Transition metals Flashcards
4 chemical properties of transition metals
-variable oxidation state
-form complex ions
-catalysis
-colour
transition metals
Transition metals = one which forms one or more stable ions which have incompletely filled d orbitals
Zn and Sc = not transition metals
complex ions
Complex ions:
1) has a metal ion at its centre
2) ligands = molecules or ions surrounding it, attached to the central ion by a dative covalent bond
ligands as donors
All ligands are lone pair donors = lewis bases. Ligands are Lewis bases - they contain at least one pair of electrons to donate to a metal atom/ion. Ligands are also called complexing agents. Metal atoms/ions are Lewis acids - they can accept pairs of electrons from Lewis bases.
ligand
Ligand = particle with a lone pair that forms a coordinate bon to a metal
complex
Complex = metal ion with ligands co-oordinately bonded to it
co-ordination number
Co-ordination number = number of co-ordinate bonds from ligands to metal ions
accepting ligands
Transition metals have empty valence orbitals which can accept pairs of electrons from ligands.
[CuCl4]2+
This molecule has a 2- charge because the Cu2+ and 4Cl- cancel out to form 2- charge overall.
The coordinate number of CuCl4 is 4 because there are 4 coordinate bonds.
copper metal vs copper ion
Copper metal = 1s2 2s2 2p6 3s2 3p6 4s1 3d10
Copper ion = 1s2 2s2 3s2 3p6 3d9
[Al(H2O)6]3+
Charge is 3+ because there is zero charge from the water molecule and 3+ from the Al.
Coordinate number = 6 bc there are 6 co-ordinate bonds
Octahedral
linear
Linear = 2 coordination number = Ag complexes e.g [Ag(NH3)2]3+
tetrahedral
Tetrahedral = 4 coordination number = large ligands like [CuCl4]2+
square planar
Square planar = Pt+ complexes e.g [PtCl4]3+
octahedral
Octahedral = most common = [Cu(H2O)2]2+
exceptions for the shape of complex ions
-Ag forms linear complexes
-Pt forms square planar complexes
types of ligands
Ligands can be unidentate (e.g. H2O, NH3 and Cl - ) which can form one coordinate bond per ligand or bidentate (e.g. NH2CH2CH2NH2 and ethanedioate ion C2O4 2- ) which have two atoms with lone pairs and can form two coordinate bonds per ligand, or multidentate (e.g. EDTA4- which can form six coordinate bonds per ligand).
unidentate ligands
-unidentate ligands = form 1 co-ordinate bonds e.g H2O, OH-, NH3, CN, Cl-
e.g [Cu(H2O)6]2+ [CuCl4)2-
bidentate ligands
-bidentate ligands = form 2 co-ordinate bonds e.g 1,2-diaminoethane (NH2CH2CH2NH2) or ethandioate (C2O4)2-
[Cr(NH2CH2CH2NH2)3]3+ or [Cr(C2O4)3]3-
NH3 charge
neutral
notes for specification
-octahedral complexes can display cis-trans complexes with monodendtate ligands and optical isomerism with bidentate ligands
cisplatin =
cis isomer
Ag+ forms linear complex used in tollens reagent
[Ag(NH3)2]+
incomplete subsitution
[Cu(NH3)4(H2O)2]2+
bidentate ligands
H2NCH2CH2NH2
C2O42-
why is carbon monoxide toxic
it is toxic bc it replaces oxygen co-ordinately bonded to Fe(II) in haemolgibin
explain the meanings of the terms multidentate and ligand with reference to the reaction of EDTA4- with [Cu(H2O)6]-
multidentate = EDTA can form 6 dative covalent bonds with central cation
ligand = lone pair can form dative bond with copper ions
hexaaquairon (II) ions react with an excess of H2NCH2CH2NH2 in a ligand subsitution reaction to form what
[Fe(NH2)6]2+
explain why water can act as a ligand
makes single dative covalent bond with metal ion
unidentate ion
name [Fe(H2O)6]2+
hexaaquairon (II)
naming monodentate ligands
Naming monodentate ligands:
-H2O = aqua
-NH3 = ammine
-OH- = hydroxo
