Coordination compounds for NEET

Coordination compounds reward a methodical student. Almost every question can be broken into the same three steps: find the metal's oxidation state and d-electron count, decide whether the ligand is strong or weak field, then read off the hybridisation, shape, number of unpaired electrons and magnetic moment. Naming and isomer-counting have their own short rule sets.

NEET questions here are usually one of these: the IUPAC name or formula of a complex, the number of geometrical or optical isomers, the hybridisation and magnetic behaviour of a complex, ions produced in solution (Werner's theory), or ordering ligands in the spectrochemical series. The chapter builds directly on d- and f-block elements.

Werner's theory and the basic terms

A double salt, such as Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O, breaks fully into its simple ions in water. A complex, such as K₄[Fe(CN)₆], keeps its complex ion intact: it gives no test for Fe²⁺ or CN⁻.

Alfred Werner proposed that a metal in a complex shows two kinds of valence:

• Primary valence: ionisable, satisfied by negative ions, equal to the oxidation state.
• Secondary valence: non-ionisable, satisfied by ligands, equal to the coordination number, and directed in space, which gives the complex its shape.

His evidence came from cobalt(III) chloride–ammonia compounds and the AgCl they precipitate:

Only chloride outside the square brackets is precipitated by AgNO₃.

Oxidation state: set the sum of charges equal to the charge on the complex ion. In K₃[Fe(C₂O₄)₃], the complex ion is 3−, so x + 3(−2) = −3 and x = +3.

IUPAC nomenclature

1. Name the cation first, then the anion, as in any salt.
2. Inside the complex, name the ligands in alphabetical order, then the metal.
3. Anionic ligands end in -o: chlorido, cyanido, hydroxido, oxalato, nitrito-N (bound through N), nitrito-O.
4. Neutral ligands keep their names, except aqua (H₂O), ammine (NH₃), carbonyl (CO) and nitrosyl (NO).
5. Use di-, tri-, tetra- for simple ligands and bis-, tris-, tetrakis- for ligands whose names already contain a number, such as ethane-1,2-diamine. Prefixes are ignored when alphabetising.
6. Give the metal's oxidation state in Roman numerals in brackets.
7. If the complex ion is an anion, the metal name ends in -ate: cobaltate, ferrate (Fe), cuprate (Cu), argentate (Ag), aurate (Au).

Isomerism

Structural isomers differ in which atoms are bonded to what:

• Ionisation: an ion swaps places between inside and outside the bracket. [Co(NH₃)₅(SO₄)]Br and [Co(NH₃)₅Br]SO₄.
• Linkage: an ambidentate ligand binds through a different atom. [Co(NH₃)₅(NO₂)]Cl₂ with N-bound or O-bound nitrite.
• Coordination: ligands swap between the cationic and anionic complexes. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆].
• Solvate (hydrate): water moves between being a ligand and being lattice water. [Cr(H₂O)₆]Cl₃ (violet) and [CrCl(H₂O)₅]Cl₂·H₂O (grey-green).

Stereoisomers have the same bonds but a different arrangement in space:

Valence bond theory

The metal uses hybrid orbitals, formed from its vacant orbitals, to accept ligand lone pairs. Strong-field ligands such as CN⁻ and CO pair up the metal's d electrons so that inner (n−1)d orbitals are free.

VBT explains shape and magnetism, but it does not explain colour and treats strong and weak ligands only qualitatively.

Crystal field theory

CFT treats ligands as point charges that repel the metal's d electrons. In an octahedral field, the five d orbitals split into a lower set t₂g (three orbitals) and a higher set e_g (two orbitals). The gap is Δo. In a tetrahedral field the order is reversed and the gap is smaller: Δt ≈ (4/9)Δo, so tetrahedral complexes are almost always high spin.

Spectrochemical series (weak field to strong field):

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NH₃ < en < CN⁻ < CO

For d⁴ to d⁷ ions, the choice between high and low spin depends on Δo and the pairing energy P:

• Δo < P (weak-field ligand): electrons spread out before pairing. d⁴ becomes t₂g³ e_g¹, high spin.
• Δo > P (strong-field ligand): electrons pair in t₂g first. d⁴ becomes t₂g⁴ e_g⁰, low spin.

Colour: the complex absorbs light of energy Δo to move an electron from t₂g to e_g, and we see the complementary colour. [Ti(H₂O)₆]³⁺ (d¹) is violet. Anhydrous CuSO₄ is white; with water as ligand, it turns blue. A complex with no d–d transition (d⁰ or d¹⁰) is colourless.

Metal carbonyls: the M–CO bond has two parts. CO donates a lone pair to the metal (σ bond), and the metal donates d electrons back into CO's empty antibonding π orbital (π back-bond). This synergic bonding strengthens the M–C bond.

