Coordination Compounds
Welcome to HSLC Guru. This article presents a complete English-medium guide for ASSEB Class 12 Chemistry Chapter 9 — Coordination Compounds. Coordination compounds are an exciting class of substances in which a central metal atom or ion is surrounded by a definite number of neutral molecules or ions called ligands. They explain the brilliant colours of transition-metal salts, the action of haemoglobin in our blood, the green colour of plants due to chlorophyll, and the catalytic action of vitamin B₁₂. This chapter walks you through Werner’s pioneering theory, the rules of IUPAC nomenclature, isomerism, bonding theories (VBT and CFT), magnetic and optical behaviour, and the practical importance of coordination compounds in biology, analysis, medicine, and industry.
The notes are written strictly according to the ASSEB Class 12 syllabus. Every concept is explained in simple language, followed by a generous bank of short and long questions, MCQs, fill-in-the-blanks, true/false statements, and a glossary. Study the summary first, attempt the questions on your own, and then check the answers. This will give you the confidence you need for HS Final Examinations as well as competitive entrance tests like NEET and JEE.
Summary
Werner’s Theory and Valencies: Alfred Werner (1893) proposed the first systematic theory of coordination compounds. According to him, a metal exhibits two types of valencies: primary valency (ionisable, satisfied by negative ions, equal to the oxidation state of the metal) and secondary valency (non-ionisable, satisfied by neutral molecules or negative ions, equal to the coordination number, and directed in space giving the complex a definite geometry). For example in [Co(NH₃)₆]Cl₃, the three Cl⁻ ions satisfy the primary valency while six NH₃ molecules satisfy the secondary valency. Ligands are the species that donate at least one lone pair of electrons to the metal. They are classified as monodentate (e.g. NH₃, H₂O, CN⁻), bidentate (e.g. ethylenediamine, oxalate), polydentate (e.g. EDTA — hexadentate), ambidentate (e.g. NO₂⁻/ONO⁻, SCN⁻/NCS⁻ — can attach through either of two atoms), and chelating ligands which form ring structures (chelates) with the metal. The coordination sphere is the central metal together with its ligands enclosed in square brackets, the coordination number (CN) is the number of donor atoms bonded to the metal, and the oxidation number is the residual charge on the metal when all ligands are removed in their normal closed-shell forms.
IUPAC Nomenclature and Isomerism: While naming, the cation is named first followed by the anion. Inside the coordination sphere, ligands are named alphabetically before the metal. Anionic ligands end in “-o” (chloro, cyano, hydroxo), neutral ligands keep their names with special cases (aqua = H₂O, ammine = NH₃, carbonyl = CO, nitrosyl = NO). Numerical prefixes di, tri, tetra are used for simple ligands and bis, tris, tetrakis for complex names. The oxidation state of the metal is given in Roman numerals in parentheses. If the complex is anionic, the metal name ends with “-ate” (ferrate, cuprate, argentate). Examples: [Co(NH₃)₆]Cl₃ — hexaamminecobalt(III) chloride; K₄[Fe(CN)₆] — potassium hexacyanidoferrate(II). Isomerism is of two broad types: Structural — (i) linkage (e.g. [Co(NH₃)₅(NO₂)]²⁺ vs [Co(NH₃)₅(ONO)]²⁺), (ii) ionisation (e.g. [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br), (iii) coordination (interchange of ligands between cation and anion), (iv) hydrate or solvate (e.g. [Cr(H₂O)₆]Cl₃ violet, [Cr(H₂O)₅Cl]Cl₂·H₂O blue-green, [Cr(H₂O)₄Cl₂]Cl·2H₂O dark green); and Stereoisomerism — (i) geometric (cis–trans, found mainly in square-planar Ma₂b₂ and octahedral Ma₄b₂ types), and (ii) optical (non-superimposable mirror images, common in [M(en)₃]ⁿ⁺ and cis-[M(en)₂Cl₂]ⁿ⁺).
