sym 6

Question 1

what do characters represent

Answer 1

what happens to vibrations // orbitals when an operation occurs

Question 2

1 =

Answer 2

unchanged orbital

Question 3

-1 =

Answer 3

reversed orbital

Question 4

random rotation =

Answer 4

cos( rotation)

Question 5

0 =

Answer 5

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

Question 6

when draeing matrices,, when trying to write down where each axis has gone,, what do u give each thing

Answer 6

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.

Question 7

basis vactor = enchanged

Answer 7

1

Question 8

bases vector changesto another in the set

Answer 8

0

Question 9

basis vector is reversed

Answer 9

-1

Question 10

for orbtials to bond they must have what

Answer 10

they must have the same symmetry.

a1 and a1

b1 and b1

Question 11

character

Answer 11

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

Question 12

classes is what

Answer 12

the operations ,, the coloumns

up down

Question 13

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

Answer 13

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.

Question 14

D4h and representations and matrices

Answer 14

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

Question 15

in D4h the xy are seen as (x,y) what does this mean

Answer 15

theyre degenerate.
they have the same energy
they have the same frequency when changing their dipoles

Question 16

looking at the point group character table for D4h,, (x,y) are in the Eu irreducible representation row,, what does that mean

Answer 16

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.

Question 17

what does Eg mean

Answer 17

gerade

when u do the inversion operator ,, i doesnt change the vectors. it doesnrt reverse or invert them.

u get a positive number

Question 18

Eu and i

Answer 18

inversion = always a negative number

Question 19

Eg and i

Answer 19

inversion = positive number

Question 20

x y and z tell u what

Answer 20

they tell u about IR // vibrational spec

Question 21

products, x^2, y^2, z^2 and xy tell us what

Answer 21

about raman spec

Question 22

so if u have a point of inversion,, what should u lok for

Answer 22

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

Question 23

when can the same irreducible reps be both raman and IR active

Answer 23

when u dont have a centre of inversion

Question 24

gerade orbitals

Answer 24

opposite lobes have the same phase.

s and d orbitals
circle and closer

Question 25

ungerade orbitals

Answer 25

opposite lobes have opposite phases

p orbitals

bumbbell shape

Question 26

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

Answer 26

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

Question 27

gc =

Answer 27

stoichiometry of the operation at the top of that column.

Question 28

xi =

Answer 28

the character we get from the point group table

Question 29

xj =

Answer 29

the character we get from the point group table from a different irreducible representation than the one from xi.

Question 30

how to find gamma ‘r’

Answer 30

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’.

Question 31

how do we do the reducable table to find what irreducable things gamma, ‘r’ is made up of.

Answer 31

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)

Question 32

describe the symmetry seen in H2O

Answer 32

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

Question 33

when to think whenever youve got a plane of symmetry

Answer 33

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

Question 34

when u draw the arrows to find gamma, ‘r’ what numbers can we give it

Answer 34

0 = changes
2 = unchanged
-1 = stays the same but reversed.

Question 35

what do we do to find out if smt is raman // ir active

Answer 35

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!!)

Question 36

whats the projection operator and how do we do it

Answer 36

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!!.

Question 37

when we think of dipole chnage what should we think of

Answer 37

we should think of the dipole arrow getting bigger // smaller

Question 38

whats supe important about the C4 axis being the principal axis

Answer 38

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’’

Question 39

when ur doing the inversion operation,, does the arrow // orbital flip aswell

Answer 39

yes,, the orbital inverts aswel!!

if its going up its gonna invert to look down