Reactions of Inorganic Compounds in Aqueous Solution

Question 1

What is lewis theory?

Answer 1

Lewis theory is the description of acids and bases in terms of whether they accept or donate lone pairs. It is generally applied in the formation of coordinate covalent bonds.

Question 2

What is a lewis acid?

Answer 2

An electron pair acceptor

Question 3

What is a lewis base?

Answer 3

An electron pair donor

Question 4

How do metal ions exist in aqueous solution free from other ions?

Answer 4

When in aqueous solution, without the presence of other ions, metal ions exist as ‘metal aqua ions’ – they have a central metal ion with six water ligands.
-Metal ions act as Lewis acids in aqueous solutions, as they accept electron pairs from the surrounding water molecules.
[M(H2O)6]2+ M=Fe, Co or Cu
[M(H2O)6]3+ M=Al, Cr or Fe

Question 5

What type of reactions do metal aqua ions undergo?

Answer 5

• Hydrolysis (loss of H+ from H2O ligand)
• Substitution (replacement of H2O by other ligands)
• Redox (metal changes in oxidation state)

Question 6

What is the hydrolysis of metal aqua ions?

Answer 6

When in a solution, there’s a reaction between the metal aqua ion and the water – this is a hydrolysis or acidity reaction. An H+ ion is lost.
-The easier the H+ ion is lost, the more acidic the metal aqua ion solution is.

Question 7

What happens when [M(H2O)6]2+ is hydrolysed?

Answer 7

[M(H2O)6]2+ will dissociate in water to form [M(H2O)5(OH)]+

• [M(H2O)6]2+(aq) + H2O(l) > [M(H2O)5(OH)]+(aq) + H3O+(aq)
• [M(H2O)6]2+(aq) > [M(H2O)5(OH)]+(aq) + H+

Question 8

What happens when [M(H2O)6]3+ is hydrolysed?

Answer 8

[M(H2O)6]3+ will dissociate in water to form [M(H2O)5(OH)]2+

• [M(H2O)6]3+(aq) + H2O(l) > [M(H2O)5(OH)]2+(aq) + H3O+(aq)
• [M(H2O)6]3+(aq) > [M(H2O)5(OH)]2+(aq) + H+

Question 9

Why are solutions of M3+ more acidic than M2+?

Answer 9

The M3+ ions is both smaller and more highly charged, meaning it has a high charge density, making it more highly polarising.
-In the [M(H2O)6]3+, the M3+ strongly attracts electrons from the oxygen atom of the water ligands, thus weakening the O-H bonds in the water ligands. This complex will then more readily release an H+, making the solution more acidic.

Question 10

What is further hydrolysis?

Answer 10

It is possible for further hydrolysis to occur by additional water molecules.

• When the metal complex ion is no longer charged, the complex ion will be a solid precipitate.
• Further hydrolysis can continue until the water molecules are completed changed to OH.

Question 11

What happens if a base is added to a metal aqua ion?

Answer 11

If a base is added to a metal aqua ion, hydrolysis will take place.
-The base removes a H+, shifting the equilibrium to the right.

Question 12

If NaOH is added, how does [M(H2O)6]2+ react?

Answer 12

[M(H2O)6]2+(aq) + 2OH-(aq)  M(H2O)4(OH)2 + 2H2O(l)
M2+(aq) + 2OH-(aq)  M(OH)2(s)
Fe: pale green sol  green gelatinous ppt
Co: pink sol  blue ppt
Cu: pale blue sol  pale blue ppt

Question 13

If NaOH is added, how does [M(H2O)6]3+ react?

Answer 13

[M(H2O)6]3+(aq) + 3OH-(aq)  M(H2O)3(OH)3 + 3H2O(l)
M3+(aq) + 3OH-(aq)  M(OH)3(s)
Fe: pale violet sol  brown gelatinous ppt
Al: Colourless sol  white ppt
Cr: violet sol green ppt

Question 14

If excess NaOH is added, how does [M(H2O)6]2+ react?