-CN- = cyano
oxidiation number of Fe in [Fe(CN)6]4-
+2
true or false = 3 ligands can be present but there can be 6 dative covalent bonds
true
how many ligands can fit around chloride ion
4
ethanedioate =
C2O4
explain the meaning of the term complex ion
an ion that forms a dative covalent bond and has ligands bonded to the central cation
give 2 other characteristic properties of transition metals
-form ions with different colours e.g vanadium can form VO2+ with a yellow colour yet also Vo2+ with a blue colour
-different oxidation states = VO2+ = +5 but VO+2 = +4
ligand =
atom which can donate a lone pair of electrons and form a dative covalent bond
lewis acid vs bonsted lowery acid
bronsted acid = proton donor
lewis acid = accepts electron pair and will have vacant orbitals
Lewis base = donates electron pair
explain why breathing in carbon monoxide can be fatal
CO replaces oxygen molecule and binds strongly to Fe2+
explain why titanium is a d-block element and why its a transition element
titanium has electrons in its d-orbital
last filled electrons in d-subshell + has variable oxidation state
how do colour changes arise
Colour changes arise from changes in
1. oxidation state,
2. co-ordination number
3. ligand. e.g H2O to NH3
4. identity of metal e.g Cu to Fe
UV light / visible spectroscopy
frequences at which a complex absorbs UV light can be measured with a spectrometer
UV light is passed through complex
more concentration solution = more light absorbed
colorimetry
more concentrated solution = more absorbed
colour of light is chosen that compound abosrbs
strength of absrption of a range of solutions of know conc is measured and a calibration curve is produced
vanadium catalyst
V2O5
[Ag(CN)2-
[NC- –> Ag –> CN-]- = +1 oxidiation number + linear (180) [Ag(CN)2-]
tetrahedral vs octahedral
[CuCl4]2- = tetrahedral
[Cu(H2O)6]2+ = octahedral
[Cr(NH2CH2CH2NH2)3]3+
Charge of ammonia ions = neutral
Oxidation number of pt in [Pt(NH3)Cl3]- = +2
haemoglobin
-globular protein that contains 4 Fe2+ centres each with a prorhyin ligand taking up four of the six coordination sites. Oxygen can also bond as a ligand. CN- ions and CO are better ligands than O2 so that’s why they bond easily to haemoglobin and become toxic.
cis-trans isomerism
-occurs in octahedral and square planar complexes where they are two ligands of one type different to other ligands
-special case of E-Z isomerism
optical isomerism
Optical isomersism:
-occurs in octahedral complex with bidentate ligands
[Cr(C2O4)3]3- = bidentate = 2 oxygen ligands x 3
subsitution of ligands
-ligand subsitution = one ligand is replaced by another ligand
-if the ligands are a similar size then there will be no change in co-ordination number
e.g H2O and NH3 = [Co(H2O)6]3+ + 4NH3 –> [Cu(H2O)2(NH3)4]2+ + 4H2O
-if the ligands are a different size then the co-ordination number may change
-for example Cl- ligands are significantly bigger than O on H2O so coordination number changes from 6 to 4
E.g [Co(H2O)6]2+ + 4Cl- –> [CoCl4]2- + 6H2O
chelate effect
-enthalphy change is neglibile as the sname number of the same type of similar bonds are broken and formed
-entropy is positive so gibbs is very negative and the reaction is feasible
-ligands are chelating agents as they are good at bonding to a metal ion and are very difficult to then remove. E.g EDTA4-
-bidentate and multidentate ligands replace monodentate ligands
subsitution effect
The substitution of monodentate ligand with a bidentate or a multidentate ligand leads to a more stable complex. This is called the chelate effect.
