Coordination Compounds • Topic 2 of 3

Isomerism & Bonding (VBT)

Two compounds with the same molecular formula but a different arrangement of atoms are isomers. In coordination chemistry they fall into two families: structural (constitutional) isomerism, where bonds connect different atoms, and stereoisomerism, where the connectivity is the same but the spatial arrangement differs.

Structural isomerism includes:

  • Ionisation isomerism — the counter ion and a coordinated ligand swap places, giving different ions in solution, e.g. [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4.
  • Linkage isomerism — an ambidentate ligand binds through a different donor atom, e.g. [Co(NH3)5(NO2)]2+ (nitrito-N, yellow) vs [Co(NH3)5(ONO)]2+ (nitrito-O, red).
  • Coordination isomerism — ligands are interchanged between cationic and anionic complexes, e.g. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
  • Solvate (hydrate) isomerism — water is inside or outside the coordination sphere, e.g. [Cr(H2O)6]Cl3 (violet), [Cr(H2O)5Cl]Cl2·H2O and [Cr(H2O)4Cl2]Cl·2H2O (green).

Stereoisomerism includes geometrical (cis–trans) and optical isomerism. Geometrical isomerism appears in square planar [Ma2b2] (e.g. cis- and trans-[Pt(NH3)2Cl2]) and in octahedral [Ma4b2] and [Ma3b3] (fac–mer) complexes. Tetrahedral complexes do not show cis–trans isomerism. Optical isomerism arises when a complex is non-superimposable on its mirror image (chiral), most commonly in octahedral chelates such as cis-[Co(en)2Cl2]+ and [Co(en)3]3+; the two forms are the dextro (d) and laevo (l) enantiomers.

Valence Bond Theory (VBT) treats bonding as overlap of empty hybrid orbitals of the metal ion with filled ligand orbitals (each ligand donates a lone pair). The hybridisation fixes the geometry:

  • $sp^3$ → tetrahedral (CN 4, e.g. [NiCl4]2−).
  • $dsp^2$ → square planar (CN 4, e.g. [Ni(CN)4]2−).
  • $d^2sp^3$ → octahedral using inner $(n-1)d$ orbitals (inner-orbital / low-spin, e.g. [Co(NH3)6]3+).
  • $sp^3d^2$ → octahedral using outer $nd$ orbitals (outer-orbital / high-spin, e.g. [CoF6]3−).

Whether a strong-field ligand pairs the d electrons (inner orbital) decides the number of unpaired electrons and hence the spin-only magnetic moment $\mu=\sqrt{n(n+2)}$ BM. Limitations of VBT: it does not explain colour, gives no quantitative splitting energy, cannot predict the exact magnitude of magnetic moment, and offers no satisfactory reason why some ligands are strong-field and others weak-field.

cis and trans isomers of square planar [Pt(NH3)2Cl2]cis-[Pt(NH3)2Cl2]trans-[Pt(NH3)2Cl2]PtClNH3ClNH3two Cl adjacent (90 deg)PtClClNH3NH3two Cl opposite (180 deg)
Solved Examples
1
Worked Example
Predict the hybridisation, geometry and number of unpaired electrons in [Co(NH3)6]3+ (NH3 is a strong-field ligand).
Solution
  1. Co is +3 → $d^6$ ($3d^6$).
  2. NH3 is strong field, so the six d electrons pair into three $t_{2g}$-type inner orbitals; two empty $3d$ orbitals become available.
  3. Hybridisation uses inner d: $d^2sp^3$ → octahedral, inner-orbital (low-spin).
  4. All electrons paired → $n=0$.

Answer: $d^2sp^3$, octahedral, diamagnetic ($n=0$, $\mu=0$).

2
Worked Example
For [CoF6]3− (F− weak field), predict hybridisation and spin-only magnetic moment.
Solution
  1. Co is +3 → $d^6$.
  2. F− is weak field, so no pairing; configuration $t_{2g}^4 e_g^2$ leaves the inner d filled.
  3. Outer $4d$ orbitals are used: $sp^3d^2$ → octahedral, outer-orbital (high-spin).
  4. Unpaired electrons $n=4$; $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM.

