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A2 Chemistry > 5F - Transition elements > Flashcards

5F - Transition elements Flashcards

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1
Q

Define:

D-block element

Transition metal

A

D-block element = A element that has electrons entering the d-subshell

Transition metal = A d-block element that forms at least one ion with an incomplete d-subshell

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2
Q

What type of bonding exists in d-block elements and what is their stuctural arrangement?

A

Metallic bonding

Arranged in a giant lattice structure

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3
Q

Physical properties of D-block elements:

  • Melting points / Boiling points
  • Ability to conduct electricity
  • Solubility in water

Give an example of each

A

HIGH MELTING / BOILING POINT:

  • Strong electrostatic forces of attraction between +ve metal ions and delocalised electrons ∴ require large amounts of energy to break
    • E.g. titanium to make irons

CONDUCTS ELECTRICITY BOTH MOLTEN AND SOLID:

  • Solids: delocalised electrons free to move around and act as free mobile charge carriers. Ions are fixed in place.
  • Molten: Both metal ions and delocalised electrons act as free mobile charge carriers
    • E.g. Copper (electrical wiring)

INSOLUBLE IN WATER:

  • Strong electrostatic forces of attraction between metal ions and delocalised electrons can’t be overcome by water molecules
  • BUT some metals react with water, just very slowly such as iron rusting
    • E.g. Iron used to make steel for buildings
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4
Q
  • State the 2 elements that are different when writing out electronic configuration
  • Explain how are they different and write out their electronic configuration as well
A

Chromium:

  • Takes 1 out from 4s to half fill the 3d sub-shell

Copper:

  • Takes 1 out from 4s to completely fill the 3d sub-shell

They do this to make the element more stable

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5
Q

Properties of transition metals (3)

Give 2 examples of each property and include any necessary equations

A

MULTIPLE OXIDATION STATES:

  1. ​Iron in compounds can be Fe2+ or Fe3+
  2. Copper in compounds can be Cu+, Cu2+ or Cu3+

FORM COLOURED IONS IN SOLUTION:

  1. Fe2+(aq) = Pale green and Fe3+(aq) = Yellow
  2. Cu2+(aq) = Blue and Cu+(aq) = Colourless

THE ELEMENT AND THEIR COMPOUNDS CAN ACT AS CATALYSTS:

  1. Iron used in Haber process to produce NH3(g)
    • N2(g) + 3H2(g) ⇌ 2NH3(g)
  2. V2O5 used as a catalyst for contact process
    • 2SO2(g) + O2(g) ⇌ 2SO3(g)
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6
Q

Transition elements can quite easily form different oxidation states such as Fe2+ / Fe3+ but why is this not true for Group 1 and 2 metals?

A
  • The energy difference in forming compounds of different oxidation states like Fe2+ / Fe3+ isn’t that great ∴ both can exist
  • Not the case for Group 1 and 2 metals:
    • Removing electrons involves big increases in ionisation energy because electrons are removed from shells closer to the nucleus
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7
Q

The different oxidation states of compounds formed and their stability

A

Transition elements have variable oxidation states, these won’t all have the same energy or stability so one oxidation state will be more favoured than the other

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8
Q

Transition metals form coloured ion solutions

How do we see the colours that we do? + Example

A
  1. When light passes through a coloured substance, specific wavelengths of light is absorbed
  2. We see a mixture of what’s NOT absorbed, the complementary colour to that which is absorbed

E.g. Solution that absorbs green light will appear red

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9
Q

How transition elements produce coloured ion solutions - electron promotion

How is white light produced? What transition metal would appear as white?

A
  1. Transition metal ions can absorb energy from visible light to promote electrons from lower to higher energy levels
  2. BUT that means there must be space available in d-orbital sub-shells to do this

White light = when 3d sub-shell is full or empty

  • Zinc has a full 3d subshell
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10
Q

What does the transitional property of colour in transition elements forming coloured ion solutions depend upon?

A

The metal ions having a partially flled d-subshell

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11
Q

Transition elements and catalytic behaviour: (3)

Why do they make good catalysts?

