Reactions of Inorganic Compounds in Aqueous Solution

Question 1

Explain why aqueous solution for M3+ is more acidic than M2+

Answer 1

• higher charge density
• more strongly polarising so attracts electrons from oxygen on water ligand
• weakens OH bonds
• increases tendency of complex ion to release H+

Question 2

Give an equation to show hydrolysis of [Fe(H2O)6]3+

Answer 2

[Fe(H2O)6]3+ -> [Fe(H2O)5OH]2+ + H+ OR

[Fe(H2O)6]3+ + H2O -> [Fe(H2O)5OH]2+ + H3O+

Question 3

Lewis Acid

Answer 3

electron pair acceptor

Question 4

Lewis Base

Answer 4

electron pair donor

Question 5

Colour of [Fe(H2O)6]2+

Answer 5

pale green solution

Question 6

Colour of [Fe(H2O)6]3+

Answer 6

pale purple solution

Question 7

Explain why [Fe(H2O)6]3+ can appear yellow/brown

Answer 7

• hydrolysis of [Fe(H2O)6]3+

- presence of [Fe(H2O)5OH]2+

Question 8

Describe how to distinguish between Fe2+ (aq) and Fe3+ (aq) since they are both pale in colour

Answer 8

• addition of base (NaOH)

- form coloured precipitates which are easily distinguishable

Question 9

Give equation to show addition of base to M2+ (aq) to form metal hydroxide precipitate

Answer 9

[M(H2O)6]2+ + 2OH- -> [M(H2O)4(OH)2] + 2H2O

Question 10

Give equation to show addition of base to M3+ (aq) to form metal hydroxide precipitate

Answer 10

[M(H2O)6]3+ + 3OH- -> [M(H2O)3(OH)3] + 3H2O

Question 11

Give colour of [Fe(H2O)4(OH)2] and explain why it turns brown on standing

Answer 11

• green ppt

- some Fe2+ (aq) oxidises to Fe3+ (aq) which forms brown ppt of [Fe(H2O)3(OH)3]

Question 12

Give equation to show acid-base reaction between M2+ (aq) and NH3 at low concentrations

Answer 12

[M(H2O)6]2+ + 2NH3 -> [M(H2O)4(OH)2] + 2NH4+

Question 13

Give equation to show acid-base reaction between M3+ (aq) and NH3 at low concentrations

Answer 13

[M(H2O)6]3+ + 3NH3 -> [M(H2O)3(OH)3] + 3NH4+

Question 14

Explain how acid-base reactions with CO32- can demonstrate relative acidity of M2+ (aq) and M3+ (aq)

Answer 14

• M2+ produces metal carbonate only

- M3+ is acidic enough to produce CO2 gas from CO32-

Question 15

Explain why carbonates of transition metals have an oxidation state of +2 but not +3

Answer 15

• metal with oxidation state of +3 is acidic enough to liberate CO2 from CO32- so will not form carbonate with M3+
• M2+ will form metal carbonate since it is less acidic

Question 16

Give the intermediate stage and overall equation for reaction between M3+ (aq) and CO32-

