1c

Question 1

how can we purify nickel

Answer 1

the mond process

Question 2

describe the mond process quickly

Answer 2

crude Ni + CO –> [Ni(CO)4] –> reflux to get pure Ni and CO

Question 3

the mond method is to purify metals but what metals can it be used on

Answer 3

it can be used on nickel and for Fe

Ni and Fe

Question 4

how do we synthesise metal carbonyls ,, name the 4 ways

Answer 4

• direct reaction of metal with CO
• reductive carbonylation
• heating or irridating an existing metal carbonyl
• metal vapour synthesis

Question 5

describe direct reaction of metal and CO

Answer 5

only works with Ni and Fe

metal + CO –> [Ni(CO)4]

Question 6

describe reductive carbonylation

Answer 6

metal salt is treated with reducing agent (Na) under pressure of CO

aka CrCl3 + Na + CO (200atm) –> 120 degrees –> [Cr(CO)6]

or OsO4 + CO (70atm) —> 125 degrees –> [Os3(CO)12] where CO is the reducing agent

Question 7

describe heating or irridating an existing metal carbonyl

Answer 7

heating gives off a CO ligand,, which leaves a reactive fragment that forms metal metal bond clusters

this adds CO to the complex

need uv heat and CO

Question 8

describe metal vapour synthesis, MVS or matrix isolation

Answer 8

unsuitable for large scale synthesis but can characterise otherwise unstable species

bulk metal + heat and vac –> M(g) —> CO and 40k to give [M(CO)x]

u isolate in situ

Question 9

bonding of CO in metal complexes,, give the electron config of the c and the o and describe their mo

Answer 9

C = 4e- : 1s2 2s2,,
O = 6e- : 1s2 2s2 2p2

s overlaps and u fill sigma1 and sigma2

P overlap and u fill pi and sigma 3
pi* and sigma* are empty

Question 10

describe the bonding between CO and metals

Answer 10

sigma donatiion from the C lone pair into the suitable metal vacant orbital

pi back bonding from a filled d orbital on the metal into the pi* bond on the CO

the CO bond is weakened and the M-C bond is strengthened

Question 11

what type of a ligand is a CO ligand

Answer 11

its a pi acceptor meaning it accepts e- from the metal into its pi* orbital

it takes e- density away from the M

Question 12

the sigma and pi does whattt,, aka why is it called synergic bonding

Answer 12

bc the sigma and the pi components strengthen eachother // reinforce eachother

Question 13

the bonding of M and CO changes the bond strength in the CO and the MC bond,, is there evidence for this

Answer 13

yesss
u can use xray diffraction

aka u get bond angles and lengths

the bond length increases
and as bond length increases,, the bond weakens

Question 14

if u have a CO and a NH2 in a complex how does this differ than just CO

Answer 14

NH2 is an amine and increases the electorn density of a metal meaning it can backbond int othe CO more

meaning the CO is weakened and lengthened as the MC bond is strengthened

Question 15

when backbonding occurs in CO,, what happens to the bond and explain everything

Answer 15

bond obvs is weakened
bond is lengthened
the vibrational freqeuncy is reducedddd

Question 16

when MC bond is strenghtened explain this and everything

Answer 16

bond is strengthened meaning the bond is shortened
the vibrational frequency increases

Question 17

whats the bond order for a CO thats bonded to a metal

Answer 17

the bond order is 2 and 3

Question 18

the vibrational frequency of MC and MO is sensitive to what

Answer 18

its sensitive to changes in charge and ligandssss

Question 19

if the metal is anionic aka (-) what does this mean

Answer 19

the e- density around the metal is large

meaning it can backbond a lot
meaning more CO lengthening ajd frequency reducing 1750

Question 20

if the metal is a cation what does this mean

Answer 20

theres less e- density on the metal

so theres less backbonding

meaning the vibrational frequency of CO increases 2096

Question 21

how can other ligands affect bonding in complexes

Answer 21

u have sigma donors and pi acceptors

if theres sigma donors,, theyll increase the e- density of the metal allowing more backbonding into the CO,, lengthening, weakening and reducing the vib freq of the bond

Question 22

name a good sigma donor

Answer 22

PPh3

u have PF3< P(OPh)3 < PPh3 < PMe3 < PtBu3

Question 23

a terminal M-CO has a vib freq of what in a neutral compound

Answer 23

2100-1900cm-1

Question 24

a doubly bridged M-CO (long u - co) has what vib freq and what does this look like

