What is the correct answer to the co-ordination compounds NEET 2000 cyanide-complex question?
Option C is correct: copper forms the more stable copper(I) cyanide complex and stays in solution, while cadmium precipitates as cadmium sulphide. The co-ordination compounds NEET 2000 question requires two checks: complex formula and relative stability. Remembering only that the copper complex is more stable cannot distinguish A from C.
The problem asks which cyanide complexes form when copper(II) and cadmium(II) ions are treated with potassium cyanide, and how their stability affects sulphide precipitation with hydrogen sulphide. Identify both complexes before deciding which metal stays in solution.
The supplied stem says “quantitative analysis”, but the separation in its worked solution belongs to Group II qualitative analysis. The choices below list copper first and cadmium second:
- A: Copper more stable; cadmium less stable.
- B: Copper less stable; cadmium more stable.
- C: Copper more stable; cadmium less stable.
- D: Copper less stable; cadmium more stable.
The question bank rates this item hard and provides an expected solving time of 90 seconds. These are bank labels, not measured student-performance data.
How do you derive the copper and cadmium complex formulas?
The official solution identifies highly stable cuprocyanide and comparatively less stable cadmium cyanide, respectively:
Copper is copper(I) in this cuprocyanide complex, although the starting ion is copper(II). Cyanide treatment involves reduction to copper(I), so carrying the starting oxidation state into the product gives the wrong formula.
For copper, count the charges step by step. Each cyanide ligand contributes one negative charge, so four cyanide ligands contribute four negative charges:
Copper(I) contributes one positive charge. Adding the metal and ligand charges gives:
Three potassium ions are therefore needed for a neutral salt:
For cadmium, the metal remains cadmium(II). Its two positive charges combine with the same four negative ligand charges:
Two potassium ions balance this complex:
Charge balance establishes the formulas, not the stability order. One check establishes electrical neutrality; the other explains the separation. A larger negative complex charge is not proof of greater stability.
Why does cadmium precipitate while copper stays in solution?
The copper(I) cyanide complex is much more stable than the cadmium cyanide complex in the separation described. High stability keeps copper bound in solution, preventing copper sulphide precipitation. The comparatively less stable cadmium complex can release cadmium(II) ions, which form cadmium sulphide when hydrogen sulphide is passed through the solution.
The supplied solution describes the copper sulphide precipitate that does not form as:
That wording does not change copper’s oxidation state in cuprocyanide. Keep the product-complex charge calculation separate from the supplied solution’s name for the absent precipitate; no copper sulphide ionic equation is needed.
“Less stable” does not mean the cadmium complex cannot form. It means less stable relative to the copper complex in this analytical context. Cadmium can therefore form its cyanide complex yet still give a sulphide precipitate during the separation.
- Copper: More stable; remains in solution; no copper sulphide precipitate.
- Cadmium: Comparatively less stable; cadmium sulphide precipitates.
Option C satisfies both the formula check and the stability check. This result applies to the stated separation, not every possible set of solution conditions.
Why does keeping copper as copper(II) lead to option A?
Retaining the starting copper(II) oxidation state produces the incorrect copper formula in option A. Combining that formula with the correct memory that the copper complex is more stable makes A appear to fit. The error is in identifying the product oxidation state, not in adding the charges.
The mistaken calculation is:
The arithmetic is internally consistent. However, it uses copper(II), not the copper(I) cuprocyanide species identified in the official solution.
A and C both give copper greater relative stability. Only C gives the stated copper(I) complex formula, so stability memory alone cannot distinguish them.
Identify the metal’s oxidation state in the product complex before balancing potassium counter-ions. Apply this order to other complex salts too: the starting ion’s charge is not automatically the metal’s oxidation state after treatment.
How do you apply these checks to three related coordination-compound questions?
Use the product oxidation state to calculate charge, then use relative stability to predict precipitation. These three original practice questions from the same chapter are not verified PYQs. Each answer follows its question.
Question 1: What is copper’s oxidation state in potassium cuprocyanide?
Answer: copper(I) in the salt:
The complete salt is neutral. Three potassium ions contribute three positive charges, and four cyanide ligands contribute four negative charges:
Question 2: What are the complex charge, potassium count and coordination number for cadmium(II) with four monodentate cyanide ligands?
The answers are:
- Complex-ion charge: two negative charges.
- Potassium counter-ions: two.
- Coordination number: 4.
Calculate the charge from the metal and ligands:
Each monodentate ligand binds through one donor atom. Four such ligands give coordination number 4, not 2; counter-ions balance charge but do not determine coordination number.
Question 3: Which metal gives a sulphide precipitate with hydrogen sulphide after cyanide complex formation, and why?
Answer: cadmium, as cadmium sulphide, in the stated separation. Its comparatively less stable cyanide complex can release cadmium(II) ions, while copper remains bound in its more stable complex.
Recall chain: identify the product oxidation state, balance the complex charge, compare stability, then predict precipitation.
Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).
Related on NEET JEEnius AI: Stationary Waves in Pipes NEET 2026: Counting Nodes.
Frequently asked questions
What is the correct answer to the NEET 2000 cyanide-complex question?
Option C is correct: the salts are K₃[Cu(CN)₄] and K₂[Cd(CN)₄]. In the stated separation, the copper(I) cyanide complex is more stable and keeps copper in solution, while cadmium precipitates as CdS.
Why is the copper complex K₃[Cu(CN)₄] and not K₂[Cu(CN)₄]?
Cyanide treatment involves reduction of the starting copper(II) ion to copper(I). Copper(I) and four cyanide ligands give a complex charge of +1 − 4 = −3, requiring three potassium ions. Keeping copper as copper(II) gives the incorrect formula used in option A.
Why does cadmium precipitate as CdS while copper stays in solution?
In the stated separation, the cadmium cyanide complex is comparatively less stable and can release cadmium(II) ions, which form CdS when hydrogen sulphide is passed through the solution. The more stable copper(I) cyanide complex keeps copper bound in solution. This conclusion applies to the specified separation, not every possible set of solution conditions.
What is the coordination number of cadmium in K₂[Cd(CN)₄]?
Cadmium has coordination number 4 because four monodentate cyanide ligands each bind through one donor atom. The complex-ion charge is −2, so two potassium counter-ions are needed. Counter-ions balance charge; they do not determine coordination number.