Coordination Compounds
Werner's theory, nomenclature, isomerism, VBT and crystal field theory for NEET
Structure, Nomenclature and Isomerism
Werner's Theory, Ligands and NomenclatureTopic 1
A coordination compound contains a central metal atom or ion surrounded by a fixed number of ions or molecules called ligands, joined by coordinate (dative) bonds. The metal and its ligands together form the coordination sphere, written inside square brackets — for example $[\text{Co(NH}_3)_6]\text{Cl}_3$, where $[\text{Co(NH}_3)_6]^{3+}$ is the complex ion and the three $\text{Cl}^-$ are counter (outside-sphere) ions. These compounds are central to inorganic chemistry and biology.
The behaviour of complexes was first explained by Werner's theory. Werner proposed that a metal has two types of valence: a primary valence, which is ionisable, equals the oxidation state of the metal, and is satisfied by negative ions; and a secondary valence, which is non-ionisable, equals the coordination number, is satisfied by ligands, and gives the complex its definite geometry (e.g. octahedral for coordination number 6). Distinguishing these two valences is a classic NEET starting point.
A ligand is a species that donates a lone pair of electrons to the metal, so it acts as a Lewis base. Ligands are classified by the number of donor atoms they use: monodentate (one donor, e.g. $\text{NH}_3$, $\text{Cl}^-$), bidentate (two donors, e.g. ethylenediamine, 'en'), polydentate (several donors, e.g. EDTA with six), and ambidentate (can bind through either of two different atoms, e.g. $\text{NO}_2^-$ via N or $\text{ONO}^-$ via O). Polydentate ligands form ring structures called chelates, which give extra stability — the chelate effect.
Complexes are named by the IUPAC rules, another reliable NEET question. The key rules: name the ligands first (alphabetically) then the metal; for anionic ligands use the suffix '-o' (chloro, cyano); use Greek prefixes (di, tri, tetra) for the number of simple ligands (and bis, tris for complex ones); give the metal's oxidation state in Roman numerals in brackets; and if the complex ion is an anion, add the suffix '-ate' to the metal (e.g. ferrate, cuprate). The counter ions are named like a normal salt. Mastering ligand classification, the meaning of coordination number, and these naming rules is the foundation for the rest of the chapter.
| Term | Meaning |
|---|---|
| Primary valence | ionisable = oxidation state |
| Secondary valence | non-ionisable = coordination number |
| Bidentate / polydentate | en (2 donors) / EDTA (6 donors) |
| Ambidentate | $\text{NO}_2^-$ (N) / $\text{ONO}^-$ (O) |
In $[\text{Co(NH}_3)_6]\text{Cl}_3$, identify the primary and secondary valences of cobalt.
Show solution
The primary valence (oxidation state) of Co is $+3$, satisfied by the three $\text{Cl}^-$ outside the sphere. The secondary valence (coordination number) is $6$, satisfied by the six $\text{NH}_3$ ligands, giving an octahedral geometry.
Find the oxidation state of iron and the coordination number in $\text{K}_4[\text{Fe(CN)}_6]$.
Show solution
$\text{CN}^-$ is $-1$ (six of them = $-6$) and the four $\text{K}^+$ are $+4$. For neutrality, Fe $= +2$. The coordination number is $6$ (six cyanide donors).
A ligand donates a lone pair, so it acts as a:
In Werner's theory, the secondary valence corresponds to the:
Ethylenediamine (en) is a ____ ligand.
$\text{NO}_2^-$, which can bind through N or O, is:
When the complex ion is an anion, the metal name ends in:
NEET tip: primary valence = oxidation state (counter ions); secondary valence = coordination number (ligands, geometry). Count donor atoms for coordination number. Name ligands first (alphabetical), metal last; anionic complex to '-ate'.
Isomerism in Coordination CompoundsTopic 2
Coordination compounds show rich isomerism — same formula, different arrangement — which divides into structural isomerism (different connectivity) and stereoisomerism (same connectivity, different spatial arrangement). Identifying the type from a given pair is a frequent NEET task.
