Stability of Coordination Compounds

Introduction

Not all coordination compounds are created equal! Some are rock-solid and persist in solution for years, while others fall apart the moment you make them. Why does [Fe(CN)₆]⁴⁻ remain stable in solution while [FeF₆]³⁻ readily decomposes? Understanding stability is crucial for applications ranging from medicine to metallurgy.

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Equilibrium in Complex Formation

Stepwise Formation

When ligands add to a metal ion, they add one at a time:

$$M + L \rightleftharpoons ML \quad K_1 = \frac{[ML]}{[M][L]}$$ $$ML + L \rightleftharpoons ML_2 \quad K_2 = \frac{[ML_2]}{[ML][L]}$$ $$ML_2 + L \rightleftharpoons ML_3 \quad K_3 = \frac{[ML_3]}{[ML_2][L]}$$

And so on…

K₁, K₂, K₃… are called stepwise formation constants or stability constants.

Overall Formation Constant

The overall stability constant (β_n) is the product of stepwise constants:

$$\beta_n = K_1 \times K_2 \times K_3 \times ... \times K_n$$

Interactive Demo: Visualize Equilibrium Constants

Watch how stability constants affect complex formation and dissociation in real-time.

Example: For [Cu(NH₃)₄]²⁺

$$Cu^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4]^{2+}$$ $$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$$

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Dissociation vs Formation Constants

Dissociation Constant (Instability Constant)

The reverse of formation:

$$ML_n \rightleftharpoons M + nL$$ $$K_d = \frac{[M][L]^n}{[ML_n]} = \frac{1}{\beta_n}$$

Relationship:

$$K_d \times \beta_n = 1$$

Comparison

Property	K_d (Dissociation)	β_n (Formation)
Meaning	Tendency to break apart	Tendency to form
Large value	Unstable complex	Stable complex
Small value	Stable complex	Unstable complex
Relationship	K_d = 1/β_n	β_n = 1/K_d

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Factors Affecting Stability

1. Nature of Metal Ion

Charge Density Effect

Higher charge and smaller size → Greater stability

$$\text{Charge density} = \frac{\text{Charge}}{\text{Radius}}$$

Example: Stability order for same ligand

$$M^{3+} > M^{2+} > M^+$$ $$Al^{3+} > Mg^{2+} > Na^+$$

For same group, higher period (larger size) → Lower stability

$$Ni^{2+} > Pd^{2+} > Pt^{2+}$$

(wait, this is WRONG for d⁸!)

Actually:

$$Pt^{2+} > Pd^{2+} > Ni^{2+}$$

(greater d orbital participation)

Irving-Williams Series

For first-row transition metals with the same ligand:

$$\boxed{Mn^{2+} < Fe^{2+} < Co^{2+} < Ni^{2+} < Cu^{2+} > Zn^{2+}}$$

Explanation:

• Increasing stability from Mn²⁺ to Cu²⁺ (increasing charge density, CFSE)
• Cu²⁺ maximum due to d⁹ configuration (Jahn-Teller effect adds extra stabilization)
• Zn²⁺ drops (d¹⁰, no CFSE)

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2. Nature of Ligand

Basicity of Ligand

Better Lewis base → Forms stronger coordinate bond → More stable complex

For nitrogen donors:

$$NH_3 > NH_2-NH_2 > py$$

For oxygen donors:

$$OH^- > RO^- > H_2O$$

Hard-Soft Acid-Base (HSAB) Principle

Hard acids prefer hard bases Soft acids prefer soft bases

Type	Metals	Ligands	Example
Hard	Small, high charge (Al³⁺, Fe³⁺, Cr³⁺)	F⁻, O²⁻, OH⁻, H₂O, NO₃⁻	[AlF₆]³⁻
Soft	Large, low charge (Cu⁺, Ag⁺, Pd²⁺, Pt²⁺)	I⁻, CN⁻, CO, PR₃, S²⁻	[Ag(CN)₂]⁻
Borderline	Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺	Br⁻, N₃⁻, SCN⁻	[Cu(NH₃)₄]²⁺

Examples:

• Fe³⁺ (hard) + F⁻ (hard) → Very stable [FeF₆]³⁻
• Ag⁺ (soft) + CN⁻ (soft) → Very stable [Ag(CN)₂]⁻
• Fe³⁺ (hard) + I⁻ (soft) → Unstable, rarely exists

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3. Chelate Effect

Chelates (complexes with polydentate ligands) are much more stable than analogous complexes with monodentate ligands.

