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A-Level Chemistry [OCR B Salters] (sorry for any spelling mistakes - there are many) > [Year 2] DM > Flashcards

[Year 2] DM Flashcards

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1
Q

Where are the d-block elements in the periodic table?

A

In the middle between s-block and p-block.

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2
Q

What is a transition element?

A
  • A d-block element…
  • that can form at least one stable ion…
  • with a partially filled d-orbital.
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3
Q

How many electrons can a d-orbital hold?

A

Up to 10 electrons

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4
Q

Which elements in period four are not transitioned element but are in the d-block?

Why?

A
  • Scandium (Sc) and Zinc (Zn)

- as they do not form a stable ion with a partially filled d-orbital.

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5
Q

How do electrons fill a d-orbital?

A

Singularly firsts.

Then they double up.

  • because electrons repel eachother
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6
Q

Write the shorthand electron configuration of the transition elements in period 4

Period 4 d-blocks:
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
A

Ti:
3d |↑ |↑ | | | |
4s |↑↓|
Config = [Ar] 3d² 4s²

V:
3d |↑ |↑ |↑ | | |
4s |↑↓|
Config = [Ar] 3d³ 4s²

Cr:
3d |↑ |↑ |↑ |↑ |↑ |
4s |↑ |
Config = [Ar] 3d⁵ 4s¹ (as filling all with 1 is most stable)

Mn:
3d |↑ |↑ |↑ |↑ |↑ |
4s |↑↓|
Config = [Ar] 3d⁵ 4s²

Fe:
3d |↑↓|↑ |↑ |↑ |↑ |
4s |↑↓|
Config = [Ar] 3d⁶ 4s²

Co:
3d |↑↓|↑↓|↑ |↑ |↑ |
4s |↑↓|
Config = [Ar] 3d⁷ 4s²

Ni:
3d |↑↓|↑↓|↑↓|↑ |↑ |
4s |↑↓|
Config = [Ar] 3d⁸ 4s²

Cu:
3d |↑↓|↑↓|↑↓|↑↓|↑↓|
4s |↑ |
Config = [Ar] 3d¹⁰ 4s¹ (as a full 3d orbital is more stable)

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7
Q

Show why Scandium is not a transition metal.

A

Scandium’s electron configuration is as followed…
3d |↑ | | | | |
4s |↑↓|

It only forms one stable ion, Sc³⁺, which has the configuration…
3d | | | | | |
4s | |

Because of its empty d-orbital, it cannot be a transition element.

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8
Q

Show why Zinc is not a transition metal.

A

Zinc’s electron configuration is as followed…
3d |↑↓|↑↓|↑↓|↑↓|↑↓|
4s |↑↓|

It only forms one stable ion, Zn²⁺, which has the configuration…
3d |↑↓|↑↓|↑↓|↑↓|↑↓|
4s | |

Because of its full d-orbital, it cannot be a transition element.

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9
Q

What is the electron configuration of Fe ions and Cu ions?

A

Fe²⁺ : [Ar] 3d⁶

Fe³⁺ : [Ar] 3d⁵

Cu⁺: [Ar] 3d¹⁰

Cu²⁺: [Ar] 3d⁹

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10
Q

What are the properties of transition metals?

A
  • Variable oxidation states.
  • Form coloured ions in solution.
  • able to form complex ions.
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11
Q

What colour are Fe ions in solution?

A 
Fe²⁺  = light green
Fe³⁺  = yellow
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12
Q

What colour are Cu ions in solution?

A

Cu²⁺ = Blue

Cu⁺ is unstable in solution, so will disproportionate into Cu²⁺ and Cu readily…
2Cu⁺ → Cu²⁺ + Cu

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13
Q

What is a complex ion?

A
  • Where a central transition metal ion…
  • is surrounded by ligands…
  • bonded by dative covalent bonds.
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14
Q

How do you show a complex ion?

A
  • Central ion.
  • With dative covalent bond arrows from ligands to it.
  • In square brackets.
  • With a charge.
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15
Q

What is a ligand?

A

Ions, atoms, or molecules that have at least one lone pair of electrons.

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16
Q

What are examples of monodentate ligands?

A

Water:
H₂O፡

Ammonia:
፡NH₃

Cyanide:
፡CN⁻

Chloride:
፡Cl⁻

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17
Q

What are examples of bidentate ligands?