Structural isomers come in four main kinds. Ionisation isomerism arises when the counter ion and a ligand exchange places, giving different ions in solution (e.g. $[\text{Co(NH}_3)_5\text{Br}]\text{SO}_4$ vs $[\text{Co(NH}_3)_5\text{SO}_4]\text{Br}$). Linkage isomerism occurs with ambidentate ligands that bind through different atoms (e.g. $\text{-NO}_2$ vs $\text{-ONO}$). Coordination isomerism appears when both cation and anion are complex ions and the ligands are swapped between the two metals. Hydrate (solvate) isomerism differs in how many water molecules are inside the coordination sphere versus outside as water of crystallisation (e.g. the well-known violet, green and blue chromium chloride hydrates).
Stereoisomers have the same bonds but differ in geometry. Geometrical (cis–trans) isomerism is shown by square planar complexes of the type $[\text{MA}_2\text{B}_2]$ and by octahedral complexes such as $[\text{MA}_4\text{B}_2]$ and $[\text{M(AA)}_2\text{B}_2]$: the identical ligands can sit adjacent (cis) or opposite (trans). Tetrahedral complexes do not show cis–trans isomerism because all four positions are equivalent — a common NEET trap.
Optical isomerism arises when a complex is chiral — non-superimposable on its mirror image — most commonly in octahedral complexes containing bidentate ligands, such as $[\text{Co(en)}_3]^{3+}$ and the cis form of $[\text{Co(en)}_2\text{Cl}_2]^+$. These exist as enantiomers that rotate plane-polarised light in opposite directions. Knowing which geometries permit cis–trans and which permit optical isomerism, and recognising the four structural types, equips you for the isomerism questions NEET sets in this chapter.
| Isomerism | Basis |
|---|---|
| Ionisation | counter ion and ligand exchange |
| Linkage | ambidentate ligand binds via different atom |
| Geometrical (cis/trans) | square planar / octahedral (not tetrahedral) |
| Optical | chiral, e.g. $[\text{Co(en)}_3]^{3+}$ |
What type of isomerism is shown by $[\text{Co(NH}_3)_5(\text{NO}_2)]^{2+}$ and $[\text{Co(NH}_3)_5(\text{ONO})]^{2+}$?
Show solution
The ambidentate $\text{NO}_2^-$ ligand binds through nitrogen in one and through oxygen in the other. This is linkage isomerism.
Why does a tetrahedral complex $[\text{MA}_2\text{B}_2]$ not show cis–trans isomerism?
Show solution
In a tetrahedron all four corner positions are equivalent and adjacent to one another, so there is no distinct 'opposite' position. Hence the ligands cannot be arranged as distinct cis and trans forms.
Linkage isomerism is shown by complexes containing:
Cis–trans isomerism is NOT shown by ____ complexes.
$[\text{Co(en)}_3]^{3+}$ shows which isomerism?
$[\text{Co(NH}_3)_5\text{Br}]\text{SO}_4$ and $[\text{Co(NH}_3)_5\text{SO}_4]\text{Br}$ are:
Different numbers of water molecules inside vs outside the coordination sphere give:
NEET trap: tetrahedral complexes show NO cis–trans isomerism (all positions equivalent). Optical isomerism mostly in octahedral complexes with bidentate ligands (e.g. $[\text{Co(en)}_3]^{3+}$). Linkage needs an ambidentate ligand.
Bonding Theories
Valence Bond Theory of ComplexesTopic 3
To explain the shape and magnetism of complexes, valence bond theory (VBT) treats coordinate bonding as the overlap of filled ligand orbitals (lone pairs) with empty hybrid orbitals of the metal ion. The type of hybridisation the metal adopts determines the geometry, and NEET problems lean heavily on matching hybridisation to shape.
For a coordination number of 6 (octahedral), the metal can use two kinds of hybridisation. If it uses inner $(n-1)d$ orbitals it is $d^2sp^3$ hybridised, giving an inner-orbital (low-spin) complex; if it uses outer $nd$ orbitals it is $sp^3d^2$ hybridised, giving an outer-orbital (high-spin) complex. For a coordination number of 4, $sp^3$ hybridisation gives a tetrahedral shape and $dsp^2$ gives a square planar shape. Reading these off is a routine NEET skill.
