[Fe(CN)₆]⁴⁻ is diamagnetic while [Fe(H₂O)₆]²⁺ is paramagnetic. This is because:

CN⁻ is a strong field ligand that pairs all six d electrons in t₂g

Iron has different oxidation states in the two complexes

CN⁻ is a strong field ligand that pairs all six d electrons in t₂g

The two complexes have different coordination numbers

Which complex can show cis–trans isomerism?

Square planar [Pt(NH₃)₂Cl₂]

Tetrahedral [Zn(NH₃)₂Cl₂]

Square planar [Pt(NH₃)₂Cl₂]

The IUPAC name of K₄[Fe(CN)₆] is:

Potassium hexacyanidoferrate(III)

Cisplatin is an effective anticancer drug but transplatin is not, because:

Only in the cis form are the leaving groups adjacent, letting platinum cross-link two neighbouring DNA bases

Transplatin is chemically unstable

Only in the cis form are the leaving groups adjacent, letting platinum cross-link two neighbouring DNA bases

Transplatin has a different oxidation state of platinum

Tetrahedral complexes are almost always high spin because:

Δ_t is roughly 4/9 of Δ_o and so is smaller than the pairing energy

Tetrahedral complexes have only four d orbitals

Their ligands are always strong field

They have no crystal field splitting at all

Coordination Compounds

Coordination chemistry is where inorganic questions stop being recall and start being reasoning, and it is generously represented in the paper. Nearly every question reduces to one procedure: find the oxidation state, count the d electrons, decide whether the ligand is strong or weak field, then read off geometry, magnetism and colour. Run that sequence in order and even unfamiliar complexes become tractable. Skip a step — most often the oxidation state — and everything downstream is wrong, which is why the marks in this chapter cluster so sharply between candidates who have a method and those who do not.

Werner's Insight, Ligands and Naming

Werner explained a puzzle: CoCl₃·6NH₃ gives three chloride ions to silver nitrate while CoCl₃·4NH₃ gives only one, even though both contain three chlorines. His answer was that each metal has a primary valence satisfied by ionisable groups and a fixed secondary valence — what we now call the coordination number — satisfied by directly bonded ligands. In [Co(NH₃)₄Cl₂]Cl only the outer chloride is free to precipitate; the other two are inside the coordination sphere and invisible to silver nitrate. Counting precipitated chloride is still a standard question, and the rule is simply: only what is outside the square brackets ionises.

Ligands by denticity, with names to know
Type	Donor sites	Examples
Monodentate	One	NH₃ ammine, H₂O aqua, CN⁻ cyanido, Cl⁻ chlorido, NO₂⁻ nitrito
Bidentate	Two	Ethane-1,2-diamine (en), oxalate (ox)
Hexadentate	Six	EDTA⁴⁻ — two N and four O
Ambidentate	Two possible, one used at a time	NO₂⁻ (via N or O), SCN⁻ (via S or N)
Chelating	Any polydentate forming a ring	en, ox, EDTA — five- and six-membered rings are the most stable

🧠 Naming: ligands alphabetically, then metal, then oxidation state

Ligands are named before the metal and in alphabetical order of the ligand name, ignoring the multiplying prefix — so in [Co(NH₃)₄Cl₂]Cl the "ammine" comes before "chlorido" and it is tetraamminedichloridocobalt(III) chloride. If the whole complex ion is an anion, the metal takes the Latin stem plus -ate: ferrate for iron, cuprate for copper, argentate for silver, aurate for gold, plumbate for lead. So K₄[Fe(CN)₆] is potassium hexacyanidoferrate(II). Two mistakes cost most of the marks here: alphabetising by "tetraammine" rather than "ammine", and forgetting the -ate ending when the complex is anionic.

Crystal Field Theory — Geometry, Magnetism and Colour

In an octahedral field the five d orbitals split into a lower t₂g set of three and an upper e_g set of two, separated by Δ_o. Electrons fill the lower set first; whether the fourth electron pairs there or goes up to e_g depends on which is cheaper — the pairing energy P or the gap Δ_o. If Δ_o > P the ligand is strong field and gives a low-spin complex; if Δ_o < P it is weak field and gives high-spin. In a tetrahedral field the splitting is inverted and much smaller (Δ_t ≈ 4/9 Δ_o), which is why tetrahedral complexes are essentially always high-spin.

The spectrochemical series orders ligands by field strength: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO. Everything from H₂O leftwards behaves as weak field in practice, and NH₃ onwards as strong field. Committing this one sequence to memory settles the spin state, the magnetic moment and the colour of most complexes you will be asked about.

