Ammonia forms the complex ion [Cu(NH_3)_4]^2+ with copper ions in alkaline solutions but not in acidic solutions. What is the reason for it?

In acidic solutions protons coordinate with ammonia molecules forming NH^+_4 ions and NH_3 molecules are not available

In alkaline solutions insoluble Cu(OH)_2 is precipitated which is soluble in excess of any alkali

Copper hydroxide is an amphoteric substance

In acidic solutions hydration protects copper ions.

Coordination Compounds PYQ — JEE Main 2003 · Shift 0 · Q49 with Solution

View written solutionFree

Correct answer: A

Nature of the complex formed

Copper(II) ions form the tetraammine complex in presence of free ammonia: $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$

So, for the complex to form, free $NH_3$ molecules must be available as ligands.
Behavior of ammonia in acidic solution

Ammonia is a weak base and in acidic medium it gets protonated: $NH_3 + H^+ \rightleftharpoons NH_4^+$

Thus, in acidic solution, ammonia is converted into $NH_4^+$ and is no longer available to donate its lone pair to $Cu^{2+}$ for coordination.
Behavior in alkaline solution

In alkaline medium, ammonia remains largely as $NH_3$ , so it can act as a ligand and coordinate with copper(II) ions to form: $[Cu(NH_3)_4]^{2+}$
Checking the options
- A: Correct. In acidic solution, $NH_3$ is converted to $NH_4^+$ , so free ammonia is not available for complex formation.
- B: Incorrect. $Cu(OH)_2$ is not soluble in excess of just any alkali; this statement is false.
- C: Incorrect. $Cu(OH)_2$ is not amphoteric in the usual sense relevant here.
- D: Incorrect. Hydration of $Cu^{2+}$ occurs in both acidic and alkaline aqueous solutions; it is not the reason.
Final answer

The correct reason is: $\boxed{\text{A}}$

View written solutionFree

Correct answer: A

Nature of the complex formed

Copper(II) ions form the tetraammine complex in presence of free ammonia: $Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$

So, for the complex to form, free $NH_3$ molecules must be available as ligands.
Behavior of ammonia in acidic solution

Ammonia is a weak base and in acidic medium it gets protonated: $NH_3 + H^+ \rightleftharpoons NH_4^+$

Thus, in acidic solution, ammonia is converted into $NH_4^+$ and is no longer available to donate its lone pair to $Cu^{2+}$ for coordination.
Behavior in alkaline solution

In alkaline medium, ammonia remains largely as $NH_3$ , so it can act as a ligand and coordinate with copper(II) ions to form: $[Cu(NH_3)_4]^{2+}$
Checking the options
- A: Correct. In acidic solution, $NH_3$ is converted to $NH_4^+$ , so free ammonia is not available for complex formation.
- B: Incorrect. $Cu(OH)_2$ is not soluble in excess of just any alkali; this statement is false.
- C: Incorrect. $Cu(OH)_2$ is not amphoteric in the usual sense relevant here.
- D: Incorrect. Hydration of $Cu^{2+}$ occurs in both acidic and alkaline aqueous solutions; it is not the reason.
Final answer

The correct reason is: $\boxed{\text{A}}$