VBT links directly to magnetic behaviour. Strong-field ligands force electrons to pair up, often producing inner-orbital, low-spin complexes with fewer (or no) unpaired electrons — these can be diamagnetic. Weak-field ligands leave electrons unpaired, giving outer-orbital, high-spin, paramagnetic complexes. By counting unpaired electrons you predict whether a complex is paramagnetic or diamagnetic and calculate its spin-only magnetic moment $\mu = \sqrt{n(n+2)}\ \text{BM}$, exactly as for the transition metals.
A worked logic for VBT: find the metal's oxidation state and $d$-electron count; decide (from the ligand) whether pairing occurs; arrange the $d$ electrons; identify the hybridisation needed for the coordination number; and read off geometry and magnetism. For example, $[\text{Fe(CN)}_6]^{4-}$ (Fe$^{2+}$, $d^6$, strong-field CN$^-$) is $d^2sp^3$, octahedral, low-spin and diamagnetic, whereas $[\text{FeF}_6]^{3-}$ (Fe$^{3+}$, $d^5$, weak-field F$^-$) is $sp^3d^2$, octahedral, high-spin and strongly paramagnetic. VBT is powerful but has limits — it does not explain colour or the detailed strength order of ligands, which is where crystal field theory takes over.
| Hybridisation | Geometry / type |
|---|---|
| $sp^3$ | tetrahedral (CN 4) |
| $dsp^2$ | square planar (CN 4) |
| $d^2sp^3$ | octahedral, inner-orbital (low-spin) |
| $sp^3d^2$ | octahedral, outer-orbital (high-spin) |
Predict the hybridisation, geometry and magnetism of $[\text{Fe(CN)}_6]^{4-}$.
Show solution
Fe is $+2$ ($3d^6$). $\text{CN}^-$ is a strong-field ligand, so the six $d$ electrons pair into three $d$ orbitals, freeing two inner $d$ orbitals: hybridisation $d^2sp^3$, geometry octahedral, no unpaired electrons — low-spin, diamagnetic.
What geometry corresponds to $dsp^2$ hybridisation?
Show solution
$dsp^2$ hybridisation gives a square planar geometry (coordination number 4), as in $[\text{Ni(CN)}_4]^{2-}$.
$d^2sp^3$ hybridisation gives which geometry?
A square planar complex uses ____ hybridisation.
An outer-orbital octahedral complex is:
A complex with no unpaired electrons is:
VBT cannot satisfactorily explain:
NEET tip: CN 4 → $sp^3$ (tetrahedral) or $dsp^2$ (square planar); CN 6 → $d^2sp^3$ (inner, low-spin) or $sp^3d^2$ (outer, high-spin). Count unpaired electrons for magnetism; VBT can't explain colour.
Crystal Field Theory, Colour and ApplicationsTopic 4
Crystal field theory (CFT) treats the metal–ligand bond as purely electrostatic and focuses on what the approaching ligands do to the metal's five $d$ orbitals. In a free ion these are degenerate (equal energy), but the electric field of the ligands raises their energy unequally, splitting them into sets — the central idea NEET tests repeatedly.
In an octahedral field, the five $d$ orbitals split into a lower-energy set of three ($t_{2g}$: $d_{xy}, d_{yz}, d_{zx}$) and a higher-energy set of two ($e_g$: $d_{z^2}, d_{x^2-y^2}$), separated by the crystal field splitting energy $\Delta_o$. In a tetrahedral field the order is reversed and the splitting ($\Delta_t$) is smaller (about $\tfrac{4}{9}\Delta_o$), so tetrahedral complexes are almost always high-spin. How the $d$ electrons fill these sets gives the crystal field stabilisation energy (CFSE).
The size of $\Delta_o$ depends on the ligand, captured by the spectrochemical series — an experimental order of ligands from weak-field (small $\Delta$) to strong-field (large $\Delta$): roughly $\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$. If $\Delta_o$ is larger than the pairing energy, electrons pair up in $t_{2g}$ giving a low-spin complex; if smaller, they spread out giving a high-spin complex. This neatly explains the magnetic behaviour that VBT only described.