Positive entropy change = more molecules of product than reactant e.g unidentate to multidenate (H2O –> EDTA)
naming monodentate ligands
Naming monodentate ligands:
-H2O = aqua
-NH3 = ammine
-OH- = hydroxo
-CN- = cyano
colours of transition metals
-transition metal ions form different colours. In transition metal complexes, electrons are promoted to higher energy orbital.
-For these electrons to be promoted they need to absorb light energy of a particular frequency. The frequency depends on the precise difference in energy between the d orbitals
-different colours = different wavelenghts of light
reflecting wavelengths
-different ligands split onto the obitals to a different extent creating a different energy gap. Electrons will therefore absorb a different frequency of light
More negative = further equilibrium lies to the left
Negative value = oxidation = released electrons more readily than hydrogen
More positive = reduction = gains electrons more easily than hydrogen
Ligands affect how easy it is to oxidise or reduce a transition metal
When you dissolve an aqua ion in water, water molecules pull H+ ions off H2O ligands
When you dissolve an aqua ion in water, water molecules pull H+ ions off H2O ligands
e.g [Fe(H2O)6]3+ + H2O -> [Fe(H2O)5(OH)]2+ + H3O+
Simplified = [Fe(H2O)6]3+ –> [Fe(H2O)5(OH)]2+ + H+
It is easier to oxidise transition metals in alkaline solution = greater tendency to form negative ions
effect of pH on oxidation state
Cr(+6) to Cr (3)
-CrO42- + 7H2O +3e- –> [Cr(H2O)3(OH)3] + 5OH- = more negative in alkali
-1/2Cr2O72- + 7H+ + 3e- –> [Cr(H2O)6]3+ + 31/2H”O = more positive in acid
This shows that it is much easier to reduce Cr2O7^2- in acid than CrO42- in alkali
Much easier to oxidise in alkali than acid
Overall easier to oxidise in alklaine and easier to reduce in acidic conditions
-reduction = higher oxidation state
tollens reagent
-reduction of [Ag(NH3)2]2+ tollens reagent to metallic silver is used to distiguish between aldehydes and ketones
[Ag(NH3)2]+ + e- –> Ag + 2NH3
vanadium oxidation states
-vanadium has 4 oxidation states = 5,4,3,2 = formed by the redutcion of vandanate (V) ions by zinc
-vanadium ions can be reduced using zinc in an acidic solution to forms its ions
V(+5) –> VO2^+ (yellow)
V(+4) –> VO^2+ (blue)
V(+3) –> V^3+ (green)
V(+2) –> V^2+ (purple)
From yellow to blue, green is formed in between yet is not shown in this series but should still be acknowledged
mixing ions
What ions are present in the flask between blue and yellow = mix of VO2^+ and VO^2+
Describe and explain what happens if you pour the liquid contents of flask E (purple) into another container –> solution turns green as it is oxidised by air to V(H2O)6^2+
explain what you would see during step 1
yellow –> green –> blue
when blue initially forms it reacts with the yellow unreacted to turn green (VO2+)
is scandanium coloured
Scandium is a member of the d block. Its ion
(Sc3+) hasn’t got any d electrons left to move
around. So there is not an energy transfer equal to
that of visible light.
spectrophotometry
If visible light of increasing frequency is passed through a
sample of a coloured complex ion, some of the light is
absorbed.
The amount of light absorbed is proportional to the
concentration of the absorbing species (and to the
distance travelled through the solution).
Some complexes have only pale colours and do not
absorb light strongly. In these cases a suitable ligand is
added to intensify the colour.
higher frequency
-higher frequency of light absorbed = bigger the energy level gap
-higher frequency light = closer to blue end of visible light spectrum
conversions
THz –> Hz = x10^12
Nm –> m = / 10^-9
vanadium half equations
VO2^+ + 2H+ (aq) + e- –> VO2+ (aq) + H2O (l)
VO2+ + 2H+ (aq) + e- –> V3+ (aq) + H2O (l)
V3+ + e- –> V2+ (aq)
different energy levels
Colours have different energy levels due to different wavelengths
Red –> blue = direction of increasing energy
what affects size of energy gap
-transition metal ion
-oxidation state
-ligands
absorbed colours
When white light is passed through a solution of its own ions, some of this energy in the light is used to promote an electron to another orbital
Colour is absorbed = opposite is perceived:
-red –> green
-orange –> blue
-purple –> yellow
hydrated copper
Hydrated CuSO4 = perceived (reflected) as blue and orange is absorbed.