A
  • Variable oxidation states make them good catalysts
  • Good homogeneous catalysts:
    • Because they take part in Redox reactions and form intermediates with reactants through an alternative route where the Ea in lowered
  • Good heterogeneous catalysts:
    • Because they can easily adsorb reactants, weakening bond within reactants
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12
Q

Define adsorption

A

Adsorption = The process that occurs when a gas, liquid or solute is held the the surface pf a solid where the chemical reaction takes place on

THINK ADHERE!

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13
Q

Define catalyst

A

Catalyst = A substance that increases the rate of a chemical reaction by providing the reaction with an alternative route that has a lower activation energy. It alos doesn’t get used up in the process

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14
Q

2 advantages of using catalysts

A
  • Reduced energy uasge
  • Reduced from toxicity exposure to transition metals
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15
Q

Give 5 examples of transition metals being used as catalysts:

  • Name of catalyst
  • Reaction it’s used in
  • Necessary equation
  • Homogeneous / heterogeneous
A
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16
Q

Name the 2 redox titration reactions you need to know and what they are used for

A

​Manganate (VII) titration

  • Used to find the concentration of reducing agents

Iodine / Thiosulphate titration

  • Used to find the concentration of oxidising agents
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17
Q

Give the steps for the general procedure for conducting a titration (7)

A
  1. Add standard solution of known conc into burette
  2. Use a volumetric pipette (10cm3) to pipette a sample of the solution being analysed into a conical flask (100cm3)
  3. If required, add a few drops of indicator as well as additional reactants
    • e.g. excess H+ ions
  4. Carry out a rough titration
    • Gives indication of end point
  5. Carry out accurate titrations until you have 2 or more concordant titres (± 0.1cm3)
  6. All values recorded to 2 d.p with units in a table
  7. Calculate mean average titre
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18
Q

Describe the procedure of carrying out a redox titration: manganate (VII) titration (7)

A

NOTE:

MnO4- ions are reduced ∴ other cheimcal must be oxidised (unknown is the reducing agent)

  1. Standard solution of KMnO4(aq) added to burette
  2. Use a volumetric pipette (10cm3) to pipette a sample to be analysed into a 100cm3 conical flask
  3. Add excess of dilute H2SO4(aq) to provide H+(aq) ions for the reduction of MnO4-(aq) ions.
    • No indicator is needed
  4. Titration is complete when you add 1 extra drop of KMnO4(aq) and the solution produces a colour change colourless → pale pink
    • ​Indicates an excess of MnO4-(aq) ions as they should decolourise as they react to produce a colourless solution
  5. Carry out 1 rough titiration then accurate titrations afterwards.
  6. Repeat experiment until 2 or more concordant titres are achieved (± 0.10 cm3)
  7. Take the mean average titre for calculations
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19
Q

For a redox titration of manganate (VII):

  • Why is reactions with Fe2+(aq) not a suitable redox titration at very high concentrations?
  • What is (COOH)2 . xH2O?
    • How is the value of x determined?
    • What happens to xH2O when (COOH)2 . xH2O is dissolved in water?
A
  • Reaction with Fe2+(aq) ions is not really a suitable redox titration at very high concentrations because Fe3+(aq) is ORANGE ∴ it would be hard to see the end point
  • (COOH)2 . xH2O = solid ethanoic acid where x is determined through a redox titration with MnO4-(aq) ions
  • xH2O becomes a part of the solvent when (COOH)2 . xH2O is dissolved in water
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20
Q

Question:

Redox calculation

A
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21
Q

Question:

Redox titration

A
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22
Q

Question:

Redox titration

A
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23
Q

Analysing the percentage purity of an iron (II) compound from a redox titration

A
24
Q

Describe an experiment to analyse oxidising agents through a redox titration (8)

A

Redox titration of iodine / thiosulphate:

  1. Standard solution of Na2S2O3(aq) added to burette
  2. Volumetric pipette (10cm3) to pipette sample to be analysed into 100cm3 conical flask
  3. Add excess KI(aq) as unknown oxidising agent reacts with KI(aq) to produce I2(aq) which turns solution yellow-brown
  4. Titrate solution with Na2S2O3(aq). During titration, I2(aq) is also reduced to 2I-(aq) causing the yellow-brown colour to fase to a pale-straw colour
  5. Difficult to decide end point ∴ once near end point add a few drops of starch indicator. Colour change pale straw → blue black
  6. Continue dripping Na2S2O3(aq), end point is when one extra drop of Na2S2O3(aq) is added and solution changes from blue-black → colourless
  7. Repeat titration until 2 concordant titires are achieved (± 0.10 cm3)
  8. Calculate mean avergae titre
25
Q

In the redox titration of iodine / thiosulphate why is starch added only once close to the end point and not right at the start?

