5.3- Transition Elements Flashcards Preview

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Flashcards in 5.3- Transition Elements Deck (47)
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0
Q

Define a transition element

A

Transition element is a d-block element that forms an ion with an incomplete d sub-shell

1
Q

Which 2 d-block elements are not classified as transition elements.

A

Zinc and scandium (which explains why their compounds are white, rather than coloured).

2
Q

Which 2 d-block elements do not fill their sub-shells as expected

A

Chromium (4s1,3d5)

Copper (4s1,3d10)

3
Q

What are the three main transition metal properties

A
  1. Variable oxidation states (eg. Fe2+/Fe3+ & Cu+/Cu2+)
  2. Form coloured ions
  3. Show catalytic behaviour
4
Q

Why do transition metals have variable oxidation states

A

It is largely because 4s and 3d sub-shells are so close in energy meaning that a 3d electron can be lost as well as 4s and still form a stable ion

5
Q

Why do transition metals form coloured ions

A

Most coloured inorganic compounds contain a transition element and the colour is related to partially filled d-orbitals.

6
Q

Why do transition metals show catalytic behaviour

A

They provide an alternate reaction pathway with a lower activation energy by either providing a surface for the reaction to take place (reactants adsorb to the surface) or use their variable oxidation states to bind to the reactants and for intermediates as part of an alternate pathway.

7
Q

What is a precipitation reaction

A

A precipitation reaction is one in which soluble ions, in separate solutions, are mixed together to produce an insoluble compound, which settles out of the solution as a solid. This insoluble compound is called the precipitate.

8
Q

Write the equation for the precipitation of Cu2+ with NaOH and to give the colour change

A

Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

Pale blue solution → Pale blue precipitate

9
Q

Write the equation for the precipitation of Co2+ with NaOH and to give the colour change

A

Co2+(aq) + 2OH-(aq) → Co(OH)2(s)

Pink solution → Blue precipitate (turns beige in air)

10
Q

Write the equation for the precipitation of Fe2+ with NaOH and to give the colour change

A

Fe2+(aq) + 2OH-(aq) → Fe(OH)2(s)

Pale green solution → Green precipitate (turns brown on contact with air as Fe2+ ions oxidise to Fe3+)

11
Q

Write the equation for the precipitation of Fe3+ with NaOH and to give the colour change

A

Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)

Pale yellow solution → Brown precipitate

12
Q

Define a complex ion

A

A transition metal ion bonded to one or more ligands by coordinate bonds (dative covalent bonds)

13
Q

Define a Ligand

A

A molecule or ion and that can donate a pair of electrons to a transition metal ion to form a coordinate bond.

14
Q

Define coordination number

A

The total number of coordinate bonds formed between a central metal ion in and it’s ligands.

15
Q

What do you call a ligand that donates one pair of electrons to form one coordinate bond

A

Monodentate

16
Q

What do you call a ligand that donates 2 pairs of electrons to form 2 coordinate bonds

A

Bidentate

The most common example is ethanediamine (‘en’; or NH2CH2CH2NH2).

17
Q

What do you call a ligand that donates many pairs of electrons to form many coordinate bonds

A

Multi-dentate

18
Q

What do you call the structures of a complex ion with 4 ligands

A

tetrahedral, 109.5°

square planar, 90°

19
Q

What do you call the structure of a complex ion with 6 ligands

A

octahedral, 90°

e.g. hexaaquacobalt(II) [Co(H2O)6]2+

20
Q

What is a counterion

A

An ion that accompanies a ionic species (complex ion) to maintain a neutral charge.
e.g. for K3[Fe(CN)6] the counter ions are 3K+, the complex ion is [Fe(CN)6]3- with 6CN- ligands and a central Fe3+ ion (compound name: potassium hexacyanoiron(III)

21
Q

Defined a stereoisomer

A

Species with the same structural formula but with different arrangements of the atoms in space

22
Q

What are the two types of stereoisomerism in transition element chemistry

A

Cis-Trans isomerism and optical isomerism

23
Q

Describe a stereoisomer that has a specific use

A

Square planar Pt(NH3)2Cl2 complex, cis- platin is an anticancer drug that works by binding to DNA in cancer cells, preventing cell division.

24
Q

What three situations can cause Cis-Trans isomerism in complexes

A
  1. Octahedral- 4 ligands of one type, 2 of another
  2. Square planar- 2 or each type
  3. Octahedral- 2 Bidentate ligands and 2 Monodentate ligands
25
Q

What are optical isomers

A

Non-superimposable mirror images

26
Q

What are the three possible requirements for an optical isomer for complexes

A

(All octahedral)

  1. 3 bidentate ligands
  2. 2 bidentate ligands and 2 Monodentate ligands
  3. 1 hexadentate ligand
27
Q

What is a ligand substitution reaction

A

When one ligand in a complex ion is replaced by another ligand

28
Q

Give the equation for the ligand substitution reaction between aqueous copper (II) ions and ammonia and give the colour change and explain the shape of the product.

