9.3 Coordination Compounds & Complex Ion Formation

9.3 Coordination Compounds & Complex Ion Formation

Quick Summary: Coordination compounds (or complex ions) form when a central transition metal cation, functioning as a Lewis acid, binds to surrounding neutral molecules or anions, known as ligands, functioning as Lewis bases. The resulting linkages are coordinate covalent bonds. Polydentate ligands wrap around the central metal to form chelates of remarkable thermodynamic stability driven by a large gain in entropy ($\Delta S^\circ > 0$). Complex ion formation equilibria, quantified by massive formation constants ($K_f$), can couple with sparingly soluble salts to dissolve precipitates and account for the amphoteric dissolution of metal hydroxides in excess base.

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1. Anatomy of Coordination Compounds and Coordinate Covalent Bonding

A coordination compound consists of a central metal ion surrounded by an array of bound ions or molecules:

• Central Metal Cation (Lewis Acid): Transition metals or post-transition metals with vacant valence $d$, $s$, and $p$ orbitals capable of accepting electron pairs (e.g., $\text{Fe}^{2+}, \text{Fe}^{3+}, \text{Cu}^{2+}, \text{Ag}^+, \text{Co}^{3+}, \text{Pt}^{2+}$).
• Ligands (Lewis Bases): Neutral molecules or anions possessing one or more unshared lone pairs that are donated to the vacant metal orbitals.
• Coordinate Covalent (Dative) Bond: A covalent bond wherein both shared bonding electrons are supplied exclusively by one of the participating atoms (the ligand donor atom).
• Coordination Sphere: Enclosed within square brackets $[ \dots ]$, the coordination sphere encompasses the central metal ion and all directly attached ligands. Species residing outside the brackets are uncoordinated counter-ions, which dissociate completely when the compound is dissolved in water: $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3(s) \xrightarrow{\text{H}_2\text{O}} [\text{Co}(\text{NH}_3)_6]^{3+}(aq) + 3\text{Cl}^-(aq)$ In this example, the hexaamminecobalt(III) cation remains intact as a stable polyatomic ion, while three chloride counter-ions dissociate into solution.

Coordination Number and Spatial Geometries

The coordination number (CN) is the total number of donor atom bonds directly attached to the central metal cation:

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2. Ligand Taxonomy and the Chelate Effect

Ligands are classified according to their denticity (from Latin dens, tooth)—the number of donor atoms through which they can simultaneously bind to a single metal center:

1. Monodentate Ligands

Bind through a single donor atom possessing a lone pair:

• Neutral: Water ($\text{H}_2\text{O}$, IUPAC name: aqua), Ammonia ($\text{NH}_3$, IUPAC name: ammine), Carbon monoxide ($\text{CO}$, IUPAC name: carbonyl).
• Anionic: Fluoride ($\text{F}^-$, fluorido), Chloride ($\text{Cl}^-$, chlorido), Cyanide ($\text{CN}^-$, cyanido), Hydroxide ($\text{OH}^-$, hydroxido), Thiocyanate ($\text{SCN}^-$, thiocyanato). Current IUPAC names end in -ido; many textbooks and exam items still use the older forms fluoro, chloro, cyano, and hydroxo, so recognize both.

2. Bidentate and Polydentate (Chelating) Ligands

Possess two or more spatially separated donor atoms that coordinate simultaneously to the same metal ion, forming stable five- or six-membered heterocyclic rings termed chelates:

• Ethylenediamine ($en$, $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$): Bidentate; binds through two nitrogen lone pairs.
• Oxalate ion ($ox^{2-}$, $\text{C}_2\text{O}_4^{2-}$): Bidentate; binds through two carboxylate oxygen atoms.
• Ethylenediaminetetraacetate ($\text{EDTA}^{4-}$): Hexadentate; contains two amine nitrogen atoms and four carboxylate oxygen atoms, wrapping completely around a metal ion in an octahedral cage.

