JEE Main · JEE Advanced · NEET UG46 min study guide10 indexed sections
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Identify coordination spheres, ligand donor atoms and denticity.
Calculate metal oxidation states, coordination numbers and counterion counts.
Name and write common coordination compounds with modern additive ligand names.
Relate stepwise and overall formation constants and apply material balances.
Explain chelation, ligand protonation and complex-assisted dissolution without confusing stability with reaction speed.
BEFORE YOU START
Lewis acids and bases, ionic charges and oxidation states.
Equilibrium constants and solubility products.
Concentration and stoichiometric material balances.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A complex is a connected coordination entity, not the whole salt
A coordination entity contains a central atom, commonly a metal, bonded to surrounding ligands. In the simple Lewis description, ligand donor atoms supply electron pairs to the metal acceptor. The resulting bonding can include covalent and electrostatic contributions; calling a bond coordinate records the electron-pair donation description rather than creating a wholly separate kind of bond after formation.
Square brackets mark the coordination sphere in a formula such as [Co(NH₃)₅Cl]Cl₂. The chloride inside the brackets is a ligand, while the two chlorides outside are counterions. They have the same elemental identity but different connectivity and roles in the compound.
The charge on the complex follows from the central metal’s oxidation state plus the formal charges assigned to its ligands. Counterions make the overall compound electrically neutral. A neutral coordination compound can therefore contain a charged coordination entity, or it can consist of a neutral complex without counterions.
In solution, a sufficiently persistent complex may remain a recognizable entity while its counterions dissociate. Other complexes exchange ligands or change protonation state rapidly. A written solid-state formula does not guarantee that exactly the same species dominates under every solvent, pH and concentration condition.
02 / UNDERSTAND & APPLY
2. Count donor atoms, not just ligand molecules
A monodentate ligand binds through one donor atom to a particular metal, as NH₃ commonly does through nitrogen or H₂O through oxygen. A bidentate ligand can bind through two donors to the same metal; ethane-1,2-diamine, commonly abbreviated en, supplies two nitrogen donors and can form a chelate ring.
Coordination number counts directly attached donor atoms in the ordinary localized-bond examples here. Thus three bidentate en ligands give coordination number six even though only three ligand molecules are present. Oxalate can bind through two oxygens, and EDTA commonly binds through six donor atoms in its hexadentate mode.
Denticity describes the number of donor atoms from one ligand simultaneously attached to one metal. Ambidentate instead means a ligand can bind through alternative donor atoms, not necessarily both at once. Nitrite can bind through nitrogen or oxygen, and thiocyanate through nitrogen or sulfur, giving different linkages.
A bridging ligand links more than one metal center. This is different from chelation to one metal. A single ligand can have several possible modes depending on structure and conditions, so do not infer its actual denticity from the count of every lone pair in its formula.
Ligand charge must also be tracked independently of denticity. NH₃ and en are neutral, chloride and cyanide are −1, oxalate is −2, and the fully deprotonated EDTA ligand is −4. Protonated forms have different charges and may have fewer donor sites available; their abundance depends on pH.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
Local ligand-connectivity sketch, not the full coordination geometry. en is neutral and bidentate in the displayed mode. Counterion dissociation is the stated composition model.
ORIGINAL WORKED EXAMPLE
Separate ligand count from coordination number
For [Co(en)₃]Cl₃, find ligand-molecule count, coordination number, complex charge and cobalt oxidation state.
There are three en molecules in the coordination sphere.
Each en is bidentate, so there are six donor atoms and coordination number six.
Three external chlorides require the complex to have charge +3.
en is neutral, so cobalt oxidation state is +3.
Answer Three ligands; coordination number six; complex charge +3; Co(III).
03 / UNDERSTAND & APPLY
3. Charge algebra determines the metal oxidation state
For a mononuclear complex, write x+Σν_iq_i=q_complex, where x is the metal oxidation state and q_i each ligand charge. Neutral ligands contribute zero even when they donate electron pairs. Donation is a bonding description and does not mean a neutral NH₃ ligand becomes formally −2 in routine oxidation-state accounting.
