$FeSO_4$ 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?

FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1:1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution

Exercises · 5.2

FeSO_4

Lakshadweep CbseNCERTSubjective· 3mImportance★★★★★

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Concept understanding — Complex Formation

Complex Formation: The Intuition

Imagine you have a metal ion — say, a copper ion ( $\text{Cu}^{2+}$ ) — floating in water. It's positively charged, so it attracts anything negative or electron-rich nearby. Water molecules themselves have lone pairs of electrons on oxygen, so they crowd around the copper ion, each one donating a pair of electrons to form a coordinate bond. That cluster — the metal ion surrounded by water molecules — is already a complex ion: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ .

Now, suppose you add ammonia ( $\text{NH}_3$ ) to the solution. Ammonia also has a lone pair on nitrogen, and it's a stronger electron donor than water. One by one, the ammonia molecules push the water molecules aside, replacing them. You end up with a deep blue complex: $[\text{Cu}(\text{NH}_3)_4]^{2+}$ .

That process — the stepwise replacement of one set of molecules (or ions) around a central metal atom by another set — is complex formation. The central metal is the Lewis acid (electron-pair acceptor), and the molecules or ions that attach to it are ligands (Lewis bases, electron-pair donors). The resulting species is a coordination compound or complex.

Note

The word "complex" doesn't mean complicated. It just means a central atom (usually a metal) bonded to surrounding molecules or ions.

The Precise Statement

Complex formation is the reversible, stepwise reaction in which a central metal atom or ion (usually a transition metal) accepts electron pairs from one or more ligands to form a coordination entity. Each step has its own equilibrium constant, and the overall stability of the complex is measured by the formation constant ( $K_f$ ) or stability constant.

For a general reaction:

\text{M} + n\text{L} \rightleftharpoons \text{ML}_n

The overall formation constant is:

K_f = \frac{[\text{ML}_n]}{[\text{M}][\text{L}]^n}

A large $K_f$ means the complex is very stable — the ligands bind tightly and are hard to remove.

K_f = \frac{[\text{complex}]}{[\text{metal}][\text{ligand}]^n}

Key Features to Remember

Stepwise nature: Complexes don't form all at once. First one ligand binds, then another, and so on. Each step has its own constant ( $K_1, K_2, \dots$ ). The product of all stepwise constants equals the overall $K_f$ .
Coordination number: The number of ligand donor atoms directly bonded to the metal. Common values are 4 (tetrahedral or square planar) and 6 (octahedral). For $[\text{Cu}(\text{NH}_3)_4]^{2+}$ , the coordination number is 4.
Ligand denticity: A ligand can have one donor atom (monodentate, like $\text{NH}_3$ or $\text{H}_2\text{O}$ ) or multiple donor atoms (polydentate, like EDTA, which wraps around the metal with six donor atoms). Polydentate ligands form especially stable complexes — this is the chelate effect.
Reversibility: Complex formation is an equilibrium. Change the concentration of ligand, pH, or temperature, and the complex can break apart or form a different one.

Watch out

A common mistake is to think complex formation happens in one step. It doesn't. The deep blue $[\text{Cu}(\text{NH}_3)_4]^{2+}$ forms through four separate equilibria, each with its own constant. The first ammonia binds easily, the fourth one less so — because the metal is already crowded.

Why It Matters

Complex formation is everywhere in chemistry and biology:

In your blood: Hemoglobin is an iron complex that carries oxygen.
In water treatment: EDTA is added to bind metal ions and prevent scaling.
In qualitative analysis: The deep blue of copper-ammonia complex is a classic test for copper ions.
In medicine: Cisplatin, a platinum complex, is a chemotherapy drug.

The key takeaway: complex formation is about electron-pair donation from ligands to a metal, creating a stable, often colourful, coordination entity. The stability depends on the metal, the ligand, and the conditions — and it always happens step by step.

