2c Flashcards
describe [Mn(H2O)6] 2+
get ur periodic table and pen out girllll
okay so h20 is a neutral ligand so Mn’s oxidation state is +2.
its d count is d7. 7-2 = 5
so d5 + its high spin bc ur bonded to H20. which is strong field.
so u draw out ur e- on the t2g and eg energy levels.
ur going from d–>d so 🔺L is disobeyed.
ur going from t2g –> eg so symmetry is disobeyed and Oh is pure octahedral which is a centrosymmetric system bc it has a point of inversion.
ur going from up spin to down so 🔺S is also disobeyed.
all the rules are disobeyed so this transition has a very small E value (molar abs coefficient value,, meaing the probability of this trnasition occuring is low, but not impossible.)
describe [Mn(acac)3]
acac = -1
Mn = d7 - 3 = d4
high spin
- d–> d = disobeyed 🔺L rule.
- 🔺S = also disobeyed
- 🔺Symmetry in a CSS : the complex does not have pure Oh symmetry,, so the larporte rule is relaxed!!
- so 2/3 are disobeyed and one is relaxed meaning the E value is larger,, meaning the probability of a transition occuring is higher.
describe [MnO4]
O = -2
-1 - - 8 = = + 7.
d7 - 7 = d0
so no electrons.
d–> d : these transitions dont occur bc theres obvs no electrons,, but theres an intense LMCT transition.
its not a centrosymmetric system bc its Td. so the larporte rule is relaxed.
the spin rule is 🔺0 and ig this is followed bc theres no electrons.
- so all the rules are obeyed meaning the proabbiltiy of this transition occuring is high,, meaning a larger E value.
large E value =
10,000 M-1 cm-1
mideum E value
50 M-1 cm-1
small E value
0.04M-1 cm-1
what differs from the Td and Oh geometry
the d orbital geom switches
from 3 tsg being lower and 2 eg being higher for Oh.
to 3 t2 being upper and 2 e being lower for Td.
this occurs bc u need to remove the grade: bc tetrahedral geom does not have a point of inversion meaning there is no need for symmetry labels.
also bc the ligands interact with different igands. in Td: ligands approach between the axis,, so the ligands on the axes ( the 2 squared orbitals) are lower in energy,, while the 3 dx.y,z orbitals are between the axes and therefore have a larger energy due to higher repulsion. : they interact with the incoming ligand field more.
describe why the Oh geom is split in that way
bc in Oh,, the ligands approach on the axes,, meaning the squared ligands interact with the approaching ligand field more and are higher in energy.
while the ligands between the axes are lower in eneergy as they interact with the approaching ligand field less.
why is 🔺t 4/9ths of 🔺o
bc in tetrahedral complexes the d orbitals are approached by 4 ligands in total.
and in the Oh theyre approached by 6 ligands in total which means the d orbitals are interacting with more ligand fields in total!! which means more splitting!! meaning a larger 🔺o value.
comparing the UV spec of [Co(H2O)6] 2+ and [CoCl4] 2-
they have the same d electron count!!
what differs between them!!
the abs wavelength of Cl4 complex is larger. the molar abs of Cl4 complex is also larger.
okay so the main diff is that the Cl4 complex is tetrahedral,, the H2O complex is octahredral
this means that the 🔺t will be smaller,, bc it always is. meaning the tetrahedral complex will need less energy for its transition,, and less energy equals a longer wavelength of absorption. which explains that.
now the E ,, aka molar extinction coefficient depends on the larporte rules, S, L and CSS. theyre both d7.
both disobey the spin rule
both disobey the L rule.
Td isnt CSS tho so the fact its going fro e to t2 doesnt matter,, its larporte relaxed.
so its got a larger E value as its more likely to occur,, bc it disobeys less uv. rules than the Oh geometry.
in uvvis,, when comparing two complexes.. the E value os due to what
the probs of that trabsition occuring.
we need to asses which complex disobeys what rules and if they disobey the same amount or not.
this is mnormally the spin rule or the CSS rule: bc Td is not a CSS meaning this rule is relaxed for Td structures.
in uvvis spec,, when analysing it between two diff compouns,, what is the max abs wavelength due to
this is due to the enrgy gap between the energy levels that that transitions are occuring between.
the 🔺t is always lower than the 🔺o bc there are 4 incoming ligand fields instead of 6,, which obvs doesnt increase the energy of the energy levels by that much. which is why its lower, and will probs have a higher wavelength of abs.
what do we do to find the free ion terms
use the russel saunders thing
ground state term main bits
2S+ 1 at top left
L value super large and in the middle.
J vlue at the bottom
how do we do 2S+ 1
S is the total spin number.
so u add the spins that u have.
u need to respect paulis principle when it comes to this!! make sure the max number of parallel spins occur.
how do we get the largest L value
we want to put the e- in the higher value ml orbitals.
aka for d orbitals,, when drawing a transition, we use ml 2, ml1, ml0,, and try to not use ml -1 or -2 etc.
bc 1+2 = 3,, so our L value would be 3 which corresponds to an F.
what is J
the total angular momentum number after spin orbital coupling occurs
when we do the ground state term symbol,, do we want to max out the s and l values
YESSS!!
so follow the pauli principle so e- spins dont cancel eachother out.
and put the e- in the highes ml values,, not the negative ones.
how do we find L
we add the ml values
how do we find S
we add the spins
how do we find J,, aka total angular moment after spin orbital coupling occurs.
we do L-S
J = L-S
think alphabetically!! l comes before S, ,and then also stands for subtract!!
