Q.What is the coordination entity formed when excess of aqueous $KCN$ is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when $H_2S$

The key idea is complex formation: the copper ion binds strongly with cyanide ligands, changing its chemical identity and removing free Cu²⁺ from solution.

1. When excess aqueous KCN is added to CuSO₄, Cu²⁺ first reduces to Cu⁺ (cyanide acts as a reducing agent), and the Cu⁺ then coordinates with CN⁻ to form a stable, soluble complex.
2. The reaction is: $2\text{CuSO}_4 + 10\text{KCN} \rightarrow 2\text{K}_3[\text{Cu(CN)}_4] + 2\text{K}_2\text{SO}_4 + (\text{CN})_2 \uparrow$
3. The coordination entity formed is the tetracyanidocuprate(I) ion, $[\text{Cu(CN)}_4]^{3-}$. …

Why this happens — the concept of complex formation

When you add a ligand like cyanide ($CN^-$) to a metal ion in solution, the metal ion can bind several ligands to form a coordination complex. This is not just a physical mixing — it’s a chemical equilibrium that dramatically changes the metal ion’s properties. The key idea: a stable complex “locks up” the metal ion so tightly that its free concentration becomes vanishingly small.

For precipitation reactions (like forming $CuS$), you need a minimum concentration of free $Cu^{2+}$ ions. If the complex is stable enough, that free concentration falls below the threshold needed for precipitation — even though the total copper is still present.

Step-by-step reasoning

1. What happens when $KCN$ is added to $CuSO_4$ solution?

Aqueous $CuSO_4$ gives blue $Cu^{2+}$ ions. When you add $KCN$, the $CN^-$ ligand first reduces $Cu^{2+}$ to $Cu^+$ (because $CN^-$ is a reducing agent as well as a ligand), and then forms a complex with $Cu^+$.

The overall reaction is:

The cyanogen gas $(CN)_2$ escapes, and the copper ends up as the tetracyanidocuprate(I) ion, $[Cu(CN)_4]^{3-}$.

2. Why is this complex so stable?

The stability constant (formation constant) of $[Cu(CN)_4]^{3-}$ is enormous — roughly $K_f \approx 2 \times 10^{30}$. That means the equilibrium

lies almost entirely to the right. The free $Cu^+$ concentration in solution is extremely low.

3. What happens when $H_2S$ is passed through this solution?

$H_2S$ dissociates slightly in water:

$H_2S \rightleftharpoons 2H^+ + S^{2-}$ …

Method: Complex Formation & Stability Analysis

This problem is solved by understanding ligand substitution and the stability of coordination complexes.

Step 1: Identify the reaction

When excess aqueous $KCN$ is added to aqueous $CuSO_4$, the cyanide ion ($CN^-$) acts as a strong ligand and replaces water molecules around the copper ion.

• Initially, $Cu^{2+}$ in water exists as $[Cu(H_2O)_6]^{2+}$ (blue colour).
• With excess $KCN$, a complex formation occurs:

The coordination entity formed is:

Note: Copper is in +1 oxidation state here because $CN^-$ reduces $Cu^{2+}$ to $Cu^+$ during complexation (cyanide acts as both ligand and reducing agent — the equation above shows the cyanide oxidised to cyanogen, $(CN)_2$).

Step 2: Why no precipitate with $H_2S$?

Normally, $H_2S$ reacts with $Cu^{2+}$ to give black $CuS$ precipitate:

However, in the complex $[Cu(CN)_4]^{3-}$:

• The cyanide ligands are very strongly bound to copper.
• The complex has a very high stability constant ($K_f$ is extremely large). …

🧪 The Correct Chemistry First

When excess aqueous $KCN$ is added to $\text{CuSO}_4$ solution:

1. Initially, unstable copper(II) cyanide forms and rapidly decomposes to copper(I) cyanide ($\text{CuCN}$), evolving cyanogen — this is where $\text{Cu}^{2+}$ is reduced to $\text{Cu}^{+}$.
2. With excess $KCN$, the $\text{CuCN}$ dissolves to form the soluble complex:

Notice: copper is reduced from Cu²⁺ to Cu⁺ by cyanide ion (CN⁻ acts as a reducing agent here).

When $H_2S$ gas is passed through this solution, no CuS precipitate forms because:

• The complex $[\text{Cu(CN)}_4]^{3-}$ is extremely stable (high formation constant).
• The concentration of free $\text{Cu}^+$ ions is too low to exceed the solubility product of $\text{Cu}_2\text{S}$.

✗ Common Mistake #1: Writing the wrong oxidation state of copper

Mistake: Students write $[\text{Cu(CN)}_4]^{2-}$ (assuming Cu remains in +2 state).

Why it’s wrong: CN⁻ is a strong reducing agent. In excess, it reduces $\text{Cu}^{2+}$ to $\text{Cu}^+$. The complex formed is with $\text{Cu}^+$, not $\text{Cu}^{2+}$.

✓ How to avoid: Remember the redox behaviour of CN⁻ with Cu²⁺. Always check:

• CN⁻ can reduce Cu²⁺ → Cu⁺ (and itself oxidises to cyanogen, $(CN)_2$).
• The final complex has coordination number 4 and charge 3− for Cu⁺.

✗ Common Mistake #2: Forgetting the reduction step entirely

Mistake: Writing the complex as $[\text{Cu(CN)}_6]^{4-}$ or similar, ignoring the change in oxidation state.

Why it’s wrong: Cu²⁺ typically forms hexacoordinate complexes (e.g., $[\text{Cu(H}_2\text{O)}_6]^{2+}$), but with CN⁻, the reduction to Cu⁺ changes the coordination geometry to tetrahedral (CN⁻ is a strong field ligand).

✓ How to avoid:

• For Cu⁺, common coordination number is 4 (tetrahedral).
• For Cu²⁺, common coordination number is 6 (octahedral) or 4 (square planar).
• When CN⁻ is in excess, always check if reduction occurs — it does for Cu²⁺.

✗ Common Mistake #3: Saying “no precipitate because CuS is soluble”

Mistake: Claiming that CuS does not form because it is soluble in the complex solution.

Why it’s wrong: CuS is highly insoluble ($K_{sp} \approx 8 \times 10^{-37}$). The real reason is that free Cu⁺ ions are virtually absent due to the high stability of the complex.

✓ How to avoid:

• Use the concept of formation constant ($K_f$). For $[\text{Cu(CN)}_4]^{3-}$, $K_f$ is extremely large (~$10^{30}$).
• This means the equilibrium:

lies far to the left.

• So $[\text{Cu}^+]$ is negligible — insufficient to form any precipitate with $S^{2-}$.

✗ Common Mistake #4: Writing the wrong precipitate formula …