remember 1b

Question 1

CO ligand gives how many electrons

Answer 1

2

Question 2

co ligand has whag dharge

Answer 2

0 neutral

Question 3

CP ligand has what charge

Answer 3

anionic charge!!

Question 4

alkene as a ligand charge

Answer 4

0

Question 5

number of mm bonds when predicting shapes

Answer 5

NEED - HAVE / 2 !!!

Question 6

Metal carbonyl + Na/Hg

Answer 6

bc Na is positive,, the complex will be negative

u will also half the complexxx so from 2 metals to 1 and from 8 ligands to 4 etc

Question 7

negative nmr ppm means

Answer 7

hydride is formed

Question 8

a hydride has a low pka when

Answer 8

its a strong acid

mraning its conjugate base is stable

normally bc theres good pi acceptor ligands aka CO

Question 9

when can pka of a hydrode increase

Answer 9

when the conjugate base is less stable

aka theres a sigma donor here

Question 10

M2CO8,, what metal is itttt

Answer 10

Co: 2e- x 8 = 16e- for both metals therefore 8 per metal

18-8 = 10e-
so 10e- metal butttttt,, we also need to think that theres a mm bond,, so 10 is with the mm bond adding one e-,, so withut the metal metal bond there would be 9e-

look for a TM with 9e-!!!

Question 11

counting valence electrons in a complex

Answer 11

metal and contributio from ligands

Question 12

contribution of alkene

Answer 12

2 electronsss

Question 13

when asked to guess reactions what should we do, aka we get a complex and a ligand

Answer 13

count how many e- we have in the complex
ligand and metal and charge

then see if its 18e- or 16e- etccc

then look at the incoming ligand

if its a halogen its probs a SN2 type thing
if its 16e- complex then probs dissociative subs,, if its 18e- probs associative subs etc.

Question 14

can an oxidation state be neagtive

Answer 14

yus

Question 15

when finding the valence electron count what do we need to do first

Answer 15

we need to find the oxidation state of the metal

Question 16

if H is a ligand what do we need to remember

Answer 16

rememebr that its a H-

so a hydride

so it gives 2e-

and it affects the oxidation state of the metal by -1. so wed get a +1 OS.

Question 17

if our ligand is an alkene,, and our complex is 16e- and has a H as a ligand what happens

Answer 17

migratory insertion reaction

we add the alkene cis to the H then we migrate the H onto the alkene to form an alkane lowkssss

Question 18

CH2CH3 as. aligand description

Answer 18

its -1 charge therefore it affects the metals oxidation state

Question 19

what else should we look at the ligands for when were asked to guess the reaction

Answer 19

look for beta hydrogens bc beta hydride elimination may occurrrrr

one of the ligands cis to the b hydrogen needs to be removed thoo so we have a vacant site yay

Question 20

if the ligands approaching is a CO what do we need to think of

Answer 20

we need to think of migratory insertion!!!

aka Me will migrate onto a CO and then the new CO will attack the vacant site

Question 21

when we wanna rate ligands based on their pi acceptor abilities what should we look for.

Answer 21

POSITIVE CHARGE

ELECTRONEGATIVE

LARGE // DIFFUSE ORBITALS : overlap with metals d orbitals better!!

(-) charge will be very bad

the top 3 thin gs lower the energy of the Pi* which is wherre the m bb into

Question 22

5 lkgand complex undergoing oxidative addition with H2 approaching but one of the OG ligands is a Cl

Answer 22

add a H and replace Cl

Question 23

Mn2 CO10 dimer + HCl what reaction is this and what do we get as products

Answer 23

oxidative addition heterolytic version

we split them up obvs and one will get the H and the other will get the Cl

Question 24

when migratory inertion occurs where do things go

Answer 24

the migrated ligand goes from the M to the metal again, aka the migrated ligand always tries to go between the M-L of a diff liganddd

Question 25

when we see CO what should we think of

Answer 25

migratory insertion perhaps the co comes between the M and the L

Question 26

CH3 and CH2CH3,, which one will the CO migrate between

Answer 26

the CH3 bc less hinderedddd

Question 27

H and alkene cis with just reflux

Answer 27

migrate the alkene to form an alkane then kick it off via reductive elimination

Question 28

when theres a H or Cl in a complex and we are elecron counting what should we do

Answer 28

remeber H and Cl affect ur oxidation state

Question 29

when doing the 18e- rule remmeer what

Answer 29

outside = inside CH3 =-1 charge H = -1 charge halogen = -1 charge

Question 30

alkene acc charge when cwere counting e-

Answer 30

0 its neutral

Question 31

CP charge when counting e-

Answer 31

-1 anionic

Question 32

Cp e-c count when counting e-

Answer 32

5 based in the n number

Question 33

18e- but trigonal bipyramidal + H2 as an incoming ligand

Answer 33

oxidative additgion of just one H try keep trigonal bipyramidal,, maybe swap H for smt with the same e- contribution and charge aka if theres a Cl replace it with H.

Question 34

metal complex with reflux

Answer 34

- H and alkene to alkane

Question 35

ACTUAL: OA

Answer 35

- H2 or R-X ligand incoming - d8-d10 - low oxidation state - less than 18 // move to allow it ligand incoming = sigma bond

Question 36

ACTUAL: migratory insertion

Answer 36

MR CIS to CO or ALKENE CO // ALKENE migrates on smallest alkane!!!! aka ch3 over ch2ch3

Question 37

ALKENE MIGRATION

Answer 37

Alkene cis to MR or H will move and get turned into an alkane normally no ligand incoming just reflux

Question 38

B hydride elimination

Answer 38

- need beta hydride and vacant spot on the metal forms a MH bond and an alkene

Question 39

REDUCTIVE ELIMINATION

Answer 39

- 18elec trons - metal has high oxidation state aka +3 etc

Question 40

which one will have the larger IR vib freq,, what do we need to remember

Answer 40

more e- rich metal = more bb = weaker bond = lower vib frequency - negative oxidation state = more electron rich = more bb = weaker bond = lower frequency - POMe = good pi acceptor so competes against CO for e-,, less bb into CO - more sigma donors = more e- into CO = more bb = lower vib freq - less CO = more concentrated bb = weaker bond = smaller vib freq - rhs metal = more electronegative = less likely to give e- = less bb = larger vib freq

Question 41

rememeber cp = -1 charge

Question 42

if the metal is anionic,, aka - what do we do.

Answer 42

sn2 aka we atatck the C,, dont add the halogen,, then do the next step they tell us to do