A

Ethanedioate:
⁻፡OOC=COO፡⁻

Ethylenediamine (diamino ethane):
H₂N̈C₂H₄N̈H₂

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18
Q

What are examples of polydentate ligands?

A

EDTA⁴⁻

6 lone pairs

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19
Q

What factors affect the shape of a complex ion?

A
  • Size of ligands

- Coordination number

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20
Q

What is the coordination number?

A

The number of coordinate bonds in a complex.

not number of ligands as ligands can have multiple coordinate (dative) bonds

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21
Q

Which ligands are small?

How many can fit around a central metal ion?

What shape do these form?

A

Water:
H₂O፡

Ammonia:
፡NH₃

Cyanide:
፡CN⁻

  • 6 can fit around one central ion.
  • This forms an octahedral shape.
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22
Q

Which ligands are medium?

How many can fit around a central metal ion?

A

Chloride:
፡Cl⁻

4 can fit around one central ion.
- This forms tetrahedral shapes
and square planar for Pt[(NH₃)₂(Cl)₂)]₍ₐᵩ₎ (cis platin, an anti-cancer drug).

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23
Q

Which ligands are large?

A

Ethanedioate:
⁻፡OOC=COO፡⁻

Ethylenediamine (diamino ethane):
H₂N̈C₂H₄N̈H₂

Normally 3 can fit around one ion.

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24
Q

What happens to the d-orbital when a ligands bond?

A

The d-orbital splits.

Before:
3d |↑↓|↑↓|↑ |↑ |↑ |

After:
|↑ |↑ |
(this energy gap is ΔE)
|↑↓|↑↓|↑ |

This is known as the ground state

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25
Q

Why are complex ions coloured?

A
  • Complex ions are made up of ligands and transition metals.
  • Ligands split the d-orbital of the transition element, as shown…
    |↑ |↑ |
    (this energy gap is ΔE)
    |↑↓|↑↓|↑ |
  • This is the ground state.
  • When electrons absorb light energy some electrons from the ground state are excited to a higher energy level (an excited state), as shown…
    |↑↓|↑ |

|↑↓|↑ |↑ |

  • This light energy has energy = ΔE
  • and so has a specific frequency.
  • The frequencies that are not absorbed are reflected or transmitted.
  • And so we see the complementary colour to what was absorbed.
26
Q

What effects ΔE during d-orbital splitting?

A
  • The central metal ion and its oxidation state.
  • The type of ligand.
  • The coordination number.
27
Q

Show what happens sodium hydroxide is added to Cu²⁺₍ₐᵩ₎.

A

[Cu(H₂O)₆]²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → [Cu(OH)₂(H₂O)₄]₍ₛ₎ + 2H₂O₍ₗ₎

This forms a pale blue precipitate.

  • It precipitates out because the product is not charged.
28
Q

Show what happens excess sodium hydroxide is added to Cu²⁺₍ₐᵩ₎.

A
  • insoluble in excess NaOH.
29
Q

Show what happens ammonia₍ₐᵩ₎ is added to Cu²⁺₍ₐᵩ₎.

A

Cu(H₂O)₆]²⁺₍ₐᵩ₎ + 2NH₃₍ₐᵩ₎ → [Cu(OH)₂(H₂O)₄]₍ₛ₎ + 2NH₄⁺₍ₐᵩ₎

This forms a pale blue precipitate.

  • It precipitates out because the product is not charged.
30
Q

Show what happens excess ammonia₍ₐᵩ₎ is added to Cu²⁺₍ₐᵩ₎.

A

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

This forms a Dark blue solution
- and is known as a part ligand substitution.

31
Q

Show what happens sodium hydroxide is added to Fe²⁺₍ₐᵩ₎.

A

[Fe(H₂O)₆]²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → [Fe(OH)₂(H₂O)₄]₍ₛ₎ + 2H₂O₍ₗ₎

This forms a darker green precipitate.

  • It precipitates out because the product is not charged.
32
Q

Show what happens excess sodium hydroxide is added to Fe²⁺₍ₐᵩ₎.

A
  • insoluble in excess NaOH.
33
Q

Show what happens ammonia₍ₐᵩ is added to Fe²⁺₍ₐᵩ₎.

A

[Fe(H₂O)₆]²⁺₍ₐᵩ₎ + 2NH₃₍ₐᵩ₎ → [Fe(OH)₂(H₂O)₄]₍ₛ₎ + 2NH₄⁺₍ₐᵩ₎

This forms a darker green precipitate.

