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Chemistry > (1) Transition Metals > Flashcards

(1) Transition Metals Flashcards

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1
Q

Define a Transition Element

A

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

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2
Q

What are the key features of Transition Elements?

A
  • Form complex ions
  • Coloured ions
  • Catalytic properties
  • Variable oxidation states
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3
Q

How is the electron configuration of a Transition Metal ion determined?

A

FIRST IN - FIRST OUT

Electrons fill up 4s subshell before 3d so electrons are also removed from 4s subshell before 3d.

e.g.
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

Define a Coordinate Bond.

A

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

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5
Q

Define a Ligand.

A

Forms a Dative Bond with a Transition Metal Ion.

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6
Q

Define Coordination Number.

A

Number of Coordinate Bonds formed with Transition Metal.

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7
Q

Define a Complex Ion.

A

Transition Metal Ion surrounded by Ligands.

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8
Q

Define a Monodentate Ligand.

A

Has 1 lone pair and forms a Dative Bond with Transition Metal Ion.

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9
Q

Define a Bidentate Ligand.

A

Has 2 lone pairs and forms 2 Dative Bonds with Transition Metal Ion.

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10
Q

Define a Multidentate Ligand.

A

Has 2 or more lone pairs and forms 2 or more Dative Bonds with Transition Metal Ion.

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11
Q

Which complexes usually form a Linear Shape?

A

Ag+ complexes

e.g.
[Ag(NH3)2]+

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12
Q

Which complexes form a Tetrahedral shape?

A

Large Ligands

e.g. Cl-

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13
Q

Which complexes form a Square Planar shape?

A

Pt2+ complex

e.g.
[PtCl4]2-

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14
Q

Which complexes form a Octahedral shape?

A

Most complexes

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

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15
Q

Give an example of a Hexadentate Ligand.

A

EDTA4-

EDTA can form 6 Coordinate Bonds as it has 6 lone pairs.
2x on N, 4x on O

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16
Q

What (observable) changes are made when:

[Cu(H2O)6]2+ + 4Cl- –> [CuCl4]2- + 6H2O

A

Blue Solution –> Yellow Solution
Octahedral –> Tetrahedral
(Coordination Number) 6 –> 4

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17
Q

What happens to the feasibility of a reaction when 6 Monodentate Ligands are replaced by 3 Bidentate Ligands?

A

FEASIBILITY INCREASES

Large increase in entropy (4 moles of reactants becomes 7 moles of products)

(CHELATE EFFECT)

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18
Q

Explain why this reaction is 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 ∆G value will be negative.

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19
Q

What is an example of a Transition Metal Complex found in the blood?

A

HAEMOGLOBIN

(6 Coordinate Bonds)

  • 4 from N to Fe2+ (haem)
  • 1 from O2 to Fe2+
  • 1 from globin to Fe2+
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20
Q

Why is Carbon Monoxide (CO) toxic?

A

CO is toxic because CO bonds more strongly to the Fe2+ in haemoglobin.

This prevents O2 from bonding to the Fe2+, causing suffocation.

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21
Q

Define Stereoisomerism. What is the form of Stereoisomerism sometimes displayed by Ligands?

A

Same Structural Formula, atoms occupy different positions in space. Cis-Trans Isomerism.

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22
Q

What is a Cis Isomer?

A

Alike Ligands are next to each other.

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23
Q

What is a Trans Isomer?

A

Alike Ligands are opposite one another.

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24
Q

What is Cisplatin?

A

A Cancer Drug.

Binds to DNA, preventing DNA replication.
The 2 Cl- ions are substituted for 2 N atoms on adjacent Guanine bases.

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25
Q

How can damage to healthy cells be minimized when using Cisplatin?

A
  • Target cancerous cells.

- Low dosage

26
Q

What is Optical Isomerism? When does it occur?

A

A form of stereoisomerism where the complex forms non-superimposable mirror images of itself.

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

27
Q

How can Optical Isomers be distinguished?

A
  • Shine plane polarised light at the mixture

- Isomers rotate light in opposite directions

28
Q

How can the colours of light be remembered?

A

ROY G BIV

29
Q

Which colours of light are low in energy?

A

ROY

  • Red
  • Orange
  • Yellow
30
Q

Which colours of light are high in energy?

A

BIV

  • Blue
  • Indigo
  • Violet
31
Q

Why are Transition Metal compounds coloured?

A
  • Partially filled d-subshell
  • Certain colours (energies) of light are absorbed while the others are transmitted/reflected
  • d-subshell electrons absory energy matching the difference between the ground and excited state
  • Electrons are excited
32
Q

Is light transmitted or reflected from a solution?

A

Transmitted.

33
Q

What does a TM compound need to absorb light?

A

A partially filled d-subshell.

34
Q

What causes a TM compound to appear Red/Orange?

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

35
Q

What causes a TM compound to appear Blue/Purple?

A

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

36
Q

What can affect the colour of a TM compound?