The Chelate Effect: Thermodynamic and Entropic Driving Force

Coordination complexes containing polydentate chelating ligands exhibit markedly greater thermodynamic stability (higher formation constants $K_f$) than analogous complexes containing equivalent monodentate ligands. Consider the ligand displacement equilibrium: $[\text{Ni}(\text{NH}_3)_6]^{2+}(aq) + 3\text{en}(aq) \rightleftharpoons [\text{Ni}(\text{en})_3]^{2+}(aq) + 6\text{NH}_3(aq)$

• Equilibrium constant: Tabulated overall formation constants are about $10^{8.61}$ for $[\text{Ni}(\text{NH}_3)_6]^{2+}$ and $10^{18.28}$ for $[\text{Ni}(\text{en})_3]^{2+}$, so $K$ for the exchange is about $10^{9.67}$ (roughly $5 \times 10^9$) and $\Delta G^\circ = -RT\ln K \approx -54\text{ kJ/mol}$ at 298 K.
• Entropy ($\Delta S^\circ$): On the reactant side, $1$ complex ion reacts with $3$ $en$ molecules ($4$ solute particles total). On the product side, $1$ chelate complex ion releases $6$ free $\text{NH}_3$ molecules ($7$ solute particles total). The net creation of independent particles ($\Delta n = +3$) gives a favorable entropy change of about $+88\text{ J/(mol}\cdot\text{K)}$, so $-T\Delta S^\circ \approx -25\text{ kJ/mol}$.
• Enthalpy ($\Delta H^\circ$): Both complexes have six Ni–N bonds, yet $\Delta H^\circ$ is still about $-29\text{ kJ/mol}$ in the measured data. Older textbooks quoted $\Delta H^\circ \approx -12\text{ kJ/mol}$ and called the chelate effect purely entropic; the measured breakdown shows the entropy and enthalpy terms are comparable. For exam purposes, remember that the gain in free particles (entropy) is the distinctive reason chelates beat monodentate analogues.

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3. Systematic Nomenclature of Coordination Compounds

IUPAC nomenclature follows rigorous sequential conventions:

1. Cation precedes Anion: In any ionic salt, the cation is named before the anion.
2. Within the Coordination Entity: Ligands are listed in alphabetical order by their chemical name (ignoring multiplying prefixes like di-, tri-, tetra-), followed immediately by the central metal.
3. Ligand Names:
   • Anionic ligands end in -o (e.g., $\text{Cl}^-$ is chlorido, $\text{CN}^-$ is cyanido, $\text{OH}^-$ is hydroxido, $\text{C}_2\text{O}_4^{2-}$ is oxalato; the older names chloro, cyano, and hydroxo are still widely used).
   • Neutral ligands retain standard names, with four historical exceptions: $\text{H}_2\text{O}$ (aqua), $\text{NH}_3$ (ammine with two m's), $\text{CO}$ (carbonyl), and $\text{NO}$ (nitrosyl).
4. Multiplying Prefixes: Use di-, tri-, tetra-, penta-, hexa- for simple ligands. For polydentate or complex ligands containing existing numerical prefixes (e.g., ethylenediamine), use bis-, tris-, tetrakis-, enclosing the ligand name in parentheses.
5. Metal Suffix and Oxidation State:
   • If the complex is a cation or neutral molecule, the standard English name of the metal is used (e.g., cobalt, platinum, chromium).
   • If the complex is an anion, the metal name takes the Latinate suffix -ate (e.g., iron $\to$ ferrate, copper $\to$ cuprate, silver $\to$ argentate, tin $\to$ stannate, lead $\to$ plumbate, gold $\to$ aurate).
   • The formal oxidation state of the metal is indicated by a Roman numeral in parentheses immediately following the metal name.

Examples:

• $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$: hexaamminecobalt(III) chloride
• $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$: diamminedichloridoplatinum(II)
• $\text{K}_4[\text{Fe}(\text{CN})_6]$: potassium hexacyanoferrate(II)
• $[\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl}$: tetraaquadichloridochromium(III) chloride

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4. Complex Ion Equilibria: Formation Constants ($K_f$)

The formation of a complex ion in aqueous solution occurs through stepwise ligand substitution replacing coordinated solvent water molecules: $\text{M}^{m+}(aq) + n\text{L}(aq) \rightleftharpoons [\text{ML}_n]^{m+}(aq)$

The equilibrium expression is governed by the formation constant ($K_f$) (or stability constant): $K_f = \frac{[[\text{ML}_n]^{m+}]}{[\text{M}^{m+}][\text{L}]^n}$ The inverse of the formation constant is the dissociation constant ($K_d = 1 / K_f$). Because $K_f$ values for transition metal complexes are exceptionally large ($10^7$ to $10^{35}$), the position of equilibrium lies overwhelmingly to the right.

