5. Coordination Compounds

What you will learn

• Identify sphere, counterions, donor atoms and denticity.
• Apply oxidation-state and naming rules.
• Draw geometric and optical arrangements in 2D and inspect the same connectivity in 3D.
• Compare valence-bond and crystal-field descriptions.
• Predict simple spin and colour behaviour with stated assumptions.
• Explain synergic metal–carbonyl bonding and coordination applications.

Big idea: the arrangement around a metal can matter as much as the identity of the metal.

A molecular formula counts species, but it may not show which atoms touch the centre or where they sit. Brackets, donor atoms and spatial labels supply the missing structure. This is why two samples with the same overall composition can behave differently.

Use the paired diagrams actively. First copy the 2D arrangement and identify its labels. Then rotate the matching model. Finish by redrawing the structure without the screen. A beautiful model is useful only if you can translate it into a clear exam diagram.

Werner explained observations that simple valence counting could not organise: counterion behaviour, fixed coordination numbers and isomers with distinct spatial arrangements. His coordination theory separated what is directly bound from what only balances charge, making structure testable through chemical evidence.

A coordination compound asks you to distinguish atoms attached directly to a metal from ions that merely balance the compound’s overall charge.

A coordination entity contains a central atom or ion bonded to surrounding ligands. A ligand supplies an electron pair through a donor atom. The metal–ligand arrangement can remain recognisable in solution, although some complexes exchange ligands readily and others do so slowly.

Werner distinguished a primary valence, now connected with oxidation state, from a secondary valence, now connected with coordination number. The surrounding ligands have definite spatial arrangements. This explains why equal overall compositions can produce different numbers of free counterions and different isomers.

For $[\mathrm{Cr(H_2O)_4Cl_2}]\mathrm{Cl}\cdot2\mathrm{H_2O}$, identify the coordination sphere, counterion and uncoordinated water.

Only the species inside brackets belong to the written coordination sphere. The chloride after the brackets balances charge. The dotted water is water of crystallisation in this formula.

The four water donors and two chlorido donors give six direct donor atoms. Chromium’s oxidation state follows from the bracket charge and ligand charges.

Result: The entity is $[\mathrm{Cr(H_2O)_4Cl_2}]^+$, there is one outside chloride per formula unit, and two waters lie outside the written sphere. The coordination number is six and chromium is in oxidation state +3.

A double salt typically gives its ordinary constituent ions on dissolving. A complex salt can give complex ions as identifiable entities as well as counterions. The distinction does not claim that every complex survives forever or that no ligand equilibrium exists; it concerns the species relevant under the stated solution conditions.

A precipitation observation can help distinguish outer chloride from bound chlorido, but the timescale and ligand-exchange conditions matter. Do not equate total chlorine content with instantly free chloride.

The brackets are therefore chemical information. Removing them to make a formula shorter can erase the distinction that the question is testing.

1. Does water of crystallisation automatically coordinate to the metal?
2. What did Werner’s secondary valence help describe?

Answers

1. No. Its placement and structural information must be checked.
2. The number and spatial arrangement of groups directly coordinated to the centre.

Learn the vocabulary by pointing to a specific complex. Definitions are easier when every term has a visible object.

A monodentate ligand binds through one donor atom. A bidentate ligand binds through two donor atoms to the same centre. A polydentate ligand can supply more. When a ligand binds through several donors and forms a ring involving the metal, the structure is chelated.

Ambidentate means a ligand has alternative donor atoms and can bind through one or the other in different structures. It does not mean both necessarily bind together. Nitrite and thiocyanate provide useful examples. Denticity and ambidentate behaviour answer different questions.

Find oxidation state, d count and coordination number for $[\mathrm{Co(CN)_6}]^{3-}$.

Each cyanido ligand has formal charge minus one. Let the metal oxidation state be $x$: $x+6(-1)=-3$. Each ligand is monodentate in this formula.

Use the first-row group-number method for the formal d count.

Result: Co is +3, its formal count is $d^{6}$, and its coordination number is 6.

A metal binds three oxalato ligands, each using two oxygen donors. How many donor atoms are directly attached?

Multiply the number of ligands by the denticity. The formula contains three ligands, not three donor atoms.

Result: 6 donor atoms; the coordination number is six.

