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chemistry ~ inorganic chmistry (3.2) > Transition Metals > Flashcards

Transition Metals Flashcards

1
Q

characteristics of TM metals (Ti-Cu) 5

A
  • variable oxidation states
  • formation o different coloured compounds
  • catalytic activity
  • formation of complex ions
  • standard metallic properties
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2
Q

what causes the characteristics of TMs?

A

incomplete d sub-level. the interactions with electrons and ligands are unique to TM ions because of the partially filled nature of the d subshell

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3
Q

what are TMs?

A

metals found in the 3d block on the periodic table that contain an incomplete d sub-level in their atomic OR ionic form. this results in their characteristics

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4
Q

ligand

A

molecule or ion that forms a co-ordinate bond with a TM. ligands donate a pair of e- and results in the formation of a TM ion complex

they act as lone pair donors.

they have 1+ lone pair/s of e- to form the coord bond

co ordinate bond: covalent bond where both electrons in it come from the same atom.

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5
Q

(TM ion) complex

A

central metal atom or ion surrounded by ligands

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6
Q

what is a lewis acid

A

lone pair acceptor - the TM acts as a L acid

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7
Q

what is a lewis base

A

lone pair donor - the ligand acts as a L base

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8
Q

co ordination number

A

the number of co-ordinate bonds from ligands to a TM atom/ion

it is the number of coord bonds present NOT the number of ligands present. you can have co-ord number bigger than when you have bi/multi dentate ligands

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9
Q

uni or monodentate ligand

examples

A

form one coord bond. therefore they only have one lone pair to donate

Cl- ion, NH3, H2O, CN- ion

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10
Q

what will happen if there is a substitution reacton with H2O and NH3

or with Cl- and either of those

A

the water and ammonia ligands are similar sized and uncharged. because of this the coord number of a TM ion complex is unchanged when they are substituted for one another

the Cl- ligand is larger than the uncharged ligands above. so exchanges of these could result in a change in coord number. this is seen in Co2+, Cu2+, and Fe3+

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11
Q

bidentate ligands

examples

A

form two coord bonds. they have 2 lone pairs available to donate, one pair from two different donor atoms

1,2-diaminoethane (NH2CH2CH2NH2 - Ns have lone pair, no charge)

ethanedioate ion C2O4 (ethanedioic acid without the Hs so those Os have lone pairs and negative charge)

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12
Q

multidentate ligands

examples

A

form more than 2 coord bonds

EDTA forms 6 bonds.
- NCH2CH2N main chain
- then the Ns each have two (CH2COO-) groups bonded to them
- 4 negatively charged Os and the 2 Ns have lone pairs on them available to donate to the TM

porphyrin rings form 4 bonds

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13
Q

what biological protein incorporates porphyrin rings

A

haemoglobin

porphyrin forms 4 coord bonds with the central Fe(ii) ion, which can have 6 bonds. the fifth is the globin (protein bit) and the sixth is a coordinate bond with H2O when O2 is not bonded to it.

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14
Q

why is CO and CN- respiratory inhibitors?

A

they are toxic because they bind more strongly to Fe than oxygen so reduces oxygen carrying capacity. for the oxygen that does bind (cooperative binding) they are more strongly held (left shifted - increased loading) so they don’t get released into tissues

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15
Q

what does the size of the ligand affect

A

how many can fit around the central TM, so the coord number and consequently the shape of the complex

the same TM ion can show different complex ion geometries

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16
Q

size of ligand affects coordination number. this is whatgives the metal ion complex…

A

its distinct shape

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17
Q

shape, bond angle and usual occurrences for a coordination number of 2

A

linear

180

Ag+ complexes - eg [Ag(NH3)2]+ which is the active particle in Tollen’s Reagent.

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18
Q

Tollen’s reagent - an example of uses of variable oxidation states.

the colour changes of organic test tube reactions can be explained by the variable ox states.

what test is Tollen’s for and explain?

