year 3 coordination 🔆 > remember remember > Flashcards

remember remember Flashcards

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1
Q

when we’re doing the cp and bz metallocene and arene questions what do we need to remember about the energy levels

A

for merallocenes,, u fill the first 3 as if they were degenerate

with arenes, u fill the 2 and the 1 separate

the order: 2,1,2

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2
Q

when asked to describe a uvvis or to predict one what should we think of

A

think of the larporte rules and if any of them are broken

then think of the splitting things

the peaks most intense will be the ones that obey the rules

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3
Q

increase in oxidation state of metal does what to splitting

A

makes the splitting larger

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4
Q

T term has what

A

T term has a fooc
is temp dependent

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5
Q

what is fooc

A

first order orbital coupling

happens when an e- can choose diff orbitals to fit into bc theyre all degen and the same shape

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6
Q

are eg the same shape

A

no

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7
Q

d8 ground state term

A

3F

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8
Q

3 F has what energy ;eve;s

A

3A2g​+3T1g​(F)+3T2g​+³T₁g(P)

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9
Q

whats B again

A

B is the measure of e- e- repulsion in a free ion

this value is reduced // decreases when a complex is formed

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10
Q

why does B get reduced whe na complex is formed

A

bc theres more covalency bc the e- are spread out

bc interelectron repulsion has been reduced bc the d electron clouds have been expanded somehow

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11
Q

the greater the reduction in B value the what

A

the greater the reduction in B value,, the larger the degree of covalency between M and L

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12
Q

FOR d2 config and therefore d7 config too: if we get 2 uv transitions,, how can we find B and delta

A

if they have B values in them u need to first divide them by B to remove the B parts.

we need to do the frst transition divided by the second transition

u then see a plot of V1/V2 against delta/B,, look at the x axis of this graph to find the value that delta/B corresponds to

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13
Q

when does the first transition equal delta

A

the first transition equals delta if u have a d3 or a d8 bc they are equal

bc dn + 5 = dn

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14
Q

d3 and d6 term and d8

A

A term!!!

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15
Q

paramagnetic means

A

attracted to magnetic field

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16
Q

number of spin allowed transition for d1-d9 all high spin exceot d6

A

1,2,3,3,0,0,3,3,1

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17
Q

anything about the assumption of uvvis and it not being correct =

A

could undergo jahn teller distortion

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18
Q

positive waveength

A

d5-d9

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19
Q

expected magnetic moment,, everything on it

A

u expected = u so (1-a h / delta)

Aa = 4
Ea = 2

h = wavelength given in cm-1

v = must be in cm-1 too ( 10^7/nm)

do the division,, then subtract from 1. then multiply by u.so!!!