Oxidation state is formal electron bookkeeping. It is not the same as the overall complex charge and does not claim that a covalent bond places every electron completely on one atom.

1. Can a complex have three ligands and coordination number six?
2. Are complex charge and metal oxidation state generally equal?

Answers

1. Yes, if all three ligands bind through two donor atoms each.
2. No. Ligand charges contribute to the complex charge.

A coordination name should let another reader rebuild the formula without guessing the metal’s charge or the ligand count.

Write the central metal first inside brackets, followed by ligand formulas. State the overall entity charge when the ion is written alone. Counterions make a complete salt electrically neutral. Use parentheses around a multi-atom ligand when the count needs grouping.

Name ligands before the metal. Alphabetise ligand names without letting ordinary multiplicative prefixes decide the order. Use di-, tri- and related prefixes for simple counts; use bis-, tris- and related forms with suitable parentheses for more complicated ligand names. The metal oxidation state appears in Roman numerals.

Ammonia is named ammine, with a double m. Water is aqua. Common anionic ligand names include chlorido, cyanido and hydroxido. A negative coordination entity uses an anionic metal name, often ending in -ate; some metals have established Latin-derived roots. A positive or neutral entity generally uses the usual metal name.

Name $\mathrm{Na_3[Cr(C_2O_4)_3]}$.

The three sodium ions require a complex charge of minus three. Three oxalato ligands contribute minus six, so chromium is plus three. The negative entity takes the metal ending -ate.

The multi-atom ligand name is grouped with tris.

Result: Sodium tris(oxalato)chromate(III). The chromium state is +3.

Write the formula of tetraamminecopper(II) sulfate.

Ammine is neutral, so the copper complex is doubly positive. Sulfate is doubly negative. One of each gives a neutral salt.

Result: $[\mathrm{Cu(NH_3)_4}]\mathrm{SO_4}$.

A systematic name is a compact structural description, but not always a complete three-dimensional structure. A name may also require a stereochemical prefix such as cis or trans to distinguish isomers with the same basic composition.

1. Why does “ammine” have different spelling from the organic class “amine”?
2. When is an -ate metal name used?

Answers

1. It is the established ligand name for coordinated ammonia and should not be confused with the organic functional-group class.
2. When naming an anionic coordination entity, following the applicable established name.

If two compounds share a formula, ask two separate questions: are the connections the same, and is the arrangement in space the same?

Structural isomers differ in which species are connected or where species lie inside and outside the coordination sphere. Stereoisomers have the same connectivity but a different spatial arrangement. A redrawing or a rotation of the entire molecule does not automatically create a new isomer.

In a square-planar arrangement with two ligands of each of two types, identical ligands may be adjacent, called cis, or opposite, called trans. A tetrahedral arrangement of the same simple ligand pattern does not produce that cis/trans pair, because all four corners are equivalent under rotation.

Octahedral arrangements can also show cis and trans for two identical ligands among four of another type. For three ligands of each type, fac puts one set on a triangular face; mer puts them along a meridional arrangement including an opposite pair. These descriptions refer to donor positions, not to whether the printed formula looks symmetrical.

A chiral complex and its mirror image cannot be superposed by rotating the complete object. Three bidentate ligands around an octahedral centre can generate this kind of handed arrangement. A mirror image is not a rotation. Optical activity also depends on sample composition: an equal mixture of opposite enantiomers can show no net rotation.

An ambidentate ligand can attach through different donor atoms. Nitrite may bind through nitrogen or oxygen. The ligand formula is unchanged, but the donor-to-metal connection differs. Specify the donor rather than guessing it from an unqualified condensed formula.

When both a cation and an anion are coordination entities, exchanging the ligand distribution between the two centres can give coordination isomers. Keep total composition and overall salt neutrality the same while identifying which ligands surround each metal.

A coordinated ligand and an outer counterion can exchange roles, producing different ions on dissolution. Distinct precipitation tests can follow because the freely available counterions differ. Again, the observation depends on relevant exchange conditions.

Solvent molecules can occupy different positions inside and outside the coordination sphere. When the solvent is water, the special name hydrate isomerism is often used. Counting all waters in the overall composition does not by itself identify which waters are coordinated.

A square-planar complex has two ammine and two chlorido ligands. How many cis/trans arrangements are possible?