A

Tollens reagent, the test for aldehydes. a colour change, allowed by the variable ox state of Ag, allows for identification of organic compound

Ag+ is reduced to AG(0), its elemental form. it exists as a solid therefore a silver mirror forms on the test tube as solid is deposited

aldehyde is oxidised to carboxylic acid

2Ag+ + 2e- –> 2Ag
RCHO + H2O –> RCOOH + 2H+ + 2e-

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19
Q

shapes, bond angles and usual occurrences for a coordination number of 4

A

1 - TETRAHEDRAL
109.5
large ligans eg Cl-

2 - SQUARE PLANAR
90
Pt2+ complexes (eg cisplatin)

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20
Q

shape, bond angle and usual occurrences for a coordination number of 6

A

octahedral

90

commonest - if not silver, large ligand, platinum, then its this

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21
Q

what shapes can exhibit geometric isomerism

A

square planar or octahedral

when there are only two ligands of 1 type, that is different to other ligands

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22
Q

geometric isomerism

A

same m/s formula, different 3d spatial arrangement of atoms

normally due to restructed rotation of a C=C. obv not the case now

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23
Q

cis/z isomers of complex ions showing g isomerism will have

A

the two ligands of one type next to eachother

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24
Q

trans/e isomers of complex ions showing g isomerism will have

A

the ligands opposite eachother

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25
Q

what shape can exhibit optical isomerism

A

octahedral
when there are 3 bidentate ligands.

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26
Q

optical isomerism

A

when molecules are mirror images of the other and therefore non-superimposable due to a chiral centre. the TM ion acts as the chiral centre, when would usually be a C

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27
Q

to determine the structure of a TM ion and determine isomerism, what to do?

A
  • coord number / the central TM (Ag, Pt)
  • will inform shape. so structure determined
  • is octahedral/square planar?
  • if yes could exhibit isomerism, now look at the ligands
  • if there are two of 1 and the rest are different, then geometric
  • if octahedral and there are 3 bidentate ligands bonded, then optical
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28
Q

a substitution reaction is

A

when ligand/s in a TM ion complex is replaced by another

it can be incomplete.
it can cause a change in co-ordination number if the ligands are of different sizes/charges

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29
Q

how do you do a ligand sub reaction

A

introduce excess of the ligand you want to substitute in

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30
Q

the chelate effect

A

bidentate and multidentate ligands replace monodentate ligands from complexes. they are better at forming TM ion complexes, to say that they form more stable complexes

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31
Q

entropy

A

measure of disorder. matter tends toward higher entropy because there is then a higher energy dispersal. spontaneous reactions increase entropy

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32
Q

enthalpy

A

heat content of a system, measure of the total energy of a system in kj per mole

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33
Q

how can entropy and enthalpy explain the chelate effect

A

enthalpy change could be negligible because number c. bonds is unchanged.
however entropy increases because there are more particles in the products than reactants. so more thermodynamically favourable bc creates more disorder

34
Q

the ppq answer for explaining that entropy energy thing

A

b

35
Q

ex of a chelating agent

A

EDTA 4-

good for binding harmful metals into nonharmful complexes

36
Q

what factors cause a colour change in TM ion complexes

A

changes in
1. ox state (variable ox state therefore can change)

  1. coord number (could be changed by ligand, would change shape)
  2. ligand (known as a subsitution reaction)
  3. tm itself
37
Q

what is colour

A

TM ions absorb some wavelengths of visible light and transmits others (in solution) / reflects others (in solids)

colour depends on the absorption and transmission. colour is colour transmitted

colour is determined by the gap in energy between low and high energy d orbitals which are split (in terms of energy value) by ligand bonding. freq of light absorbed is this energy gap

