Pre- & Post- Transition Metals

Question 1

transition metal

Answer 1

an element from the d-block that normally forms complexes with an incomplete d-shell

Question 2

pre-transition metals

Answer 2

metals before central d-block

A group metals
- in group 1, group 2, group 3 + Lanthanides
-also includes Al

group A elements: largely hard (but with exceptions)

Question 3

post-transition metals

Answer 3

B group transition metals
- elements in group 11-12
- metals in group 13-16

Group B elements: largely soft

Question 4

hard and soft acids and bases

Answer 4

hard prefers hard
soft prefers soft

in this definition, acid and base refer to lewis acid and lewis base

hard and soft refers to the polarisability of a metal ion or species
- hard ion: does not deform much when brought up to ion of opposite charge
- soft ion: readily deforms when brought up to ion of opposite charge

Question 5

lewis acid

Answer 5

electron pair acceptor (Na+)

Question 6

lewis base

Answer 6

electron pair donor (F-, NH3)

Question 7

hard (for lewis acids)

Answer 7

~ pre-transition metals are hard
~ transition metals in high oxidation states are hard

a high oxidation state would be a charge of +3 or higher
- however, for 2nd and 3rd row transition metals, sometimes higher oxidation states are needed
- Pt4+ is soft

Question 8

soft (for lewis acids)

Answer 8

~ post-transition metals are soft
~ transition metals in low oxidation states are soft

Question 9

hard (for lewis base)

Answer 9

F, N, O donors are typically all hard bases

F-, -OH, NH3, RNH2, ROH

Question 10

soft (for lewis base)

Answer 10

all other donors are soft bases
S, Se, Te, As, P donors are typically all soft bases

H2S, PR3, H3R3, TeR2

Question 11

determine if transition metal ion is hard/soft

Answer 11

look at the stability of the corresponding halide

if we have: F- > Cl- > Br- > I-
metal cation is hard

if we have: F- < Cl- < Br- < I-
metal cation is soft

stability of complex determined by looking at the stability constant (Kn, Bn)

Question 12

Kn - stability constant

Answer 12

for a metal ion (cation) combining with a ligand (lewis base) the equilibrium constant is called the stability constant or formation constant

addition of ligand occurs stepwise
K1 - first ligand addition
K2 - second ligand addition

Question 13

βn - overall stability constant

Answer 13

considers adding all the ligands in one go (rather than stepwise)

βn = K1 x K2 x K3 … Kn

Question 14

trends in stability constants

Answer 14

if stability constant for a reaction is high, then that complex will be readily formed and is encompassed by our definition of a strong complex via HSAB principle

it is commonly observed that the stepwise stability constants lie in the order:
K1 > K2 > K3 > K4 … Kn

Question 15

stepwise reduction in Kn values

Answer 15

consider reactions in water and the number of sites available for a displacement reaction

K1 - 6 sites
K2 - 5 sites
K3 - 4 sites
K4 - 3 sites
K5 - 2 sites
K6 - 1 sites

less sites available for L - therefore, statistically less likely to happen

Question 16

exceptions to stepwise reduction in Kn values

Answer 16

occur when there is a structural change at the metal (i.e. 6 to 4 coordination)

occur due to variation in high spin - low spin

occur due to changes associated wth the John-Teller distortion (Cu2+ ~ John-Teller ion)

Question 17

changes to stability constant when structural change occurs at metal

Answer 17

stability constant for Cd with Br-
-logK1 = 1.58
-logK2 = 0.54
-logK3 = 0.06
-logK4 = 0.37

increase in 4th stability constant - this anomaly suggests a structural change at the metal
[Cd(H2O)6]2+ is a 6-coordinate complex but Br- is a larger ligand than H2O

stability constant K4 relates to :
[CdBr3(OH)3]- + Br- ⇌ [CdBr4] + 3H2O
OCTAHEDRAL ⇌ TETRAHEDRAL

addition of 4th Br- ligand is favourable (increase in entropy ~ 2 to 4 molecules) - as Br- is so large, there is not enough space for H2O molecules when 4th ligand is added

Question 18

the chelate effect

Answer 18

K1 for a bidentate ligand (en) compared to β2 of corresponding di-ligand complex (diamine)

logK1 > logβ2

essentially, the same two Cu-N bonds are formed ~ but the chelated complex is more favoured

ΔS is large and positive for the en addition
ΔG more negative

Question 19

reattachment of chelate ring

Answer 19

loss of one arm of the chelate ring - can easily reattach due to high local concentration of the end of the molecule

reattachment best for smaller ring sizes

Question 20

chelate effect maximised for smaller ring sizes

Answer 20

chelate effect maxmised for ring sizes of 5 and 6 and it virtually non existant above a ring size of 9 atoms