Answer 24

M - M - - CO,, triangle type shape

1900-1760cm-1 bc the one CO can experience bb from 2 metals
2 bond order

Question 25

triply bridging (long u3- CO) has a vib freq of what and what does rhis look like

Answer 25

triangular based pyramid type shape bond order is less than 2 1750-1600 bc theres 3 metals that can bb into CO

Question 26

what metals do CO not like bridging to

Answer 26

the left of the transition metals the ones at the bottom of the transition series

Question 27

when do CO like bridging to metals

Answer 27

when they have a short radius so they can bridge well aka kinda like electronegativity,, we like rhs and up of the transition metals

Question 28

elusive meaning

Answer 28

hard to capture or achieve

Question 29

whats a metal alkyl

Answer 29

a metal bonded to a C via sigma bond

Question 30

are metal alkyls thermodynamically stable

Answer 30

app noooo

Question 31

whats more stable,, metal alkyls on rhs or lhs of the periodic tablw

Answer 31

metal alkyls on the LHS are more stable than the RHS

Question 32

a larger BDE means what

Answer 32

it means the bond is strongerrr // more stable

Question 33

apart from thermodynamic reasons,, why else are metal alkyls so elusive

Answer 33

theres stability issues due to kinetics

Question 34

what kinetics affect the elusiveness of meta alkyls

Answer 34

the MC bond in metal alkyls is strong,, but they have a low energy reaction pathway for decomposition aka this makes it very easy for them to be decomposed they also have beta hydride elimination

Question 35

whats beta hydride elimination

Answer 35

its when a metal alkyl turns into a metal hydride and an alkene attached to it

Question 36

what do u need for a beta hydride elimination to occur,, this affects how elusive metal alkyls are

Answer 36

- need a vacant coordination site on the metal - need a beta hydrogen the metal will take the beta hydrogen and form a little transition state bonded to it ,, there will be a dotted line between the M and the alpha C this then forma an alkene and the H is moved onto the metalll

Question 37

so why are metal alkyls so elusive aka hard to capture

Answer 37

- low thermodynamic stability - kinetics: decomposition pathway is low in energy and if theres a beta hydrogen the complex will form a metal hydride and an alkene instead

Question 38

how do we stabilise metal alkyls

Answer 38

we need to prevent beta hydride elimination!!! or block the vacant coordination site // maintain the 18e- count use group that cannot form // its difficult to form planar M C C H units (due to blocking beta hydrides) or use groups that cannot form stable products after elimination (if metal is bonded to a double bond,, this cannot be stable once it forms a triple bond)

Question 39

how do we prevent beta hydride elimination

Answer 39

we use alkyls in metal alkyls that don't have a beta hydride

Question 40

what metal alkyls can be used to prevent beta hydride elimination to decompose metal hydrides

Answer 40

M-Me M-CH2- Benzene (benzyl) M-CH2-T-Bu (neopentyl) M-CH2-Si,Me3 (trimethylsilylmethyl)

Question 41

even if beta hydride elimination occurs,, what other elimination pathway can occur

Answer 41

alpha elimination pathways can occur this is where M-CH3 turns into M=CH2 radical pathways

Question 42

metal alkyls are also very reactive to what

Answer 42

oxygen and water : need air sensitive experimental technique to handle metal alkyls in the lab

Question 43

how do we make transition metal alkyls

Answer 43

- alkylation ofmetal hydrides with main group M-alkyl reagent - insertion of alkene into metal hydride - oxidative addition of aryl-X or alkyl-X to nucleophilic metal complex

Question 44

if we have a nucleophilic metal,, what C do we need

Answer 44

we need an electrophilic C

Question 45

if we have an electrophilic metal,, what C do we need

Answer 45

we need a nucleophilic C

Question 46

describe how we make metal alkyls through the alkylation of metal hydrides with main group M - alkyl reagent

Answer 46

WCl6 + 6MeLi --> WMe6 + 6LiCl WMe6 is the metal alkyl

Question 47

describe how we make metal alkyls via insertion of alkene into metal hydride

Answer 47

Cp2Zr(H)Cl + RCH=CH2 --> Cp2ZrCl(CH2 CH2 R)

Question 48

describe how we make metal alkyls via oxidative addition of aryl-X or alkyl-X to a nucleophilic metal complex

Answer 48

Ir(CO)(Cl)(PPh3)2 + Me-I ---> Ir(Me)(CO)(I)(Cl)(PPh3)2 Pd(t-Bu3P)2 + Ph-Br --> Pd(Ph)(Br) (t-Bu3P)2