38
Q

how does colour arise in TM ion complexes and why can it change

A

t metals have partially filled d orbitals

orbitals have specific energy values

when ligands bond to the TM the orbitals become different energy values. the bonding splits them into different energy values

when visible light strikes the complex, the e- absorb wavelengths and are excited to higher energy d orbitals

the light transmitted is what we see

39
Q

how does changing the coord number or the ligand change the TM ion colour

A

alter the energy split between d orbitals. the difference in energy is the energy needed/absorbed by the e- to become in an excited state. the energy required is supplied by the UV/visible light wavelengths.

the energy gap determines colour

40
Q

ΔE = hv

ΔE = (hc) / wavelength of light absorbed (m)

A

e = energy difference
h = plancks constant 6.63x10^-34
v = freq of light absorbed Hz
c = speed of light 3x10^8 ms^-1

41
Q

spectroscopy

??

A

measure of how much light is absorbed by a chemical substance and at what intensity light passes through it

42
Q

how is the absorption of vis light used in spectroscopy

??

A

by measuring the absorbance/transmittance frequencies of a substance, you can identify it

the more conced your solution, the more it absorbs. this is the beer-lambert law. thus you can determine [] of a substance

43
Q

what does a colourimeter determine

A

conc of coloured ions in solution

44
Q

how does a colourimeter determine the conc of coloured ions in solution

A

by measuring absorbance. absorbance is related to conc because if more conced then will absorb more light (colour is fixed wavelengths observable only in the visible light spectrum.) so the colourimeter absorbance reading will be higher

45
Q

how can you use a colourimeter to determine conc of a coloured ion solution

A

create a series of known concs and measure their absorbance. this allows you to create a calibration curve ([] on x, absorbance on y)

the line created can be read off to find what the [] is of an unknown substance once you have its absorbance reading

46
Q

why will TM ions of the same element have different colours?

A

variable ox states

colour is determined by this (one of the factors)

47
Q

colour of Fe2+

of Fe3+

A

pale green

brown

48
Q

colour of Mn7+ (MnO4-)

of Mn2+

A

purple

colourless

49
Q

colour of Cu1+ (Cu2O)

of Cu2+

A

brick red ppt is copper 1 oxide

blue solution

50
Q

what are Vanadium’s different oxidation states

A

+5
VO2 with a +1 charge

+4
VO with a 2+ charge

[[above are vanadate ions]]

+3
V3+ - its most stable form

+2
V2+

51
Q

colour of the different vanadium oxidation states

A

you better get vanadium

VO2 + yellow

VO2+ blue

V3+ green

V2+ violet

52
Q

how can vanadate (V) ions be reduced to form vanadium species at ox states IV, III, AND II

A

reduction of vanadate (V) ions with zinc in acidic solution, with HCl

test tube with cotton wool in it.
Zn + HCl –> H2 + ZnCl

the H2 is the reducing agent, it gets oxidised to water. Is h2 or zn the reducing agent?

53
Q

the half equation for the reduction of vanadate V

colour change

A

VO2 + + 2H+ + e- —> VO2+ + H2O

yellow -> green because mixing primary colours as VO2+ gets formed -> blue

54
Q

half eq for reduction of vanadate IV

colour change

A

VO2+ + 2H+ + 2e- —> V3+ + H2O

blue -> green

55
Q

half eq for reduction of V3+

A

V3+ + e- —> V2+

green -> violet

56
Q

how would you oxidise vanadium species up to vanadate V

A

oxidising, so going from lower to higher ox state

  1. take out cotton wool
  2. O2 in air acts as oxidising agent
  3. alkaline conditions
57
Q

what is the reactive species in Fehling’s solution

what does it test for

what happens

A

a colour change, allowed by the variable ox state of Cu, allows for identification of aldehydes

fehlings is blue solution containing Cu2+. when reduced to Cu(I) it forms brick red ppt CuO with O2 in air i presume

red ppt formed = aldehyde

58
Q

what reagent is a way to test for alcohols
reduction half eq

(and also aldehydes or ketones cause ketones wont undergo further oxidation)