A

∆E between d-orbitals can be altered by a:

  • Change in ligands
  • Change in metal oxidation state
  • Change in complex’s coordination number
  • Change in complex’s shape
37
Q

How is ∆E calculated? (equation)

A

∆E = hc/λ

  • ∆E is change in energy
  • h is Planck Constant (6.626 x 10-34 J·s)
  • c is speed of light (3 x 108 m/s)
  • λ is wavelength of light
38
Q

How can the concentration of a TM solution be determinded using Colorimetry?

A
  • An appropriate ligand (such as -SCN) is added to the solution in order to intensify the colour.
  • A range of solutions of the same complex ion are made of known concentrations.
  • One at a time these are tested in a colorimeter and the transmission (or absorbance) is measured.
  • A graph is plotted of Conc vs Transmission (or absorbance) and a line of best fit drawn.
  • The transmittance/absorbance of the unknown solution is measured in a colorimeter, and its concentration is determined by reading off the calibration curve.
39
Q

What should be done before taking a reading during Colorimetry? Why?

A

Place a cuvette filled with water (solvent) into the colorimeter and set to zero
(Reason) Negates/removes absorbance due to solvent and ensures accuracy

40
Q

What are other key experimental points of Colorimetry?

A
  • Length of solution’s container affects the amount of absorption as distance light travels is affected.
  • A filter is used to only allow one colour through the sample (use opposite light to colour of solution so light is strongly absorbed by sample)
41
Q

What is a Heterogenous Catalyst?

A

Catalyst and Reactants are in different phases.

42
Q

What is a Homogenous Catalyst?

A

Catalyst and Reactants are in the same phase.

43
Q

Outline Heterogenous Catalysis.

A
  • Reactants adsorb onto catalyst surface (active site of catalyst)
  • Reactant bonds break then product bonds form
  • Product’s molecules desorb from catalyst’s surface
44
Q

How are catalysts made more efficient?

A

INCREASE SURFACE AREA

  • by using a fine powder
  • by using honeycomb design/support then spread catalyst on it
45
Q

Why don’t catalysts last forever?

A

Poisoning can occur. Poison strongly adsorbs to catalyst’s active site and blocks it, reactant can’t displace it.

46
Q

How can the Haber Process by poisoned?

A

Sulphur impurities can be present in CH4 that generates H2 gas, poisoning Fe catalyst.

47
Q

Outline the Haber Process.

A

N2(g) + 3N2(g) 2NH3(g

48
Q

Outline the Contact Process.

A

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)

OVERALL REACTION:
2SO2(g) + O2(g) –>2SO3(g)

Sulfuric Acid is then formed by reacting SO3 with H2O

49
Q

Outline how Methanol is manufactured.

A

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)

50
Q

What is Autocatalysis?

A

One of the products of the reaction actually catalyses the reaction as it proceeds further.

51
Q

Outline the overall equation of oxidation of Ethanedioic Acid by Manganate (VII) Ions, and name the Autocatalyst.

A

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

Mn2+(aq) acts as the autocatalyst.

52
Q

How does rate of reaction change during Autocatalysis?

A

Initially the rate is slow (usually due to repulsion between negative ions), but as more Catalyst is produced the rate increases (name the catalyst).

53
Q

What is the first step of Autocatalysis (ethanedioic acid and manganate ions)?

A

4Mn2+(aq) + MnO4-(aq) + 8H+(aq) –> 5Mn3+(aq) + 4H2O(l)

In the first step the Mn2+ is oxidised to MnO4-.

54
Q

What is the second step of Autocatalysis (ethanedioic acid and manganate ions)?

A

2Mn3+(aq) + C2O42-(aq) –> 2CO2(g) + 2Mn2+(aq)

In the second step the Mn3+ is reduced by C2O42-

55
Q

Is Autocatalysis an example of Heterogenous Catalysis or Homogenous Catalysis?

A

Homogenous Catalysis.

56
Q

Why can’t Group 1 Metals act as Catalysts?

A

They only have 1 oxidation state.

57
Q

Why does this reaction have a high Ea?

S2O82-(aq) + 2I-(aq) –> 2SO42-(aq) + I2(aq)

A

The 2 negative ions repel each other.

58
Q

Give an example of Homogenous Catalysis (reaction). Outline both steps.

A

S2O82-(aq) + 2I-(aq) –> 2SO42-(aq) + I2(aq)

STEP 1: S2O82-(aq) + 2Fe2+(aq) –> 2SO42-(aq) + 2Fe3+(aq)

STEP 2: 2I-(aq) + 2Fe3+(aq) –> I2(aq) + 2Fe2+(aq)

CATALYST: Fe2+

59
Q

How does colour change in this equation?:

Fe2+ –> Fe3+

A

Green to Brown.

60
Q

How does colour change in this equation?

MnO4- –> Mn2+

A

Purple to Colourless.

61
Q

How does colour change in this equation?

Cr2O72- –> Cr3+

A

Orange to Green

Chemistry (29 decks)

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