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Aqa a level chemistry > Transition metals > Flashcards

Transition metals Flashcards

1
Q

4 chemical properties of transition metals

A

-variable oxidation state

-form complex ions

-catalysis

-colour

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2
Q

transition metals

A

Transition metals = one which forms one or more stable ions which have incompletely filled d orbitals

Zn and Sc = not transition metals

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3
Q

complex ions

A

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

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4
Q

ligands as donors

A

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.

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5
Q

ligand

A

Ligand = particle with a lone pair that forms a coordinate bon to a metal

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6
Q

complex

A

Complex = metal ion with ligands co-oordinately bonded to it

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7
Q

co-ordination number

A

Co-ordination number = number of co-ordinate bonds from ligands to metal ions

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8
Q

accepting ligands

A

Transition metals have empty valence orbitals which can accept pairs of electrons from ligands.

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9
Q

[CuCl4]2+

A

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.

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10
Q

copper metal vs copper ion

A

Copper metal = 1s2 2s2 2p6 3s2 3p6 4s1 3d10

Copper ion = 1s2 2s2 3s2 3p6 3d9

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11
Q

[Al(H2O)6]3+

A

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

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12
Q

linear

A

Linear = 2 coordination number = Ag complexes e.g [Ag(NH3)2]3+

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13
Q

tetrahedral

A

Tetrahedral = 4 coordination number = large ligands like [CuCl4]2+

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14
Q

square planar

A

Square planar = Pt+ complexes e.g [PtCl4]3+

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15
Q

octahedral

A

Octahedral = most common = [Cu(H2O)2]2+

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16
Q

exceptions for the shape of complex ions

A

-Ag forms linear complexes

-Pt forms square planar complexes

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17
Q

types of ligands

A

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).

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18
Q

unidentate ligands

A

-unidentate ligands = form 1 co-ordinate bonds e.g H2O, OH-, NH3, CN, Cl-

e.g [Cu(H2O)6]2+ [CuCl4)2-

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19
Q

bidentate ligands

A

-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-

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20
Q

NH3 charge

A

neutral

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21
Q

notes for specification

A

-octahedral complexes can display cis-trans complexes with monodendtate ligands and optical isomerism with bidentate ligands

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22
Q

cisplatin =

A

cis isomer

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23
Q

Ag+ forms linear complex used in tollens reagent

A

[Ag(NH3)2]+

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24
Q

incomplete subsitution

A

[Cu(NH3)4(H2O)2]2+

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25
Q

bidentate ligands

A

H2NCH2CH2NH2

C2O42-

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26
Q

why is carbon monoxide toxic

A

it is toxic bc it replaces oxygen co-ordinately bonded to Fe(II) in haemolgibin

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27
Q

explain the meanings of the terms multidentate and ligand with reference to the reaction of EDTA4- with [Cu(H2O)6]-

A

multidentate = EDTA can form 6 dative covalent bonds with central cation

ligand = lone pair can form dative bond with copper ions

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28
Q

hexaaquairon (II) ions react with an excess of H2NCH2CH2NH2 in a ligand subsitution reaction to form what

A

[Fe(NH2)6]2+

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29
Q

explain why water can act as a ligand

A

makes single dative covalent bond with metal ion
unidentate ion

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30
Q

name [Fe(H2O)6]2+

A

hexaaquairon (II)

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31
Q

naming monodentate ligands

A

Naming monodentate ligands:

-H2O = aqua

-NH3 = ammine

-OH- = hydroxo

-CN- = cyano

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32
Q

oxidiation number of Fe in [Fe(CN)6]4-

A

+2

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33
Q

true or false = 3 ligands can be present but there can be 6 dative covalent bonds

A

true

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34
Q

how many ligands can fit around chloride ion

A

4

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35
Q

ethanedioate =

A

C2O4

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36
Q

explain the meaning of the term complex ion

A

an ion that forms a dative covalent bond and has ligands bonded to the central cation

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37
Q

give 2 other characteristic properties of transition metals

A

-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

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38
Q

ligand =

A

atom which can donate a lone pair of electrons and form a dative covalent bond

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39
Q

lewis acid vs bonsted lowery acid

A

bronsted acid = proton donor
lewis acid = accepts electron pair and will have vacant orbitals
Lewis base = donates electron pair