why are transition metal complexes usually coloured (5 marks)
-transition metals have part-filled d orbitals
-in a compound these d-orbitals have slights different energies
-energy equal to the difference between these d-orbitals can be used to excite an electron
-this energy is related to the frequency of light. This determines the colour so one colour of light is absorbed
-the rest of light is transmitted and we see this combination of colours
equations to recall
∆E = hv = hc / λ = energy difference
Λ = c (speed of light) / f
planks constant
Planks constant = 6.63 x 10^-34 Js
velocity of light
Velocity of light c = 3.00 x 10^8 ms^-1
frequency of light
Frequency of light is related to energy difference
5 d orbitals = don’t all have the same energy = gap corresponds to energy of visible light
equation for iron and mangante
Iron + Manganate:
MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) –> Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)
(purple –> colourless)
A 2.41 g nail made from an alloy containing iron is dissolved in 100 cm3 acid. The solution formed contains Fe(II) ions. 10cm3 portions of this solution are titrated with potassium manganate (VII) solution of 0.02 mol dm-3 . 9.80cm3 of KMnO4 were needed to react with the solution containing the iron. Calculate the percentage of iron by mass in the nail.
1) Write out equation = MnO4 - (aq) + 8H+ (aq) + 5Fe2+ –> Mn2+ (aq) + 4H2O + 5Fe3+
2) 0.02 x 0.0098 = 1.96 x 10^-4
3) bc iron is 5Fe in balanced equation multiply previous answer by 5 = 9.8 x 10^-4
4) Find moles of Fe in 100cm3 = 9.8 x 10^-4 x 10 = 9.8 x 10^-3
5) ANS x mr = 0.547g
6) Find percentage by mass = 0.547 / 2.41 x 100 = 22.6%
iron and ethandioate
With iron (II) ethanedioate both the Fe2+ and the C2O4 2- react with the MnO4 - 1 MnO4 - reacts with 5 Fe2+ and 2 MnO4 - reacts with 5C2O4 2- MnO4 -(aq) + 8H+ (aq) + 5Fe2+ –> Mn2+ (aq) + 4H2O + 5Fe3+
Mn and Fe
2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2- –> 10CO2 + 2Mn2+ (aq) + 8H2O
So overall 3MnO4 -(aq) + 24H+ (aq) + 5FeC2O4 –> 10CO2 + 3Mn2+ (aq) + 5Fe3+ + 12H2O So overall the ratio is 3 MnO4 - to 5
A 1.412 g sample of impure FeC2O4 .2H2O was dissolved in an excess of dilute sulfuric acid and made up to 250 cm3 of solution. 25.0 cm3 of this solution decolourised 23.45 cm3 of a 0.0189 mol dm–3 solution of potassium manganate(VII). Calculate the percentage by mass of FeC2O4 .2H2O in the original sample.
1) 0.0189 / 0.002345 = 4.34 x 10^-4
2) Write balanced equation = KMnO4 = 5/3 x ANS = 7.39 x 1
Step 3 : find moles FeC2O4 .2H2O in 250 cm3 = 7.39x10-4 mol x 10 = 7.39x10-3 mol
Step 4 : find mass of FeC2O4 .2H2O in 7.39x10-3 mol mass= moles x Mr = 7.39x10-3 x 179.8 = 1.33g
Step 5 : find % mass %mass = 1.33/1.412 x100 = 94.1%
H2O2 + MnO4
Ox H2O2 –> O2 + 2H+ + 2e-
Red MnO4 -(aq) + 8H+ (aq) + 5e- –> Mn2+ (aq) + 4H2O
Overall 2MnO4 -(aq) + 6H+ (aq) + 5H2O2 -> 5O2 + 2Mn2+ (aq) + 8H2O
ethandioate + mangante
Ox C2O4 2- –> 2CO2 + 2e-
Red MnO4 -(aq) + 8H+ (aq) + 5e- –> Mn2+ (aq) + 4H2O
Overall 2MnO4 -(aq) + 16H+ (aq) + 5C2O4 2-(aq) –> 10CO2 (g) + 2Mn2+(aq) + 8H2O(l)
Copper(II) sulfate,
CuSO4 is dissolved in water. The solution contains a complex ion that acts as an acid in water. Give the equilibrium equation for this acid-base reaction.