Bonding Theories — VBT and CFT: Valence Bond Theory (VBT) proposed by Pauling explains bonding through hybridisation. The metal provides empty hybrid orbitals (sp³, dsp², sp³d², d²sp³) which overlap with filled ligand orbitals to form coordinate bonds. It accounts for geometry and magnetic properties (inner-orbital/low-spin vs outer-orbital/high-spin) but fails to explain colour, the relative strengths of ligands, and quantitative magnetic data. Crystal Field Theory (CFT) treats metal–ligand interaction as purely electrostatic. In an octahedral field the five degenerate d-orbitals split into a lower triply degenerate t₂g set (dxy, dyz, dxz) and a higher doubly degenerate eg set (dz², dx²-y²) with energy gap Δ₀ (10 Dq). In a tetrahedral field the order is reversed (e below t₂) and the splitting Δt is smaller.
$\Delta_t = \frac{4}{9}\Delta_o$
The Crystal Field Stabilisation Energy (CFSE) for an octahedral complex is given by:
$\text{CFSE} = (-0.4 n_{t_{2g}} + 0.6 n_{e_g})\Delta_o$
If Δ₀ > pairing energy (P), electrons pair up giving a low-spin complex (strong-field ligand); if Δ₀ < P, electrons remain unpaired giving a high-spin complex (weak-field ligand). The spectrochemical series arranges ligands in increasing field strength: I⁻ < Br⁻ < SCN⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Tetrahedral complexes are nearly always high-spin because Δt is small. Colour arises due to d–d electronic transitions; the complex absorbs a wavelength corresponding to Δ and transmits the complementary colour. Magnetic moment is calculated using the spin-only formula:
$\mu = \sqrt{n(n+2)}\,\text{BM}$
Importance of Coordination Compounds: Biological — haemoglobin (Fe²⁺ porphyrin complex transports O₂), chlorophyll (Mg²⁺ porphyrin, photosynthesis), vitamin B₁₂ or cyanocobalamin (Co³⁺ corrin complex), carboxypeptidase-A (Zn²⁺ enzyme). Analytical — EDTA estimates Ca²⁺ and Mg²⁺ hardness of water; DMG detects Ni²⁺ as a rosy-red precipitate; K₄[Fe(CN)₆] tests for Fe³⁺ (Prussian blue). Industrial / Medicinal — extraction of silver and gold by cyanide leaching, electroplating using [Ag(CN)₂]⁻ and [Au(CN)₂]⁻, Wilkinson’s catalyst [RhCl(PPh₃)₃] for hydrogenation, cisplatin cis-[PtCl₂(NH₃)₂] in cancer chemotherapy, EDTA in lead-poisoning therapy.
Question and Answers
1-Mark Questions
Q1. Define a coordination compound.
Answer: A coordination compound is a compound in which a central metal atom or ion is bonded to a definite number of neutral molecules or negative ions (ligands) through coordinate bonds, retaining its identity in solution.
Q2. What is meant by coordination number?
Answer: The coordination number of a metal in a complex is the total number of donor atoms (ligand sites) directly attached to the central metal ion through coordinate bonds.
Q3. What is the oxidation number of Fe in K₄[Fe(CN)₆]?
Answer: Let oxidation state of Fe = x. 4(+1) + x + 6(-1) = 0, hence x = +2.
Q4. Give one example of an ambidentate ligand.
Answer: NO₂⁻ (nitrito-N or nitrito-O) and SCN⁻ (thiocyanato-S or thiocyanato-N) are ambidentate ligands.
Q5. What is a chelate?
Answer: A chelate is a closed-ring complex formed when a polydentate ligand attaches to the central metal through two or more donor atoms, producing a stable ring structure.
Q6. Name the metal present in chlorophyll.
Answer: Magnesium (Mg²⁺) is the central metal ion present in chlorophyll.
Q7. Write the IUPAC name of [Cu(NH₃)₄]SO₄.
Answer: Tetraamminecopper(II) sulphate.
Q8. What is the geometry of [Ni(CN)₄]²⁻?
Answer: Square planar (dsp² hybridisation, diamagnetic).
Q9. Which type of isomerism is shown by [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂?
Answer: Linkage isomerism.
Q10. What is CFSE?
Answer: Crystal Field Stabilisation Energy is the net lowering of energy of the d-electrons of a metal ion when placed in a crystal field of ligands, compared with the energy of the same electrons in a spherical (degenerate) field.
