Required Practical 11 – Testing Transition Metal Ions

Question 1

What precipitate does Cu²⁺ form with NaOH and with excess NH3?

Answer 1

With NaOH: blue precipitate of Cu(OH)2(H2O)4. With excess NH3: precipitate dissolves to give deep blue solution of [Cu(NH3)4(H2O)2]²⁺.

Question 2

What precipitate does Fe²⁺ form with NaOH and with excess NH3?

Answer 2

With NaOH: green precipitate of Fe(OH)2(H2O)4. With excess NH3: no further change — precipitate does not dissolve.

Question 3

What precipitate does Fe³⁺ form with NaOH?

Answer 3

Brown precipitate of Fe(OH)3(H2O)3. Does not dissolve in excess NaOH or excess NH3.

Question 4

What precipitate does Al³⁺ form with NaOH, and what happens in excess?

Answer 4

White precipitate of Al(OH)3. Dissolves in excess NaOH to form colourless [Al(OH)4]⁻. Al(OH)3 is amphoteric — dissolves in both acid and base.

Question 5

What is observed when Cu²⁺, Fe²⁺, and Fe³⁺ react with Na2CO3 solution?

Answer 5

Cu²⁺ → blue/green CuCO3 precipitate (precipitation reaction). Fe²⁺ → green FeCO3 precipitate. Fe³⁺ → brown Fe(OH)3 precipitate + CO2 gas evolved (acidity reaction).

Question 6

Why do 3+ ions react differently with CO3²⁻ compared to 2+ ions?

Answer 6

The higher charge density of 3+ ions gives greater polarising power, causing M2(CO3)3 to be unstable. Instead, M(OH)3 precipitate forms with evolution of CO2.

Question 7

Give the ionic equation for Cu²⁺ with limited NH3.

Answer 7

[Cu(H2O)6]²⁺(aq) + 2NH3(aq) → Cu(H2O)4(OH)2(s) + 2NH4⁺(aq). NH3 acts as a Bronsted-Lowry base.

Question 8

Give the ionic equation for Cu²⁺ with excess NH3.

Answer 8

Cu(OH)2(H2O)4(s) + 4NH3(aq) → [Cu(NH3)4(H2O)2]²⁺(aq) + 2H2O(l) + 2OH⁻(aq). NH3 acts as a Lewis base/ligand.