5.3- Transition Elements

Question 0

Define a transition element

Answer 0

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

Question 1

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

Answer 1

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

Question 2

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

Answer 2

Chromium (4s1,3d5)

Copper (4s1,3d10)

Question 3

What are the three main transition metal properties

Answer 3

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

Question 4

Why do transition metals have variable oxidation states

Answer 4

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

Question 5

Why do transition metals form coloured ions

Answer 5

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

Question 6

Why do transition metals show catalytic behaviour

Answer 6

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.

Question 7

What is a precipitation reaction

Answer 7

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.

Question 8

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

Answer 8

Cu2+(aq) + 2OH-(aq) → Cu(OH)2(s)

Pale blue solution → Pale blue precipitate

Question 9

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

Answer 9

Co2+(aq) + 2OH-(aq) → Co(OH)2(s)

Pink solution → Blue precipitate (turns beige in air)

Question 10

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

Answer 10

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+)

Question 11

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

Answer 11

Fe3+(aq) + 3OH-(aq) → Fe(OH)3(s)

Pale yellow solution → Brown precipitate

Question 12

Define a complex ion

Answer 12

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

Question 13

Define a Ligand

Answer 13

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

Question 14

Define coordination number

Answer 14

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

Question 15

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

Answer 15

Monodentate

Question 16

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

Answer 16

Bidentate

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

Question 17

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

Answer 17

Multi-dentate

Question 18

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

Answer 18

tetrahedral, 109.5°

square planar, 90°

Question 19

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

Answer 19

octahedral, 90°

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

Question 20

What is a counterion

Answer 20

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)

Question 21

Defined a stereoisomer

Answer 21

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

Question 22

What are the two types of stereoisomerism in transition element chemistry

Answer 22

Cis-Trans isomerism and optical isomerism

Question 23

Describe a stereoisomer that has a specific use

Answer 23

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

Question 24

What three situations can cause Cis-Trans isomerism in complexes

Answer 24

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

Question 25

What are optical isomers

Answer 25

Non-superimposable mirror images

Question 26

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

Answer 26

(All octahedral)

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

Question 27

What is a ligand substitution reaction

Answer 27

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

Question 28

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.

Answer 28

[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))

Question 29

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

Answer 29

[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)

Question 30

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

Answer 30

[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)

Question 31

Describe what haemoglobin is and how it transports oxygen

Answer 31

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.

Question 32

Explain how and why carbon monoxide is so dangerous for haemoglobin

Answer 32

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.

Question 33

Define the stability constant

Answer 33

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

Question 34

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

Answer 34

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).

Question 35

How many molecules of oxygen can one molecule of haemoglobin carry

Answer 35

Four, one in each of the haem groups

Question 36

What does a higher Kstab value tell you

Answer 36

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.

Question 37

Why do bidentate and multidentate ligands form stable complex ions

Answer 37

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

Question 38

How can you use Kstab values explain ligand substitutions

Answer 38

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.

Question 39

Give the colour change at the end point of MnO4-

Answer 39

MnO4- -> Mn2+

Purple Colourless

Question 40

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

Answer 40

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

Question 41

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.

Answer 41

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).

Question 42

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

Answer 42

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

Question 43

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

Answer 43

2(S2O3)2- + I2→ (S4O6)2- + 2I-

Question 44

What do you call (S2O3)2-

Answer 44

Thiosulfate ion

Question 45

What do you call (S4O6)2-

Answer 45

Tetrathionate ion

Question 46

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

Answer 46

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).