2-3 Marks Questions
Q11. State the postulates of Werner’s theory.
Answer: (i) Metals exhibit two types of valencies — primary (ionisable, equal to oxidation state, satisfied by anions) and secondary (non-ionisable, equal to coordination number, satisfied by ligands). (ii) Every metal has a fixed coordination number. (iii) Secondary valencies are directed towards fixed positions in space, giving each complex a definite geometry such as octahedral, tetrahedral, or square planar.
Q12. Distinguish between double salt and complex compound.
Answer: A double salt (e.g. Mohr’s salt FeSO₄·(NH₄)₂SO₄·6H₂O, alum) dissociates completely in water giving all simple ions, and loses its identity in solution. A complex compound (e.g. K₄[Fe(CN)₆]) gives the complex ion as such in solution and retains its identity; the ligands are tightly bound to the metal through coordinate bonds.
Q13. Write IUPAC names: (a) [Co(en)₃]Cl₃, (b) Na₃[Co(NO₂)₆], (c) [Pt(NH₃)₂Cl₂].
Answer: (a) Tris(ethane-1,2-diamine)cobalt(III) chloride. (b) Sodium hexanitrito-N-cobaltate(III). (c) Diamminedichloridoplatinum(II).
Q14. Differentiate between geometric and optical isomerism with one example each.
Answer: Geometric (cis–trans) isomerism arises from different spatial arrangement of identical ligands; e.g. cis- and trans-[Pt(NH₃)₂Cl₂]. Optical isomerism arises when a complex and its mirror image are non-superimposable; e.g. [Co(en)₃]³⁺ exists as d- and l-forms which rotate plane-polarised light in opposite directions.
Q15. What are strong-field and weak-field ligands? Give examples.
Answer: A strong-field ligand produces large crystal-field splitting (Δ₀ > P), forces electron pairing and gives low-spin complexes; e.g. CN⁻, CO, NO₂⁻, en. A weak-field ligand produces small splitting (Δ₀ < P), gives high-spin complexes; e.g. I⁻, Br⁻, Cl⁻, F⁻, H₂O.
Q16. Why is [Fe(CN)₆]³⁻ low-spin while [FeF₆]³⁻ is high-spin?
Answer: CN⁻ is a strong-field ligand causing Δ₀ > P; therefore the five d-electrons of Fe(III) pair up in t₂g giving one unpaired electron (low-spin, μ ≈ 1.73 BM, d²sp³). F⁻ is a weak-field ligand giving Δ₀ < P; the d-electrons remain unpaired across t₂g and eg (high-spin, μ ≈ 5.92 BM, sp³d²).
5-7 Marks Questions
Q17. Discuss the salient features of Crystal Field Theory and explain octahedral splitting with the help of a diagram. Calculate CFSE of [Cr(H₂O)₆]³⁺.
Answer: CFT (Bethe and Van Vleck) considers the metal–ligand interaction to be purely electrostatic. The ligands are treated as point negative charges (or dipoles) which approach the metal ion. In an isolated gaseous ion the five d-orbitals are degenerate. When six ligands approach along ±x, ±y, ±z axes (octahedral arrangement), the d-orbitals pointing along the axes (dz² and dx²-y² → eg set) experience greater repulsion and are raised in energy, while orbitals pointing between axes (dxy, dyz, dxz → t₂g set) are lowered. The energy gap is Δ₀ (10 Dq); t₂g lies 0.4 Δ₀ below and eg lies 0.6 Δ₀ above the average (barycentre). Diagrammatically:
average d-level —— splits into —— t₂g (-0.4 Δ₀, three orbitals) below and eg (+0.6 Δ₀, two orbitals) above. For [Cr(H₂O)₆]³⁺, Cr³⁺ is d³ → t₂g³ eg⁰. CFSE = (-0.4 × 3 + 0.6 × 0) Δ₀ = -1.2 Δ₀. The negative sign indicates stabilisation. CFT successfully explains colour, magnetic behaviour, and order of stability of complexes.