A

If it was added to the start then it’s impossible to see when the end point is approaching since starch is blue-black

26
Q

Name 3 different oxidising agents that can be determined through a redox titration with iodine / thiosulphate

A
  • ClO- content in household bleach
  • Cu2+ content in copper (II) compounds
  • Cu content in copper alloys
27
Q

Describe the procedure to analyse household bleach (5 + calculation)

Results = Mean titre of 24.20cm3

A
  1. Use volumetric pipette (10cm3) to pipette a sample of bleach into a volumetric flask and add water to prepare 250cm3 of solution
  2. Using pipette, measure 25cm3 of this solution into a conical flask. Add 10cm3 of 1moldm-3 KI(aq) followed by sufficient 1moldm-3 HCl(aq) to acidify the solution
    • HCl provides H+ ions for the reaction
  3. Titrate this solution against a standard 0.05moldm-3 of Na2S2O3(aq)
  4. Repeat titration until you get 2 or more concordant results (± 0.10 cm3)
  5. Finally analyse results to determine the concentration of chlorate(I) ions in the bleach
28
Q

Define

Complex ion

A

Complex ion = A transition metal bonded to ligands by coordinate bonds

29
Q

Define co-ordinate bond

A

Co-ordinate bond = A shared pair of electrons in which the bonded pair has been provided by one of the bonded atoms only

(Same definition as a dative covalent bond)

30
Q

Define Ligand

A

A species that can donate pair of electrons to the transition metal ion

31
Q

Rewrite Fe3+(aq) in complex ion form

A

Fe3+(aq) is the same as [Fe(H2O)6]3+

32
Q

Explain using VSEPR theory explain explain how the number of co-ordinate bonds will influence the shape of the complex ion

A

VSEPR theory = valence shell electron pair repulsion theory

  • VSEPR theory states that electron pairs repel each other and try to be as far apart as possible at equal distances
  • No’ of bonded electron pairs in co-ordinate bond will influence the amount of repulsion between electrons ∴ influencing the shape of the complex ion
33
Q

Define

Monodentate ligand

Bidentate ligand

Polydentate ligand

A

Monodentate ligand = A ligand using 1 pair of electrons to form a single co-ordinate bond to a central metal ion

Bidentate ligand = A ligand using 2 pairs of electrons to form 2 co-ordinate bond to a central metal ion

Polydentate ligand = A ligand using several pairs of electrons to form several co-ordinate bonds to a central metal ion

34
Q

For 5 of the most common monodentate ligands:

  • Name
  • Formula
  • Charge on ligand
A
35
Q

Draw the complex ion [Cu(C2O4)3]4-

What type of ligand is this?

A
36
Q

Define

Stereoisomers

Optical isomers

Enantiomers

A

Stereoisomers = Compounds with the same structural formula but a differemt arrangement of atoms in space

Optical isomers = Transition metal surrounded by ligands such that if reflected in a mirror will form 2 non superimposable images

Enantiomers = A pair of optical isomers

37
Q

Name the 5 different shapes for complex ions, the first and second ligands as well as steroisomers formed

A
38
Q

Question:

[PtCl2(NH3)2] has a square planar shape

  • Draw the possible stereoisomers that can form and name them
  • Bond angle = ?
A
  • 4 x monodentate
  • Cis-trans isomerism (2 stereoisomers)
  • Bond angle = 180º
39
Q

When do square planar complexes form?