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40
Q

explain why breathing in carbon monoxide can be fatal

A

CO replaces oxygen molecule and binds strongly to Fe2+

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41
Q

explain why titanium is a d-block element and why its a transition element

A

titanium has electrons in its d-orbital

last filled electrons in d-subshell + has variable oxidation state

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42
Q

how do colour changes arise

A

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

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43
Q

UV light / visible spectroscopy

A

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

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44
Q

colorimetry

A

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

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45
Q

vanadium catalyst

A

V2O5

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46
Q

[Ag(CN)2-

A

[NC- –> Ag –> CN-]- = +1 oxidiation number + linear (180) [Ag(CN)2-]

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47
Q

tetrahedral vs octahedral

A

[CuCl4]2- = tetrahedral

[Cu(H2O)6]2+ = octahedral

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48
Q

[Cr(NH2CH2CH2NH2)3]3+

A

Charge of ammonia ions = neutral

Oxidation number of pt in [Pt(NH3)Cl3]- = +2

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49
Q

haemoglobin

A

-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.

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50
Q

cis-trans isomerism

A

-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

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51
Q

optical isomerism

A

Optical isomersism:

-occurs in octahedral complex with bidentate ligands

[Cr(C2O4)3]3- = bidentate = 2 oxygen ligands x 3

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52
Q

subsitution of ligands

A

-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

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53
Q

chelate effect

A

-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

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54
Q

subsitution effect

A

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)

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55
Q

naming monodentate ligands

A

Naming monodentate ligands:

-H2O = aqua

-NH3 = ammine

-OH- = hydroxo

-CN- = cyano

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56
Q

colours of transition metals

A

-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

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57
Q

reflecting wavelengths

A

-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

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58
Q

When you dissolve an aqua ion in water, water molecules pull H+ ions off H2O ligands

A

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

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59
Q

effect of pH on oxidation state

A

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

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60
Q

tollens reagent

A

-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

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61
Q

vanadium oxidation states

A

-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

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62
Q

mixing ions

A

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+

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63
Q

explain what you would see during step 1

A

yellow –> green –> blue

when blue initially forms it reacts with the yellow unreacted to turn green (VO2+)

64
Q

is scandanium coloured

A

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.

65
Q

spectrophotometry

A

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.

66
Q

higher frequency

A

-higher frequency of light absorbed = bigger the energy level gap

-higher frequency light = closer to blue end of visible light spectrum

67
Q

conversions

A

THz –> Hz = x10^12

Nm –> m = / 10^-9

68
Q

vanadium half equations

A

VO2^+ + 2H+ (aq) + e- –> VO2+ (aq) + H2O (l)

VO2+ + 2H+ (aq) + e- –> V3+ (aq) + H2O (l)

V3+ + e- –> V2+ (aq)

69
Q

different energy levels

A

Colours have different energy levels due to different wavelengths

Red –> blue = direction of increasing energy

70
Q

what affects size of energy gap

A

-transition metal ion

-oxidation state

-ligands

71
Q

absorbed colours

A

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

72
Q

hydrated copper

A

Hydrated CuSO4 = perceived (reflected) as blue and orange is absorbed.

73
Q

why are transition metal complexes usually coloured (5 marks)

A

-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

74
Q

equations to recall

A

∆E = hv = hc / λ = energy difference

Λ = c (speed of light) / f

75
Q

planks constant

A

Planks constant = 6.63 x 10^-34 Js

76
Q

velocity of light

A

Velocity of light c = 3.00 x 10^8 ms^-1

77
Q

frequency of light

A

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

78
Q

equation for iron and mangante

A

Iron + Manganate:

MnO4 -(aq) + 8H+ (aq) + 5Fe2+ (aq) –> Mn2+ (aq) + 4H2O (l) + 5Fe3+ (aq)

(purple –> colourless)

79
Q

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.

A

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%

79
Q

iron and ethandioate

A

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+

80
Q

Mn and Fe

A

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

81
Q

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.