Why coordination compounds matter

• Qualitative analysis: Ni²⁺ gives a red precipitate with dimethylglyoxime.
• Extraction of metals: silver and gold dissolve as cyanide complexes such as [Au(CN)₂]⁻ and are then recovered.
• Biological systems: chlorophyll contains magnesium, haemoglobin contains iron, and vitamin B₁₂ (cyanocobalamin) contains cobalt.
• Analysis and medicine: EDTA is used to estimate hardness of water (Ca²⁺ and Mg²⁺), and cisplatin is used against cancer.

Worked NEET-style questions

Example 1: Werner's theory

One mole of CoCl₃·5NH₃ is treated with excess AgNO₃. How many moles of AgCl form, and how many ions does the compound give in water?

• Co is +3 with coordination number 6: five NH₃ and one Cl⁻ inside, [CoCl(NH₃)₅]Cl₂.
• Two Cl⁻ are outside: 2 mol AgCl.
• Ions: one [CoCl(NH₃)₅]²⁺ and two Cl⁻, so 3 ions.

Example 2: oxidation state and coordination number

For [Co(en)₂Cl₂]⁺, find both.

• en is neutral; each Cl⁻ is −1: x − 2 = +1, so Co is +3.
• en is bidentate: 2 × 2 + 2 = 6.

Example 3: magnetism by CFT

Compare [Fe(CN)₆]⁴⁻ and [Fe(H₂O)₆]²⁺.

• Both contain Fe²⁺, which is 3d⁶.
• CN⁻ is strong field: t₂g⁶ e_g⁰, 0 unpaired, diamagnetic, d²sp³.
• H₂O is weak field: t₂g⁴ e_g², 4 unpaired, μ ≈ 4.90 BM.

Example 4: shape from ligand strength

Why is [Ni(CN)₄]²⁻ diamagnetic but [NiCl₄]²⁻ paramagnetic?

• Ni²⁺ is 3d⁸. Strong-field CN⁻ pairs the eight electrons into four d orbitals, leaving one 3d free for dsp²: square planar, 0 unpaired.
• Weak-field Cl⁻ cannot force pairing, so Ni uses 4s and 4p: sp³, tetrahedral, 2 unpaired, μ ≈ 2.83 BM.

Practice MCQs

1. Which is an ambidentate ligand? (a) NH₃ (b) en (c) NO₂⁻ (d) EDTA⁴⁻
2. The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is: (a) pentaamminechloridocobalt(III) chloride (b) chloridopentaamminecobalt(II) chloride (c) pentaamminechloridocobaltate(III) chloride (d) pentaamminecobalt(III) trichloride
3. The number of ions [Co(NH₃)₆]Cl₃ gives in water is: (a) 2 (b) 3 (c) 4 (d) 6
4. The number of geometrical isomers of square planar [Pt(NH₃)₂Cl₂] is: (a) 0 (b) 2 (c) 3 (d) 4
5. The hybridisation and magnetic nature of [Ni(CO)₄] are: (a) dsp², diamagnetic (b) sp³, paramagnetic (c) sp³, diamagnetic (d) d²sp³, diamagnetic
6. Which is optically active? (a) [Co(NH₃)₆]³⁺ (b) [Co(en)₃]³⁺ (c) trans-[CoCl₂(en)₂]⁺ (d) [Ni(CN)₄]²⁻
7. The strongest field ligand among these is: (a) Cl⁻ (b) H₂O (c) NH₃ (d) CO
8. The number of unpaired electrons in [Fe(CN)₆]³⁻ is: (a) 0 (b) 1 (c) 3 (d) 5

Answers

1. (c) Nitrite can bind through N or O.
2. (a) Ammine before chlorido alphabetically; cationic complex, so no -ate.
3. (c) [Co(NH₃)₆]³⁺ plus three Cl⁻.
4. (b) cis and trans.
5. (c) Ni(0) becomes 3d¹⁰ with CO; 4s and 4p give sp³.
6. (b) Three chelate rings make a non-superimposable mirror image.
7. (d) CO is at the strong end of the series.
8. (b) Fe³⁺ is d⁵; strong-field CN⁻ gives t₂g⁵, one unpaired electron.

What to do next

• Write the IUPAC names of ten complexes and then reverse the exercise, formula from name.
• Learn the isomer-count table and draw cis, trans, fac and mer shapes once by hand.
• For each of d³ to d⁸, write the high-spin and low-spin octahedral configurations and unpaired electrons.
• Revise spin-only moments in d- and f-block elements and hybridisation shapes in chemical bonding.