What elements does this apply to? (3)

A

When you have 8 d-electrons in the 3-d sub-shell

  • Pt2+
  • Pa2+
  • Au2+
40
Q

​​Question:

[Co(NH3)4Cl2]+

  • Shape of complex ion
  • Type of ligands present
  • Draw the possible stereoisomers that can form and name them
  • Bond angle = ?
A
  • Shape = octahederal
  • 4 x monodentate and 2 x monodentate
  • Cis-trans isomerism (3 stereoisomers as there’s 2 forms of the cis isomer)
  • Bond angle = 90º
41
Q

​​Question:

[Cr(C2O4)2(H2O)2]-

  • Type of ligands present
  • Draw the possible stereoisomers that can form and name them
  • Bond angle = ?
A
  • Shape = octahederal
  • 2 x Bidentate and 2 x Monodentate
  • Cis-trans isomerism (2 stereoisomers)
  • Bond angle = 90º
42
Q

Ligand substitution reactions with Cu2+(aq) ions and Cl- ions

  • Reagents / observations
  • Equations / formulae
  • What is particular about the observation?
  • Shape of final complex ion / it’s bond angle
  • Why has the co-ordination no’ changed?
A

Cu2+(aq) = [Cu(H2O)6]2+ complex ion

  • Pale blue in solution

Cu2+(aq) AND Cl- IONS:

​[Cu(H2O)6]2+ + 4Cl- ⇌ [CuCl4]2- + 6H2O

  • R = Conc HCl as source of Cl- ions
  • O = Colour change in solution from pale blue → yellow (in theory but we tend to see green instead since yellow mixed with blue gives green)
  • [CuCl4]2- is tetreahedral (bond angle = 109.5º)
  • Co-ordination no’ has changed from 6 to 4
    • Because Cl- ligands take up more space than H2O ligands ∴ they repel more
43
Q

Ligand substitution reactions with Cu2+(aq) and ammonia solution:

  • Reagents / observations
  • Equations / formulae
  • What is particular about the observation?
  • Shape of final complex ion / it’s bond angle
A

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

Pale blue solution → Pale blue precipitate → Dark blue solution

44
Q

Ligand substitution reactions with Chromium (III) ions and ammonia solution:

  • Reagents / observations
  • Equations / formulae
  • Why does the coloured precipitate form?
A

Cr3+(aq) = [Cr(H2O)6]3+ complex ion

[Cr(H2O)6]3+ → Cr(OH)3(s) → [Cr(NH3)6]3+

Violet solution → Grey green precipitate → Purple solution

  • Grey green precipitate forms because dilute NH3(aq) is an alkali, it contains OH- ions
    • NH3 + H2O ⇌ NH4+ + OH-
45
Q

What is Cis-platin?

  • Formula
  • Shape of molecule
  • Overall charge
  • Use
  • Why is it toxic to humans? (2)
A

Cis-platin = [Pt(NH3)2]

  • Square planar with no overall charge
  • Used in chemotherapy for cancer treatments

Toxic because:

  • Binds to DNA and prevents it from unravelling ∴ preventing DNA replication
  • Pt is a very heavy metal (NOTE that heavy metals are rare) ∴ not used to having them in the body
    • Damage kidney’s and causes nausea
46
Q

Once cis-platin has crossed the cell surface membrane what ligand substitution reaction takes place?

Give an equation

A

A chloride ligand is replaced with a water ligand

[Pt(NH3)2Cl2] + H2O ⇌ [Pt(NH3)2(H2O)Cl]+ + Cl-

  • Reaction occurs ine quilibrium and happens more readily once cis-platin has crossed CSM
  • Conc of Cl- ions within cell is less than outside cell in bloodstream
  • Equili shift to favour right side of reaction
47
Q

How is cis-platin formed from potassium tetrachloroplatinate?

  • How does this process ensure that only cis-platin (no trans-platin) is produced
A

STAGE 1:

[PtCl4]2- + NH3 ⇌ [Pt(NH3)Cl3]- + Cl-

STAGE 2:

[Pt(NH3)Cl3]- + NH3 ⇌ [Pt(NH3)2Cl2]- + Cl-

  • ​NH3 has a weaker trans effect than chloride ∴ the first NH3 tha tbinds is ineffective at pushing off the chloride ligand trans to it
  • NH3 wants to be next to each other (more favourable)
48
Q

What is carboplatin?