A

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%

82
Q

H2O2 + MnO4

A

Ox H2O2 –> O2 + 2H+ + 2e-

Red MnO4 -(aq) + 8H+ (aq) + 5e- –> Mn2+ (aq) + 4H2O

Overall 2MnO4 -(aq) + 6H+ (aq) + 5H2O2 -> 5O2 + 2Mn2+ (aq) + 8H2O

83
Q

ethandioate + mangante

A

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)

84
Q

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.

A

[Cu(H2O)6]2+ + H2O –> [Cu(OH)(H2O)5]+ + H3O+

85
Q

complex ions reacts wth water to form

A

X(OH)(H2O)5 + H3O+

86
Q

why is aluminium colourless not iron or copper

A

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.

87
Q

complex ion + NaOH

(2+ metal ion)

A

Complex + 2OH- –> X(OH)2(H2O)4 + 2H2O

88
Q

complex ion + NaOH

(3+ metal aqua ion)

A

Complex + 3OH- –> X(OH)3(H2O)3 + 3H2O

89
Q

Al and Fe3+ in NaOH colours

A

Fe3+ = brown ppt
Al = white ppt

90
Q

in air Fe(OH)2(H2O)4 is converted (oxidised to)

A

Fe(OH)3(H2O)3

91
Q

return back to complex aqua ion =

A

add acid e.g H3O+

92
Q

ammonia + copper hydroxide

A

ligand subsitution reaction

Copper hydroxide + 4NH3 –> complex ion + 2OH- + 2H2O

93
Q

test for Al3+

A

add excess NaOH

94
Q

explain why the conical flask had to be heated during this titration

A

reaction between 2 negative ions is slow

95
Q

explain why after a while the manganate ions did decolourise straight away

A

Mn2+ produced in the reaction acts as a catalyst to reduce rate

96
Q

finding X of crystalled molecule

A

find regular moles
mass / moles to get mr

ANS - mr of molecule excluding water

ANS / mr of water

97
Q

Lewis acid and Lewis base

A

Lewis acid: electron pair acceptor
Lewis base: electron pair donator

98
Q

NH3 + OH

A

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)

99
Q

excess NH3

A

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-

100
Q

iron and ethanedioate ratios to Mn

A

Mn and Fe = 1:5 mol ratio

Mn and C2O4 = 2:5 mol ratio

101
Q

MnO4- equations (Fe and C2O4)

A

MnO4- + 8H+ + 5Fe2+ –> Mn2+ + 4H2O + 5Fe3+

or

2MnO4- + 16H+ + 5C2O4^2- –> 2Mn2+ + 8H2O + 10CO2

102
Q

mass of 0.12mol Fe2+ in 50cm3 of 250cm3 solution

A

50/250 = 5

0.12 x 5 = 0.6

0.6 x mr = mass

103
Q

manganate reduction half equation

A

MnO4- + 8H+ + 5e- –> Mn2+ + 4H2O

104
Q

oxidation of Fe2+

A

Fe2+ –> Fe3+ + e-

105
Q

oxidation of C2O4^2-

A

C2O4^2- –> 2CO2 + 2e-

106
Q

ammonia + colour

A

Ammonia = more splitting of d orbitals = higher frequency of light absorbed

107
Q

charge of ethanedioate

A

2-

108
Q

why is there a change in coordination numbers between chloride and NH2

A

Change in coordination number as chloride ligands are larger than NH2

109
Q

increase in entropy

A

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

110
Q

potassium manganate

A

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+)

111
Q

acidity of 3+ vs 2+ aqua metal ions

A

-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

112
Q

[Fe(H2O)6]3+ reversible reaction

A

[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)

113
Q

complex ion bonding

A

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.

114
Q

increase conc of H+

A

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.

115
Q

is Fe(H2O)3(OH)3

A

[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

116
Q

metal aqua ion + ammonia

A

[Metal(H2O)6]3+ + 3NH3 ⇌ [Metal (H2O)4(OH)2]+ + 3NH4+

[Metal(H2O)6]2+ + 2NH3 ⇌ [Metal (H2O)4(OH)2]+ + 2NH4+

117
Q

metal aqua ion + OH-

A

[Metal (H2O)6]3+ + OH- –> [Fe(H2O)5(OH)2+ + H2O

Lower charged e.g 2+ then 1 less water molecule and lower charge

118
Q

[Cu(H2O)6]2+

A

blue

119
Q

[Fe(H2O)6]2+

A

green

120
Q

[Fe(H2O)6]3+

A

pale violet

121
Q

[Al(H2O)6]3+

A

colourless

122
Q

yellow vs violet

A

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.