[Cu(H2O)6]2+ + H2O –> [Cu(OH)(H2O)5]+ + H3O+
complex ions reacts wth water to form
X(OH)(H2O)5 + H3O+
why is aluminium colourless not iron or copper
Aluminium is the only non-transition metal, as it does not have partially-filled d orbitals. There are no available electrons to excite into a higher-energy
d orbital. Therefore, the aluminium complex cannot absorb or reflect visible light, so the solution is colourless.
complex ion + NaOH
(2+ metal ion)
Complex + 2OH- –> X(OH)2(H2O)4 + 2H2O
complex ion + NaOH
(3+ metal aqua ion)
Complex + 3OH- –> X(OH)3(H2O)3 + 3H2O
Al and Fe3+ in NaOH colours
Fe3+ = brown ppt
Al = white ppt
in air Fe(OH)2(H2O)4 is converted (oxidised to)
Fe(OH)3(H2O)3
return back to complex aqua ion =
add acid e.g H3O+
ammonia + copper hydroxide
ligand subsitution reaction
Copper hydroxide + 4NH3 –> complex ion + 2OH- + 2H2O
test for Al3+
add excess NaOH
explain why the conical flask had to be heated during this titration
reaction between 2 negative ions is slow
explain why after a while the manganate ions did decolourise straight away
Mn2+ produced in the reaction acts as a catalyst to reduce rate
finding X of crystalled molecule
find regular moles
mass / moles to get mr
ANS - mr of molecule excluding water
ANS / mr of water
Lewis acid and Lewis base
Lewis acid: electron pair acceptor
Lewis base: electron pair donator
NH3 + OH
Here the NH3 and OHions are acting as
Bronsted-Lowry bases accepting a proton
[Al(H2O)6]3+(aq) + 3OH- (aq) —> Al(H2O)3(OH)3 (s) + 3H2O (l)
excess NH3
With excess NH3 a ligand substitution reaction occurs with Cu and its precipitate dissolves to form a deep blue
solution.
Cu(OH)2(H2O)4(s) + 4NH3 (aq) –> [Cu(NH3)4(H2O)2]2+(aq) + 2H2O (l) + 2OH-
iron and ethanedioate ratios to Mn
Mn and Fe = 1:5 mol ratio
Mn and C2O4 = 2:5 mol ratio
MnO4- equations (Fe and C2O4)
MnO4- + 8H+ + 5Fe2+ –> Mn2+ + 4H2O + 5Fe3+
or
2MnO4- + 16H+ + 5C2O4^2- –> 2Mn2+ + 8H2O + 10CO2
mass of 0.12mol Fe2+ in 50cm3 of 250cm3 solution
50/250 = 5
0.12 x 5 = 0.6
0.6 x mr = mass
manganate reduction half equation
MnO4- + 8H+ + 5e- –> Mn2+ + 4H2O
oxidation of Fe2+
Fe2+ –> Fe3+ + e-
oxidation of C2O4^2-
C2O4^2- –> 2CO2 + 2e-
ammonia + colour
Ammonia = more splitting of d orbitals = higher frequency of light absorbed
charge of ethanedioate
2-
why is there a change in coordination numbers between chloride and NH2
Change in coordination number as chloride ligands are larger than NH2
increase in entropy
Entropy increases if there are more molecules in the products than the reactants
If 6 coordination bonds in reactants + products = the same so enthalphy is zero but entropy is positive meaning gibbs is negative so the reaction is thermodynamically feasible
potassium manganate
MNO4- + 8H+ + 5e- –> Mn2+ + 4H2O = oxidising agent = needs to be used in acidic conditions (dilute sulfuric acid)
HCl = cannot be used with MnO4- as it would oxidised Cl- to Cl2
Concentrated sulfuric acid cannot be used as they are oxidisng agents themselves
Ethanoic acid cannot be used as it is a weak acid and would not provide enough protons (H+)
acidity of 3+ vs 2+ aqua metal ions
-Acidity of [M(H2O)6]3+ is greater than [M(H2O)6]2+ bc 3+ has greater charge density but smaller ionic radius.
