[Year 2] DM Flashcards
Where are the d-block elements in the periodic table?
In the middle between s-block and p-block.
What is a transition element?
- A d-block element…
- that can form at least one stable ion…
- with a partially filled d-orbital.
How many electrons can a d-orbital hold?
Up to 10 electrons
Which elements in period four are not transitioned element but are in the d-block?
Why?
- Scandium (Sc) and Zinc (Zn)
- as they do not form a stable ion with a partially filled d-orbital.
How do electrons fill a d-orbital?
Singularly firsts.
Then they double up.
- because electrons repel eachother
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
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)
Show why Scandium is not a transition metal.
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.
Show why Zinc is not a transition metal.
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.
What is the electron configuration of Fe ions and Cu ions?
Fe²⁺ : [Ar] 3d⁶
Fe³⁺ : [Ar] 3d⁵
Cu⁺: [Ar] 3d¹⁰
Cu²⁺: [Ar] 3d⁹
What are the properties of transition metals?
- Variable oxidation states.
- Form coloured ions in solution.
- able to form complex ions.
What colour are Fe ions in solution?
Fe²⁺ = light green Fe³⁺ = yellow
What colour are Cu ions in solution?
Cu²⁺ = Blue
Cu⁺ is unstable in solution, so will disproportionate into Cu²⁺ and Cu readily…
2Cu⁺ → Cu²⁺ + Cu
What is a complex ion?
- Where a central transition metal ion…
- is surrounded by ligands…
- bonded by dative covalent bonds.
How do you show a complex ion?
- Central ion.
- With dative covalent bond arrows from ligands to it.
- In square brackets.
- With a charge.
What is a ligand?
Ions, atoms, or molecules that have at least one lone pair of electrons.
What are examples of monodentate ligands?
Water:
H₂O፡
Ammonia:
፡NH₃
Cyanide:
፡CN⁻
Chloride:
፡Cl⁻
What are examples of bidentate ligands?
Ethanedioate:
⁻፡OOC=COO፡⁻
Ethylenediamine (diamino ethane):
H₂N̈C₂H₄N̈H₂
What are examples of polydentate ligands?
EDTA⁴⁻
6 lone pairs
What factors affect the shape of a complex ion?
- Size of ligands
- Coordination number
What is the coordination number?
The number of coordinate bonds in a complex.
not number of ligands as ligands can have multiple coordinate (dative) bonds
Which ligands are small?
How many can fit around a central metal ion?
What shape do these form?
Water:
H₂O፡
Ammonia:
፡NH₃
Cyanide:
፡CN⁻
- 6 can fit around one central ion.
- This forms an octahedral shape.
Which ligands are medium?
How many can fit around a central metal ion?
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).
Which ligands are large?
Ethanedioate:
⁻፡OOC=COO፡⁻
Ethylenediamine (diamino ethane):
H₂N̈C₂H₄N̈H₂
Normally 3 can fit around one ion.
What happens to the d-orbital when a ligands bond?
The d-orbital splits.
Before:
3d |↑↓|↑↓|↑ |↑ |↑ |
After:
|↑ |↑ |
(this energy gap is ΔE)
|↑↓|↑↓|↑ |
This is known as the ground state
Why are complex ions coloured?
- 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.
What effects ΔE during d-orbital splitting?
- The central metal ion and its oxidation state.
- The type of ligand.
- The coordination number.
Show what happens sodium hydroxide is added to Cu²⁺₍ₐᵩ₎.
[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.
Show what happens excess sodium hydroxide is added to Cu²⁺₍ₐᵩ₎.
- insoluble in excess NaOH.
Show what happens ammonia₍ₐᵩ₎ is added to Cu²⁺₍ₐᵩ₎.
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.
Show what happens excess ammonia₍ₐᵩ₎ is added to Cu²⁺₍ₐᵩ₎.
[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.
Show what happens sodium hydroxide is added to Fe²⁺₍ₐᵩ₎.
[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.
Show what happens excess sodium hydroxide is added to Fe²⁺₍ₐᵩ₎.
- insoluble in excess NaOH.
Show what happens ammonia₍ₐᵩ is added to Fe²⁺₍ₐᵩ₎.
[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.
