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Chemistry > Transition Metals > Flashcards

Transition Metals Flashcards

1
Q

Transition Element definition

A

one which forms at least one stable ion with a partially filled d sub-shell

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2
Q

Which d block elements are not transition metals?

A

Scandium has no electrons in a d sub-shell and so therefore it is not a transition element.

Zinc has a completely full d sub-shell and so therefore it is not a transition element

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3
Q

Transition Metal Ions Configuration

A

First In – First Out

V = 1s2 2s2 2p6 3s2 3p6 4s2 3d3
V+ = 1s2 2s2 2p6 3s2 3p6 4s1 3d3
V2+ = 1s2 2s2 2p6 3s2 3p6 3d3

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4
Q

What features do transition metals display?

A

(1) Transition Elements form Complex Ions
(2) Coloured Ions
(3) Catalytic Properties
(4) Variable Oxidation States

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5
Q

Coordinate Bond

A

a shared pair of electrons which have both come from the same atom.

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6
Q

Ligand

A

an ion or molecule with a lone pair of electrons which can form a coordinate bond with a transition metal ion.

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7
Q

Coordination Number

A

the number of coordinate bonds a transition metal ion forms.

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8
Q

Complex Ion

A

central atom or ion surrounded by ligands.

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9
Q

Different ligand definitions

A

Monodentate Ligands –Each ligand forms 1 coordinate bond

Bidentate Ligands – Each ligand forms 2 coordinate bonds

Multidentate Ligands – Each ligand forms 2 or more coordinate bonds

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10
Q

monodentate ligands examples

A

water = :OH2
ammonia = :NH3
chloride Cl:-
cyanide = :-CN

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11
Q

bidentate ligands examples

A
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12
Q

shapes which transition metal complexes form

A
  • Octahedral
  • Tetrahedral
  • Square Planar
  • Linear
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13
Q

Octahedral shape

A

Octahedral
[Cu(H2O)6]2+

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14
Q

Tetrahedral shape

A

Tetrahedral
[CoCl4]2-
chloride ions are relatively large, only 4 can fit around transition metal ions, meaning the cannot be octahedral

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15
Q

Square planar

A

Square Planar
[PtCl2(NH3)2]
Cisplatin

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16
Q

Linear

A

Linear
[Ag(NH3)2]+
active part of Tollen’s Reagent which can be used to distinguish between Aldehydes and Ketones

COH + [O] → COOH
Ag+ + e- → Ag(s) - silver mirror

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17
Q

Bidentate comple ion shape

A
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18
Q

EDTA4-

A

EDTA4- can form six coordinate bonds; two coordinate bonds form from the N atoms, and four coordinate bonds form the O- atoms.

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19
Q

Ligand substitution reactions
forming:
[Cu(H2O)2(NH3)4]
[Cu(H2O)5CN]
[CuCl4]

A

[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(H2O)2(NH3)4]2+ + 4H2O

[Cu(H2O)6]2+ + CN- ⇌ [Cu(H2O)5CN]+ + H2O

[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O
Blue solution Yellow solution
Coordination No = 6 Coordination No = 4
Octahedral Tetrahedral

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20
Q

Chelate Effect Definition

A

Whenever ligand substitution reactions occur which involve many ligands being replaced by fewer multidentate ligands, there is always an increase in entropy

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21
Q

The Chelate Effect
[Cu(H2O)6] replaced by ethane-1,2-diamine
[Co(NH3)6] replaced by EDTA4-

A

[Cu(H2O)6]2+ + 3H2NCH2CH2NH2 ⇌ [Cu(H2NCH2CH2NH2)3]2+ + 6H2O

4 molecules on the LEFT turn into 7 molecules on the RIGHT
large increase in entropy
thermodynamically very favourable.

[Co(NH3)6]2+ + EDTA4- ⇌ [CoEDTA]2- + 6NH3

2 molecules on the LEFT of the equation turn into 7
even larger increase in entropy

22
Q

Explain why is this reaction feasible?