A

K2Cr2O7
potassium dichromate

primary and secondary alcohols are oxidised to aldehydes and ketones respectively

Cr2O7 2- ions are reduced to Cr3+, the way chromium exists

ORANGE TO GREEN

Cr2O7 2- + 6e- + 14H+ –> 2Cr3+ + 7H2O

59
Q

what is a redox titration

A
  • Redox titration is used in chemistry to determine and quantify substances based on their oxidation-reduction reactions.
  • relies on the transfer of electrons between reactants, where one compound loses electrons and one gains electrons.
  • concentration of the unknown can be calculated by measuring amount of reagent needed for reaction to go to completion
60
Q

how could the iodine clock reaction be an ex of a redox titration

A

2 S2O32- + I2 → S4O62- + 2 I–

Sodium thiosulfate (Na2S2O3) is titrated with iodine (I2). the iodine oxidises thiosulfate ions to tetrathionate ions. the disappearance of the iodine solution’s blue-black colour signifies the reaction’s endpoint; self indicating

61
Q

what is the redox titration with Fe2+ and MnO4-

A

5Fe2+ + MnO4– + 8 H+ → 5Fe3+ + Mn2+ + 4 H2O

red: MnO4- + 8H+ —> Mn2+ + 4H2O + 5e-
ox: 5Fe2+ —> 5Fe3+ + 5e-

  • need to use dilute sulfuric acid
  • manganate ion is purple, so if in burette is hard to see bottom of meniscus. when in burette colour change is colourless to purple. bc as soon as purple then thats when the reaction has just gone to completion and now MnO4- is in excess
62
Q

whats the redox titration with ethanedioate ion and MnO4-

A

red: 5e- + MnO4– + 8 H+ → Mn2+ + 4 H2O
ox: C2O4^2- → 2CO2 + 2e-

overall:
5C2O4^2- + 2MnO4– + 16H+ → 10CO2 + 2Mn2+ + 8H2O

  • the reaction is slow because it is between two negative ions, so the conical flask must be heated to 60C to increase
63
Q

catalyst

A

substance that speeds up the ROR without undergoing a permanent chemical change. they are involved in the reaction but get regenerted

it provides an alternative reaction route with a lower Ea

they increase the no successful collisions bc more particles have Ea bc it is reduced.

64
Q

why do TM ions make good catalysts

A

variable ox state, so can act as oxidising and reducing agents (get reduced and oxidised), which is the basis for their catalytic activity

they can easily change their ox states whereas other metals have 1 fixed ox state they can’t easily revert back and forth from

65
Q

heterogenous catalyst

A

catalyst in different phase to reactants

solid catalyst catalysing gaseous/in solution reaction

66
Q

how does het catalysis catalyse a reaction

A

has active sites, locations on surface of catalyst where adsorption can occur

  • adsorption is a chemical process, due to chemical bonding.
  • when the reactants adsorb and desorb(??) onto the surface, the local conc reactants on the surface of the catalyst is increased so freq successful collisions is increased.
  • it also puts strain on the bonds and weakens them so break more easily
  • puts molecules in optimal orientation
67
Q

why must adsorption strength be strong but not too strong

A

STRONG ENOUGH TO
adsorb reactants and hold them in optimal orientation so can weaken bonds, lowering Ea, and increase local conc so increased success of collisions

WEAK ENOUGH TO
desorb products, allow reactants to desorb and adsorb so move along surface
so active sites not occupied and unavailable/blocked

68
Q

what is a support medium and why is it used

A

a foundation that catalyst is spread around eg honeycomb for catalyst in catalytic converters

increased surface area will improve effectiveness/activity of catalyst because more active sites exposed