Answer 14

-None of the M2+ ions react further when excess NaOH is added.
Fe: remains green gelatinous ppt
Co: remains blue ppt
Cu: remains pale blue ppt

Question 15

If excess NaOH is added, how does [M(H2O)6]3+ react?

Answer 15

-Fe3+ does not react further, remains brown gelatinous ppt

Al and Cr both do react further.

Question 16

If excess NaOH is added to [Al(H2O)3(OH)3], how does it react?

Answer 16

Al(H2O)3(OH)3 + OH-(aq)  [Al(H2O)2(OH)4]-(aq) + H2O(l)
Al(OH)3(s) + OH-(aq)  Al(OH)4(aq)
-White ppt colourless sol

Question 17

If excess NaOH is added to [Cr(H2O)3(OH)3], how does it react?

Answer 17

Cr(H2O)3(OH)3 + 3OH-(aq)  [Cr(OH)6]3-(aq) + 3H2O(l)
Cr(OH)3(s) + 3OH-(aq)  Cr(OH)6(aq)
-green ppt green sol

Question 18

How can the addition of OH- be explained using equilibrium?

Answer 18

In water, metal-aqua 3+ ions form the equilibrium:
-M(H2O)6 + H2O(l) > [M(H2O)5(OH)]2+(aq) + H3O+(aq).
If you add OH- ions, H3O+ ions are removed, this shifts the equilibrium to the right. Another equilibrium is set up:
-[M(H2O)5(OH)]2+(aq) + H2O(l) > [M(H2O)4(OH)2]+(aq) + H3O+(aq).
When this continues, you are left with the insole uncharged metal hydroxide.

Question 19

If Na2CO3 is added, how does [M(H2O)6]2+ react?

Answer 19

Reacts to form precipitates of metal (II) carbonates in a precipitation reaction.
[M(H2O)6]2+(aq) + CO32-(aq)  MCO3(s) + 6H2O(l)
M2+(aq) +CO32-(aq) MCO3(s)
Fe: pale green sol  green ppt
Co: pink sol  pink ppt
Cu: pale blue sol  blue-green ppt

Question 20

If Na2CO3 is added, how does [M(H2O)6]3+ react?

Answer 20

Reacts to form precipitates of metal (III) hydroxides in a precipitation reaction.
2[M(H2O)6]3+(aq) + 3CO32-(aq)  2M(H2O)3(OH)3 + 3H2O(l) + 3CO2(g)
M3+(aq) + 3CO32-(aq)  M(OH)3(s) + 3H2O(l) + 3CO2(g)
Fe: pale violet sol  brown gelatinous ppt + bubbles of CO2
Al: colourless sol  white ppt + bubbles of CO2
Cr: violet sol  green ppt + bubbles of CO2

Question 21

What happens when ammonia is added to metal aqua ions?

Answer 21

When ammonia dissolves in water, it can accept protons from the water molecules, to form NH4+ ions and OH- ligands. This gives the same result as adding NaOH.

Question 22

If NH3 is added, how does [M(H2O)6]2+ react?

Answer 22

[M(H2O)6]2+(aq) + 2NH3(aq)  M(H2O)4(OH)2 + 2NH4+(aq)
M2+(aq) + 2OH-(aq)  M(OH)2(s)
Fe: pale green sol  green gelatinous ppt
Co: pink sol  blue ppt
Cu: pale blue sol  pale blue ppt

Question 23

If NH3 is added, how does [M(H2O)6]3+ react?

Answer 23

[M(H2O)6]3+(aq) + 3NH3(aq)  M(H2O)3(OH)3 + 3NH4+(aq)
M3+(aq) + 3OH-(aq)  M(OH)3(s)
Fe: pale violet sol  brown gelatinous ppt
Al: Colourless sol  white ppt
Cr: violet sol green ppt

Question 24

What are ligand substitution reactions?