Draw one arrangement with chlorido ligands adjacent and another with them opposite. Check whether a rotation can turn one into the other without exchanging positions.

Result: 2 arrangements, cis and trans. A formula without spatial information does not distinguish them.

1. What must change in a linkage isomer?
2. What is the key distinction in hydrate isomerism?

Answers

1. The donor atom through which an ambidentate ligand attaches.
2. Which water molecules are coordinated and which are outside the coordination sphere.

Bonding models help answer different questions. Use the model that addresses geometry, spin or colour, and keep its limitations visible.

In the traditional valence-bond description, suitable metal orbitals accept ligand electron pairs. Common school-level labels are $sp^3$ for tetrahedral, $dsp^2$ for square planar, and $d^2sp^3$ or $sp^3d^2$ for octahedral arrangements. These labels organise geometry and pairing pictures; they do not replace a modern molecular-orbital treatment of the bonds.

Find oxidation state and d count first. Then determine how the geometry and ligand field affect occupation. Counting unpaired electrons from a free ion without considering the actual complex can give the wrong answer.

Compare the usual school-level pictures of $[\mathrm{NiCl_4}]^{2-}$ and $[\mathrm{Ni(CN)_4}]^{2-}$.

Both have nickel(II), hence a formal d count of eight. Chlorido commonly gives the tetrahedral paramagnetic complex; cyanido gives the square-planar diamagnetic complex in this familiar comparison.

Result: The tetrahedral model has two unpaired electrons and spin-only moment about $2.828\,\mu_B$. The square-planar model has all electrons paired.

The approach can summarise chosen geometry and magnetic behaviour, but it does not quantitatively predict absorption energies or by itself establish which ligands cause a particular splitting. Treating hybridisation labels as a complete cause of colour overstates the model.

In an octahedral field, ligands approach along the coordinate axes. The orbitals directed most strongly along those axes, $d_{x^2-y^2}$ and $d_{z^2}$, form the higher $e_g$ set. The other three form the lower $t_{2g}$ set. The energy separation is $\Delta_o$.

$$E(t_{2g})=-0.4\Delta_o,\qquad E(e_g)=+0.6\Delta_o$$

Whether electrons pair in the lower set or enter the upper set depends on splitting relative to pairing cost. A weak field can favour a high-spin arrangement. A sufficiently strong field can favour a low-spin one for d counts where the alternatives exist. Not every d count has a distinct high-spin/low-spin choice.

A useful selected weak-to-strong field sequence is $\mathrm{I^-<Br^-<Cl^-<F^-<H_2O<NH_3<en<CN^-<CO}$. It guides appropriate comparisons rather than ignoring metal, oxidation state or donor atom. Linkage through a different atom can change the field behaviour.

Compare the usual octahedral models of $[\mathrm{CoF_6}]^{3-}$ and $[\mathrm{Co(CN)_6}]^{3-}$.

Both give Co(III), hence $d^6$. The fluoride case is commonly high spin, $t_{2g}^4e_g^2$; the cyanido case is low spin, $t_{2g}^6$.

Result: The models have 4 and 0 unpaired electrons respectively. Splitting and pairing change, not the formal metal state.

A tetrahedral field reverses the ordering: the $e$ set is lower and the $t_2$ set higher. For comparable metal–ligand situations, $\Delta_t$ is roughly $4\Delta_o/9$. The smaller splitting helps explain why tetrahedral complexes are commonly high spin.

An absorbed photon can promote an electron between appropriate states if the energy matches the gap. The observed colour relates to the light not absorbed. Metal oxidation state, ligand identity and geometry can all change the absorption. The same dopant ion can give different colours in different crystal environments; chromium-containing ruby and emerald illustrate environment dependence.

A pure electrostatic model treats ligands as charges or dipoles and does not describe the full covalent bonding. It does not explain every ligand trend without additional ideas. Ligand-field and molecular-orbital treatments extend the picture. Charge-transfer absorption is another reason not to reduce every colour to a single d–d gap.

1. Which octahedral d-orbital set lies higher?
2. Does a zero field-stabilisation contribution prove a complex cannot form?

Answers

1. The eg set.
2. No. Many other bonding and energetic contributions exist.

A carbonyl ligand can donate electron density to a metal and accept electron density back. The two interactions reinforce one another.