Question 21

macrocylic ligand

Answer 21

ligand that has a ring or a caged structure

Question 22

macrocyclic effect

Answer 22

the enhanced stability of complexes with macrocyclic ligands compared to their open chain analogues

arises from changes in configurational entropy
- the open chain analogue loses a lot of configurational entropy (rotation around bonds) on complexation
- the macrocyclic ligand loses less configurational entropy as it is already in ring form

Question 23

there is a cavity size effect for macrocyclic ligands

Answer 23

K+ fits best in the cavity for 18 crown 6
Rb+ fits best in cavity for valimaycin

the size selection properties of macrocyclic ligands can be used in the separation of various metal ions

macrocylic ligands that are three dimensional are known as cryptands

Question 24

solubilities of ionic substances

Answer 24

solubility of ionic salts related to four main factors

relating to change in ΔH
1. lattice energy lose on dissolution
2. solvation energy gain on dissolving

relating to change in ΔS
3. the entropy gain by the ions moving in solution
4. the entropy lost by the solvent molecules

Question 25

solubility of NaCl

Answer 25

Na+ and Cl- massively gain entropy - there is very little comparative entropy in the rigid NaCl lattice - on dissolving the anion and cation are free to move in solution H2O molecules lose entropy - attracted to ions

Question 26

solvation energy is given by:

Answer 26

the solvation energies and solubilities increase with the dielectric constant of the solvents (dielectric constant - measure of how easy it is to sepatate ions) the solubilities of the salts generally increase with size of cation and anion -entropy change accompanying dissolution is less favourable for small ions solvation energy inverserly proportional to ionic radius

Question 27

lattice energy is given by:

Answer 27

lattice energy inversely proportional to the ionic radius there is a charged effect - the lattice energy increases faster for multiply charged ions lattice energy is higher for ions of similar size as they pack well in the lattice - salts are likely to be more soluble if the cation and anion differ in size

Question 28

lattice energy higher for ions of similar size - less soluble salt

Answer 28

NaF 1.0 mol dm-3 (less soluble) NaCl 6.2 mol dm-3 NaBr 8.8 mol dm-3 NaI 11.9 mol dm-3 (more soluble) Na+ 0.95 Å F- 1.36 Å Cl- 1.81 Å Br- 1.95 Å I- 2.21 Å - Na+ is closer in size to F- - NaF has good packing in a lattice and high lattice energy - much more difficult to dissolve - in NaI, there is a big size difference between the anion and cation - low lattice energy, less energy is needed to dissolve NaI

Question 29

multiply charged ions are less souble than singly charged ions

Answer 29

require more solvent ions to go around them NaI (singly charged) more soluble than MgSO4 (muliply charged)

Question 30

oxidation and reduction

Answer 30

oxidised: species lost an electron A → A+ + e- reduced: species gained an electron A + e- → A- oxidising agent: species that removed electrons for oxidation reducing agent: species that supplies electrons for reduction

Question 31

REDOX REACTION: Zn + 2H+ → Zn+ + H2

Answer 31

can be viewed as the sum of two half reactions - each with its own ΔG 2H+ 2e- → H2 Zn → Zn2+ + 2e-

Question 32

ΔG from half reactions

Answer 32

ΔG = -nFE ΔG - change in Gibbs Free energy n - number of electrons transferred in reaction F - Fariday constant E - standard reduction potential if E > 0, ΔG will be negative for reaction to be spontaneous, E must be positive

Question 33

Latimer diagram

Answer 33

applies thermodynamics to solution inorganic chemistry. it is in the form: Ox → Red enables us to calculate a non-adjacent couple ~ for example the E value for: Cu2+ → Cu Latimer diagrams are normally shown for pH = 0, but they are also tabulated for pH = 14

Question 34

Latimer diagram (example)

Answer 34

(number of electrons) x (potential E for stepwise reduction) = (number of electrons transferred) x (potential of overall reaction)

Question 35

Latimer diagram - strongest oxidising agent

Answer 35

look at the highest potential for any couple ~ oxidising agent is on the LHS of the couple

Question 36

Latimer diagram - strongest reducing agent

Answer 36

look at the lowest potential for any couple ~ reducing agent is on the RHS of the couple

Question 37

Latimer diagram - species likely to undergo disproportionation

Answer 37

M2+ → M+ + M3+ species goes from one oxidation state to a higher and lower oxidation state a species is unstable to disproportionation if the potential on the left is lower than the potential on the right