[Cu(NH3)6]2+ + 3H2NCH2CH2NH2 ⇌ [Cu(H2NCH2CH2NH2)3]2+ + 6NH3

A

The enthalpy change is zero
Because the same number of Cu-N bonds are broken and formed

There are 7 molecules produced from 4 molecules
So there is a large increase in entropy

The ΔG value will be negative

23
Q

Haemoglobin

A
  • 4 coordinate bonds between Fe2+ and the N atoms in the haem structure, enabling oxygen to be transported in the blood.
  • 1 coordinate bond between the Fe2+ and the protein globin.
  • 1 coordinate bond between the Fe2+ and the O2 molecule
24
Q

danger of carbon monoxide poisoning

A

CO is toxic because CO bonds more strongly to the Fe2+ in haemoglobin.
This prevents O2 from bonding to the Fe2+, causing suffocation

25
Q

Cis-Trans Isomerism

A

Cis= 90
trans = 180

26
Q

Cis platin

A

Cis platin binds to the DNA in cancer cells and stops cell replication.
The two Cl- ions on Cisplatin are substituted for two N atoms on adjacent Guanine bases.

27
Q

Optical Isomerism

A

This only occurs when complexes form with at least two bidentate ligands.

28
Q

Colours

A

ROY light = low energy
BIV light = high energy

29
Q

Colours key points

A
  • Transition Metal compounds are coloured because they have partially filled d-sub shells.
  • In a pure transition metal atom all d-orbitals are of equal energy. This is known as the ground state.
  • In transition metal compounds, the presence of other atoms causes the d-orbitals to have slightly different energies.
  • This enables electrons to be excited from one d orbital to another.
  • When an electron moves from a lower energy d orbital to a higher energy d orbital (called the excited state) the energy needed to make the transition is taken from white light.
  • That colour of the light which is absorbed is then missing from the light which reflects from the substance. The colours which are reflected give the compound its colour.

The difference in energy between the d sub-shells can be called ∆E

30
Q

State the electron configuration of a Ti(III) ion and that of a Ti(IV) ion. Explain, in terms of electron configurations and electron transitions, why Ti(III) compounds are usually coloured but Ti(IV) compounds are colourless

A

Ti(IV) = 1s2 2s2 2p6 3s2 3p6

Ti(III) = 1s2 2s2 2p6 3s2 3p6 3d1

Ti(III) has a d electron that can be excited to a higher level
Absorbs one colour of light from white light

Ti(IV) has no d electron so no electron transition with
energy equal to that of visible light

31
Q

Explain why this electron transition causes a solution containing the transition metal ion to be coloured.

A

One colour of light is absorbed
The remaining colour are transmitted (through the solution)

32
Q

∆E equations

A

∆E = hv
∆E = Change in Energy
h = Planck Constant 6.626x10-34
v = Frequency of Light (Hz)

∆E = h x c/λ
∆E = Change in Energy
h = Planck Constant 6.626x10-34
c = speed of light
λ = wavelenght of Light

33
Q

BIG E BIV

A

If a transition metal compound has LARGE ∆E between d sub-shells:
- High energy light (i.e. Violet, Indigo and Blue)
- will be absorbed to excite the electrons
- Red, Orange, Yellow light will be reflected
- This means the compound will look Red/Orange

34
Q

Small E ROY

A

If a transition metal compound has SMALL ∆E between d sub-shells:

  • Low energy light (i.e. Red, Orange, Yellow)
  • will be absorbed to excite the electrons
  • Blue, Indigo, Violet) will be reflected by the compound
  • This means the compound will look Blue/Purple
35
Q

What can changes in ∆E cause

A

LOCo
- Change in ligands
- Change in oxidation state of the metal
- Change in the coordination number of the complex

36
Q

Colorimetry

A

Suggest why it is important that the container for each sample has the same dimensions.
- Absorption depends on path length

Suggest why the coloured filter is used.
- To select the colour that is most strongly absorbed
( blue solution = red filter

Suggest one reason why a colorimetric method might be chosen in preference to titration.
- Quicker to analyse extracted samples than by titration / uses smaller volumes of solution

1.0 = All the light passed through the sample, and none was absorbed.

37
Q

colorimetry key points

A
  • measure absorbance of known conc.
  • plot graph of absorbance v conc.
  • read value of conc. for measured absorbance for this graph
38
Q

What is a catalyst?

A
  • A substance which increases the rate of a chemical reaction
  • Without being used up
39
Q

How do catalysts work?