In an ordinary metal carbonyl, carbon monoxide typically coordinates through carbon. Donation from a carbon-based lone pair provides a sigma component towards the metal. Occupied metal orbitals can also donate into an appropriate antibonding orbital of the ligand, producing a pi back-bonding component.

Back-bonding strengthens the metal–carbon interaction while weakening the internal carbon–oxygen bond relative to a simple isolated-CO comparison. That is why vibrational information can reveal changes in bonding. The idea is synergic: one component can help support the other.

A neutral carbonyl ligand carries zero formal ligand charge. In a neutral homoleptic carbonyl such as $\mathrm{Ni(CO)_4}$, formal oxidation-state bookkeeping therefore assigns the metal state zero. This does not mean the metal and ligand share no electron density.

Assign the formal metal oxidation state in neutral $\mathrm{Cr(CO)_6}$.

Every CO ligand is formally neutral. The sum of metal state and ligand charges must equal the overall neutral charge.

Result: Chromium is in formal oxidation state $0$.

In a labelled exam sketch, use one arrow from CO towards the metal for sigma donation and another from the metal towards a ligand antibonding orbital for pi back-donation. State which internal bond becomes weaker. Do not draw the back-donation arrow as if it simply added another C–O bond.

1. Through which atom does CO usually bind in these elementary carbonyl examples?
2. What happens qualitatively to the internal C–O bond when metal-to-CO back-bonding increases?

Answers

1. Carbon.
2. Its bond strength is reduced relative to the comparison without that back-bonding.

Coordination changes what a metal can do. That makes ligand binding useful in analysis, extraction, catalysis and living systems.

A ligand can selectively bind an ion, change its colour, alter its solubility or shift its redox behaviour. Analytical methods use such changes to identify or determine metal ions. A chelating reagent can form a predictable complex, allowing a titration or another measurement under controlled conditions.

Metal extraction and purification can use selective complex formation to move a metal into a solution or to separate it from another species. Catalysts use metal centres and ligands to control which reactants bind and how they react. The ligand environment is part of the catalytic design, not decorative surrounding material.

Biological coordination centres include iron-containing oxygen-transport systems, magnesium-containing light-harvesting pigments and cobalt-containing vitamin structures. These examples show that metal identity alone is insufficient: the coordination environment gives the centre its specific role.

Some therapeutic substances contain coordination compounds, but biological effects depend on exact structure, dose and context. Learning a cis/trans distinction explains a structural difference; it does not justify medical use or substituting one compound for another.

A reagent gives a different optical signal when it binds a metal ion. What must be established before using the signal to measure that ion?

The relevant complex must form predictably under the selected conditions. Competing ions, pH, concentration and the calibration relation can affect the response.

Result: A selective, calibrated relationship between signal and the intended ion under controlled chemical conditions.

A good application answer connects three parts: the ligand or complex involved, the chemical effect it creates, and the practical task that effect enables. This is more informative than listing several metal-containing substances without explanation.

1. Why can changing a ligand alter a metal catalyst?
2. What makes chelation useful in an analytical measurement?

Answers

1. It changes binding, electronic structure and the accessible reaction pathways.
2. Controlled multidentate binding can provide a defined, selective complex and a measurable response.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

