Coordination Compounds (Class 12): CFT, Colour, Magnetism and Isomerism Made Clear
Why [Fe(CN)6]4− is diamagnetic while [Fe(H2O)6]2+ carries four unpaired electrons — and how one idea, crystal field splitting, answers colour, magnetism and geometry together.
Werner’s theory: the idea the chapter rests on
Werner proposed that a metal shows two kinds of valency. The primary valency is ionisable, satisfied by negative ions, and corresponds to the oxidation state. The secondary valency is non-ionisable, satisfied by ligands, and corresponds to the coordination number, which fixes the geometry.
This is why CoCl3·6NH3 gives three moles of AgCl with silver nitrate but CoCl3·4NH3 gives only one. The chloride inside the coordination sphere is not free to precipitate.
Naming a complex without losing marks
| Rule | What it means in practice |
|---|---|
| Cation before anion | Name the positive part first, whether or not it is the complex |
| Ligands alphabetically, then metal | Ignore the multiplying prefix when alphabetising |
| Anionic ligands end in -o | chlorido, cyanido, hydroxido, oxalato, sulphato |
| Neutral ligands keep special names | aqua (H2O), ammine (NH3), carbonyl (CO), nitrosyl (NO) |
| bis / tris / tetrakis | Used when the ligand name already contains di, tri or tetra |
| Oxidation state in Roman numerals | Written in parentheses right after the metal |
| Anionic complex takes -ate | ferrate, cuprate, argentate, stannate, plumbate, aurate |
Applied to the standard set:
- [Co(NH3)6]Cl3 → hexaamminecobalt(III) chloride
- K4[Fe(CN)6] → potassium hexacyanidoferrate(II)
- [Pt(NH3)2Cl2] → diamminedichloridoplatinum(II)
- [Cr(en)3]Cl3 → tris(ethane-1,2-diamine)chromium(III) chloride
- [Ni(CO)4] → tetracarbonylnickel(0)
Isomerism: two families, six types
Structural isomerism
- Ionisation. [Co(NH3)5SO4]Br and [Co(NH3)5Br]SO4 — they exchange an ion between inside and outside the sphere, so they give different precipitation tests.
- Hydrate. [Cr(H2O)6]Cl3 is violet, [Cr(H2O)5Cl]Cl2·H2O is blue-green, [Cr(H2O)4Cl2]Cl·2H2O is dark green.
- Linkage. An ambidentate ligand binds through either donor atom: nitrito-N (−NO2) or nitrito-O (−ONO); similarly thiocyanato-S and thiocyanato-N.
- Coordination. [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6] swap the ligand sets between two complex ions.
Stereoisomerism
- Geometrical. cis- and trans-[Pt(NH3)2Cl2], cis- and trans-[Co(NH3)4Cl2]+. Tetrahedral complexes do not show it, because all four positions are adjacent.
- Optical. [Co(en)3]3+ exists as a non-superimposable pair. Note the classic trap: cis-[Co(en)2Cl2]+ is optically active, but the trans form has a plane of symmetry and is not.
Crystal field theory: one diagram that explains the chapter
CFT treats the metal–ligand interaction as purely electrostatic. As ligands approach, d-orbitals pointing straight at them are raised in energy and those pointing between them are lowered.
In a tetrahedral field the pattern inverts — the e set lies below t2 — and the gap is much smaller:
That single relation carries a large consequence: because Δt is always smaller than the pairing energy in practice, tetrahedral complexes are essentially always high spin. You will not be asked to choose.
Strong field or weak field? Compare Δ with P
Electrons fill the lower set first. Whether the fourth electron pairs up there or jumps to the upper set is decided by a straight comparison:
- Δ > P → pairing is cheaper → low spin (strong field ligand)
- Δ < P → promotion is cheaper → high spin (weak field ligand)
The spectrochemical series ranks ligands by the Δ they produce, weakest first:
Magnetic behaviour: the spin-only formula
Count unpaired electrons (n), then apply:
| Complex | Config | Field | Unpaired e− | μ (BM) |
|---|---|---|---|---|
| [Fe(CN)6]4− | Fe2+, d6 | Strong, low spin | 0 | 0 (diamagnetic) |
| [Fe(H2O)6]2+ | Fe2+, d6 | Weak, high spin | 4 | 4.90 |
| [Co(NH3)6]3+ | Co3+, d6 | Strong, low spin | 0 | 0 |
| [CoF6]3− | Co3+, d6 | Weak, high spin | 4 | 4.90 |
| [Ni(CN)4]2− | Ni2+, d8 | Square planar, dsp2 | 0 | 0 |
| [NiCl4]2− | Ni2+, d8 | Tetrahedral, sp3 | 2 | 2.83 |
Useful values to hold in memory: n = 1 → 1.73, n = 2 → 2.83, n = 3 → 3.87, n = 4 → 4.90, n = 5 → 5.92 BM.