Complex formation through ligand electron-pair donation is a central theme of the NCERT/CBSE Class 12 Chemistry chapter on Coordination Compounds, and ‘complex formation important questions’ or ‘coordination complex formation examples’ are commonly searched for board exams, JEE Main and NEET. Real-world applications like EDTA and cisplatin also make this a favourite for application-based competitive-exam questions.

Why this formula?

Complex Formation: Understanding the Why Behind the Key Formulas

Complex formation is a fundamental concept in coordination chemistry and equilibrium. Let's build the reasoning step-by-step, starting from the simplest idea.

1. What is Complex Formation?

A complex forms when a central metal ion (Lewis acid) accepts electron pairs from surrounding molecules or ions called ligands (Lewis bases).

Example:

$Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$

The key question: Why do we get a specific formula for the equilibrium constant?

2. The Stepwise Formation (The Core Reason)

Complex formation does not happen in one giant leap. It occurs in successive, reversible steps — each step adding one ligand.

For a metal $M$ and ligand $L$ :

Step 1:

$M + L \rightleftharpoons ML$

Equilibrium constant: $K_1 = \frac{[ML]}{[M][L]}$

Step 2:

$ML + L \rightleftharpoons ML_2$

$K_2 = \frac{[ML_2]}{[ML][L]}$

Step 3:

$ML_2 + L \rightleftharpoons ML_3$

$K_3 = \frac{[ML_3]}{[ML_2][L]}$

... and so on up to $ML_n$ .

Each $K_i$ is called a stepwise formation constant.

3. The Overall Formation Constant (The Key Formula)

Now, what if we want the equilibrium constant for the overall reaction:

$M + nL \rightleftharpoons ML_n$

We can multiply the stepwise equilibria (because when you add reactions, you multiply their equilibrium constants):

$K_f = K_1 \times K_2 \times K_3 \times \dots \times K_n$

So:

\boxed{K_f = \frac{[ML_n]}{[M][L]^n}}

Why this form?

Because each step contributes one $[L]$ in the denominator and one $[ML_i]$ in the numerator, but intermediate species cancel out when multiplied.

4. Why the Denominator Has $[L]^n$ (Not $n[L]$ )

This is a common confusion. Let's derive it explicitly:

From step 1: $[ML] = K_1[M][L]$

From step 2: $[ML_2] = K_2[ML][L] = K_1K_2[M][L]^2$

From step 3: $[ML_3] = K_3[ML_2][L] = K_1K_2K_3[M][L]^3$

Continuing:

$[ML_n] = (K_1K_2\dots K_n)[M][L]^n$

Therefore:

\frac{[ML_n]}{[M][L]^n} = K_1K_2\dots K_n = K_f

The exponent $n$ comes from repeated multiplication, not addition. Each ligand adds one factor of $[L]$ in the denominator.

5. The Stability Connection

A larger $K_f$ means:

The complex is more stable
Equilibrium lies far to the right (products favoured)
The metal ion is more effectively "tied up" by the ligand

This is why $K_f$ is also called the stability constant.

6. Quick Summary — The "Why" in One Line

The formula $K_f = \frac{[ML_n]}{[M][L]^n}$ arises because complex formation is a series of n sequential equilibria, each adding one ligand, and multiplying the stepwise constants gives the overall constant with $[L]^n$ in the denominator.

7. Exam Tip

Stepwise constants ( $K_1, K_2, \dots$ ) are usually not equal — often $K_1 > K_2 > K_3$ due to steric hindrance and statistical factors.
The overall constant $K_f$ is what you use for direct calculations involving the fully formed complex.
Always check: Is the ligand monodentate or polydentate? For polydentate ligands (like EDTA), the formula still holds — $n$ is the number of donor atoms that bind.

Final thought: The beauty of this derivation is that it shows how a complex multi-step process collapses into a single, elegant equilibrium expression — as long as you remember the stepwise logic behind it.