Q18. Using VBT, explain the geometry, hybridisation and magnetic nature of (a) [Co(NH₃)₆]³⁺, (b) [NiCl₄]²⁻, (c) [Ni(CN)₄]²⁻.
Answer: (a) Co³⁺ = 3d⁶. NH₃ is a strong-field ligand → all six 3d electrons pair up in three orbitals leaving two 3d orbitals empty. These two 3d, one 4s and three 4p orbitals form six d²sp³ hybrid orbitals → octahedral geometry, no unpaired electron, diamagnetic, inner-orbital low-spin complex. (b) Ni²⁺ = 3d⁸. Cl⁻ is weak-field → no pairing. 4s, three 4p and one 4d? No — actually Ni²⁺ with weak field uses sp³ hybridisation (one 4s + three 4p) → tetrahedral, two unpaired electrons (μ ≈ 2.83 BM), paramagnetic. (c) Ni²⁺ = 3d⁸. CN⁻ is strong-field → the two unpaired 3d electrons pair up giving one empty 3d orbital. dsp² hybridisation (one 3d + 4s + two 4p) → square planar, diamagnetic.
Q19. Discuss different types of structural isomerism in coordination compounds with examples.
Answer: (i) Linkage isomerism — occurs with ambidentate ligands; e.g. [Co(NH₃)₅NO₂]²⁺ (nitro, N-bonded, yellow) and [Co(NH₃)₅ONO]²⁺ (nitrito, O-bonded, red). (ii) Ionisation isomerism — interchange of ligand and counter-ion; e.g. [Co(NH₃)₅Br]SO₄ (gives BaSO₄ with BaCl₂) and [Co(NH₃)₅SO₄]Br (gives AgBr with AgNO₃). (iii) Coordination isomerism — exchange of ligands between cation and anion of complex salts; e.g. [Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]. (iv) Hydrate / solvate isomerism — different number of water molecules inside and outside the coordination sphere; e.g. [Cr(H₂O)₆]Cl₃ violet, [Cr(H₂O)₅Cl]Cl₂·H₂O blue-green, [Cr(H₂O)₄Cl₂]Cl·2H₂O dark green.
Q20. Explain why coordination compounds are coloured. How is colour related to the spectrochemical series?
Answer: Most transition-metal complexes possess unfilled d-orbitals split by a crystal field. Visible light excites electrons from the lower (t₂g) to upper (eg) set — a d–d transition. The complex absorbs the wavelength whose energy equals Δ₀ and transmits the complementary colour, which we observe. The magnitude of Δ₀ depends upon the ligand: weak-field ligands give small Δ₀ (absorb red, transmit green/blue) and strong-field ligands give large Δ₀ (absorb blue/UV, transmit yellow/red). The spectrochemical series arranges ligands in increasing order of Δ₀: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Hence [Ti(H₂O)₆]³⁺ is purple, [Cu(H₂O)₄]²⁺ blue, [Cu(NH₃)₄]²⁺ deep-violet. Complexes of d⁰ (Sc³⁺) and d¹⁰ (Zn²⁺) are colourless because no d–d transition is possible.
Q21. Describe the importance of coordination compounds in biology, analysis and industry.
Answer: Biological: (i) Haemoglobin — Fe²⁺ porphyrin complex carries O₂ from lungs to tissues. (ii) Chlorophyll — Mg²⁺ porphyrin complex traps sunlight for photosynthesis. (iii) Vitamin B₁₂ (cyanocobalamin) — Co³⁺ corrin complex essential for erythropoiesis. (iv) Many metalloenzymes — carbonic anhydrase (Zn²⁺), cytochromes (Fe). Analytical: (i) EDTA — quantitative estimation of Ca²⁺ and Mg²⁺ (water hardness). (ii) DMG — detection of Ni²⁺ as rosy-red precipitate. (iii) K₄[Fe(CN)₆] — confirmatory test of Fe³⁺ (Prussian-blue). Industrial / Medicinal: (i) Extraction of silver and gold by cyanide leaching forming [Ag(CN)₂]⁻, [Au(CN)₂]⁻. (ii) Electroplating of silver and gold using cyanide complexes for smooth coatings. (iii) Wilkinson’s catalyst [RhCl(PPh₃)₃] for hydrogenation. (iv) Cisplatin cis-[Pt(NH₃)₂Cl₂] — anti-cancer drug. (v) EDTA — antidote in lead and mercury poisoning. (vi) Photographic fixer — Na₃[Ag(S₂O₃)₂].