In [Fe(CN)₆]⁴⁻, six cyanido ligands contribute −6, so x−6=−4 and iron is +2. The coordination number is six in the usual carbon-bound monodentate description. Neither the −4 complex charge nor the six donor atoms is the iron oxidation state.
When the compound formula includes counterions, first derive the complex charge by overall electroneutrality. For K₂[PtCl₆], two K⁺ ions imply a 2− complex; six Cl⁻ ligands then give platinum +4. Counterions are not counted in coordination number unless the stated structure places them in the coordination sphere.
Composition can constrain but not always uniquely determine structure. A coordination number of four may correspond to tetrahedral or square-planar geometry depending on metal and ligands. Geometry, isomerism and ligand-field electronic structure require additional evidence and are developed in separate lessons.
metal oxidation state + Σ(ligand count × ligand charge)=complex charge
Keep ligand charge, denticity and coordination number as separate quantities.
ORIGINAL WORKED EXAMPLE
Analyze an anionic complex salt
For K₃[Cr(C₂O₄)₃], assume each oxalate is bidentate. Find chromium oxidation state and coordination number.
Three K⁺ counterions require complex charge −3.
Three oxalate ligands contribute 3×(−2)=−6.
x−6=−3 gives chromium +3.
Each of three oxalates has two donors, so coordination number is six.
Answer Cr(III), coordination number six, complex charge −3.
04 / UNDERSTAND & APPLY
4. Build the name from ligands, metal and charge information
For an ionic coordination compound, name the cation before the anion. Within a coordination entity, ligand names precede the metal name and are ordered alphabetically by ligand name, ignoring ordinary multiplicative prefixes such as di, tri and tetra for that ordering. Use multiplicative prefixes to show how many identical ligands are present.
Common neutral ligand names include aqua for H₂O, ammine with two m letters for NH₃, and carbonyl for carbon-bound CO. Modern additive names for halide ligands include fluorido, chlorido, bromido and iodido; cyanide is cyanido. Older chloro and cyano forms still appear in historical teaching materials, but this lesson uses the modern forms.
Anionic ligand names generally change ide to ido, ate to ato, and ite to ito in additive nomenclature. More complicated ligand names, especially those already containing numerical prefixes or locants, use bis, tris or tetrakis with parentheses to avoid ambiguity. Thus three ethane-1,2-diamine ligands are described with tris(ethane-1,2-diamine).
The metal name is followed by its oxidation state in Roman numerals in parentheses in the common oxidation-state naming system used here. If the coordination entity is an anion, the metal name takes an ate ending, sometimes with a conventional stem such as ferrate for iron. Cationic and neutral complexes retain the ordinary metal name.
Donor connectivity can be specified with κ notation where needed, such as nitrito-κN versus nitrito-κO. Geometrical descriptors such as cis/trans and fac/mer add information beyond the composition-only name when a definite stereoisomer is intended. A formula that does not specify geometry should not be assigned an arbitrary stereodescriptor.
To write a formula from a name, identify the metal oxidation state, ligand charges and counts, calculate the coordination-entity charge, then supply enough counterions for neutrality. The Roman numeral is the metal oxidation state, not the total complex charge. Check both after assembling the formula.
ORIGINAL WORKED EXAMPLE
Name a cationic complex salt
Name [Co(NH₃)₄Cl₂]Cl using the oxidation-state system, without assigning geometry.
The outside chloride means the coordination entity has charge +1.
Inside, four ammines are neutral and two chlorido ligands contribute −2, so cobalt is +3.
Alphabetical order places ammine before chlorido; counts give tetraammine and dichlorido.
The complex is cationic, so use cobalt rather than cobaltate, followed by external chloride.
Answer Tetraamminedichloridocobalt(III) chloride.
ORIGINAL WORKED EXAMPLE
Write a formula from a name
Write the formula for potassium hexacyanidoferrate(II).
Iron is +2 and six cyanido ligands contribute −6.