A

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

Pale blue solution -> Dark blue solution

(Octahedral product as ammonia and water molecules are of a similar size. The two water molecules are located above and below the Cu2+, with the four ammonia molecules in the same plane (horizontal). NB Ammonia is a stronger ligand than water because it can donate its lone pair more easily (since nitrogen is less electronegative than oxygen))

29
Q

Give the equation for the ligand substitution reaction between aqueous copper (II) ions and hydrochloric acid and give the colour change and explain the shape of the product

A

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

Pale blue solution -> Yellow solution

(Tetrahedral product as chloride ions are larger than water ions so fewer of them can fit around the central metal ion)

30
Q

Give the equation for the ligand substitution reaction between aqueous cobalt (II) ions and concentrated hydrochloric acid and give the colour change and explain the shape of the product

A

[Co(H2O)6]2+(aq) + 4Cl- ⇌ [CoCl4]2-(aq) + 6H2O

Pink solution -> Blue solution

(Tetrahedral product as chloride ions are larger than water ions so fewer of them can fit around the central metal ion)

31
Q

Describe what haemoglobin is and how it transports oxygen

A

It is a complex protein found in red blood cells that transports oxygen. Each protein chain contains four non-protein components called haem groups. Each haem group has an Fe2+ ion at its centre which can bind with oxygen. Each Fe2+ ion has coordinate bonds between four nitrogen atoms in the haem structure, one to the protein globin and one to the oxygen molecule.

32
Q

Explain how and why carbon monoxide is so dangerous for haemoglobin

A

In the presence of carbon monoxide, less oxygen is transported in the blood because the stability constant for the haem complex with carbon monoxide bonded is greater than for that with oxygen (the co-ordinate bond is stronger with CO than O2). This is an example of ligand substitution, with oxygen replaced (irreversibly) by carbon monoxide, starving vital organs.

33
Q

Define the stability constant

A

Exam definition: The equilibrium constant for the formation of the complex ion in a solvent from its constituent ions

Can also be thought as: The equilibrium constant for an equilibrium between a transition metal ion surrounded by water ligands and the complex formed when the same ion has undergone a ligand substitution

34
Q

What is the only difference when writing Kstab equations compared to writing Kc equations

A

Water is left out of the expression (all the species are dissolved in water, which is in large excess so that its concentration is virtually constant).

35
Q

How many molecules of oxygen can one molecule of haemoglobin carry

A

Four, one in each of the haem groups

36
Q

What does a higher Kstab value tell you

A

Kstab values are usually referenced against the complex with water ligands. A large value of Kstab indicates that the equilibrium lies over to the right. Complex ions with high stability constants are more stable than those with lower values and are more easily formed.

37
Q

Why do bidentate and multidentate ligands form stable complex ions

A

Bidentate and multidentate ligands form particularly stable complex ions, because their formation results in an increase in the number of moles in the system and therefore an increase in entropy

38
Q

How can you use Kstab values explain ligand substitutions

A

Ligand substitution reactions can be explained by comparing values of Kstab, where a stronger ligand (higher Kstab) is able to replace a weaker ligand. However, don’t forget that ligand substitution reactions are equilibria systems and so, according to Le Chatelier’s principle, adding a higher concentration of a weaker ligand (e.g. adding water to excess) would shift the equilibrium in the reverse direction. Changes in temperature would also cause shifts e.g. increase temperature shifts in the endothermic direction.

39
Q

Give the colour change at the end point of MnO4-

A

MnO4- -> Mn2+

Purple Colourless

40
Q

What acid cannot be used in titrations with MnO4-and why

A

Hydrochloric acid cannot be used because it reacts with MnO4- producing chlorine gas. Therefore the manganate(VII) solution must be acidified using sulphuric acid

41
Q

It’s the colour change at the end point of the titration when the manganate (VII) solution is in the burette and the acid/metal (Fe2+) solution in the conical flask.

A

Colourless to purple if the manganate is in the burette (once all of the Fe2+ ions in the conical flask have been oxidised, one more drop of manganate added from the burette gives the first permanent pink/purple colour in the flask because MnO4- is now in excess).

42
Q

Give the overall equation for the titration between MnO4- and Fe2+

A

MnO4- + 8H+ + 5Fe2+ → Mn2+ + 5Fe3+ + 4H2O

43
Q

Give the overall equation for the titration between I2 and (S2O3)2-

A

2(S2O3)2- + I2→ (S4O6)2- + 2I-

44
Q

What do you call (S2O3)2-

A

Thiosulfate ion

45
Q

What do you call (S4O6)2-

A

Tetrathionate ion

46
Q

What is the colour change at the end of the iodine thiosulfate titration

A

Colour change at the end point: the brown iodine solution becomes a pale straw colour, at which point starch is added. Adding starch gives a blue/black colour in the presence of iodine, which disappears sharply at the end point to give a colourless solution (this is much more obvious than pale yellow to colourless seen for iodine in the absence of starch).

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