Question

Identify 'P' and 'Q' in the following reaction

• A. P = K$_2$[Cd(CN)$_4$], Q = CdS
• B. P = CdS, Q = K$_2$[Cd(CN)$_4$]
• C. P = Cd(NO$_3$)$_2$, Q = CdSO$_4$
• D. P = [Cd(OH$_2$)$_4$](NO$_3$)$_2$, Q = [Cd(NO$_3$)$_4$](NO$_3$)$_2$

Hint:

Using a simple frame or just bolding for the box Key Points:
Ligand Substitution: Stronger ligands (like CN$^-$) displace weaker ligands (like NH$_3$) from complex ions.
Cd$^{2+$ forms a stable tetracyano complex: [Cd(CN)$_4$]$^{2-$.
Precipitation: H$_2$S is used to precipitate metal sulfides. CdS is highly insoluble and yellow.
Even stable complexes can be decomposed if a very insoluble precipitate can be formed (like CdS).

Answer 1

The Correct Option is A

Let's analyze the reaction sequence step by step.
(A) Formation of P: The starting material is tetraamminecadmium(II) nitrate, [Cd(NH₃)₄](NO₃)₂. It reacts with potassium cyanide (KCN). Cyanide ions (CN^-) are stronger ligands than ammonia (NH₃) for Cd²⁺ and will replace the ammonia ligands. Cadmium(II) typically forms a stable tetracyano complex. The reaction (balancing the cyanide) is:
[Cd(NH₃)₄](NO₃)₂ + 4KCN → K₂[Cd(CN)₄] + 4NH₃ + 2KNO₃
The ammonia released reacts with water to form ammonium hydroxide (NH₄OH, or more accurately NH₃ in equilibrium with NH₄⁺ + OH^-).
Therefore, the major cadmium-containing product P is potassium tetracyanocadmate(II), K₂[Cd(CN)₄]. (B) Formation of Q: Product P (K₂[Cd(CN)₄]) is treated with hydrogen sulfide (H₂S). H₂S is a source of sulfide ions (S^2-). Cadmium sulfide (CdS) is a highly insoluble yellow precipitate. Even though Cd²⁺ is complexed in [Cd(CN)₄]^2-, the sulfide ion concentration from H₂S is sufficient to precipitate CdS, especially because CdS is very stable and has a very low solubility product (K_sp). The cyanide complex may decompose, particularly if the solution becomes acidic due to H₂S. The net reaction leads to the precipitation of CdS.
K₂[Cd(CN)₄] + H₂S → CdS ↓ + 2KCN + 2HCN
Therefore, Q is Cadmium Sulfide (CdS). Matching the identified products P = K₂[Cd(CN)₄] and Q = CdS with the options, we find that option (A) is correct.