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5. Dissolution of Insoluble Precipitates via Coupled Equilibria

A sparingly soluble salt can be induced to dissolve by introducing a Lewis base ligand that sequesters the free aqueous metal cation into a highly stable complex ion, shifting the solubility equilibrium to the right in accordance with Le Chatelier's principle.

Case Study: Dissolving Insoluble Silver Halides

Solid silver chloride is sparingly soluble in water: $\text{AgCl}(s) \rightleftharpoons \text{Ag}^+(aq) + \text{Cl}^-(aq) \quad K_{sp} = 1.8 \times 10^{-10}$

When aqueous ammonia is added, free $\text{Ag}^+$ ions combine with $\text{NH}_3$ to produce the diamminesilver(I) complex ion: $\text{Ag}^+(aq) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) \quad K_f = 1.7 \times 10^7$

Adding these two chemical equations yields the overall dissolution reaction: $\text{AgCl}(s) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$

The equilibrium constant for the coupled process ($K_{net}$) is the product of the individual equilibrium constants: $K_{net} = K_{sp} \times K_f = (1.8 \times 10^{-10})(1.7 \times 10^7) = 3.1 \times 10^{-3}$

Selective Halide Differentiation

While $K_{net} = 3.1 \times 10^{-3}$ is sufficiently large to dissolve substantial amounts of $\text{AgCl}$ in $1\text{ M } \text{NH}_3$, silver iodide ($\text{AgI}$) has a far smaller solubility product ($K_{sp} = 8.5 \times 10^{-17}$): $K_{net(\text{AgI})} = (8.5 \times 10^{-17})(1.7 \times 10^7) = 1.4 \times 10^{-9}$ Because $K_{net}$ is negligible, $\text{AgI}$ does not dissolve in aqueous $\text{NH}_3$. However, $\text{AgI}$ dissolves readily in aqueous sodium thiosulfate ($\text{Na}_2\text{S}_2\text{O}_3$, photographic "hypo") or potassium cyanide ($\text{KCN}$), because their respective formation constants ($K_f = 2.9 \times 10^{13}$ and $1.0 \times 10^{21}$) are massive enough to overcome $\text{AgI}$'s tiny $K_{sp}$.

Worked Coupled Equilibrium Calculation

Problem: Calculate the molar solubility of silver chloride ($\text{AgCl}$) in a $1.00\text{ M}$ aqueous ammonia solution at 25 °C. Compare this with its solubility in pure water ($s = \sqrt{K_{sp}} = 1.34 \times 10^{-5}\text{ M}$).

Solution:

1. Set up an equilibrium analysis for the coupled reaction: $\text{AgCl}(s) + 2\text{NH}_3(aq) \rightleftharpoons [\text{Ag}(\text{NH}_3)_2]^+(aq) + \text{Cl}^-(aq)$
   • Initial concentrations: $[\text{NH}_3] = 1.00\text{ M}$, $[[\text{Ag}(\text{NH}_3)_2]^+] = 0$, $[\text{Cl}^-] = 0$
   • Equilibrium concentrations: $[\text{NH}_3] = (1.00 - 2s)\text{ M}$, $[[\text{Ag}(\text{NH}_3)_2]^+] = s\text{ M}$, $[\text{Cl}^-] = s\text{ M}$
2. Substitute equilibrium expressions into $K_{net}$: $K_{net} = \frac{[[\text{Ag}(\text{NH}_3)_2]^+][\text{Cl}^-]}{[\text{NH}_3]^2} = \frac{s \times s}{(1.00 - 2s)^2} = \frac{s^2}{(1.00 - 2s)^2} = 3.1 \times 10^{-3}$
3. Take the square root of both sides: $\frac{s}{1.00 - 2s} = \sqrt{3.1 \times 10^{-3}} = 0.0557$
4. Solve for molar solubility $s$: $s = 0.0557(1.00 - 2s) = 0.0557 - 0.1114s$ $1.1114s = 0.0557 \implies s = \frac{0.0557}{1.1114} = 0.0501\text{ M}$

In $1.00\text{ M } \text{NH}_3$, the solubility of $\text{AgCl}$ is $0.0501\text{ M}$ ($5.01 \times 10^{-2}\text{ M}$), representing a roughly 3,700-fold increase in solubility relative to pure water ($1.34 \times 10^{-5}\text{ M}$).