Multiple Choice Questions (MCQ)
Q1. The coordination number of cobalt in [Co(en)₃]Cl₃ is —
(a) 3 (b) 4 (c) 6 (d) 9
Answer: (c) 6 (en is bidentate, three en × 2 = 6).
Q2. Which of the following is an ambidentate ligand?
(a) NH₃ (b) en (c) SCN⁻ (d) C₂O₄²⁻
Answer: (c) SCN⁻.
Q3. The IUPAC name of K₃[Fe(CN)₆] is —
(a) Potassium ferricyanide (b) Potassium hexacyanidoferrate(III) (c) Potassium hexacyanoiron(III) (d) Tripotassium hexacyanoferrate
Answer: (b).
Q4. Which complex is diamagnetic?
(a) [NiCl₄]²⁻ (b) [Ni(CN)₄]²⁻ (c) [CoF₆]³⁻ (d) [Fe(H₂O)₆]³⁺
Answer: (b) [Ni(CN)₄]²⁻ (square planar, dsp², no unpaired electron).
Q5. The geometry of [Ni(CO)₄] is —
(a) Square planar (b) Octahedral (c) Tetrahedral (d) Linear
Answer: (c) Tetrahedral (sp³, diamagnetic).
Q6. Which one is the strongest field ligand?
(a) F⁻ (b) Cl⁻ (c) H₂O (d) CO
Answer: (d) CO.
Q7. CFSE for d⁵ high-spin octahedral complex is —
(a) 0 (b) -0.4 Δ₀ (c) -2.0 Δ₀ (d) -1.6 Δ₀
Answer: (a) 0 Δ₀ (t₂g³ eg² → -0.4×3 + 0.6×2 = 0).
Q8. The metal present in vitamin B₁₂ is —
(a) Fe (b) Mg (c) Zn (d) Co
Answer: (d) Co.
Q9. Which pair shows ionisation isomerism?
(a) [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br (b) [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O (c) cis- and trans-[Pt(NH₃)₂Cl₂] (d) d- and l-[Co(en)₃]³⁺
Answer: (a).
Q10. The magnetic moment of [Fe(CN)₆]³⁻ (low-spin, d⁵, n = 1) is approximately —
(a) 1.73 BM (b) 2.83 BM (c) 4.90 BM (d) 5.92 BM
Answer: (a) 1.73 BM.
Fill in the Blanks
Q1. The geometry of [Cu(NH₃)₄]²⁺ is __________.
Answer: Square planar.
Q2. The number of unpaired electrons in [CoF₆]³⁻ is __________.
Answer: Four.
Q3. EDTA is a __________ ligand.
Answer: Hexadentate.
Q4. The relationship between Δt and Δ₀ is __________.
Answer: Δt = (4/9) Δ₀.
Q5. The central metal in haemoglobin is __________.
Answer: Iron (Fe²⁺).
True or False
Q1. Primary valency is non-ionisable. — False (it is ionisable; secondary is non-ionisable).
Q2. [Ni(CN)₄]²⁻ is square planar and diamagnetic. — True.
Q3. Tetrahedral complexes are usually low-spin. — False (Δt is small, so high-spin).
Q4. Cisplatin is used as an anti-cancer drug. — True.
Q5. CO is a weaker-field ligand than F⁻. — False (CO is one of the strongest-field ligands).
