Q.$FeSO_4$ solution mixed with $(NH_4)_2SO_4$ solution in 1:1 molar ratio gives the test of $Fe^{2+}$ ion but $CuSO_4$ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of $Cu^{2+}$ ion. Explain why?

The key idea is Complex Formation — the difference arises because one mixture forms a stable complex that masks the metal ion, while the other does not.

Reasoning:

1. $FeSO_4 + (NH_4)_2SO_4$ (1:1): This simply forms a double salt, Mohr's salt, $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$. In solution, it dissociates completely into $Fe^{2+}$, $NH_4^+$, and $SO_4^{2-}$ ions. The $Fe^{2+}$ ion remains free and uncomplexed, so it gives its characteristic tests (e.g., the deep blue Turnbull's blue precipitate with $K_3[Fe(CN)_6]$).

2. $CuSO_4 + NH_3(aq)$ (1:4): Ammonia acts as a ligand. $Cu^{2+}$ initially forms a pale blue precipitate of $Cu(OH)_2$, which then dissolves in excess ammonia to form the deep blue tetraamminecopper(II) complex:

This complex is very stable, so the concentration of free $Cu^{2+}$ ions becomes negligible. Hence, tests that rely on free $Cu^{2+}$ (e.g., the chocolate-brown precipitate of $Cu_2[Fe(CN)_6]$ with $K_4[Fe(CN)_6]$) fail.

The question contrasts two seemingly similar mixtures — but the chemistry of the two metal ions is fundamentally different. Let’s see why.

1. What does “test of the ion” mean?

When we say a solution “gives the test” of $Fe^{2+}$ or $Cu^{2+}$, we mean that the free, hydrated metal ion is present in sufficient concentration to react with a specific reagent (like $K_3[Fe(CN)_6]$ for $Fe^{2+}$, or $NH_3$ itself for $Cu^{2+}$) to produce a characteristic colour or precipitate. If the metal ion is tightly bound in a complex, it may not be available for that test.

2. The iron(II) case: $FeSO_4 + (NH_4)_2SO_4$ (1:1 molar ratio)

$(NH_4)_2SO_4$ provides $NH_4^+$ and $SO_4^{2-}$ ions. It does not provide free ammonia ($NH_3$) in significant amount — ammonium ion is a weak acid ($pK_a \approx 9.25$), so in neutral solution it does not release enough $NH_3$ to form ammine complexes with $Fe^{2+}$.

$Fe^{2+}$ does form weak complexes with sulfate ($[FeSO_4]^0$ ion pair) and possibly with water, but these are labile — they dissociate instantly. The $Fe^{2+}$ remains essentially as the hexaaqua ion $[Fe(H_2O)_6]^{2+}$ in solution.

So when you add a test reagent like potassium ferricyanide, you get the deep blue Turnbull’s blue precipitate:

The test works because free $Fe^{2+}$ is abundant.

3. The copper(II) case: $CuSO_4 + \text{aqueous ammonia}$ (1:4 molar ratio)

Here, aqueous ammonia ($NH_3$ in water) is a strong ligand and is present in excess (4 moles $NH_3$ per mole $Cu^{2+}$). $Cu^{2+}$ has a strong tendency to form ammine complexes. The reaction proceeds stepwise:

The overall formation constant for the tetraammine complex is very large:

With 1:4 stoichiometry and such a high $K_f$, essentially all $Cu^{2+}$ ions are converted to the deep blue $[Cu(NH_3)_4]^{2+}$ complex. The concentration of free $Cu^{2+}$ drops to astronomically low levels (on the order of $10^{-12}$ M or less).

4. The critical difference: stability and lability

• $Fe^{2+}$ with $(NH_4)_2SO_4$: No strong complex forms; free $Fe^{2+}$ remains.
• $Cu^{2+}$ with $NH_3$: A thermodynamically very stable complex forms ($K_f \approx 10^{12}$), so the equilibrium leaves virtually no free $Cu^{2+}$ — the masking comes from this huge formation constant. (Cu(II) complexes are actually kinetically labile — ligands exchange fast — but the equilibrium position keeps free $Cu^{2+}$ negligible.)

5. Why doesn’t the same happen with $Fe^{2+}$ and $(NH_4)_2SO_4$?

Even if we replaced $(NH_4)_2SO_4$ with actual $NH_3$ solution, $Fe^{2+}$ forms much weaker ammine complexes ($K_f$ for $[Fe(NH_3)_6]^{2+}$ is only about $10^{2.2}$ — negligible compared to copper). Plus, $(NH_4)_2SO_4$ doesn’t even provide free $NH_3$ in the first place.

Concept: Complex Formation

Method: Ligand Substitution & Coordination Number Analysis

This method explains why certain metal ions become "hidden" from their characteristic tests due to the formation of stable coordination complexes.

Step-by-Step Explanation

Step 1: Identify the reactants and their molar ratios

• Case 1: $FeSO_4$ + $(NH_4)_2SO_4$ in 1:1 molar ratio

  • $Fe^{2+}$ is present, and $(NH_4)_2SO_4$ provides $NH_4^+$ and $SO_4^{2-}$ ions — no strong ligand like $NH_3$ is available in significant amount.