Question 38

Latimer diagram - species likely to undergo comproportionation

Answer 38

M+ + M3+ → M2+ a higher and lower oxidation state converge into a single species (opposite of disproportionation)

Question 39

Frost diagram

Answer 39

pictorial representation of a Latimer diagram (constructed from Latimer diagram) plot of electropotential against oxidation state tells us about: -comproportionation/disproportionation -most thermodynamically stable species -strongest oxidising/reducing agent

Question 40

Frost diagram - comproportionation/disproportionation

Answer 40

Question 41

Frost diagram - most thermodynamically stable species

Answer 41

species which is lowest on the diagram

Question 42

Frost diagram - strongest oxidising/reducing agent

Answer 42

strongest oxidising agent: species which lies at the top of the most positive slope - [O] agent is on the RHS of the most positive slope strongest reducing agent: species which lies at the top of the most negative slope - [R] agent is on the LHS of the most negative slope

Question 43

Frost and Latimer diagrams

Answer 43

tell us about solution thermodynamics and not about kinetics but an overpotential of 0.6V normally means the reaction occurs kinetically fast overpotential for oxidation: 0.6V greater than that required for the minimum thermodynamic amount overpotential for reduction: 0.6V less than than required for the minimum thermodynamic amount

Question 44

stability field of water

Answer 44

region 1: any oxidising agent can oxidise water rapidly - will kinetically and thermodynamically oxidise water region 2: any species can thermodynamically oxidise water, but the reaction is kinetically slow region 3: water is stable to oxidation and reduction region 4: species can thermodynamically reduce water, but the reaction is kinetically slow region 5: any reducing agent can reduce water rapidly - will kinetically and thermodynamically reduce water oxidation line relates to oxidising water. lines slope downwards to higher pH - easier to oxidise species at high pH - easier to reduce species at low pH

Question 45

thermodynamic and kinetic oxidation of water

Answer 45

thermodynamic oxidation - at pH=0, a species with a potential greater than 1.23V required - at pH=10, a species with a potential greater than 0.63V required kinetic oxidation an overpotential of 0.6V is required for reaction to proceed quickly - at pH=0, a species with a potential greater than 1.83V - at pH=10, a species with a potential greater than 1.23V required

Question 46

thermodynamic and kinetic reduction of water

Answer 46

thermodynamic reduction - at pH=0, species with a potential less than 0V required - at pH=10 species with a potential less than -0.6V required kinetic reduction an overpotential of 0.6V is required for reaction to proceed quickly - at pH=0, species with a potential of less than -0.6V required - at pH=10, species with a potential of less than -1.2V required

Question 47

Pourboix diagram

Answer 47

phase diagrams which plot electropotential against pH produce a form of phase diagram that enable us to predict the form of a species at a particular potential and pH value

Question 48

Pourboix diagram example

Answer 48

strongest oxidising agent: at the top of the diagram ~ FeO4 2- strongest reducing agent: at the bottom of the diagram ~ Fe

Question 49

areas in the Pourbaix diagram

Answer 49

mark regions where a single species is stable - more stable species tend to occupy larger regions

Question 50

lines in the Pourbaix diagram

Answer 50

mark places where two species exist in equilibrium horizontal lines: pure REDOX reactions - reactions are not pH-dependent - no acid-base equilibria verticle lines: pure acid-base reactions - reactions do not depend on potential downward slope of -0.0592 V/pH: reactions that are both acid-base and redox - acid/base exchange is an important part of process

Question 51

acidity of cations in solution

Answer 51

dissolution of ionic salts in water can cause changes to solution pH - due to metal ion polarising the water molecules and producing H+ in solution smaller, multiply charged ions tend to cause more distortion of the water molecules (related to the charge to radius ratio of the ion)

Question 52

pH

Answer 52

-log[H+] lower pKa value more acidic

Question 53

pH ~ Henderson Hasselbalch equation

Answer 53

pH = pKa + log10 ([base]/[acid])

Question 54

metal is able to polarise the M-O bond towards itself

Answer 54

the bond gets elongated and easier to break dependent on the q/r ratio of the cation ions of high charge/small radius are more polarising - more chance of breaking the O-H bond and forming H+ in solution lower pKa values indicates stronger acids metal aquo ions act as Bronsted acids ~ H+ donors

Question 55

metal aqua ions - acidity

Answer 55

1. smaller ions have higher acidity 2. higher charged metal ions have higher acidity 3. electronegativity of ion also greatly affects pKa and can override other effects most electronegative metal: gold