Glossary
| Term | Meaning |
|---|---|
| Coordination compound | Compound with central metal bonded to ligands by coordinate bonds. |
| Ligand | Ion or molecule donating lone pair to a metal. |
| Coordination number | Number of donor atoms attached to the central metal. |
| Coordination sphere | The metal and its ligands enclosed within square brackets. |
| Primary valency | Ionisable valency, equal to oxidation state of metal. |
| Secondary valency | Non-ionisable valency, equal to coordination number. |
| Chelate | Closed-ring complex formed by a polydentate ligand. |
| Ambidentate ligand | Ligand that can bond through either of two donor atoms. |
| Werner’s theory | First systematic theory of complexes (1893). |
| VBT | Valence Bond Theory — bonding via hybrid orbitals. |
| CFT | Crystal Field Theory — electrostatic interaction model. |
| Δ₀ | Octahedral crystal-field splitting energy (10 Dq). |
| Δt | Tetrahedral splitting; equals (4/9) Δ₀. |
| CFSE | Crystal Field Stabilisation Energy of d-electrons in a field. |
| High-spin complex | Weak-field ligand, max unpaired electrons (Δ < P). |
| Low-spin complex | Strong-field ligand, electrons paired (Δ > P). |
| Spectrochemical series | Order of ligands in increasing field strength. |
| Linkage isomerism | Isomerism due to ambidentate ligands. |
| Ionisation isomerism | Exchange of ligand and counter-ion. |
| Geometric isomerism | cis–trans isomerism due to spatial arrangement. |
| Optical isomerism | Existence as non-superimposable mirror images. |
| Magnetic moment | μ = √[n(n+2)] BM where n is unpaired electrons. |
| Haemoglobin | Iron-porphyrin complex, oxygen carrier in blood. |
| Chlorophyll | Magnesium-porphyrin complex, photosynthetic pigment. |
| Vitamin B₁₂ | Cobalt-corrin complex (cyanocobalamin). |
| EDTA | Hexadentate chelating ligand for Ca²⁺ / Mg²⁺ estimation. |
| DMG | Dimethyl-glyoxime, used to detect Ni²⁺. |
| Cisplatin | cis-[Pt(NH₃)₂Cl₂], anti-cancer drug. |
| Wilkinson’s catalyst | [RhCl(PPh₃)₃], hydrogenation catalyst. |
| Prussian blue | Fe₄[Fe(CN)₆]₃, confirmatory test for Fe³⁺. |
Important Solved Numericals and Reasoning Questions
Q1. Calculate the magnetic moment of [Mn(H₂O)₆]²⁺ (high-spin, d⁵).
Answer: Mn²⁺ = 3d⁵. H₂O is a weak-field ligand → all five electrons remain unpaired (n = 5). Using the spin-only formula μ = √[5(5+2)] = √35 = 5.92 BM. So [Mn(H₂O)₆]²⁺ is paramagnetic with five unpaired electrons.
Q2. A metal complex has the formula CrCl₃·6H₂O. Aqueous AgNO₃ precipitates only one Cl⁻ per formula unit. Write the structural formula.
Answer: If only one Cl⁻ is ionisable, the other two must be inside the coordination sphere. Therefore, the structural formula is [Cr(H₂O)₄Cl₂]Cl·2H₂O. This is the dark-green hydrate isomer of chromium(III) chloride hexahydrate.
Q3. Why is [Co(NH₃)₆]³⁺ diamagnetic but [CoF₆]³⁻ paramagnetic, even though both contain Co³⁺ (d⁶)?
Answer: NH₃ is a strong-field ligand: Δ₀ > pairing energy P. All six d-electrons pair in the t₂g level (t₂g⁶ eg⁰); n = 0 → diamagnetic, d²sp³, inner-orbital, low-spin. F⁻ is a weak-field ligand: Δ₀ < P. The electrons remain unpaired (t₂g⁴ eg²); n = 4 → paramagnetic, sp³d², outer-orbital, high-spin (μ ≈ 4.90 BM).
Q4. Compute the CFSE of (a) d⁴ low-spin, (b) d⁷ high-spin octahedral complexes.
Answer: (a) d⁴ low-spin: t₂g⁴ eg⁰ → CFSE = (-0.4 × 4 + 0.6 × 0) Δ₀ = -1.6 Δ₀ + P (one extra pairing). (b) d⁷ high-spin: t₂g⁵ eg² → CFSE = (-0.4 × 5 + 0.6 × 2) Δ₀ = -0.8 Δ₀.
Q5. [Ti(H₂O)₆]³⁺ absorbs at 500 nm and appears violet. Explain.