Answer 24

What are ligand substitution reactions? The water molecules that act as ligands in metal aqua ions can be replaced by other ligands – either because the other ligands form stronger coordinate bonds (are better Lewis bases) or because they are present in higher concentration and an equilibrium is displaced.
-Replacement may be complete or partial.

Question 25

What are the different types of ligand substitution reactions?

Answer 25

• Replacement by other neutral ligands, e.g. NH3.
• Replacement by negatively charged ligands, e.g. Cl-
• Replacement by bi- or multidentate ligands, e.g. EDTA. This is known as chelation.

Question 26

What happens in a substitution reaction with excess NH3?

Answer 26

The N atom of the NH3 ligand is a similar size to the O of the H2O ligand. When ligands of similar size are substituted, the co-ordination number stays the same.

• When added in excess, NH3 ligands will displace the H2O and OH- ligands. This is because ammonia is a better ligand than water and is in a higher concentration, displacing the equilibrium to the right.
• This happens with Cu2+, Co2+ and Cr3+.

Question 27

Why is ammonia a better ligand than water?

Answer 27

Ammonia is a better ligand than water because the lone pair on the nitrogen atom is less strongly held than that on the more electronegative oxygen atom. It is therefore more readily donated.

Question 28

If excess NH3 is added to [Cu(H2O)6]2+, how does it react?

Answer 28

[Cu(H2O)6]2+(aq) + 4NH3(aq) > [Cu(NH3)4(H2O)2]2+(aq) + 4H2O(l)
Pale blue sol  pale blue ppt deep blue sol
In this instance, ligand substitution is only partial.
The ammonia first acts as a base, removing protons from two of the water molecules to form Cu(OH)2(H2O)4. As we add more of the concentrated ammonia, the pale blue precipitate dissolved to form a deep blue solutions of [Cu(NH3)4(H2O)2]2+.

Question 29

If excess NH3 is added to [Co(H2O)6]2+, how does it react?

Answer 29

[Co(H2O)6]2+(aq) + 6NH3(aq) > [Co(NH3)6]2+(aq) + 6H2O(l)
Pink sol  blue ppt pale yellow sol –[O]-> brown sol
Cobalt first form a blue precipitate of hydrated cobalt (II) hydroxide, produced by the loss of a proton from each of two of the six water ligands. Here NH3 is acting as a base.
-[Co(H2O)6]2+(aq) + 2NH3(aq)  Co(H2O)4(OH)2 + 2NH4+(aq)
If we add more of the concentrated ammonia, all ligands are replaced by NH3. The blue precipitate dissolves to form a pale yellow solution (oxidised by oxygen in the air to a brown mixture containing Co(II).
-Co(H2O)4(OH)2 + 6NH3(aq) > [Co(NH3)6]2+(aq) + 4H2O(l) + 2OH-(aq).

Question 30

If excess NH3 is added to [Cr(H2O)6]3+, how does it react?

Answer 30

If excess NH3 is added to [Cr(H2O)6]3+, how does it react? [Cr(H2O)6]3+(aq) + 6NH3(aq) > [Cr(NH3)6]3+(aq) + 6H2O(l)
Violet sol  green ppt purple sol

Question 31

If excess NHS is added to [Fe(H2O)6]2+ or [Fe(H2O)6]3+, how will they react?

Answer 31

Both of these reactions are acid-base reactions only. We will form the green and orange/brown precipitates only. These are insoluble in excess, so there are not ligand substitution reactions.

Question 32

If excess NH3 is added to [Al(H2O)6]3+, how does it react?

Answer 32

The white precipitate of [Al(H2O)3(OH)3] is insoluble in excess so this is not a ligand substitution reaction.

Question 33

What happens in a substitution reaction with HCl?

Answer 33

When H2O is replaced by a larger ligand, less ligands fit around the metal ion, this also brings about a change in coordination number.