The coordination entity therefore has charge −4: [Fe(CN)₆]⁴⁻.
Four potassium ions are needed to balance that charge.
Each cyanido ligand is counted once in the formula; its −1 charge is distinct from its donor bond.
Answer K₄[Fe(CN)₆].
05 / UNDERSTAND & APPLY
5. Counterion tests probe connectivity only under stated kinetic conditions
Werner-type composition questions compare total elemental analysis with ions released into solution. In the ideal model, [Co(NH₃)₆]Cl₃ releases three external chloride ions per formula unit, [Co(NH₃)₅Cl]Cl₂ releases two, and [Co(NH₃)₄Cl₂]Cl releases one. Their coordinated chloride counts differ even though related gross compositions may look similar.
Adding Ag⁺ can precipitate freely available chloride as AgCl. Interpreting the precipitate as an external-chloride count assumes a suitable short observation interval and a complex whose coordinated chlorides do not substantially dissociate or react during the test. Over longer times or with labile complexes, this simple inference can fail.
In an ideal complete-dissociation model, [Co(NH₃)₅Cl]Cl₂ produces one complex cation plus two chloride anions, giving three solute ions per formula unit. This particle count can be linked to ideal colligative behavior, but real ion pairing and incomplete dissociation can make measured particle factors differ.
Double salts and coordination compounds illustrate another distinction. A double salt may dissociate into its constituent simple ions under the stated conditions, whereas a sufficiently stable coordination entity can persist as a complex ion. Neither classification is a substitute for examining the actual aqueous equilibria.
ORIGINAL WORKED EXAMPLE
Use a rapid external-chloride test
A 0.0100 mol sample of [Co(NH₃)₅Cl]Cl₂ is tested under ideal conditions that precipitate only external chloride. How many moles AgCl form, and how many moles of ions result from ideal salt dissociation?
There are two external chlorides per formula unit.
AgCl amount is 2×0.0100=0.0200 mol.
Each formula unit yields one complex cation plus two chloride anions, three ions in total.
Total ion amount is 3×0.0100=0.0300 mol, assuming ideal complete dissociation and persistent complex.
Answer 0.0200 mol AgCl; 0.0300 mol dissolved ions in the stated model.
06 / UNDERSTAND & APPLY
6. Stepwise constants multiply to give an overall constant
Complex formation can be written in stages: M+L⇌ML with K₁, then ML+L⇌ML₂ with K₂, and so on. Multiplying equilibrium expressions cancels intermediate complexes. The overall constant for M+nL⇌ML_n is β_n=K₁K₂…K_n.
Thermodynamic constants use dimensionless activities. In dilute classroom calculations, concentration ratios with an implicit standard-state convention are commonly used: β_n≈[ML_n]/([M][L]ⁿ). The free ligand concentration belongs in this expression, not automatically the total amount of ligand added.
The dissociation constant for exactly the reverse overall reaction is 1/β_n. It is not the reciprocal of one arbitrary stepwise constant unless that step is the reaction being reversed. Likewise, β₂ is not K₁+K₂: equilibrium constants multiply when reaction equations are added.
A large formation constant means thermodynamic favorability relative to the specified free species and conditions. It does not mean ligand substitution is necessarily fast. A thermodynamically stable complex can be kinetically labile, while an inert complex can persist because its ligand-exchange barrier is high.
β_n=ΠK_i≈[ML_n]/([M][L]ⁿ)
Overall formation; concentration form assumes the stated dilute standard-state convention.
K_d,overall=1/β_n
Only for the exactly reversed overall formation reaction.
ORIGINAL WORKED EXAMPLE
Combine two ligand-binding steps
For a hypothetical system, K₁=2.00×10⁵ and K₂=5.00×10³ in a consistent convention. Find β₂ and the reverse overall dissociation constant.
Add M+L⇌ML and ML+L⇌ML₂ to obtain M+2L⇌ML₂.
β₂=K₁K₂=(2.00×10⁵)(5.00×10³)=1.00×10⁹.
