sym 6 Flashcards
(39 cards)
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