Reactions of Transition Metal Elements (Edexcel A Level Chemistry): Flashcards

True.

The interconversion between chromate(VI) and dichromate(VI) is an acid-base reaction, not a redox reaction. No change in chromium oxidation state occurs — it remains +6 in both ions.

Adding acid to chromate(VI) (yellow) shifts the equilibrium towards dichromate(VI) (orange):

2CrO₄^2- (aq) + 2H⁺ (aq) ⇌ Cr₂O₇^2- (aq) + H₂O (l)

The colour changes from yellow to orange. Adding alkali reverses this.

Cr(OH)₃ can be oxidised to CrO₄^2- using H₂O₂ as the oxidising agent under alkaline conditions.

The oxidation of Cr(OH)₃ to CrO₄^2- by H₂O₂ is:

2Cr(OH)₃ (aq) + 4OH^- (aq) + 3H₂O₂ (aq) → 2CrO₄^2- (aq) + 8H₂O (l)

This requires alkaline conditions and raises chromium's oxidation state from +3 to +6.

An amphoteric hydroxide is a metal hydroxide that can act as both an acid and a base.

Chromium(III) hydroxide, [Cr(H₂O)₃(OH)₃], is an example: it dissolves in excess acid to give [Cr(H₂O)₆]³⁺ and in excess alkali to give [Cr(OH)₆]^3-.

The reduction of Cr₂O₇^2- to Cr³⁺ by zinc is:

Cr₂O₇^2- (aq) + 14H⁺ (aq) + 3Zn (s) → 2Cr³⁺ (aq) + 7H₂O (l) + 3Zn²⁺ (aq)

This requires acidic conditions; chromium's oxidation state falls from +6 to +3.

True.

Chromium(III) hydroxide is amphoteric — it acts as an acid with excess OH^-, undergoing further deprotonation to form the soluble [Cr(OH)₆]^3- ion:

[Cr(H₂O)₃(OH)₃] (s) + 3OH^- (aq) → [Cr(OH)₆]^3- (aq) + 3H₂O (l)

The reduction of Cr³⁺ to Cr²⁺ by zinc is:

2Cr³⁺ (aq) + Zn (s) → 2Cr²⁺ (aq) + Zn²⁺ (aq)

This also requires acidic conditions and is a further step from the reduction of dichromate(VI).

The precipitate colours formed with NaOH(aq) are:

• Fe²⁺: green [Fe(H₂O)₄(OH)₂]

• Fe³⁺: brown [Fe(H₂O)₃(OH)₃]

• Cu²⁺: blue [Cu(H₂O)₄(OH)₂]

• Co²⁺: blue/green precipitate

• Cr³⁺: grey-green [Cr(H₂O)₃(OH)₃]

The ionic equation for [Fe(H₂O)₆]³⁺ with limited NaOH is:

[Fe(H₂O)₆]³⁺ (aq) + 3OH^- (aq) → [Fe(H₂O)₃(OH)₃] (s) + 3H₂O (l)

This is a deprotonation reaction — the OH^- removes H⁺ from water ligands, converting them to hydroxide ligands.

False.

Although two water ligands appear to be replaced by hydroxide ions, this is actually a deprotonation reaction. The OH^- ions remove H⁺ from water ligands, converting them into hydroxide ligands — no ligand substitution occurs.

When excess NH₃(aq) is added to the blue Cu²⁺ hydroxide precipitate, it dissolves to form deep blue solution containing the complex ion [Cu(NH₃)₄(H₂O)₂]²⁺.

The ionic equation for dissolution of the copper hydroxide precipitate in excess NH₃ is:

[Cu(H₂O)₄(OH)₂] (s) + 4NH₃ (aq) → [Cu(NH₃)₄(H₂O)₂]²⁺ (aq) + 2H₂O (l) + 2OH^- (aq)

This is ligand substitution — NH₃ replaces two water molecules and two hydroxide ligands.

In excess NH₃, Cu²⁺, Co²⁺ and Cr³⁺ precipitates dissolve via ligand substitution.

Fe²⁺ and Fe³⁺ precipitates do not dissolve in excess ammonia.

False.

Iron(III) hydroxide does not dissolve in excess NH₃. Only Cu²⁺, Co²⁺ and Cr³⁺ hydroxide precipitates undergo ligand substitution with excess ammonia.

When [Cu(H₂O)₆]²⁺ reacts with excess concentrated HCl:

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

The coordination number changes from 6 to 4 and the geometry changes from octahedral to tetrahedral because Cl^- is larger than H₂O.

With limited NaOH, a grey-green precipitate forms:

[Cr(H₂O)₆]³⁺ (aq) + 3OH^- (aq) → [Cr(H₂O)₃(OH)₃] (s) + 3H₂O (l)

With excess NaOH, the precipitate dissolves via further deprotonation:

[Cr(H₂O)₃(OH)₃] (s) + 3OH^- (aq) → [Cr(OH)₆]^3- (aq) + 3H₂O (l)

Ligand exchange (also called ligand substitution) is when one or more ligands in a complex ion are replaced by other ligands, forming a new complex that is more stable than the original.

If the incoming ligand is a similar size to the outgoing ligand, the coordination number and geometry are unchanged.

Chelation by EDTA is feasible because ΔG^ð is negative.

Although the enthalpy change is small, the entropy change is large and positive: many monodentate ligands are released as one multidentate EDTA^4- ligand coordinates, giving a net increase in the number of particles. This large positive ΔS makes –TΔS^ð strongly negative, ensuring ΔG^ð = ΔH^ð – TΔS^ð < 0.

False.

If the incoming ligand is a different size to the outgoing ligand, the coordination number and geometry will change. For example, replacing H₂O (small) with Cl^- (large) changes coordination number from 6 to 4 and shape from octahedral to tetrahedral.

The ligand exchange of [Co(H₂O)₆]²⁺ with Cl^- is:

[Co(H₂O)₆]²⁺ (aq) + 4Cl^- (aq) ⇌ [CoCl₄]^2- (aq) + 6H₂O (l)

The coordination number changes from 6 to 4 and the solution turns blue. Note the copper(II) ion is not involved here.

Chelate effect is the replacement of monodentate ligands in a complex ion by bidentate or multidentate ligands, forming more stable complexes.

The reaction is feasible (ΔG^ð < 0) because of the large positive entropy change resulting from a net increase in the number of particles.

The ligand exchange of [Cu(H₂O)₆]²⁺ with Cl^- is:

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

The coordination number changes from 6 to 4. The complex changes from blue to yellow, with a green intermediate mixture.

When concentrated HCl is added slowly to CuSO₄(aq), the colour changes from blue to green and finally to yellow as [Cu(H₂O)₆]²⁺ is converted to [CuCl₄]^2-.

The chelation of [Co(H₂O)₆]²⁺ by EDTA^4- is:

[Co(H₂O)₆]²⁺ (aq) + EDTA^4- (aq) → [CoEDTA]^2- (aq) + 6H₂O (l)

Two reactant particles become seven product particles, giving a large positive entropy change that drives the reaction.

True.

In chelation, a multidentate ligand replaces several monodentate ligands, increasing the total number of particles in solution. This produces a net increase in disorder, so ΔS^ð is always positive, making chelation thermodynamically feasible.