Chemistry A2 > Transition Metals > Flashcards

Transition Metals Flashcards

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1
Q

Transition Metal

A
  • any d-block element that forms at least one ion with a partially filled d SUB-shell of electrons
  • found between Sc and Zn
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2
Q

Which subshell are electrons removed from first to form transition metal ions

A

4s then 3d

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3
Q

State chemical properties of transition metals

A
  • variable oxidation states
  • characteristic colour of complexes
  • catalysts
  • complex formation
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4
Q

Suggest why transition metals have a variable oxidation state

A

partially filled d shell is able to gain or lose electrons

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5
Q

Give reasons why transition metals act as good catalysts

A
  • able to donate and accept electrons
  • variable oxidation states
  • form complexes
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6
Q

Ligand

A
  • electron pair donor

- forms coordinate bond with central metal atom/ion

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7
Q

Complex

A

central metal atom or ion surrounded by ligands

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8
Q

Coordination Number

A

number of co-ordinate bonds to central metal atom or ion

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9
Q

Monodentate ligand

A

donate a single pair of electrons from a single atom to central metal atom or ion

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10
Q

Multidentate ligand

A

donate two or more electron pairs from two or more atoms to central metal atom or ion per ligand

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11
Q

Examples of ionic and neutral monodentate ligands

A

ionic - Cl-, F-, CN-, H-

neutral - :CO, :NH3, H2O:

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12
Q

Examples of ionic and neutral bidentate ligands

A

ionic - ethanedioate (C2O42-)
neutral - 1,2-diaminoethane (H2NCH2CH2NH2),
benzene-1,2-diol (C6H4(OH)2)

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13
Q

Example of hexadentate ligand

A

EDTA4-

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14
Q

How to determine the shape of a complex based on its ligands

A
  • small ligands (e.g. H2O/NH3) = octahedral
  • large ligands (e.g. Cl-) = tetrahedral
  • Ag+ forms linear e.g. [Ag(NH3)2]+ (Tollen’s reagent)
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15
Q

How to determine the shape of a complex based on coordination number of metal

A 
6 = octahedral 
4 = tetrahedral (or square planar for Ni2+ and Pt2+)
2 = linear
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16
Q

Ligand Substitution

A

ligands in a complex are exchanged for ligands which bind more strongly to central metal atom or ion

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17
Q

Types of ligand substitution reactions

A
  • similar size ligands like NH3 and H2O occurs without
    change of co-ordination number
  • different size ligands like H2O by Cl– can involve a change of co-ordination number
  • ligand substitution may be incomplete
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18
Q

Give an example of incomplete ligand substitution

A
  • replacing water in hexaaqua copper(II) ion
  • only four of six water molecules replaced by NH3
  • [Cu(NH3)4(H2O)2]2+
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19
Q

Explain how haem is a transition metal complex and how ligand substitution occurs in carbon monoxide poisoning

A
  • (iron(II) complex with multidentate ligand)
  • oxygen forms a co-ordinate bond to Fe2+ in haemoglobin
  • carbon monoxide has greater affinity for Fe2+/ bonds more strongly than oxygen so displaces it
  • toxic since oxygen no longer transported in blood
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20
Q

Chelation

A

multidentate ligands act as chelating agents by replacing several monodentate ligands from complexes

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21
Q

Sequestering Agent

A
  • (chelating agent)
  • remove (unwanted) metal ions form a solution
  • monodentate (water ligands replaced in aqua ions) by multidentate ligand so a precipitate forms
  • forms complex ion that does not have the chemical reactions of the ion that is removed
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22
Q

Chelate Effect

A
  • chelate complex more stable than monodentate
  • due to increase in entropy during chelation
  • few species replaced by several species
  • S = positive, G = negative so reaction favourable
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23
Q