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20
Q

d3, 6 ,8 are what

A

A terms

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21
Q

d1, 2, 5 are

A

T terms
everythiung else is E

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22
Q

eg set

A

cross bc largest energy

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23
Q

t2g set

A

pirate cross bc lower in energy

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24
Q

eg use what orbital

A

pz orbitsl

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25
eg what do we do
put 4 e- in see if 1st and 4rth are same or not
26
antiferromagnetism
when e- point in diff ways
27
ferromagnetism
when e- point in the same way spins are parallel
28
for 180 anti /ferro what do we use
we only use 1 p orbital
29
for 90 antiferro // ferro what do we use
we use 2 p orbitsls and we fill he 2 and 3rd electron the same wayy y
30
how do we know if smt is eg or t2g
draw out the molecular orbital diagram to see which orbital the last electron goes into.
31
is ferro or anti ferrow lower in energuy
antiferro is lowest in energy
32
more negative J means whgat
more antiferro more stable more lower in energyt
33
whats ferrimage=netism
when u still have net magnetism but the arrows are still opposite facing. aka 50:50 ratio of up and down but the down arrows will be shorter etc.
34
weak but multiple sharp dd transitions is what d config
thats d5!!
35
2 diff temps
low spin = low temp high spin = high temp spin crossover
36
graphs of u against T
- nice smoothe S - not nice S ,, blocky S,, straight, up straight
37
TREN AND QP
TRIGONAL BIPYRAMIDAL TREN = HIGH SPIN QP = LOW SPIN 2,2,1 L TO H DXZ DYZ, DXY, DX2Y2, DZ2. BC OF TRIGONAL FACES GETTING CLOSER // FURTHER FILL 2,, THEN. 2,, THEN 1
38
somo
singly occupied moolecular orbital q
39
CP AND BZ
cp = always M^2+ bz = Mo 2,1,2 cp = all 3 bz = 2 then 1
40
diamagnetic
not attratcted to magnetic field = no unpaired e-
41
after Fe metal and BP complexes
stronger nuc charge // larger nuc charge = pulls s orbitals = weakening s bonding. after Fe,, the 2,2,1 aka the 2 and 2 switch.. from e2 the bottom to a1 the bottom
42
for d2 and d7 how do we find B and delta
v1/v2 then find delta/B from graph if delta/B= 29 then delta = 29B then fill this into the equation they give. make sure yk BxB is Bsquared and when its delta.squared u need to do 29B squared aka square 29 and B.
43
nephelauxetic effect
The nephelauxetic effect is the reduction in the Racah parameter 𝐵 B on complex formation due to expansion of the metal d-electron cloud, and it provides a quantitative measure of the covalency of the metal–ligand bond.
44
d3 and d8 delta
delta = first transition
45
sharp peaks
GS and ES are parallel as delta increases broad peaks,, GS and ES go from a point to far apart as delta increases!
46
low spin d6 delta and B
V2-V1 = 16B find B yes 4B = C V1 = delta - C
47
when is a peak broad
peak is broad due to delta aka the splitting changing as the bond vibrates,, when the ligand is close the delta is large and when the ligand is far the delta is small. this means as delta increases the ES and GS move like a cone shape,, the larger the delta,, the larger energy the ES is in. aka the ES energy is sensitive to the ligand field
48
when is the peak sharp
the peak is sharp when the ES is insensitive of the ligand field aka a larger delta doesnt affect the ES energy meaning as delta increases,, the energy of the ES and GS remain the same
49
d2 and d7 what is delta
delta is v3-v1
50
d3 and d8 the delta is
v1
51
in d6 ls what does the first transition equal and what does the second transition equal
delta - c delta + 16B-C v2-v1 = 16B C=4B
52
CFSE
-0.4 x t2g + 0.6 eg
53
Oh and d2 but bidentate ligand
it wont be a T term,, but an E or A
54
delta o if just given one nm
10^7/nm value
55
Oxidative addition is favoured by
High electron density at metal * Low π-backbonding * Low steric hindrance
56
why for oxidative addition is the yield lower when u have C-cl vs C-Br
bc C-Cl is a stronger bonddd
57
PMe₃ as a ligand
small good sigma donor makes the metal very electron rich
58
NMe₃ lower than PMe₃
N = more electronegative so a weaker sigma donor Ir is less negative so less oxidative addition
59
P(OMe)3 characteristics
pi acceptorrrrrr takes electron density away from the metal so less oxidative addition
60
why is CO bad for oxidative addition
bc such a good pi acceptor it takes electron density away from the M
61
if we know oxidative addition reaction conditions we know
we know reductive elimination reaction conditions
62
charge and backbonding ability and CO stretch
As the metal becomes progressively more positively charged from Ti²⁻ to Fe²⁺, metal→CO π-backbonding decreases, strengthening the C–O bond and therefore increasing ν(CO), giving the observed ordering.
63
whats a stronger sigma donor,, cp* or cp
cp***
64
what happens down a group with BB
BB and sigma bonding are syngeristic meaning they strengthen eachother as u go down a group,, the d orbitals increase in size and overlap better with the CO pi * orbital so more BB.
65
down a group metal affect with any ligand
down a group = larger orbitals = larger overlap = more e- for each ligand so less acidic Hs.
66
larger d orbitals
larger overlap = more e- to what its overlapping with making more bb or H- characterrr
67
d1 likes what trigonal jhan teller and how do we remmeber
t2g splits to a and e thing a = 1 and e = 2 lines if its d1 we want a then e therefore we like compressed
68
d2 and trigonal jahn teller
we like elongation bc we get e then a aka we can put 2 e- in the e set one in each
69
d4 ls and OBM
means its in trigonal jhan teller where its e then a aka elogation therefore all the 4 e- are paired up
70
alkenes onded bby n2 bond description and how this affects it and how can we exp measure thissss
In Zeise’s salt the alkene binds by σ-donation from the π orbital to Pt and π-backbonding from Pt into the π* orbital, resulting in a longer, weaker C=C bond with reduced stretching frequency, confirmed experimentally by X-ray bond lengths and IR ν(C=C) shifts.
71
descrive ZEIZEs salt
Pt 3Cl 1 alkene via n2
72
why are early tm good for olefin polymerisation
bc they have a low d electron count making them e- poor meaning theres no e- to enter the M-H sigma* bond for beta hydride elimination which temrinates polymerisation it also makes then good lewis acids bc they accept e- pairs via alkenes allowing them to continue the chain process
73
why are later Tm bad at olefin polu,erisation
bc they have a higher d eectoen count mesning theyre more e- rich,, beta hydride elimination is more likely and therefore polymerisation would stop,, so polym wouldnt work
74
more BB means what in terms of rotation
larger rotation enrgy barrier aka harder to rotate deep it,, bb weakens the bonds between the CO ligand but strengthens the M-CO bond stronger bond = less likely its gonna rotate
75
spec technique to measure bond rotation
use variable temperature nmr!! the diff environments will have different peaks when it comes to cold temp but as temp increases the rotation will increase if energy barrier is surpassed,, the peaks join and become one we can find the rotation barrier energy using the emeru of the two peaks and we get the coalescence temperature too!
76
how can a metal bb into CC alkene but not CO
bc the pi* for CC must be hella low in energy!! bc its non polar
77
what is magnetic susceptibility
a measure of how strongly a material is magentised whrn put in an external magnetic field
78
ferro mag and how its two electrons parallel when the angle is acute
bc hunds rule needs the e- pair to be paralled bc the 2 p orbitals are orthogonal
79
obtuse angles and antiferro
superexchange can occur bc both metals interact with the same p orbitals .. this stabilises the opposite spins
80
diff oxidative addition reactions
- single electron transfer - SN2 ( polar ligands) anionic metal electron rich ones - concerted reaction ( non polar ligands) , 16e- square planar ones sn2 is the selective one bc u add the. Rand then u add the X ,, it shows inversion at the carbon
81
p and t2g orbital on the super exchange pathway
put the e- in the same way arounddddddd ] bc orthogonal so dont need to follow hunds rule
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