69
Q

2 examples of het catalysis

A

V2O5 in the contact process

Fe in the haber process

70
Q

the contact process

A

catalyst V2O5 (vanadium oxide) to convert sulfur dioxide to sulfur trioxide

Overall equation : 2SO2 + O2 → 2SO3

step 1 2SO2 + 2V2O5 → SO3 + V2O4

step 2 2V2O4 + O2 → 2V2O5

71
Q

the haber process

how and why can it undergo catalytic posioning

A

Fe is used as a catalyst in the Haber process
N2 + 3H2 <=> 2NH3

sulfur impurities in natural gas, where the hydrogen comes from, binds and occupies active sites. sulfur is in natural gas naturally or put in deliberately so gas leaks smell

72
Q

catalytic poisoning

cost implication

A

when impurities/contaminants bind to active sites of catalyst surface more strongly than reactants so they occupy and block active sites

catalysts have reduced efficiency because less sites available, so ror deceases and yield decreases. also difficult to remove from catalyst, and removal damages catalyst, so must replace

73
Q

how can catalytic converters be poisoned

A

the plat/pallad/rhodium catalyst can be poisond by lead. when occurs must replace converter because otherwise more pollutants escape into atmosphere

recall that
2CO + 2NO → 2CO2 + N2
and
unburnt HCs and N2 → CO2 + N2 + H2O

74
Q

homogenous catalyst

A

in the same phase as reactants, eg all aq

75
Q

how does hom catalysis catalyse a reaction

A

the reaction proceeds through an intermediate species. TM ion will have different ox state in intermediate than orginally, then will revert back. this is enabled by TM ions having variable ox states due to partially filled d orbitals, which can easily gain or lose e-

76
Q

2 examples of hom catalysis

A
  1. iodide and persulfate ions to form sulfate ions and iodine
  2. autocatalysis reaction between ethanedioate and manganate ions
77
Q

autocatalysis

A

when the products of the reaction then catalyses the reaction

78
Q

iodide and persulfate ions to form sulfate ions and iodine

A

OVERALL
S2O8^2- + 2I- → 2SO4^2- + I2
two opply charged ions so very high Ea

CATALYSIS ROUTE
stage 1
S2O8^2- + 2Fe2+ —> 2SO4^2- + 2Fe3+

stage2
2I- + 2Fe3+ —> 2Fe2+ + I2

steps can occur in either order so Fe3+ can also act as a catalyst

79
Q

autocatalysis reaction between ethanedioate and manganate ions

A

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

  • slow because repulsion and big number of particles need to collide
  • speeds up when products start to get formed because the Mn2+ acts as catalyst for reaction.

Step 1
4Mn2+ + MnO4- + 8 H+—> 5Mn3+ + 4 H2O

Step 2
2Mn3+ + C2O4^2- —> 2Mn2+ + 2 CO2

Mn2+ reacts with manganate ion reactant to form Mn3+ intermediate. then Mn3+ intermediate reacts with ethanedioate ion reactant to form Mn2+ again and CO2

80
Q

why is the conc time graph weird for autocatalysed reactions

A

slow to begin with because uncatalysed

as products have ben formed, ror increases to rate of decreasing [reactant] increases, rapidly decreasing

rate slows again because less reactants; getting used up

plateaus because reactant/s depleted

81
Q

how can you figure out the conc of MnO4- and therefore the ror using a colourimeter

??

A

is purple so more purple, more absorbance, more conced

  1. take samples at intervals during exp
  2. absorbance value is related to [MnO4-]
  3. see how absorbance value changes over time
  4. can have absorbance on graph over time
82
Q

how can you figure out the conc of MnO4- and therefore the ror using titration

??

A
  1. take samples at intervals during exp
  2. titrate them. when reacted with reducing agent, end point being colourless, the moles of that reducing agent can be found (n=cv ; c known and v= titre)
  3. determine molar ratio reacting in, redox eq maybes
  4. so find moles MnO4- therefore in sample
  5. sample has certain vol so c=n/v

chemistry ~ inorganic chmistry (3.2) (6 decks)

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