The same d⁶ ion, two different ligands
Feature	[Fe(H₂O)₆]²⁺	[Fe(CN)₆]⁴⁻
Oxidation state and d count	Fe(II), d⁶	Fe(II), d⁶ — the same
Ligand field	Weak	Strong
Configuration	t₂g⁴e_g² — high spin	t₂g⁶e_g⁰ — low spin
Unpaired electrons	4	0
Magnetic behaviour	Paramagnetic, μ = 4.90 BM	Diamagnetic
Hybridisation	sp³d² — outer orbital	d²sp³ — inner orbital

🎯 Why the same metal ion gives two different colours and two different magnetisms

The table above is the whole point of crystal field theory. Iron(II) is d⁶ in both complexes — nothing about the metal has changed — yet one is paramagnetic and the other diamagnetic, and their colours differ completely. The reason is that cyanide splits the d orbitals so widely that pairing all six electrons in t₂g is cheaper than promoting any to e_g, while water splits them only slightly and Hund's rule wins instead. Since the absorbed wavelength corresponds to Δ, a larger Δ means light of shorter wavelength is absorbed and a different complementary colour is seen. So the ligand, not the metal, controls colour and magnetism — and that is why a question naming the ligand has already told you the answer.

⚠️ Find the oxidation state before counting d electrons

The commonest failure in this chapter is not crystal field theory — it is arithmetic. In K₃[Fe(CN)₆] the three potassium ions carry +3, so the complex ion is 3−; six cyanides carry 6−, so iron must be +3 and the count is d⁵, not d⁶. Get that wrong and every downstream answer — spin state, magnetic moment, hybridisation, colour — follows the wrong branch. Two habits prevent it: always balance the total charge explicitly rather than by eye, and remember that neutral ligands such as NH₃, H₂O and CO contribute zero to the sum, which is why [Ni(CO)₄] has nickel in the zero oxidation state and is therefore d¹⁰, tetrahedral and diamagnetic.

Isomerism and Applications

Types of isomerism in coordination compounds
Type	What differs	Example
Ionisation	Which ion sits outside the sphere	[Co(NH₃)₅SO₄]Br and [Co(NH₃)₅Br]SO₄
Hydrate	Water inside or outside the sphere	CrCl₃·6H₂O in violet and green forms
Linkage	Which atom of an ambidentate ligand binds	Nitrito-N versus nitrito-O
Geometrical (cis/trans)	Spatial arrangement of like ligands	[Pt(NH₃)₂Cl₂] — square planar only, never tetrahedral
Optical	Non-superimposable mirror images	cis-[Co(en)₂Cl₂]⁺ is chiral; the trans form is not

🎯 Why cisplatin is a drug and transplatin is not

Cisplatin, cis-[Pt(NH₃)₂Cl₂], is one of the most widely used anticancer drugs; its trans isomer is clinically useless. The difference is purely geometric. In the cis form the two chlorides are adjacent, so after they are displaced the platinum can bind two neighbouring guanine bases on the same DNA strand, kinking the helix and blocking replication. In the trans form the leaving groups are on opposite sides and cannot reach two adjacent bases, so no such cross-link forms. This is the cleanest illustration in the whole syllabus that geometry alone can decide biological activity — and it is also a reminder that tetrahedral complexes show no cis–trans isomerism at all, since in a tetrahedron every position is adjacent to every other.

Applications recur as one-mark questions. EDTA titrations estimate water hardness and treat lead poisoning, both by chelating the metal. Cyanide complexes extract silver and gold from ores, and the same chemistry makes electroplating baths give an even deposit. Wilkinson's catalyst hydrogenates alkenes homogeneously. And nature uses the same principle throughout: haemoglobin is an iron porphyrin complex, chlorophyll a magnesium one, and vitamin B₁₂ a cobalt one — which is why carbon monoxide is lethal, since it binds haemoglobin's iron far more strongly than oxygen does and will not let go.

Key takeaways

Only groups outside the square brackets ionise — that alone answers the silver-nitrate precipitation questions.
Name ligands alphabetically ignoring prefixes, then the metal; an anionic complex takes the -ate ending.
Method: oxidation state → d count → strong or weak field → geometry, magnetism, colour. In that order.
Δ_o > P gives low spin, Δ_o < P gives high spin. Tetrahedral Δ_t is small, so tetrahedral is always high spin.
Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO.
[Fe(H₂O)₆]²⁺ is paramagnetic and [Fe(CN)₆]⁴⁻ diamagnetic though both are d⁶ — the ligand decides, not the metal.
Neutral ligands contribute zero charge, so nickel in [Ni(CO)₄] is zero-valent, d¹⁰, tetrahedral and diamagnetic.
Cisplatin works and transplatin does not, because only adjacent leaving groups can cross-link two DNA bases.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