Quantitative Comparison

Example: Ni²⁺ with ammonia vs ethylenediamine

Reaction 1: (Monodentate)

$$[Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftharpoons [Ni(NH_3)_6]^{2+} + 6H_2O$$ $$\log \beta_6 = 8.61$$

Reaction 2: (Bidentate)

$$[Ni(H_2O)_6]^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+} + 6H_2O$$ $$\log \beta_3 = 18.28$$

Difference: The bidentate complex is 10⁹·⁶⁷ ≈ 5 billion times more stable!

Why Does Chelate Effect Occur?

Entropic Advantage (Main Reason)

Monodentate replacement:

$$[Ni(H_2O)_6]^{2+} + 6NH_3 \rightleftharpoons [Ni(NH_3)_6]^{2+} + 6H_2O$$

• Reactants: 7 particles
• Products: 7 particles
• ΔS ≈ 0 (no entropy change)

Chelate replacement:

$$[Ni(H_2O)_6]^{2+} + 3en \rightleftharpoons [Ni(en)_3]^{2+} + 6H_2O$$

• Reactants: 4 particles
• Products: 7 particles
• ΔS > 0 (positive entropy change!)

More particles → Greater disorder → Favorable entropy

$$\Delta G = \Delta H - T\Delta S$$

Since ΔS is positive and large → ΔG more negative → More stable!

Statistical Advantage

Once one end of bidentate ligand binds, the other end is already nearby, making the second binding much more probable!

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4. Macrocyclic Effect

Complexes with macrocyclic ligands (ring-shaped) are even MORE stable than chelates!

Example: Porphyrin complexes (hemoglobin, chlorophyll)

Reasons:

1. Preorganization: Ligand is already in the right geometry
2. Entropic advantage: Even less reorganization needed
3. Kinetic stability: Once in, very hard to get out of the ring!

Stability order:

$$\text{Macrocyclic} > \text{Chelate} > \text{Monodentate}$$

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Thermodynamic vs Kinetic Stability

Thermodynamic Stability

How stable is the complex at equilibrium?

• Measured by formation constant (β)
• Large β = thermodynamically stable

Example: [Ni(CN)₄]²⁻ has huge β → very stable thermodynamically

Kinetic Stability (Inertness)

How fast does the complex react/exchange ligands?

• Labile: Rapidly exchanges ligands (half-life < 1 minute)
• Inert: Slowly exchanges ligands (half-life > 1 minute)

Examples

Complex	Thermodynamic	Kinetic	Explanation
[Ni(CN)₄]²⁻	Stable (large β)	Labile	d⁸, rapid exchange
[Co(NH₃)₆]³⁺	Stable	Inert	d⁶ low spin, slow exchange
[Cr(H₂O)₆]³⁺	Stable	Inert	d³, high CFSE barrier
[CrF₆]³⁻	Unstable (in water)	Inert	Slow to hydrolyze despite thermodynamic driving force

Labile vs Inert: General Rules

Labile complexes:

• d⁰ (no CFSE to lose)
• d¹⁰ (no CFSE to lose)
• d⁴-d⁷ high spin (weak field)
• Large metal ions

Inert complexes:

• d³ low spin (Cr³⁺, Co³⁺)
• d⁶ low spin (Co³⁺, Rh³⁺, Ir³⁺)
• High charge density

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Quantitative Problems

Problem Type 1: Calculating Free Metal Ion Concentration

Example: What is [Cu²⁺] in a solution containing 0.1 M [Cu(NH₃)₄]²⁺ and 1 M free NH₃?