Reverse the overall equation to obtain K_d=1/β₂=1.00×10⁻⁹.
Adding the numerical constants would not represent the combined equilibrium.
Answer β₂=1.00×10⁹; K_d=1.00×10⁻⁹.
07 / UNDERSTAND & APPLY
7. Total metal can be large while free metal is small
For a simple 1:1 system with only M, L and ML, metal balance is C_M=[M]+[ML] and ligand balance is C_L=[L]+[ML]. Together with K_f=[ML]/([M][L]), these equations determine the free concentrations. Ignoring either balance can violate conservation even if a guessed concentration satisfies the equilibrium ratio.
If ligand is present in substantial excess and binding is strong, first estimate [ML]≈C_M and [L]≈C_L−C_M. Then [M]≈[ML]/(K_f[L]). Check afterward that the free-metal amount is small relative to C_M and that the ligand approximation is self-consistent.
When free ligand is externally buffered or maintained, combine the equilibrium relation with the metal balance to obtain [M]=C_M/(1+K_f[L]) for a 1:1 complex. For several ML_i species, C_M=[M](1+β₁[L]+β₂[L]²+…), so the metal distribution depends on all relevant formation terms.
Free-ion activity can control precipitation, electrode potential and biological binding even when total dissolved metal changes little. The phrase metal concentration is therefore incomplete when speciation matters. Identify whether an analytical method measures total metal or only a particular free species.
ORIGINAL WORKED EXAMPLE
Estimate a strongly bound 1:1 system
A hypothetical solution has total metal 0.0100 M, total ligand 0.0300 M and Kf=1.00×10⁶ M⁻¹ in the concentration model. Assume only M, L and ML. Estimate free metal.
Strong binding suggests [ML]≈0.0100 M and excess free ligand≈0.0200 M.
[M]≈0.0100/(1.00×10⁶×0.0200)=5.00×10⁻⁷ M.
This is only 0.005% of total metal, supporting [ML]≈C_M.
The corresponding correction to free ligand is also tiny relative to 0.0200 M.
Answer Approximately 5.00×10⁻⁷ M free metal.
08 / UNDERSTAND & APPLY
8. Chelation and protonation modify effective binding
A chelating ligand uses multiple donor atoms to bind one metal and creates one or more rings. Relative to comparable monodentate ligands, chelation often favors complex formation through entropy and other structural contributions. The effect is not simply “two bonds are always twice as strong”: ligand geometry, ring strain, solvation and enthalpy also matter.
After one donor attaches, another donor on the same ligand is locally available to bind, which can make complete dissociation less favorable relative to comparable separate ligands. However, an unsuitable donor spacing or strained ring can reduce that advantage. Chelate stability must be considered for the actual metal and ligand.
Protonation can remove a donor lone pair from availability. For example, NH₄⁺ is not equivalent to free NH₃ as a nitrogen donor. Acidifying a metal–ammine solution can lower free NH₃ and shift complex dissociation even if the intrinsic formation constant for M+NH₃ binding has not changed.
For a ligand with several protonation states, a conditional formation constant can fold the fraction of the binding-active form into the equilibrium description at fixed pH. In a simple model where only L binds, K_cond=K_fα_L if the ligand concentration in the conditional expression means total unbound ligand across its protonation states. More complex systems can also need metal hydrolysis and competing-ligand corrections.
09 / UNDERSTAND & APPLY
9. Complexation can dissolve more solid while lowering free metal
For a sparingly soluble salt MX, dissolution gives M and X with K_sp. If ligand L binds M strongly, the free-metal activity falls and more solid can dissolve to restore the solubility-product condition. Total dissolved metal increases even though free M remains low, a distinction that resolves an apparent contradiction.
For MX(s)⇌M+X followed by M+nL⇌ML_n, the net reaction is MX(s)+nL⇌ML_n+X and its constant is K_spβ_n. Adding the reaction equations requires multiplying their constants. The solid’s activity is one while the solid phase is present.
