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

The reaction sequence is analyzed in stages.\r\n
(A) Formation of P: Tetraamminecadmium(II) nitrate, [Cd(NH₃)₄](NO₃)₂, reacts with potassium cyanide (KCN). Cyanide (CN^-) replaces ammonia (NH₃) ligands on Cd²⁺, forming a tetracyano complex. The balanced reaction is:\r\n
\r\n[Cd(NH₃)₄](NO₃)₂ + 4KCN → K₂[Cd(CN)₄] + 4NH₃ + 2KNO₃\r\n
\r\nReleased ammonia reacts with water, forming ammonium hydroxide (NH₄OH). The major cadmium-containing product P is potassium tetracyanocadmate(II), K₂[Cd(CN)₄].\r\n\r\n(B) Formation of Q: Product P (K₂[Cd(CN)₄]) reacts with hydrogen sulfide (H₂S), a source of sulfide ions (S^2-). Cadmium sulfide (CdS) precipitates as a yellow solid due to its low solubility. The cyanide complex decomposes, especially if the solution becomes acidic. The net reaction is:\r\n
\r\nK₂[Cd(CN)₄] + H₂S → CdS ↓ + 2KCN + 2HCN\r\n
\r\nTherefore, Q is Cadmium Sulfide (CdS).\r\n\r\nMatching P = K₂[Cd(CN)₄] and Q = CdS to the options, option (A) is correct.\r\n\r\n