Answer 16

[M(H2O)6]3+ + 3CO32- -> [M(H2O)3(OH)3] + 3HCO3-

2[M(H2O)6]3+ + 3CO32- -> 2[M(H2O)3(OH)3] + 3CO2 + 3H2O

Question 17

Give equation to show reaction between M2+ (aq) and CO32-

Answer 17

[M(H2O)6]2+ + CO32- -> MCO3 + 6H2O

Question 18

Colour of FeCO3 precipitate

Answer 18

green ppt

Question 19

Colour of [Al(H2O)6]3+

Answer 19

colourless solution

Question 20

Give equation to show addition of base to Al3+ (aq) and observations

Answer 20

[Al(H2O)6]3+ + 3OH- -> [Al(H2O)3(OH)3] + 3H2O

white ppt of aluminium oxide formed

Question 21

Give equations to show how [Al(H2O)3(OH)3] behaves as an amphoteric hydroxide

Answer 21

[Al(H2O)3(OH)3] + 3H+ -> [Al(H2O)6]3+

[Al(H2O)3(OH)3] + OH- -> [Al(OH)4]- + 3H2O

Question 22

Give equation to show reaction between Al3+ (aq) and NH3 in low concentrations

Answer 22

[Al(H2O)6]3+ + 3NH3 -> [Al(H2O)3(OH)3] + 3NH4+

Question 23

Explain why addition of base to Al3+ (aq) results in formation of white aluminium hydroxide precipitate but colourless solution in excess

Answer 23

• formation of [Al(H2O)3(OH)3] to begin with which is white precipitate
• [Al(H2O)3(OH)3] reacts with base to form [Al(OH)4]- which is colourless solution

Question 24

Explain why ligand substitution of H2O with NH3 occurs in transition metal complexes and why it occurs without a change in charge or oxidation number

Answer 24

• NH3 is stronger ligand than H2O
• N is more electronegative than O so more likely to donate electron pair to central metal ion
• NH3 is same size and charge to H2O

Question 25

Give equation to show ligand substitution of H2O with NH3 in transition metal complex

Answer 25

[M(H2O)6]2+ + 6NH3 -> [M(NH3)6]2+ + 6H2O

Question 26

Give equations for two stage reaction of concentrated ammonia with M2+ (aq) and explain why this occurs

Answer 26

[M(H2O)6]2+ + 2OH- -> [M(H2O)4(OH)2] + 2H2O [M(H2O)4(OH)2] + 6NH3 -> [M(NH3)6]2+ + 4H2O + 2OH- - NH3- acts as a base so M2+ (aq) reacts with OH- from OH- in NH3 (aq) solution (NH3 + H2O -> NH4+ + OH-) - excess NH3 acts as ligand and substitutes H2O

Question 27

Give equation to show (partial) ligand substitution of H2O with NH3 in Cu2+ (aq)

Answer 27

[Cu(H2O)6]2+ + 4NH3 -> [Cu(NH3)4(H2O)2]2+ + 4H2O

Question 28

Colour of [Cu(H2O)6]2+

Answer 28

pale blue solution

Question 29

Use equations to explain why Cu2+ forms a pale blue precipitate before a deep blue solution on addition of excess NH3

Answer 29

- NH3 acts as base initially [Cu(H2O)6]2+ + 2OH- -> [Cu(H2O)4(OH)2] + 2H2O - NH3 acts ligand and partially substitutes H2O and OH- in excess [Cu(H2O)4(OH)2] + 4NH3 -> [Cu(NH3)4(H2O)2]2+ + 2H2O + 2OH-

Question 30

Describe structure of complex formed when Cu2+ (aq) reacts with excess NH3

Answer 30

- distorted octahedral with 4 NH3 and 2 H2O ligands - NH3 form square planar arrangement - OH above and below plane - Cu-O bond is longer than Cu-N bond since H2O weaker ligand

Question 31

Colour of CuCO3

Answer 31

blue-green ppt

Question 32

Give equation and observations for reaction between Cu2+ (aq) and concentrated HCl

Answer 32

[Cu(H2O)6]2+ + 4Cl- -> [CuCl4]2- + 6H2O | - pale blue solution changes colour to yellow-green solution

Question 33

Explain why there is a change in charge and coordination number for complex formed from reaction between Cu2+ (aq) and concentrated HCl

Answer 33

- Cl- ligand has -1 charge where H2O ligand is neutral - Cl- ligand is larger than H2O so fewer can fit around Cu2+ ion - only four coordinate bonds form rather than six

Question 34

Explain why when NH4VO3 reacts with zinc under acidic conditions only two colour changes are observed

Answer 34

yellow>green>purple - VO2^+ (yellow) > VO2+ (blue) which appears green - remains green when reduced to V3+ - then to purple when reduced to V2+