A
  • Provide an alternative reaction pathway
  • With a lower activation energy
40
Q

Heterogeneous

A

where the catalyst is in a different phase to the reactants

41
Q

Homogeneous

A

where the catalyst is in the same phase as the reactants

42
Q

Heterogeneous Catalysis

A

The catalyst is usually a solid, and the reactants are either liquids or gases

(1) Reactants adsorb onto the surface of the catalyst on an active site.
(2) Reaction occurs on the surface of the catalyst
(3) Products desorb from the surface if the catalyst

43
Q

Making catalysts more efficient

A
  • Increase the surface area
  • Spread the catalyst over an inert support medium.
44
Q

Catalyst issues

A
  • Impurities can block the active sites.
  • This prevents the reactants from adsorbing
  • Purifying the reactants is the best way to prevent poisoning.
     
45
Q

Heterogeneous catalyst examples

A

(1) Making Ammonia in the Haber Process
Catalysed by solid IRON

N2(g) + 3H2(g) ⇌ 2NH3(g)

(2) Making Sulfuric Acid in the Contact Process
Catalysed by solid Vanadium (V) Oxide – V2O5

Step 1 – Sulfur Dioxide is oxidised to Sulfur Trioxide
- SO2(g) + V2O5(s) ⇌ SO3(g) + V2O4(s)

Step 2 – The Vanadium (IV) Oxide is then converted back to Vanadium (V) Oxide with oxygen.
- 2V2O4(s) + O2(g) ⇌ 2V2O5(s)

This means the overall reaction is:
2SO2(g) + O2(g) → 2SO3(g)

Sulfuric Acid is then formed by reacting SO3 with H2O.

(3) Manufacture of Methanol
Firstly this reaction forms a mixture known as synthesis gas
CH4(g) + H2O(g) → CO(g) + 3H2(g)

Then the following reaction is catalysed by solid Chromium (III) Oxide – Cr2O3
CO(g) + 2H2(g) → CH3OH(g)

46
Q

Why can’t group 1 metals catalyse?

A

only exist as 1 oxidation state

47
Q

Homogeneous catalyst example

A

Peroxodisulfate ions oxidise iodide ions to iodine.
The uncatalyzed reaction has a high Ea as the two negative ions repel each other:
S2O82-(aq) + 2I-(aq) → 2SO42-(aq) + I2(aq)

This reaction is catalysed by Fe2+ ions in a two-step process.
Step 1:
S2O82-(aq) + 2Fe2+(aq) → 2SO42-(aq) + 2Fe3+(aq)

Step 2:
2I-(aq) + 2Fe3+(aq) → I2(aq) + 2Fe2+(aq)

In the second step the Fe2+ catalyst is regenerated.

Fe3+ could also be used as a catalyst, as step 2 could occur before step 1.

48
Q

Autocatalysis

A

where one of the products of the reaction actually catalyses the reaction as it proceeds further

49
Q

Autocatalysis example

A

An example of this is the oxidation of ethanedioic acid by manganate (VII) ions.

The uncatalyzed reaction has a high Ea as the two negative ions repel each other:

2MnO4-(aq) + 16H+(aq) + 5C2O42-(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

The Mn2+ ions produced can catalyse the reaction in a two step process as shown below. Initially the rate is slow, but as more catalyst is produced the rate increases.

1) In the first step the Mn2+ is oxidised to MnO4-.
4Mn2+(aq) + MnO4-(aq) + 8H+(aq) → 5Mn3+(aq) + 4H2O(l)

2) In the second step the Mn3+ is reduced by C2O42-.
2Mn3+(aq) + C2O42-(aq) → 2CO2(g) + 2Mn2+(aq)

Because the MnO4- ions are a deep purple colour, the progress of this reaction can be tracked using a colorimeter. The reaction will go from purple to progressively clearer.

50
Q

MnO4- conc. time graph

A

The rate starts off slow as there is no catalyst initially.
The two negatively charged reactants collide with a very high Ea.
Then as some Mn2+ is formed the rate increases as the reaction is being increasingly catalysed.
The rate then decreases and levels off as the reactants get used up.

51
Q

Half equations

A

Chemistry (30 decks)

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