123
Q

Reactions of metal aqua ions

A

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+

124
Q

electrons pairs pulled from oxygen to metal ion

A

-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

125
Q

3+ vs 2+ ionic radius

A

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

126
Q

why is [Fe(H2O)3(OH)3 is insoluble

A

Why is [Fe(H2O)3(OH)3 insoluble = neutral molecule = not attracted to water

127
Q

formation of precipitate

A

-[Fe(H2O)6]2+ (Aq) –> Fe(H2O)5(OH)]+ (Aq)–> Fe(H2O)4(OH)2 (s) (Removal of 2 hydrogen ions)

128
Q

precipitate redissolves

A

Precipiate re-dissolves for aluminum (not transition metal)

–> Al(H2O)4(OH)2 –> [Al(H2O)3(OH)3- –> [Al(H2O2)(OH4)]2-

129
Q

summary of copper colours

A

-NaOH = blue ppt

-NH3 = blue ppt

-excess NH3 = deep blue solution

-Na2O3 = blue-green ptt

130
Q

summary of iron 2+ colours

A

-NaOH = green ppt but oxidises brown

-NH3 = same

-green ppt

131
Q

summary of iron3+ colours

A

-NaOH = brown ppt

-NH3 = brown ppt

-Na2CO3 = brown ppt and CO2 effervescence

132
Q

summary of Al3+ colours

A

NaOH = white ppt

-excess NaOH = colourless solution

-NH3 = white ppt

-Na2CO3 = white ppt = CO2 gas evolved

133
Q

aluminuium reactions

A

-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

134
Q

carbonate ions

A

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

135
Q

total oxidation state of metal =

–> remember the total oxidation state of the complex is the same as its charge

A

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

136
Q

ionic equation for forming carbonates

X = metal

A

X^+ + CO3^2- –> X2CO3

137
Q

equation to show the conversion of AgBr into its complex, using excess concentrated ammonia

A

AgBr + 2NH3 -> Ag(NH3)2^+ + Br-

138
Q

Copper(I) iodide is a white solid.
Explain why copper(I) iodide is white.

A

copper (I) has a completly filled d orbital so cannot absorb visible light

139
Q

heterogenous catalysts

A

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

140
Q

catalysts in haber process and contact process

A

contact = V2O5

haber = Fe

141
Q

reducing impact of heterogenous catalysts

A

become poisoned by impurities that block the active site = cost implication

142
Q

contact process

A

S(s)+O2(g)→SO2(g)
2SO2(g)+O2(g)⇌2SO3(g)

oxidation of SO2 to SO3 using V2O5 catalyst

143
Q

vanadium oxide is the catalyst used in the manufacture of sulfur trioxide

give 2 equations

A

V2O5 + SO2 –> V2O4 + SO3

V2O4 + 1/2O2 –> V2O5

144
Q

heating NH4VO3 produces vanadium oxide water and one other product

give an equation for the reaction

A

2NH4VO3 –> V2O5 + H2O + NH3

145
Q

[V(H2O)4Cl2]+ exists as 2 isomers give the type of isomerism

A

cis-trans isomerism

146
Q

NH3 + H2O –>

A

NH4+ + OH-

147
Q

smaller Fe3+ =

A

able to polarise water molecules more = make more hydroxide ligands that are broken

148
Q

bidentate =

A

form 2 coordinate bonds with the metal ion

149
Q

charge of ethane-1,2,diamine

NH2-CH2-CH2-NH2

A

neutral but carries lone pair on nitrogen

150
Q

one molecule of EDTA^4- can form

A

6 dative covalent bonds

151
Q

Mn to Fe ratio from equation

A

1:5

152
Q

back titration =

A

original moles - excess

153
Q

enthalphy change

A

near zero

same bonds being broken as are formed

high entropy, low enthalphy = negative gibbs = more favourable

154
Q

acidic =

A

reduction

155
Q

3+

A

dissociate more and have greater attractive power to OH-

Aqa a level chemistry (23 decks)

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