  • How is the structure different to cis-platin
  • Give 1 advantage and 1 disadvantage of using carboplatin
A

Carboplatin = Deveoped to try to work like cis-platin but with less side effects

  • The 2 chloride ligands in cis-platin is replaced with a bidentate cyclobutane-1,1-dicarboxylate group

+ve = less side effects

  • ve = causes a drop in red and white blood cell count
  • (More non lethal side effects or less side effects but they’re lethal)?*
49
Q

Haemoglobin:

  • What is it (2)
  • Draw a simplified diagram of a haem group
  • Equation for the ligand substitution reaction that haemoglobin undergoes in the lungs
  • Equation for when CO is present in lungs - why does the reaction happen?
A
  • Haemoglobin is the red pigment in RBC’s and are responsible for carrying oxygen from lungs → respiring tissues
  • It’s a conjucated protein
    • Has 4 polypeptides
    • Each polypeptide has 1 haem group which contains a Fe2+ ion in the centre

Ligand substitution of 4 x waters with 4 x oxygens in the lungs:

[Hb(H2O)4]8+ + 4O2 ⇌ [Hb(O2)4]8+ + 4H2O

Carbon monoxide is a competitive ligand with oxygen ∴ can displace oxygen as a ligand for oxyhaemoglobin:

[Hb(O2)4]8+ + 4CO ⇌ [Hb(CO)4]8+ + 4O2

50
Q

What allows H2O, O2 and CO to bind to Fe2+ in haemoglobin?

Draw the dot / cross diagrams to illustate answer

A

A pair of electrons that are not shared

51
Q

What is called when CO displaces oxygen in oxyhaemoglobin?

  • Equation
  • How is it treated?
A

Carbon monoxide poisoning

[Hb(O2)4]8+ + 4CO ⇌ [Hb(CO)4]8+ + 4O2

Treated:

  • Use 100% oxygen or hyperbaric oxygen - oxygen above normal atmospheric pressure
  • Inceasing [O2] will push equil to favour left side
  • [Hb(CO)4]8+ combines with more O2 and CO is released as a byproduct
52
Q

Precipitation reactions with NaOH(aq)__:

Cu2+ ions

  • Obervations
    • Is it solution in excess NaOH(aq)
  • Equations
A

Observations:

  • Pale blue solution → pale blue precipitate
  • Precipitate insolution in excess NaOH(aq)

Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

53
Q

Precipitation reactions with NaOH(aq):

Fe2+ ions

  • Obervations
    • Is it solution in excess NaOH(aq)
  • Equations
A

Observations:

  1. Green solution → gelatinous pale green precipitate of Fe(OH)2(s)
  2. Precipitate insoluble in excess NaOH(aq) BUT on standing precipitate becomes orange / brown on surface
  3. Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)
  4. 4Fe(OH)2(s) + O2(g) → 4Fe(OH)3(s) + 2H2O(l)
54
Q

Precipitation reactions with NaOH(aq):

Fe3+ ions

  • Obervations
    • Is it solution in excess NaOH(aq)
  • Equations
A

Obsevations:

  • Pale yellow solution → gelatinous orange / brown precipitate
  • Precipitate insoluble in excess NaOH(aq)

Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)

55
Q

Precipitation reactions with NaOH(aq):

Mn2+ ions

  • Obervations
    • Is it solution in excess NaOH(aq)
  • Equations
A

Observations:

  • Pale pink solution → light brown precipitate of Mn(OH)2(s) which darkens on standing in air
  • Precipitate insoluble in excess NaOH(aq)

Mn2+(aq) + 2OH-(aq) → Mn(OH)2(s)

56
Q

Precipitation reactions with NaOH(aq):

Cr3+ ions

  • Obervations
    • Is it solution in excess NaOH(aq)
  • Equations
A

Observations:

  1. Violet solution → grey green precipitate of Cr(OH)3(s)
  2. Precipitate soluble in excess NaOH(aq) to form a dark green solution
  3. Cr3+(aq) + 3OH-(aq) → Cr(OH)3(s)
  4. Cr(OH)3(s) + 3OH-(aq) → [Cr(OH)6]3-

A2 Chemistry (17 decks)

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