  • It precipitates out because the product is not charged.
34
Q

Show what happens excess ammonia₍ₐᵩ₎ is added to Fe²⁺₍ₐᵩ₎.

A
  • insoluble in excess NH₃.
35
Q

Show what happens sodium hydroxide is added to Fe³⁺₍ₐᵩ₎.

A

[Fe(H₂O)₆]³⁺₍ₐᵩ₎ + 3OH⁻₍ₐᵩ₎ → [Fe(OH)₃(H₂O)₃]₍ₛ₎ + 3H₂O₍ₗ₎

This forms an orange precipitate.

  • It precipitates out because the product is not charged.
36
Q

Show what happens excess sodium hydroxide is added to Fe³⁺₍ₐᵩ₎.

A
  • insoluble in excess NaOH.
37
Q

Show what happens ammonia₍ₐᵩ is added to Fe³⁺₍ₐᵩ₎.

A

[Fe(H₂O)₆]³⁺₍ₐᵩ₎ + 3NH₃₍ₐᵩ₎ → [Fe(OH)₃(H₂O)₃]₍ₛ₎ + 3NH₄⁺₍ₐᵩ₎

This forms an orange precipitate.

  • It precipitates out because the product is not charged.
38
Q

Show what happens excess ammonia₍ₐᵩ₎ is added to Fe³⁺₍ₐᵩ₎.

A
  • insoluble in excess NH₃.
39
Q

Why does part ligand substitution lead to a colour change?

A
  • Change in ΔE, as there is a different ligand.

- Change in shape if the ligand is larger/smaller.

40
Q

Why are transition metals good catalyst?

A
  • As they have variable oxidation states.
  • they are able t receive and lose electrons in their d-orbital.
  • This can lead to reactions speeding up
41
Q

How does a solid heterogeneous catalyst work?

A
  • Substances adsorb to the surface
  • This weakens bonds and breaks them to form radicals.
  • Radicals react with each other to form new substances.
  • The new substances are then desorbed.
42
Q

What makes a good solid heterogeneous catalyst?

A
  • They can attract and hold reactants long enough for them to react.
  • They do not attract products too strongly or desorption won’t occur.
  • If they are ground up, they have a larger surface area, so more substance can attach to its surface.
43
Q

What are examples of transition metal catalyst and their uses?

A
  • Iron catalysts (used in Harber process)
  • Copper sulfate (used to catalyse acid reactions with zinc)
  • Manganese dioxide (used to catalyse the decomposition of hydrogen peroxide)
44
Q

What are the risks of metal catalyst?

A
  • Long term exposure to copper can lead to liver damage.

- Long term exposure to manganese can cause psychiatric issues and physical tremors.

45
Q

Show how a homogeneous catalyst is used in the oxidation of iodide ions using peroxodisulfate (S₂O₈²⁻).

A

Normal reaction is slow…
S₂O₈²⁻₍ₐᵩ₎ + 2I⁻₍ₐᵩ₎ → I₂₍ₐᵩ₎ + 2SO₄²⁻₍ₐᵩ₎
… as like charges repel each other.

So an Iron catalyst is used (Fe²⁺)…
S₂O₈²⁻₍ₐᵩ₎+ 2Fe²⁺₍ₐᵩ₎ → 2Fe³⁺ ₍ₐᵩ₎+ 2SO₄²⁻₍ₐᵩ₎

Fe²⁺ must be reformed so…
2Fe³⁺ ₍ₐᵩ + 2I⁻₍ₐᵩ₎ → 2Fe²⁺ + I₂₍ₐᵩ₎

46
Q

What is a reducing agent?

A
  • A molecule that loses electrons and is oxidised itself.
47
Q

What is an oxidising agent?

A
  • A molecule that gains electrons and is oxidised itself.
48
Q

What rules can you follow in order to balance half equations?

A
  1. balance atoms that are not Oxygen and Hydrogen.
  2. balance any oxygens with water
  3. balance any hydrogens with H⁺ ions
  4. Balance any charges with electrons.
49
Q

What must be added in a redox titration in addition to a normal one?

A

Add excess dilute sulfuric acid to the conical flask of unknown concentration, in order to have sufficient H⁺ ions to allow the reduction of the oxidising agent.