Answer: $sp^3d^2$, octahedral, $\mu\approx4.9$ BM (paramagnetic).

3
Worked Example
Why is [Ni(CN)4]2− square planar and diamagnetic while [NiCl4]2− is tetrahedral and paramagnetic?
Solution
  1. Ni is +2 → $d^8$ in both.
  2. CN− is strong field: it pairs the d electrons, freeing one $3d$ orbital → $dsp^2$ → square planar, all paired (diamagnetic).
  3. Cl− is weak field: no pairing; $sp^3$ → tetrahedral with 2 unpaired electrons (paramagnetic).

Answer: Strong-field CN− → $dsp^2$ square planar, diamagnetic; weak-field Cl− → $sp^3$ tetrahedral, $\mu=\sqrt{2(2+2)}=2.83$ BM.

4
Worked Example
How many geometrical isomers does the octahedral complex [Co(NH3)4Cl2]+ show? Name them.
Solution
  1. Type [Ma4b2] octahedral.
  2. The two Cl can be adjacent (90°) or opposite (180°).

Answer: 2 geometrical isomers — cis (Cl adjacent) and trans (Cl opposite).

5
Worked Example
Identify the type of isomerism in the pair [Co(NH3)5(NO2)]Cl2 and [Co(NH3)5(ONO)]Cl2.
Solution
  1. Same atoms; only the donor atom of the ambidentate NO2− changes (N-bonded vs O-bonded).
  2. This is a structural isomerism arising from the mode of attachment.

Answer: Linkage isomerism (nitrito-N vs nitrito-O).

6
Worked Example
Does cis-[Co(en)2Cl2]+ show optical isomerism? Explain.
Solution
  1. The cis form has no plane of symmetry; the two chelating en rings make it chiral.
  2. Its mirror image is non-superimposable → two enantiomers (d and l).
  3. The trans form has a plane of symmetry and is optically inactive.

Answer: Yes — cis-[Co(en)2Cl2]+ is chiral and exists as d and l optical isomers; the trans isomer is achiral.

Key Points

  • Structural isomerism: ionisation, linkage, coordination and solvate (hydrate) types — different atoms are bonded.
  • Stereoisomerism: geometrical (cis–trans; fac–mer in [Ma3b3]) and optical (chiral, non-superimposable d/l enantiomers, common in octahedral chelates); tetrahedral complexes show no cis–trans isomerism.
  • VBT: ligand lone pairs fill empty metal hybrid orbitals — $sp^3$ tetrahedral, $dsp^2$ square planar, $d^2sp^3$ inner-orbital octahedral, $sp^3d^2$ outer-orbital octahedral.
  • Strong-field ligands pair electrons (inner-orbital, low-spin); weak-field ligands do not (outer-orbital, high-spin); $\mu=\sqrt{n(n+2)}$ BM gives the spin-only moment.
  • VBT limitations: it cannot explain colour, gives no $\Delta$ value, does not quantify magnetic moments, and does not explain the spectrochemical order of ligands.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The pair [Co(NH3)5Br]SO4 and [Co(NH3)5SO4]Br are:
Explanation: The counter ion and a coordinated ligand have swapped, giving different ions in solution — ionisation isomerism.
Q2.The hybridisation of nickel in the square planar [Ni(CN)4]2− is:
Explanation: Square planar geometry corresponds to $dsp^2$ hybridisation.
Q3.Which complex can show optical isomerism?
Explanation: The cis-bis(en) octahedral complex is chiral (no symmetry plane) and exists as d/l enantiomers.
Q4.The spin-only magnetic moment of an outer-orbital $d^6$ octahedral complex (4 unpaired e−) is about:
Explanation: $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.90$ BM.
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