-Electrons in OH bond are pulled towards the oxygen atom. Depending on the charge of the ion the strength at which the pull on electrons away from hydrogen atom is decided. Hydrogen has a greater partial positive charge
[Fe(H2O)6]3+ reversible reaction
[Fe(H2O)6]3+ + H2O ⇌ [Fe(H2O)5(OH)]2+ + H3O+
Or
[Fe(H2O)6]3+ (aq) ⇌ [Fe(H2O)5(OH)]2+ (aq) + H+ (aq)
complex ion bonding
Complex ion = dative covalent bond using empty orbital of iron ion
Fe3+ is more acidic because of charge and ionic radius = better electron acceptor as electrons get pulled to the oxygen atom more.
increase conc of H+
Increased concentration of H+ ions = lower pH
Less distortion on H bond = hydrogen is not as positive = wont be lost easily. Less hydroxyonium (H3O+) ions formed.
is Fe(H2O)3(OH)3
[Fe(H2O)3(OH)3] = insoluble in water as it is a neutral molecule so not attracted to another water molecule and instead form prectipiates
metal aqua ion + ammonia
[Metal(H2O)6]3+ + 3NH3 ⇌ [Metal (H2O)4(OH)2]+ + 3NH4+
[Metal(H2O)6]2+ + 2NH3 ⇌ [Metal (H2O)4(OH)2]+ + 2NH4+
metal aqua ion + OH-
[Metal (H2O)6]3+ + OH- –> [Fe(H2O)5(OH)2+ + H2O
Lower charged e.g 2+ then 1 less water molecule and lower charge
[Cu(H2O)6]2+
blue
[Fe(H2O)6]2+
green
[Fe(H2O)6]3+
pale violet
[Al(H2O)6]3+
colourless
yellow vs violet
When in aqueous solution metal ions exist as metal aqua ions.
Sometimes solutions containing [Fe(H2O)6]3+ appear yellow rather than violet due to small amounts of [Fe(H2O)5(OH)]2+ formed by hydrolysis.