Answer: Ti³⁺ is a 3d¹ ion. The single d-electron in t₂g is excited to eg by absorbing a photon of wavelength 500 nm (yellow-green region). The complementary colour transmitted is violet, hence the observed colour. The energy of the photon equals Δ₀ for the complex.
Q6. Why is Zn²⁺ complex like [Zn(NH₃)₄]²⁺ colourless?
Answer: Zn²⁺ has a 3d¹⁰ configuration. All five d-orbitals are completely filled, so there is no vacant d-orbital available for d–d transition. Therefore, no visible-light absorption takes place and the complex is colourless. Similarly Sc³⁺ (d⁰) and Cu⁺ (d¹⁰) complexes are colourless.
Q7. Why are square-planar Pt(II) complexes more inert than the corresponding tetrahedral Ni(II) complexes?
Answer: Pt(II) is a 5d⁸ ion with very large CFSE in a square-planar geometry; the strong dsp² bonding and the large 5d-orbital extension make ligand substitution kinetically slow. Ni(II) tetrahedral complexes have small Δt and weaker bonding, and so undergo rapid ligand exchange.
Q8. Give the formula and IUPAC name of (a) Wilkinson’s catalyst, (b) cisplatin, (c) Zeise’s salt.
Answer: (a) [RhCl(PPh₃)₃] — chloridotris(triphenylphosphine)rhodium(I). (b) cis-[Pt(NH₃)₂Cl₂] — cis-diamminedichloridoplatinum(II). (c) K[PtCl₃(η²-C₂H₄)] — potassium trichlorido(η²-ethene)platinate(II).
Quick Revision Points
- Werner’s primary valency = oxidation state; secondary valency = coordination number.
- EDTA is hexadentate; en is bidentate; oxalate is bidentate; DMG is bidentate.
- Coordination sphere is written within square brackets [].
- While naming, ligands come alphabetically before the metal; anionic complex ends with “-ate”.
- Strong-field ligands → low-spin; weak-field → high-spin (octahedral only).
- Tetrahedral complexes are nearly always high-spin because Δt = (4/9) Δ₀.
- Spin-only μ = √[n(n+2)] BM; n is unpaired electrons.
- Colour of complex = complementary of the absorbed wavelength.
- Cisplatin is anti-cancer; Wilkinson’s catalyst hydrogenates alkenes; Zeise’s salt has Pt–C₂H₄ π-bond.
- Haemoglobin (Fe), chlorophyll (Mg), vitamin B₁₂ (Co) — three biological coordination compounds to remember.
Common Examples — Geometry, Hybridisation, Magnetic Nature
| Complex | Geometry | Hybridisation | Magnetic Nature |
|---|---|---|---|
| [Co(NH₃)₆]³⁺ | Octahedral | d²sp³ | Diamagnetic |
| [CoF₆]³⁻ | Octahedral | sp³d² | Paramagnetic (4 e⁻) |
| [Fe(CN)₆]³⁻ | Octahedral | d²sp³ | Paramagnetic (1 e⁻) |
| [Fe(CN)₆]⁴⁻ | Octahedral | d²sp³ | Diamagnetic |
| [FeF₆]³⁻ | Octahedral | sp³d² | Paramagnetic (5 e⁻) |
| [Ni(CN)₄]²⁻ | Square planar | dsp² | Diamagnetic |
| [NiCl₄]²⁻ | Tetrahedral | sp³ | Paramagnetic (2 e⁻) |
| [Ni(CO)₄] | Tetrahedral | sp³ | Diamagnetic |
| [Cu(NH₃)₄]²⁺ | Square planar | dsp² | Paramagnetic (1 e⁻) |
| [Pt(NH₃)₂Cl₂] | Square planar | dsp² | Diamagnetic |
| [Cr(H₂O)₆]³⁺ | Octahedral | d²sp³ | Paramagnetic (3 e⁻) |
| [Mn(H₂O)₆]²⁺ | Octahedral | sp³d² | Paramagnetic (5 e⁻) |
Keep visiting HSLC Guru for more ASSEB Class 12 Chemistry chapter notes, MCQs, and previous-year solved papers in English medium. Best wishes for your HS Final Examination.