50
Q

How would you work out the concentration after you have done a redox titration?

A
  1. write out balenced equation.
  2. work out the number of moles of known concentration substance by doing…
    mol = conc x vol (in dm³)
  3. Use molar ratios to work out moles of desired product/reactant.
  4. With the known volme and moles now work out its concentration…
    conc= mol/vol (in dm³)
51
Q

What is a half cell?

A
  • half of an electrochemical cell.
  • made from metal in its aqueous ion.
  • or platinum with 2 aqueous ions in solution.
52
Q

Describe the method of setting up an electrochemical cell.

A
  • Wear gloves for safety and to prevent contamination
  • obtain metals under investigation and clean with sandpaper/emery paper to ensure surface is free of impurities.
  • Wash surface with propanone to remove grease
  • place each metal electrode into a solution containing the ions of the same metal. (if you are using an oxidising agent then add acid).
  • make a salt bridge from filter paper in saturated KNO₃ or KCl linking the beakers. Do not touch the electrode with this though.
  • Connect the electrodes with wires and crocodile clips to a voltmeter.
  • The reading on the voltmeter is the E⁰cell.
53
Q

How do you work out which half cell is undergoing reduction and oxidation?

A

Compare electrode potentials (E⁰) from an electrochemical series.

  • The larger E⁰ will be reduced.
54
Q

What is the standard electrode potential compared to?

A

the standard electrode potential of hydrogen

at 298K
100kPa
with a conc of 1 moldm³ of ions.

55
Q

Which half cells in the electrochemical series are stronger oxidising agents?

A

The more positive ones.

56
Q

How do you work out E⁰cell of an electrochemical cell?

A

E⁰cell = E⁰reduced - E⁰oxidised.

57
Q

How do you show if a reaction is feasible using the standard electrode potentials??

A
  • write out full equation.
  • identify half equations and their E⁰.
  • identify which will be reduced and which will be oxidised.
  • work out Ecell.
  • if Ecell is positive the reaction is feasible.
58
Q

why might a positive E⁰cell not be feasible?

A
  • non-standard conditions, as the electrochemical series assumes standard conditions.
  • Change in concentration of reactants, as the electrochemical series assumes it is 1 moldm³.
  • Reaction may not be kinetically favourable, so will react too slow.
  • Reaction may have high activation energy so it may stop before the reaction has happened.
59
Q

What reactions take place during rusting?

A

Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎
O₂₍𝓰₎ + 2H₂O₍ₗ₎ + 4e⁻ ⇌ 4OH⁻₍ₐᵩ₎
1. 2Fe₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ ⇌ Fe²⁺₍ₐᵩ₎ + 4OH⁻₍ₐᵩ₎

  1. Fe²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → Fe(OH)₂₍ₛ₎
  2. 4Fe(OH)₂₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ → 4Fe(OH)₃₍ₛ₎
  3. overtime Fe(OH)₃₍ₛ₎ becomes hydrated and becomes Fe₂O₃xH₂O
60
Q

Why is rust less likely to form in alkaline conditions?

A
  1. Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎
  2. O₂₍𝓰₎ + 2H₂O₍ₗ₎ + 4e⁻ ⇌ 4OH⁻₍ₐᵩ₎
  • in alkaline conditions, the conc of OH⁻ ions are increased.
  • This shifts the equilibrium in equation 2 to the left to remove the excess OH⁻.
  • This means there are more e⁻ available so equilibrium in equation 1 shifts to the right using up the electrons.
  • This forms more iron from ion ions so it is less likely for rust to occur.
61
Q

What forms of protection can be used to prevent rusting?

A

Barrier methods:
- Oil/Grease act as a physical barrier to prevent water from accessing the Iron/

  • Painting.

Sacrificial methods:

  • galvanisation…
  • Zn/Zn²⁺ half-reaction has a more negative E⁰ compared to Iron.
  • This means it is a better reducing agent, and so oxidises before the iron does.
62
Q

List the oxidation states od vanadium and their colours

A

VO₂⁺
Oxidation state: +5
Colour: Yellow.

VO²⁺
Oxidation state: +4
Colour: Blue.

V³⁺
Oxidation state: +3
Colour: Green.

V²⁺
Oxidation state: +2
Colour: Purple.

A-Level Chemistry [OCR B Salters] (sorry for any spelling mistakes - there are many) (14 decks)

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