Why coordination compounds are coloured
An electron in the lower set absorbs a photon of exactly Δ and jumps to the upper set — a d–d transition. The compound transmits what it does not absorb, so the colour you see is complementary to the colour absorbed.
[Ti(H2O)6]3+ is the textbook case: a single d electron, absorption near 500 nm in the green-yellow, and a violet solution as a result.
Two situations give no colour at all, and both follow from the same reasoning: d0 has no electron to promote (Sc3+, Ti4+) and d10 has no vacancy to promote into (Zn2+, Cu+, Ag+). This is exactly why zinc salts are white while copper(II) salts are blue.
Crystal field stabilisation energy
CFSE measures how much the splitting stabilises the ion relative to an unsplit set:
For low-spin d6, all six electrons sit in t2g: CFSE = −2.4Δo, the largest value available in an octahedral field. That is the quantitative reason [Co(NH3)6]3+ and [Fe(CN)6]4− are so notably stable.
Where students actually lose marks
- Forgetting ligand charges when working out oxidation state — in K4[Fe(CN)6], six cyanido ligands contribute −6, so Fe must be +2.
- Using the octahedral splitting order for a tetrahedral complex. The pattern inverts.
- Debating high spin versus low spin for d3 or d8 octahedral complexes, where no choice exists.
- Calling trans-[Co(en)2Cl2]+ optically active. It has a plane of symmetry.
- Writing "chloro" where current IUPAC practice, followed by NCERT, requires "chlorido".
- Quoting μ without the unit. It is Bohr magneton, BM.
Quick revision checklist
- Work out the oxidation state and the dn configuration first, before anything else.
- Read the ligand off the spectrochemical series and label it strong or weak field.
- Fill the split orbitals; count unpaired electrons.
- Apply μ = √[n(n+2)] for magnetic moment.
- Use Δ size to reason about the colour absorbed, then take the complement.
- Check the geometry for cis/trans and for a plane of symmetry before claiming optical activity.
FAQs
Why is [Ni(CN)4]2− diamagnetic but [NiCl4]2− paramagnetic?
Both are Ni2+ (d8), but CN− is a strong field ligand and forces pairing, giving a square planar dsp2 complex with no unpaired electrons. Cl− is weak field, so the complex stays tetrahedral (sp3) with two unpaired electrons and μ = 2.83 BM.
What is the difference between Δo and Δt?
Δo is the octahedral splitting, with t2g below eg. Δt is the tetrahedral splitting, with the order inverted and the gap smaller: Δt = (4/9)Δo. Because that gap rarely exceeds the pairing energy, tetrahedral complexes are high spin.
Why are zinc(II) compounds colourless?
Zn2+ is d10. The lower and upper sets are both full, so no d–d transition is possible and nothing in the visible range is absorbed.
Should I learn VBT or CFT for the board exam?
Both. VBT is asked for hybridisation and inner/outer orbital classification; CFT is asked for colour, magnetic behaviour and the high spin versus low spin argument. CFT is also the one that carries forward to JEE Advanced and to postgraduate entrance papers.
How do I decide high spin or low spin quickly?
Check the d-count first. If it is not between d4 and d7, there is no choice to make. If it is, place the ligand on the spectrochemical series: H2O and anything to its left behaves as weak field, NH3 and anything to its right as strong field.
Struggling with Inorganic Chemistry?
ABC Chemistry runs Class 11 and 12 Chemistry coaching in Dwarka, Delhi, along with GATE, CSIR-NET and IIT-JAM Chemistry batches at the centre and through live online classes.