The amount dissolved depends on free ligand, counterion concentration, finite ligand inventory and possible competing equilibria. A large β alone does not guarantee complete dissolution of any amount of solid. If binding consumes most ligand, free ligand falls and the driving force changes.
A complete analysis may combine solubility product, formation constants, acid–base equilibria, metal and ligand balances, and charge balance. Begin with a minimal justified model and check its assumptions rather than writing a large formation constant as a substitute for the balances.
ORIGINAL WORKED EXAMPLE
Combine dissolution and binding
A hypothetical 1:1 salt MX has Ksp=2.00×10⁻¹⁰. Its metal forms ML₂ with β₂=5.00×10⁸ in a compatible activity convention. Find the constant for MX(s)+2L⇌ML₂+X.
Write the dissolution and complex-formation equations and cancel free M.
The product is dimensionless in the specified activity convention.
The value alone is not a solubility; free ligand and product activities still enter the equilibrium expression.
Answer K_net=0.100.
10 / UNDERSTAND & APPLY
10. Eight solved composition and naming checks
Keep charge, coordination number, ligand count and thermodynamic concentration distinct in every answer.
Count donor atoms for coordination number.
Do not confuse metal oxidation state with complex charge.
Use free ligand in the formation expression.
Multiply stepwise constants; do not add them.
State the kinetic assumption behind external-chloride tests.
Modern chlorido/cyanido names differ from common older textbook forms.
Chelation and pH effects require the actual ligand form and geometry.
ORIGINAL PRACTICE · 1
Find the metal oxidation state in [Cu(NH₃)₄]²⁺.
Reveal answer & reasoning
Cu(II).
All four NH₃ ligands are neutral, so the metal oxidation state equals the +2 complex charge.
ORIGINAL PRACTICE · 2
Find coordination number in [Cr(C₂O₄)₂(H₂O)₂]⁻ if oxalate is bidentate.
Reveal answer & reasoning
Six.
Two oxalates provide four donors and two waters provide two more. The number of ligand molecules is four, not six.
ORIGINAL PRACTICE · 3
Find chromium oxidation state in that same complex.
Reveal answer & reasoning
+3.
Two oxalates contribute −4 and waters zero; x−4=−1 gives x=+3.
ORIGINAL PRACTICE · 4
Name K₂[PtCl₆] in the oxidation-state system.
Reveal answer & reasoning
Potassium hexachloridoplatinate(IV).
The complex has charge −2; six chlorido ligands give Pt +4. Anionic coordination entities use the ate metal form.
ORIGINAL PRACTICE · 5
Is an ambidentate ligand automatically bidentate?
Reveal answer & reasoning
No.
Ambidentate means alternative donor sites can bind; bidentate means two donors bind simultaneously to the same metal in the specified mode.
ORIGINAL PRACTICE · 6
If β₃=10¹² for M+3L⇌ML₃, what is the constant for ML₃⇌M+3L?
Reveal answer & reasoning
10⁻¹².
The exactly reversed reaction has the reciprocal equilibrium constant.
ORIGINAL PRACTICE · 7
At fixed free [L], a 1:1 binding system has Kf[L]=99. What fraction of total metal is free?
Reveal answer & reasoning
1.00%.
[M]/C_M=1/(1+Kf[L])=1/100=0.0100, assuming only free M and ML.
ORIGINAL PRACTICE · 8
Does a large formation constant prove rapid ligand exchange?
Reveal answer & reasoning
No.
Formation constants describe equilibrium thermodynamics; substitution rates depend on activation barriers. Stability and kinetic inertness are different concepts.
KEEP THE BIG IDEAS
Quick revision
A coordination sphere and its external counterions have different structural roles.
Ligand charge and denticity are independent properties.
Naming requires ligand identity/count, metal oxidation state and complex charge.
Overall binding constants multiply from stepwise constants.
Material balances determine free and total concentrations.
Complex formation can enhance total solubility while suppressing free metal.
Thermodynamic stability is not the same as kinetic inertness.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Official technical-report HTML inspected for additive names, chlorido/ido endings, multiplicative prefixes and connectivity notation; no PDF downloaded.