Type of isomerism in monodentate ligand complexes

A
  • octahedral / square planar / tetrahedral

- cis-trans isomerism

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24
Q

Type of isomerism in bidentate/multidentate ligand complexes

A
  • octahedral

- optical isomerism

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25
Describe method for redox titration with Fe2+ or C2O42- with MnO4-
- rinse and then fill burette with MnO4- using funnel - rinse with then transfer standard volume of Fe2+ or C2O42- to conical flask using volumetric pipette - add excess sulfuric acid to conical flask - add MnO4- to solution in conical flask and swirl - dropwise near end point - end point indicated by permanent colour change to pink observed against white tile
26
Explain why Fe2+ and C2O42- are acidified during redox titration of MnO4-
provides H+ to reduce MnO4-
27
Explain why the colour of KMnO4 purple before the titration and after the end point it is pink
- burette contains MnO4- (Mn7+) = purple - when added to Fe2+/C2O42- redox reaction occurs so MnO4- is reduced to Mn2+ = colourless - once fully reacted at end point, additional MnO4- remains in oxidation state +7 but concentration is only small so appears pink
28
Explain how the redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by pH
- metal ions often exist in an equilibrium with H+ | - so their tendency to undergo oxidation or reduction depends on pH
29
Explain how the redox potential for a transition metal ion changing from a higher to a lower oxidation state is influenced by ligands it bonds with
- some ligands stabilise higher oxidation states more than lower ones and other ligands, visa versa - different ligands change the ability of the transition metal ion to alter its oxidation state (redox potential)
30
Explain how a yellow solution of vanadium (V) can be reduced to form a purple solution
- reducing agent zinc in acidic conditions - VO2^+ + 4H+ + 3e- -> V2^+ + 2H2O - Zn -> Zn2+ + 2e- - yellow (VO2^+) to blue (VO^2+) to green (V^3+) to purple (V2^+ )
31
Explain in terms of electrons why transition metal complexes are coloured
- certain wavelengths of visible light are absorbed - electrons in d orbitals are excited to higher energy levels - remaining wavelengths transmitted are within the visible range
32
Explain why the complex ion formed after chelation is more stable
- entropy increases | - several species formed from a few
33
Suggest factors which cause transition metal complexes to change colour
- change in oxidation state - change in coordination number - change in TYPE of ligand
34
Explain why ligand substitution reactions cause colour changes in transition metal complexes
- in different complexes d orbital splitting will be different / d orbital electrons have different energies - energy absorbed causes electrons in d orbitals to excite - different wavelengths/frequencies of absorbed and transmitted depending on complex
35
Equation for energy difference between ground and excited state of d electrons
E = hv = hc/wavelength
36
Colorimeter
measures intensity of light of a particular wavelength passing through a coloured solution
37
Describe how colorimetry is used to determine the transition metal ion concentration in an aqueous sample
- make up standard solutions of known metal ion concentrations by serial dilution (lighter and darker than unknown sample) - choose a suitable filter (light most absorbed by coloured solution) - calibrate colorimeter using distilled water in cuvette to zero - measure absorbable reading of different known concentrations - plot calibration curve of absorbance against concentrations - measure absorbance reading of unknown concentration - interpolate concentration from calibration curve
38
Describe how to determine the formula of a transition metal complex using colorimetry
- make upon standard solutions of different ratios of metal ions and ligand solutions - plot graph of absorbance against ratios of metal ion to ligand concentrations - mixture with maximum absorbance has same ratio of ligands to metal ions as in complex
39
Homogenous Catalyst
catalyst in same phase as reactant and intermediate species formed
40
Heterogenous Catalyst
catalyst in different phase to reactants and provides a surface for reaction
41
Suggest advantages of heterogenous catalyst over homogenous catalyst
- catalyst not lost | - catalyst does not need to be separated from products
42
Describe how to increase efficiency of heterogenous catalyst
- increase surface area | - impregnate onto inert support medium to increase surface area to mass ratio
43