Crucially, CFT explains the colour of complexes: a $d$ electron absorbs a photon of energy equal to $\Delta_o$ and jumps from $t_{2g}$ to $e_g$ (a $d$–$d$ transition), and the complementary colour is seen. A larger $\Delta_o$ (stronger ligand) means higher-energy light is absorbed and a different colour appears. Coordination compounds are also enormously important in practice and in life: haemoglobin (Fe), chlorophyll (Mg) and vitamin B$_{12}$ (Co) are all coordination compounds, cisplatin is an anticancer drug, EDTA is used in titrations and to treat metal poisoning, and complexes are vital in metallurgy and electroplating. Knowing the $t_{2g}/e_g$ splitting, the spectrochemical series, the high-spin/low-spin rule, and these biological/medical examples covers the CFT questions NEET asks.
| Idea | Key point |
|---|---|
| Octahedral splitting | $t_{2g}$ (lower) and $e_g$ (higher), gap $\Delta_o$ |
| Tetrahedral splitting | reversed, $\Delta_t \approx \tfrac{4}{9}\Delta_o$ (usually high-spin) |
| Spectrochemical series | $\text{I}^- < \ldots < \text{H}_2\text{O} < \text{NH}_3 < \text{CN}^- < \text{CO}$ |
| Colour | $d$–$d$ transition across $\Delta_o$ |
Why is $[\text{Co(CN)}_6]^{3-}$ low-spin while $[\text{CoF}_6]^{3-}$ is high-spin?
Show solution
$\text{CN}^-$ is a strong-field ligand: $\Delta_o$ exceeds the pairing energy, so the $d^6$ electrons pair in $t_{2g}$ — low-spin. $\text{F}^-$ is weak-field: $\Delta_o$ is small, electrons stay unpaired across $t_{2g}$ and $e_g$ — high-spin.
How does crystal field theory explain the colour of $[\text{Ti(H}_2\text{O})_6]^{3+}$?
Show solution
Ti$^{3+}$ is $d^1$. Its single $d$ electron absorbs visible light of energy equal to $\Delta_o$ and is promoted from $t_{2g}$ to $e_g$ (a $d$–$d$ transition). The complementary colour of the absorbed light gives the ion its characteristic purple colour.
In an octahedral field, the $d$ orbitals split into:
The strongest-field ligand in the spectrochemical series is:
A complex is low-spin when $\Delta_o$ is:
The colour of a complex arises from an electronic transition between:
Which biomolecule is a cobalt coordination compound?
NEET tip: octahedral $t_{2g}$ (low) + $e_g$ (high), gap $\Delta_o$; large $\Delta_o$ (strong field, e.g. CN⁻/CO) to low-spin. Colour = $d$–$d$ transition across $\Delta_o$. Bio: Fe-haemoglobin, Mg-chlorophyll, Co-B₁₂.
Quick Revision — Coordination Compounds
- Coordination compound: a central metal atom/ion bonded to surrounding ligands by coordinate bonds, written in square brackets, e.g. $[\text{Co(NH}_3)_6]^{3+}$.
- Werner's theory: primary valence (ionisable, = oxidation state) and secondary valence (non-ionisable, = coordination number, fixed geometry).
- Ligands: monodentate, bidentate (en), polydentate (EDTA), ambidentate (NO₂⁻/ONO⁻); donate a lone pair (Lewis bases).
- Coordination number = number of donor atoms bonded to the metal (commonly 4 — tetrahedral/square planar — or 6 — octahedral).
- Isomerism: structural (ionisation, linkage, coordination, hydrate) and stereo (geometrical cis/trans, optical).
- Valence Bond Theory: metal hybridises ($d^2sp^3$ inner / $sp^3d^2$ outer for octahedral); predicts geometry & magnetism.
- Crystal Field Theory: ligands split the $d$ orbitals ($t_{2g}/e_g$); splitting $\Delta$ governs colour, magnetism, and strong vs weak field (spectrochemical series).
- Vital in biology: haemoglobin (Fe), chlorophyll (Mg), vitamin $\text{B}_{12}$ (Co).
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