5.1 Distinguish the coordination sphere from counterions.

5.2 Find metal state in $[\mathrm{Fe(H_2O)_6}]^{2+}$.

5.3 Find metal state in $[\mathrm{CrCl_6}]^{3-}$.

5.4 Six water ligands bind a centre. What is its coordination number?

5.5 Three bidentate ligands bind through both donors. What is the coordination number?

5.6 Explain denticity without using charge.

5.7 Explain why nitrite can give linkage isomers.

5.8 Is a complex with only one ligand type necessarily monodentate?

5.9 Name $[\mathrm{Ag(NH_3)_2}]\mathrm{NO_3}$.

5.10 Write the formula of hexaamminecobalt(III) chloride.

5.11 Find formal d count in $[\mathrm{Fe(CN)_6}]^{4-}$.

5.12 Why is ammine alphabetised by its ligand name rather than its numerical prefix?

5.13 What distinction is encoded by cis and trans?

5.14 Can a rotation of a molecule alone create its distinct geometric isomer?

5.15 Why does a simple tetrahedral MA₂B₂ pattern lack a cis/trans pair?

5.16 Compare fac and mer for octahedral MA₃B₃.

5.17 Define a pair of enantiomers.

5.18 Can an equal mixture of enantiomers show zero net optical rotation?

5.19 How do ionisation isomers differ?

5.20 How do hydrate isomers differ?

5.21 Find the formal metal state in a neutral complex containing only CO ligands.

5.22 Identify the higher orbital set in an octahedral crystal field.

5.23 Calculate the field contribution for octahedral $t_{2g}^2e_g^1$ occupancy.

5.24 For $\Delta_o=18000\,\mathrm{cm^{-1}}$, estimate $\Delta_t$ using the usual approximate ratio.

5.25 Why can ligand strength change magnetism?

5.26 Does crystal-field theory fully describe metal–ligand covalency?

5.27 Why may the same metal ion show different colours in different complexes?

5.28 Describe sigma donation in a metal carbonyl.

5.29 Describe pi back-donation in the same system.

5.30 What is the qualitative effect of increased back-donation on the internal CO bond?

5.31 Plan a complete solution to a coordination-magnetism question.

Answers

5.1 The sphere contains the centre and directly bound ligands; counterions outside it balance entity charge.

5.2 $+2$.

5.3 $+3$.

5.4 $6$.

5.5 $6$.

5.6 Denticity counts donor atoms through which one ligand binds to the same centre.

5.7 It has alternative donor atoms, so the metal can connect through a different atom of the same ligand.

5.8 No. Homoleptic describes ligand types, not their denticity.

5.9 Diamminesilver(I) nitrate.

5.10 $[\mathrm{Co(NH_3)_6}]\mathrm{Cl_3}$.

5.11 Iron is +2, hence $d^{6}$.

5.12 Multiplicative prefixes are ignored for the ordinary alphabetical ordering of ligand names.

5.13 Whether selected identical ligands occupy adjacent or opposite positions in the stated geometry.

5.14 No. A whole-object rotation preserves the arrangement.

5.15 The tetrahedral positions can be interchanged by suitable whole-object rotations; there is no distinct adjacent/opposite pairing.

5.16 Fac places the three A ligands on a face; mer includes an opposite pair of A positions.

5.17 Nonsuperposable mirror-image stereoisomers.

5.18 Yes. Opposite contributions can cancel.

5.19 An inside ligand and an outside counterion exchange roles, changing the available ions on dissolution.

5.20 They differ in which water molecules are coordinated and which are outside the sphere.

5.21 Zero, because each CO ligand is formally neutral.

5.22 The eg set, with orbitals oriented strongly towards the axial ligand directions.

5.23 $-0.2\Delta_o$, before pairing-energy terms.

5.24 $8000\,\mathrm{cm^{-1}}$, for the stipulated comparable situation.

5.25 Changing splitting relative to pairing cost can change orbital occupancy and the unpaired-electron count.

5.26 No. Its pure form is electrostatic; more complete bonding models include covalency explicitly.

5.27 The ligand environment, geometry and transitions available can change.

5.28 A carbon-based CO lone pair donates towards a suitable metal acceptor orbital.

5.29 Occupied metal orbitals donate towards suitable ligand antibonding orbitals.

5.30 It weakens that internal bond relative to an otherwise suitable comparison.

5.31 Determine entity charge, ligand charges and metal state; count d electrons; identify geometry and splitting; assign occupancy; count unpaired electrons; use a justified magnetic approximation.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

Why may the same metal ion show different colours in different complexes?

The ligand environment, geometry and transitions available can change.

Describe sigma donation in a metal carbonyl.

A carbon-based CO lone pair donates towards a suitable metal acceptor orbital.

Describe pi back-donation in the same system.

Occupied metal orbitals donate towards suitable ligand antibonding orbitals.

What is the qualitative effect of increased back-donation on the internal CO bond?

It weakens that internal bond relative to an otherwise suitable comparison.

Plan a complete solution to a coordination-magnetism question.

Determine entity charge, ligand charges and metal state; count d electrons; identify geometry and splitting; assign occupancy; count unpaired electrons; use a justified magnetic approximation.