Show what happens excess ammonia₍ₐᵩ₎ is added to Fe²⁺₍ₐᵩ₎.
- insoluble in excess NH₃.
Show what happens sodium hydroxide is added to Fe³⁺₍ₐᵩ₎.
[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.
Show what happens excess sodium hydroxide is added to Fe³⁺₍ₐᵩ₎.
- insoluble in excess NaOH.
Show what happens ammonia₍ₐᵩ is added to Fe³⁺₍ₐᵩ₎.
[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.
Show what happens excess ammonia₍ₐᵩ₎ is added to Fe³⁺₍ₐᵩ₎.
- insoluble in excess NH₃.
Why does part ligand substitution lead to a colour change?
- Change in ΔE, as there is a different ligand.
- Change in shape if the ligand is larger/smaller.
Why are transition metals good catalyst?
- 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
How does a solid heterogeneous catalyst work?
- 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.
What makes a good solid heterogeneous catalyst?
- 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.
What are examples of transition metal catalyst and their uses?
- 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)
What are the risks of metal catalyst?
- Long term exposure to copper can lead to liver damage.
- Long term exposure to manganese can cause psychiatric issues and physical tremors.
Show how a homogeneous catalyst is used in the oxidation of iodide ions using peroxodisulfate (S₂O₈²⁻).
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₎
What is a reducing agent?
- A molecule that loses electrons and is oxidised itself.
What is an oxidising agent?
- A molecule that gains electrons and is oxidised itself.
What rules can you follow in order to balance half equations?
- balance atoms that are not Oxygen and Hydrogen.
- balance any oxygens with water
- balance any hydrogens with H⁺ ions
- Balance any charges with electrons.
What must be added in a redox titration in addition to a normal one?
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.
How would you work out the concentration after you have done a redox titration?
- write out balenced equation.
- work out the number of moles of known concentration substance by doing…
mol = conc x vol (in dm³) - Use molar ratios to work out moles of desired product/reactant.
- With the known volme and moles now work out its concentration…
conc= mol/vol (in dm³)
What is a half cell?
- half of an electrochemical cell.
- made from metal in its aqueous ion.
- or platinum with 2 aqueous ions in solution.
Describe the method of setting up an electrochemical cell.
- 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.
How do you work out which half cell is undergoing reduction and oxidation?
Compare electrode potentials (E⁰) from an electrochemical series.
- The larger E⁰ will be reduced.
What is the standard electrode potential compared to?
the standard electrode potential of hydrogen
at 298K
100kPa
with a conc of 1 moldm³ of ions.
Which half cells in the electrochemical series are stronger oxidising agents?
The more positive ones.
How do you work out E⁰cell of an electrochemical cell?
E⁰cell = E⁰reduced - E⁰oxidised.
How do you show if a reaction is feasible using the standard electrode potentials??
- 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.
why might a positive E⁰cell not be feasible?
- 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.
What reactions take place during rusting?
Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎
O₂₍𝓰₎ + 2H₂O₍ₗ₎ + 4e⁻ ⇌ 4OH⁻₍ₐᵩ₎
1. 2Fe₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ ⇌ Fe²⁺₍ₐᵩ₎ + 4OH⁻₍ₐᵩ₎
- Fe²⁺₍ₐᵩ₎ + 2OH⁻₍ₐᵩ₎ → Fe(OH)₂₍ₛ₎
- 4Fe(OH)₂₍ₛ₎ + O₂₍𝓰₎ + 2H₂O₍ₗ₎ → 4Fe(OH)₃₍ₛ₎
- overtime Fe(OH)₃₍ₛ₎ becomes hydrated and becomes Fe₂O₃xH₂O
Why is rust less likely to form in alkaline conditions?
- Fe²⁺₍ₐᵩ₎ + 2e⁻ ⇌ Fe₍ₛ₎
- 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.
What forms of protection can be used to prevent rusting?
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.
List the oxidation states od vanadium and their colours
VO₂⁺
Oxidation state: +5
Colour: Yellow.
VO²⁺
Oxidation state: +4
Colour: Blue.
V³⁺
Oxidation state: +3
Colour: Green.
V²⁺
Oxidation state: +2
Colour: Purple.