Given: β₄ = 1.1 × 10¹³

Solution:

$$\beta_4 = \frac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}][NH_3]^4}$$ $$1.1 \times 10^{13} = \frac{0.1}{[Cu^{2+}](1)^4}$$ $$[Cu^{2+}] = \frac{0.1}{1.1 \times 10^{13}} = 9.1 \times 10^{-15} \text{ M}$$

Interpretation: Almost all Cu²⁺ is complexed!

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Problem Type 2: Comparing Stabilities

Example: Which is more stable: [Ni(NH₃)₆]²⁺ (β₆ = 10⁸) or [Ni(en)₃]²⁺ (β₃ = 10¹⁸)?

Solution: log β[Ni(en)₃]²⁺ = 18 log β[Ni(NH₃)₆]²⁺ = 8

Difference: 10¹⁰ times more stable!

Answer: [Ni(en)₃]²⁺ is vastly more stable (chelate effect)

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Applications of Stability Concepts

1. Qualitative Analysis

Example: Detection of Fe³⁺ using SCN⁻

$$Fe^{3+} + SCN^- \rightleftharpoons [Fe(SCN)]^{2+}$$

(blood-red color)

But adding F⁻:

$$[Fe(SCN)]^{2+} + 6F^- \rightleftharpoons [FeF_6]^{3-} + SCN^-$$

Red color disappears because [FeF₆]³⁻ is MORE stable (hard-hard combination)!

2. Metallurgy

Extraction of gold:

$$4Au + 8CN^- + O_2 + 2H_2O \rightarrow 4[Au(CN)_2]^- + 4OH^-$$

Very stable [Au(CN)₂]⁻ (β ≈ 10³⁸) allows gold to be leached from ore.

3. Medicine

EDTA for heavy metal poisoning:

• Forms very stable complexes with Pb²⁺, Hg²⁺, Cd²⁺
• β values > 10¹⁶
• Complexes are water-soluble and excreted

4. Water Softening

Hard water contains Ca²⁺ and Mg²⁺. Adding EDTA:

$$Ca^{2+} + EDTA^{4-} \rightarrow [Ca(EDTA)]^{2-}$$

Stability constant: β ≈ 10¹⁰ This “ties up” Ca²⁺, preventing soap scum formation.

5. Photography

Fixing agent: Sodium thiosulfate (hypo)

$$AgBr(s) + 2S_2O_3^{2-} \rightarrow [Ag(S_2O_3)_2]^{3-} + Br^-$$

The stable thiosulfate complex removes unexposed AgBr from film.

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Memory Tricks

Stability Factors - “CMCL”

Charge density: Higher → More stable Metal hardness: Match with ligand (HSAB) Chelate effect: Rings → More stable Ligand basicity: Better base → More stable

Irving-Williams Order

“Most Friendly Cats Need Careful Zookeepers”

Mn < Fe < Co < Ni < Cu > Zn

Labile vs Inert

“d³ and d⁶ low spin are SLOW” (inert) “d⁰ and d¹⁰ are GO-GO” (labile)

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Common Mistakes

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Practice Problems

Level 1: Basic Concepts

Q1. Arrange in increasing order of stability: a) [Ni(NH₃)₆]²⁺, [Ni(en)₃]²⁺, [Ni(EDTA)]²⁻ b) [FeF₆]³⁻, [FeCl₆]³⁻, [Fe(CN)₆]³⁻

Q2. If β₄ for [Cu(NH₃)₄]²⁺ is 10¹³, what is its dissociation constant K_d?

Q3. Why is [Co(NH₃)₆]³⁺ kinetically inert despite being thermodynamically stable?

Level 2: Application

Q4. Given: K₁ = 10⁴, K₂ = 10³, K₃ = 10² Calculate β₃ for the complex.

Q5. A solution contains 0.01 M [Ag(NH₃)₂]⁺ and 0.1 M free NH₃. Calculate [Ag⁺] if β₂ = 10⁸.