How many moles of AgCl are precipitated when excess AgNO₃ is added to one mole of [Co(NH₃)₄Cl₂]Cl?
1. 3
2. 2
3. 1
4. 0
Show answer
Answer: C — 1
Only the chloride outside the square brackets is ionisable, so just one mole of free Cl⁻ is available to precipitate. The two chlorides bonded directly to cobalt are inside the coordination sphere and silver nitrate cannot reach them. The answer 3 is what you get by counting every chlorine in the formula, which is exactly the confusion Werner's primary and secondary valences resolved.
[Fe(CN)₆]⁴⁻ is diamagnetic while [Fe(H₂O)₆]²⁺ is paramagnetic. This is because:
1. Iron has different oxidation states in the two complexes
2. CN⁻ is a strong field ligand that pairs all six d electrons in t₂g
3. Water is a bidentate ligand
4. The two complexes have different coordination numbers
Show answer
Answer: B — CN⁻ is a strong field ligand that pairs all six d electrons in t₂g
Iron is Fe(II) and d⁶ in both, and both are six-coordinate octahedral, so neither oxidation state nor coordination number can be the discriminator. Cyanide splits the d orbitals so widely that Δ_o exceeds the pairing energy, so all six electrons crowd into t₂g and none is left unpaired; water gives a small Δ_o, so Hund's rule wins and four electrons stay unpaired. Water is monodentate, donating through its single oxygen.
The oxidation state of nickel in [Ni(CO)₄] is:
1. +2
2. +4
3. 0
4. −4
Show answer
Answer: C — 0
Carbon monoxide is a neutral ligand and contributes nothing to the charge, and the complex itself is neutral, so nickel must be zero-valent. That makes it 3d¹⁰, which is why the complex is tetrahedral and diamagnetic rather than octahedral. Assuming +2 out of habit for nickel is the standard error, and it leads to the wrong geometry and the wrong magnetic prediction.
Which complex can show cis–trans isomerism?
1. Tetrahedral [Zn(NH₃)₂Cl₂]
2. Square planar [Pt(NH₃)₂Cl₂]
3. Linear [Ag(NH₃)₂]⁺
4. Octahedral [Co(NH₃)₆]³⁺
Show answer
Answer: B — Square planar [Pt(NH₃)₂Cl₂]
In a square plane the two chlorides can be adjacent or opposite, giving distinct cis and trans forms — the very difference that makes cisplatin a drug and transplatin useless. A tetrahedron has every position adjacent to every other, so no cis–trans distinction exists; a linear complex has no choice of arrangement; and [Co(NH₃)₆]³⁺ has six identical ligands, so all arrangements are the same molecule.
The IUPAC name of K₄[Fe(CN)₆] is:
1. Potassium hexacyanidoiron(II)
2. Potassium hexacyanidoferrate(II)
3. Potassium hexacyanidoferrate(III)
4. Tetrapotassium ironhexacyanide
Show answer
Answer: B — Potassium hexacyanidoferrate(II)
Four potassium ions give +4, so the complex ion is 4−; six cyanides give 6−, leaving iron at +2. Because the complex is an anion, the metal takes its Latin stem with the -ate ending, giving ferrate rather than iron. Both the -ate ending and the oxidation state have to be right, which is why the two near-miss options differ from the answer by exactly one of those details.
Which ligand produces the largest crystal field splitting?
1. I⁻
2. H₂O
3. NH₃
4. CN⁻
Show answer
Answer: D — CN⁻
The spectrochemical series runs I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO, so cyanide is at the strong-field end and iodide at the weak. That single ordering determines whether a complex is high or low spin, and hence its magnetic moment and its colour, which is why it is worth committing to memory rather than reasoning out each time.
Cisplatin is an effective anticancer drug but transplatin is not, because:
1. Transplatin is chemically unstable
2. Only in the cis form are the leaving groups adjacent, letting platinum cross-link two neighbouring DNA bases
3. Transplatin has a different oxidation state of platinum
4. Cisplatin is optically active
Show answer
Answer: B — Only in the cis form are the leaving groups adjacent, letting platinum cross-link two neighbouring DNA bases
Both isomers have platinum(II) in a square plane with the same ligands, so composition and oxidation state are identical and neither is unstable. The difference is purely spatial: adjacent chlorides can both be displaced and let platinum bind two guanines on the same strand, kinking the DNA and blocking replication, whereas trans chlorides point in opposite directions and cannot reach two adjacent bases. Neither isomer is chiral, since both possess a plane of symmetry.
Tetrahedral complexes are almost always high spin because:
1. Δ_t is roughly 4/9 of Δ_o and so is smaller than the pairing energy
2. Tetrahedral complexes have only four d orbitals
3. Their ligands are always strong field
4. They have no crystal field splitting at all
Show answer
Answer: A — Δ_t is roughly 4/9 of Δ_o and so is smaller than the pairing energy
With only four ligands, and none of them pointing directly at a d orbital, the field is much weaker — Δ_t is about four-ninths of the octahedral value. Since Δ_t is then almost always smaller than the pairing energy, electrons spread out to keep spins parallel and the complex is high spin. There are still five d orbitals, split into a lower e set of two and an upper t₂ set of three, so splitting does occur and is simply inverted relative to octahedral.