The key is complex formation: $Fe^{2+}$ forms a weak, labile complex with $(NH_4)_2SO_4$ that still releases free $Fe^{2+}$ for its characteristic test, whereas $Cu^{2+}$ forms a very stable deep-blue tetraammine complex $[Cu(NH_3)_4]^{2+}$ that locks up virtually all $Cu^{2+}$ ions, preventing them from giving the usual $Cu^{2+}$ test.

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.

Tip

Even if a tiny amount of $NH_3$ were present, $Fe^{2+}$ forms only weak ammine complexes (unlike $Fe^{3+}$ or $Cu^{2+}$ ). The equilibrium heavily favours free $Fe^{2+}$ .

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

3Fe^{2+} + 2[Fe(CN)_6]^{3-} \rightarrow Fe_3[Fe(CN)_6]_2 \downarrow

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:

[Cu(H_2O)_6]^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)(H_2O)_5]^{2+} + H_2O

\vdots

[Cu(NH_3)_3(H_2O)_3]^{2+} + NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + H_2O

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

K_f = \frac{[[Cu(NH_3)_4]^{2+}]}{[Cu^{2+}][NH_3]^4} \approx 10^{12} \text{ to } 10^{13}

Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+} \quad K_f \approx 2 \times 10^{12}

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).

Watch out

A common mistake is to think that because the solution is blue, it still contains $Cu^{2+}$ ions. The blue colour is from the complex $[Cu(NH_3)_4]^{2+}$ , not from free $Cu^{2+}$ (which is pale blue). The test for $Cu^{2+}$ (e.g., with $K_4[Fe(CN)_6]$ to give a chocolate brown precipitate of $Cu_2[Fe(CN)_6]$ ) requires free $Cu^{2+}$ — which is virtually absent.

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.)

Important

The 1:4 molar ratio for $Cu^{2+}:NH_3$ is exactly the stoichiometry needed to form the tetraammine complex. If you used less $NH_3$ , some free $Cu^{2+}$ would remain and the test would partially work. But at 1:4, the complexation is essentially complete.

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.

›Proof

Why $(NH_4)_2SO_4$ doesn’t release $NH_3$ significantly:

$NH_4^+ \rightleftharpoons NH_3 + H^+$ has $K_a = 5.6 \times 10^{-10}$ . In a neutral solution ( $pH \approx 7$ ), the ratio $[NH_3]/[NH_4^+] = K_a/[H^+] \approx 5.6 \times 10^{-3}$ . So less than 1% of ammonium is present as $NH_3$ — far too little to complex $Fe^{2+}$ even if it wanted to.

✓Final answer

FeSO_4

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:

Cu^{2+} + 4NH_3 \rightarrow [Cu(NH_3)_4]^{2+}

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:

Mixture	Product	In water, it gives
$FeSO_4 + (NH_4)_2SO_4$	Mohr's salt (double salt)	$Fe^{2+}$ , $NH_4^+$ , $SO_4^{2-}$ — all free ions
$CuSO_4 + 4NH_3$	$[Cu(NH_3)_4]SO_4$ (complex)	$[Cu(NH_3)_4]^{2+}$ , $SO_4^{2-}$ — no free $Cu^{2+}$

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

Concept	Double Salt	Coordination Complex
Bonding	Ionic (electrostatic)	Coordinate covalent
In solution	Dissociates completely	Complex ion stays intact
Metal ion test	Positive	Negative
Example	Mohr's salt	$[Cu(NH_3)_4]SO_4$

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.

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Q.FeSO4FeSO_4FeSO4​ solution mixed with (NH4)2SO4(NH_4)_2SO_4(NH4​)2​SO4​ solution in 1:1 molar ratio gives the test of Fe2+Fe^{2+}Fe2+ ion but CuSO4CuSO_4CuSO4​ solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+Cu^{2+}Cu2+ ion. Explain why?

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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?

Unlock everything free for 14 days