• Case 2: $CuSO_4$ + aqueous ammonia in 1:4 molar ratio

  • $Cu^{2+}$ is present, and $NH_3$ (a strong ligand) is available in excess (4 moles per mole of $Cu^{2+}$).

Step 2: Check coordination tendency of each metal ion

• $Fe^{2+}$ has a coordination number of 6, but here the ligand is only $SO_4^{2-}$ (a weak ligand) and water. No strong ligand like $NH_3$ is present in sufficient quantity to form a stable complex that masks $Fe^{2+}$.

  • Result: $Fe^{2+}$ remains as free ions → gives its characteristic test.

• $Cu^{2+}$ has a coordination number of 4 (common for tetrahedral or square planar complexes with $NH_3$).

  • With 4 moles of $NH_3$ per mole of $Cu^{2+}$, the following complex forms:

This is a deep blue, stable complex.

Step 3: Consequence for ion detection

• In the $FeSO_4$ case: No stable complex forms → $Fe^{2+}$ ions are free → test positive.
• In the $CuSO_4$ case: All $Cu^{2+}$ ions are bound in $[Cu(NH_3)_4]^{2+}$ → no free $Cu^{2+}$ → test for $Cu^{2+}$ (e.g., with $K_4[Fe(CN)_6]$ or $NaOH$) fails because the ligand is already occupying all coordination sites.

Final Answer

$FeSO_4$ with $(NH_4)_2SO_4$ does not form a stable complex (no strong ligand in excess), so $Fe^{2+}$ remains free and gives its test.

$CuSO_4$ with $NH_3$ in 1:4 ratio forms $[Cu(NH_3)_4]^{2+}$, a stable complex that masks $Cu^{2+}$ ions, so no test for $Cu^{2+}$ is obtained.

Key concept: Excess of a strong ligand can completely complex a metal ion, removing it from solution as a free ion.

Common Mistakes Students Make on This Question

This is a classic coordination chemistry question from JEE/NEET. Here are the most frequent errors and how to avoid each:

Mistake 1: Thinking Both Reactions Are Similar

The error: Students assume both mixtures form simple salts or double salts, and wonder why only one gives the cation test.

Why it's wrong:

• $FeSO_4 + (NH_4)_2SO_4$ forms a double salt — Mohr's salt $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$
• $CuSO_4 + NH_3$ forms a coordination complex — $[Cu(NH_3)_4]SO_4$

How to avoid:

• Double salts dissociate completely in water into simple ions.
• Coordination complexes bind the metal ion tightly — the metal is no longer free to give its characteristic tests.

Mistake 2: Forgetting the Key Difference — Dissociation vs. Complexation

The error: Students write both as "salt formation" and miss the core concept.

Correct understanding:

How to avoid:

• Remember: Double salts exist only in solid state; in solution they break into constituent ions.
• Complexes have coordinate bonds that survive in solution — the metal is "masked."

Mistake 3: Ignoring the Molar Ratio Given

The error: Students ignore the "1:1" and "1:4" ratios and treat the problem generically.

Why it matters:

• 1:1 ratio of $FeSO_4 : (NH_4)_2SO_4$ is exactly the stoichiometry for Mohr's salt — no excess ammonia to form a complex.
• 1:4 ratio of $CuSO_4 : NH_3$ is exactly the stoichiometry for the tetraamminecopper(II) complex — all $Cu^{2+}$ gets bound.

How to avoid:

• Always check if the given ratio matches the formula of a known complex or double salt.
• If excess ligand is present, complex formation is likely.

Mistake 4: Confusing the Tests for $Fe^{2+}$ and $Cu^{2+}$

The error: Students think both tests fail or both succeed.

Correct:

• $Fe^{2+}$ test (e.g., with $K_3[Fe(CN)_6]$ giving a dark blue precipitate) works because free $Fe^{2+}$ exists.
• $Cu^{2+}$ test (e.g., with $K_4[Fe(CN)_6]$ giving a chocolate brown precipitate) fails because $Cu^{2+}$ is locked in $[Cu(NH_3)_4]^{2+}$.

How to avoid:

• Know the specific tests for each ion.
• Understand that a complexed metal ion does not respond to simple precipitation tests for the free ion.

Mistake 5: Writing Incorrect Formulae

The error: Students write $FeSO_4 \cdot (NH_4)_2SO_4$ as a complex or $[Cu(NH_3)_4]SO_4$ as a double salt.

How to avoid:

• Double salt formula: $FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$ — written with a dot, not brackets.
• Complex formula: $[Cu(NH_3)_4]SO_4$ — square brackets indicate the coordination sphere.

Quick Summary to Remember

Final takeaway:

The $Fe^{2+}$ test works because Mohr's salt is a double salt that breaks apart. The $Cu^{2+}$ test fails because $Cu^{2+}$ is tightly bound in a complex ion — no free $Cu^{2+}$ remains to react.