Explain how heterogenous catalysts work
- transition metal ions have partially filled d sub-shell so from weak chemical bonds with reactants when adsorbed - weaken bonds so new bonds from or - bring reactants in closer proximity so increase concentration in particular area or - absorbed in correct orientation for reaction
44
Explain why d orbital splitting occurs in transition metal complexes
- ligand forms dative covalent bonds with metal ion - electrons in some of the orbitals will experience a greater repulsion than electrons in other orbitals - d sub-shell splits to higher and lower energy levels
45
Catalyst Poisoning (with examples)
- substances adsorbed too strongly to active sites of catalyst - so active sites blocked - lead poisoning to catalytic converters - sulfur poisoning of Fe in Haber process
46
Give equations to show how iron(II) acts as a homogenous catalyst in reaction of iodide ions with peroxodisulfate (S2O82-) and explain specific need for catalyst
S2O82- + 2Fe2+ -> 2SO42- + 2Fe3+ 2Fe3+ + 2I- -> 2Fe2+ + I2 opposite charges repel so reaction between S2O82- and I- has high activation energy
47
Give equations to show how vanadium pentoxide acts as a heterogenous catalyst in contact process and explain specific need for catalyst
SO2 + V2O5 -> SO3 + V2O4 2V2O4 + O2 -> 2V2O5 gases are far part so vanadium pentoxide provides surface for reaction by bringing reactants in close proximity and in correct orientation
48
Autocatalysis
product of reaction is catalyst for reaction
49
Give the half equations and overall equation for oxidation of ethanedioate ions by manganate (VII) ions to form carbon dioxide
MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O C2O42- -> 2CO2 + 2e- 5C2O42- + 2MnO4- + 16H+ -> 2Mn2+ + 8H2O + 10CO2
50
Give equations to show how Mn2+ behaves as an autocatalyst in oxidation of ethanedioate ions by manganate (VII) ions to form carbon dioxide
MnO4- + 4Mn2+ + 8H+ -> 5Mn3+ + 4H2O 2Mn3+ + C2O42- -> 2Mn2+ + 2CO2 (no electrons so balance charges using Mn3+ / Mn2+) - construct ionic equation from MnO4- + Mn2+ -> Mn3+ - MnO4- + Mn2+ + 8H+ -> 2Mn3+ + 4H2O (charge LHS = 9+, RHS=6+) - 3Mn2+ -> 3Mn2+ (charge LHS = 6+, RHS=9+) - combine equations so charges balance
51
Examples of square planner complex
Pt(NH3)2Cl2 or PtCl42- or NiCl42-
52
When is H positive for ligand substitution
more coordinate bonds are broken than made
53
When is H negative for ligand substitution
more coordinate bonds are made than broken
54
Suggest why NH3 is a better ligand compared to H2O
- N less electronegative than O | - more willing to donate electron pair to central metal ion
55
Suggest methods by which concentrations of transition metal ion solutions can be determined
- titration | - colorimetry
56
Suggest how filters can be tested to see which is appropriate complementary colour to use in colorimeter when determining metal ion concentration
- place solution in cuvette and measure absorbance reading using each filter - filter which gives highest absorbance is complementary colour
57
Explain why mixture with same ratio of ligands to metal ions as transition metal complex has maximum absorbance
maximum amount of complex precipitate formed
58
Explain why s block elements are not good catalysts
exist only in one oxidation state/ do not have a variable oxidation state
59
Explain why Fe3+ are as effective as Fe2+ at catalysing reaction between I- and S2O82-
reactions can occur in any order
60
Explain graph for autocatalysis reaction
- rate of reaction increases as catalyst formed | - rate of reaction decreases as reactants are used up
61
Suggest a property of an impurity that could cause calculated mass of FeSO4.7H2O to be greater than actual mass in redox titration of FeSO4.7H2O solution against K2Cr2O7
- impurity reduces Cr2O72- | - impurity reacts with more Cr2O72- for equal masses of impurity and FeSO4.7H2O
62
Outline a plan of an experiment to determine the percentage of iron present as iron(III) in a solution containing both Fe2+ (aq) and Fe3+ (aq) ions - provided with Zn (reduces Fe3+) and K2Cr2O7/H+
6Fe2+ + Cr2O72- + 14H+ -> Fe3+ + 2Cr3+ + 7H2O Zn + 2Fe3+ -> Zn2+ + 2Fe2+ - titrate measured volume of solution against K2Cr2O7 - reduce same volume with excess zinc - filter off excess zinc - titrate remaining solution using K2Cr2O7 - % Fe3+ = (titre 2 - titre 1)/titre 2
63
Suggest how selecting the wrong filter will reduce accuracy of absorbance reading of colorimeter
absorbance reading will be lower than actual absorbance since not all light absorbed by the sample regardless of concentration
Chemistry A2
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