24.4 Ligand Substitution Of Copper (II) Ions

Question 1

Reaction with water Cu

Answer 1

1. PALE BLUE complex forms when copper (II) sulfate is dissolved in water.

CuSo4 + 6H2O -> [Cu(H2O)6]2+ So42-

Question 2

Reaction with excess of aqueous ammonia Cu

Answer 2

1. In first stage of reaction, PALE BLUE SOLUTION -> PALE BLUE PRECIPITATE of Cu(OH)2 is formed.

Cu2+(aq) + OH-(aq) -> Cu(OH)2(s)

2. Cu(OH)2 precipitate then dissolves in excess ammonia to form a DARK BLUE SOLUTION.

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

Question 3

Reaction with chloride ions

Answer 3

1. Addition of excess CONCENTRATED hydrochloride acid changes the solution colour from PALE BLUE to YELLOW.

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

2. GREEN solution forms as a result of the colours yellow and blue mixing.
3. If WATER is added to the YELLOW solution, it will be a more DILUTE and PALER BLUE than the original blue solution.

Question 4

Why is there a change in coordination number from [Cu(H2O)6]2+(aq) to [CuCl4]2-(aq)?

Answer 4

Chloride ligands are larger in size than water molecules, so fewer fit around the central Cu2+ ion.