Q6. Explain why: a) [Cr(H₂O)₆]³⁺ exchanges water very slowly b) [Cu(H₂O)₆]²⁺ exchanges water very rapidly

Q7. Which is more stable: [CaEDTA]²⁻ or [Ca(H₂O)₆]²⁺? Explain using the chelate effect.

Level 3: JEE Advanced

Q8. The stability constant for [Cu(NH₃)₄]²⁺ is 10¹³. If excess KCN is added to this solution:

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

Given β for [Cu(CN)₄]²⁻ = 10²⁸, calculate the equilibrium constant for the above reaction.

Q9. Explain the Irving-Williams series using: a) CFSE arguments b) Charge density arguments c) Why Zn²⁺ deviates from the trend

Q10. A student adds Fe³⁺ solution to SCN⁻ and observes blood-red color. When F⁻ is added, the color disappears. But when the student adds EDTA to the colorless solution, the red color DOES NOT return. Explain all observations.

Q11. Calculate ΔG° for the formation of [Ni(en)₃]²⁺ at 298 K if β₃ = 10¹⁸. (R = 8.314 J/mol·K)

Q12. A complex has β = 10⁵ and exchanges its ligands in 10⁻⁶ seconds. Another has β = 10¹⁰ but exchanges ligands in 10⁴ seconds. Which is: a) More thermodynamically stable? b) More kinetically inert? c) Better for long-term storage? d) Better for rapid catalysis?

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Solutions to Selected Problems

Q1. a) [Ni(NH₃)₆]²⁺ < [Ni(en)₃]²⁺ < [Ni(EDTA)]²⁻ (monodentate < bidentate < hexadentate) b) [FeCl₆]³⁻ < [FeF₆]³⁻ < [Fe(CN)₆]³⁻ (based on spectrochemical series and HSAB)

Q2. K_d = 1/β₄ = 10⁻¹³

Q4. β₃ = K₁ × K₂ × K₃ = 10⁴ × 10³ × 10² = 10⁹

Q5.

$$\beta_2 = \frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$$ $$10^8 = \frac{0.01}{[Ag^+](0.1)^2}$$ $$[Ag^+] = 10^{-9} \text{ M}$$

Q8.

$$K_{eq} = \frac{\beta_{[Cu(CN)_4]^{2-}}}{\beta_{[Cu(NH_3)_4]^{2+}}} = \frac{10^{28}}{10^{13}} = 10^{15}$$

Reaction goes essentially to completion!

Q11.

$$\Delta G° = -RT \ln \beta = -8.314 \times 298 \times \ln(10^{18})$$ $$= -8.314 \times 298 \times 18 \times 2.303$$ $$= -103 \text{ kJ/mol}$$

Q12. a) Second (β = 10¹⁰) b) Second (slow exchange = inert) c) Second (stable and doesn’t decompose) d) First (rapid exchange allows catalytic turnover)

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Summary Table

Factor	Effect on Stability	Example
Higher charge	↑ Stability	Al³⁺ > Mg²⁺
Smaller size	↑ Stability	Ni²⁺ > Sr²⁺
Better Lewis base	↑ Stability	CN⁻ > NH₃ > H₂O
Hard-Hard match	↑ Stability	Fe³⁺-F⁻
Soft-Soft match	↑ Stability	Ag⁺-CN⁻
Chelate effect	↑↑ Stability	en > 2NH₃
Macrocyclic	↑↑↑ Stability	Porphyrin
Higher CFSE	↑ Stability	d³, d⁶ (low spin)

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Related Topics

Within Coordination Compounds

• Werner’s Theory — Foundation of coordination chemistry
• Bonding Theories — CFSE and stability
• Crystal Field Theory — CFSE calculations
• Applications — Real-world uses of stable complexes

Cross-Chapter Connections

• Chemical Equilibrium — Equilibrium constants
• Thermodynamics — ΔG, ΔH, ΔS
• d-Block Elements — Metal properties
• Chemical Kinetics — Labile vs inert

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