve got a plane of symmetry
where is my principal axis
does this plane have a z component in it?
is it parallel to the z axis or is it perpendicular to it.
parallel = sigma v
perpendicular = sigma h
when u draw the arrows to find gamma, ‘r’ what numbers can we give it
0 = changes
2 = unchanged
-1 = stays the same but reversed.
what do we do to find out if smt is raman // ir active
we find what gamma is equal to by doing the whole thing.
then we look at the charts we are given.
if r = A1
we look at the A1 row. the lhs = IR active
rhs = raman active
then if ur given a spec,, draw lines from each peak. if the line grows through both raman + ir peak = theres an orbital whos vibration makes it both raman and ir active.
if its just raman or just ir, aka the line only goes through either peak,, theres an orbital whos vibration makes it only ir // raman active.
basically just match what u see with what the diff symmetry irreducible reps are (if theyre both active or just one active!!)
whats the projection operator and how do we do it
figuring out if smt is ir active and what the bonds are doing!! aka if there is a chnage in dipole (dipole arrow getting bigger//smaller)
u first find the irreducible representations by filling out the whole chart.
u draw the molecule and label the H’s. u then pick H1 to be r1 and go through the irreducible symmetry operations and see what r1 turns into after the operation. aka r1 o r3,, we write r3. u do this for every operation.
then u multiply it by the characters in the irreducible rep rows.
then u add then together!! noramlly: 2(r1 + r2 + r3 + r4) etc!! if theyre all positive. u put - if theyre negative.
the u draw the arrows on the molecule. + theyre pointing away. - theyre pointing inside.
if opposite arrows strengthen eachother, its IR active as there is a dipole change. u look at the chart fir that irreducible representation in the lhs IR column to see in what direction does the dipole change in!!.
when we think of dipole chnage what should we think of
we should think of the dipole arrow getting bigger // smaller
whats supe important about the C4 axis being the principal axis
if u have a C4 axis u also immediately have a C2 axis in the same place!!
meaning horizontal C2 axis need a C2’ and C2’’
when ur doing the inversion operation,, does the arrow // orbital flip aswell
yes,, the orbital inverts aswel!!
if its going up its gonna invert to look down