• When H2O ligands are replaced by larger Cl- ligands, only four Cl- ligands can fit.
• Cu2+ and CO2+ both react with HCl.

Question 34

If HCl is added to [Cu(H2O)6]2+, how does it react?

Answer 34

[Cu(H2O)6]2+(aq) + 4Cl-(aq) > [CuCl4]2-(aq) + 6H2O(l)
Pale blue sol yellow sol
(The yellow solution is likely to appear green as some blue [Cu(H2O)6]2+ will remain).
Coord no = 6  4 (tetrahedral shape)

Question 35

If HCl is added to [Co(H2O)6]2+, how does it react?

Answer 35

[Co(H2O)6]2+(aq) + 4Cl-(aq) > [CoCl4]2-(aq) + 6H2O(l)
Pink sol  blue sol
Coord no = 6  4 (tetrahedral shape)

Question 36

What is chelation?

Answer 36

The formation of complexes with multidentate ligands.

Question 37

What happens in a substitution reaction with multidentate ligands?

Answer 37

Multidentate ligands are able to displace unidentate ligands, e.g. ethylene diamine (en) and EDTA.
[Cu(H2O)6]2+(aq) + 3en(aq)  [Cu(en)3]2+(aq) + 6H2O(l)
[Cu(H2O)6]2+(aq) + EDTA4-(aq)  [Cu(EDTA)]2-(aq) + 6H2O(l)

Question 38

Why do multidentate ligands form more stable complexes than unidentate ligands?

Answer 38

This can be explained using ∆G=∆H-T∆S.
∆H: The enthalpy change for a ligand substitution reaction is usually very small so ∆H≈0Jmol-1.
∆S: The number of particles always increases during a reaction with multidentate ligands, so ∆S will be positive.
∆G: When unidentate ligands are substituted for multidentate ligands, ∆G will be negative (0-T(+ve)). This means the reaction is always feasible, forming a more stable ion.

Question 39

What does amphoteric mean?

Answer 39

Amphoteric means showing both acidic and basic properties.

-Aluminium hydroxide and chromium hydroxide are amphoteric.

Question 40

How is aluminium hydroxide amphoteric?

Answer 40

We would expect a metal hydroxide to act as a base and react with an acid:
-Al(H2O)3(OH)3 + 3HCl(aq)  [Al(H2O)6]3+(aq) + 3Cl-(aq)
Aluminium hydroxide also shows acidic properties by reacting with sodium hydroxide to give a colourless solution of tetrahydroxoaluminate
-Al(H2O)3(OH)3 + OH-(aq)  [Al(OH)4]-(aq) + 3H2O(l)

Question 41

How is chromium hydroxide amphoteric?

Answer 41

• Cr(H2O)3(OH)3 + 3H3O+(aq)  [Cr(H2O)6]3+(aq) + 3H2O(l)
• Cr(H2O)3(OH)3 + 3OH-(aq)  [Cr(OH)6]3-(aq) + 3H2O(l)

Question 42

In high oxidation states, in what form do some metals exist?

Answer 42

• Cr(H2O)3(OH)3 + 3H3O+(aq)  [Cr(H2O)6]3+(aq) + 3H2O(l)
• Cr(H2O)3(OH)3 + 3OH-(aq)  [Cr(OH)6]3-(aq) + 3H2O(l)

Question 43

In high oxidation states, in what form do some metals exist?

Answer 43

In most transition metal compounds, the metal exists in the cationic form, but in high oxidation states, some metals exist in anionic form, e.g. MnO4-, CrO4 2- and Cr2O7 2-.
-The dichromate ion has an oxygen atom that forms a bridge between two chromiums. The orange Cr2O7 2-ion is stable in acid solution and the yellow CrO4 2- in alkali.
2CrO42-(aq) + 2H+(aq) > Cr2O72-(aq) + H2O(l)
Yellow > orange.