Reactions of metal aqua ions
1) Hydrolysis (loss of H+ from H2O ligands as OH bond breaks)
2) Substitution (replacement of H2O by other ligands)
3) Redox (metal changes oxidation state)
g [Fe(H2O)6]2+ ⇌ [Fe(H2O)5(OH)+ + H+
electrons pairs pulled from oxygen to metal ion
-Electron pairs pulled away from oxygen to metal ion
-causes electron pairs in OH bond to be pulled even closed to the oxygen atom
-hydrogen atoms have a greater partial positive charge
-more attracted to water molecules in solution so bonds more easily lost/broken
3+ vs 2+ ionic radius
Does 3+ or 2+ have smaller ionic radius:
-3+ ion = same nuclear charge but 2+ charge is spread over larger area so there is less distortion of the OH bond
3+ = [Fe(H2O)5(OH)]2+ + H2O
2+ = [Fe(H2O)4(OH)2]+ + H2O
why is [Fe(H2O)3(OH)3 is insoluble
Why is [Fe(H2O)3(OH)3 insoluble = neutral molecule = not attracted to water
formation of precipitate
-[Fe(H2O)6]2+ (Aq) –> Fe(H2O)5(OH)]+ (Aq)–> Fe(H2O)4(OH)2 (s) (Removal of 2 hydrogen ions)
precipitate redissolves
Precipiate re-dissolves for aluminum (not transition metal)
–> Al(H2O)4(OH)2 –> [Al(H2O)3(OH)3- –> [Al(H2O2)(OH4)]2-
summary of copper colours
-NaOH = blue ppt
-NH3 = blue ppt
-excess NH3 = deep blue solution
-Na2O3 = blue-green ptt
summary of iron 2+ colours
-NaOH = green ppt but oxidises brown
-NH3 = same
-green ppt
summary of iron3+ colours
-NaOH = brown ppt
-NH3 = brown ppt
-Na2CO3 = brown ppt and CO2 effervescence
summary of Al3+ colours
NaOH = white ppt
-excess NaOH = colourless solution
-NH3 = white ppt
-Na2CO3 = white ppt = CO2 gas evolved
aluminuium reactions
-NaOH = [Al(H2O)6]3+ + OH- –> [Al(H2O)5(OH)]2+ + H2O
[Al(H2O)3(OH)3 + OH- –> [Al(OH)4]- + 3H2O
-Here Al acts as an acid as H+ to form H2O is donated
Al + HCl = basic
Al + OH = acidic
So it is amphoteric
carbonate ions
Carbonate (CO3^2-) ions are weak bases. They are able to remove protons from 3+ metal aqua ions but not from 2+ metal aqua ions. (Cu, Al, Fe)
e.g [Fe(H2O)6]2+ + CO3^2- –> FeCO3 + 6H2O
total oxidation state of metal =
–> remember the total oxidation state of the complex is the same as its charge
total oxidation state of complex - oxidation state of ligands
e.g [Co(Cl)4]2-
total = 2-
ligands = -4 x 1 = -4
-2–4 = +2
ionic equation for forming carbonates
X = metal
X^+ + CO3^2- –> X2CO3
equation to show the conversion of AgBr into its complex, using excess concentrated ammonia
AgBr + 2NH3 -> Ag(NH3)2^+ + Br-
Copper(I) iodide is a white solid.
Explain why copper(I) iodide is white.
copper (I) has a completly filled d orbital so cannot absorb visible light
heterogenous catalysts
catalyst is in a different phase from the reactants. Reaction occurs on the active sites of the surface
use of support medium maxmizes surface area and minimum cost
catalysts in haber process and contact process
contact = V2O5
haber = Fe
reducing impact of heterogenous catalysts
become poisoned by impurities that block the active site = cost implication
contact process
S(s)+O2(g)→SO2(g)
2SO2(g)+O2(g)⇌2SO3(g)
oxidation of SO2 to SO3 using V2O5 catalyst
vanadium oxide is the catalyst used in the manufacture of sulfur trioxide
give 2 equations
V2O5 + SO2 –> V2O4 + SO3
V2O4 + 1/2O2 –> V2O5
heating NH4VO3 produces vanadium oxide water and one other product
give an equation for the reaction
2NH4VO3 –> V2O5 + H2O + NH3
[V(H2O)4Cl2]+ exists as 2 isomers give the type of isomerism
cis-trans isomerism
NH3 + H2O –>
NH4+ + OH-
smaller Fe3+ =
able to polarise water molecules more = make more hydroxide ligands that are broken
bidentate =
form 2 coordinate bonds with the metal ion
charge of ethane-1,2,diamine
NH2-CH2-CH2-NH2
neutral but carries lone pair on nitrogen
one molecule of EDTA^4- can form
6 dative covalent bonds
Mn to Fe ratio from equation
1:5
back titration =
original moles - excess
enthalphy change
near zero
same bonds being broken as are formed
high entropy, low enthalphy = negative gibbs = more favourable
acidic =
reduction
3+
dissociate more and have greater attractive power to OH-
state the meaning of the term transition metal complex
central metal ion surrounded by ligands via dative covalent bonds
