Chapter 5 — Coordination Chemistry Exercise Solutions
This chapter is highly scoring if IUPAC naming, oxidation state calculation, and VBT/CFT concepts are clear.
Key Concepts Quick Recap
| Term | Meaning |
|---|
| Ligand | Ion or molecule that donates an electron pair to the central metal |
| Coordination number (CN) | Number of donor atoms directly bonded to the central metal |
| Chelate | Ring complex formed by a polydentate ligand; chelation gives extra stability |
| Ambidentate ligand | Ligand that can bond through different atoms (e.g. SCN⁻ via S or N) |
| Monodentate | One donor atom per ligand (e.g. Cl⁻, NH₃, H₂O) |
| Bidentate | Two donor atoms (e.g. en = ethylenediamine, ox²⁻ = oxalate) |
| EDTA | Hexadentate ligand — forms 5-membered rings |
| Effective Atomic Number (EAN) | Z + electrons from ligands; stable when = nearest noble gas |
| VBT | Valence Bond Theory — uses hybrid orbitals; explains shape and magnetism |
| CFT | Crystal Field Theory — explains colour and magnetic behaviour using d orbital splitting |
| CFSE | Crystal Field Stabilisation Energy — extra stability from d electron arrangement |
I. Choose the Best Answer (MCQs)
1. The coordination number of Co in [Co(NH₃)₆]³⁺ is: (a) 3 (b) 4 (c) 6 (d) 12
Answer(c) 6 NH₃ is monodentate — one donor atom each. Six NH₃ ligands → coordination number = 6.
2. The IUPAC name of [Cu(NH₃)₄]²⁺ is: (a) Copper(II) tetraammine (b) Tetraamminecopper(II) ion (c) Ammoniacopper(IV) ion (d) Cupric ammoniate
Answer(b) Tetraamminecopper(II) ion Naming: ligands first (alphabetical) → metal → OS in brackets. tetra + ammine (NH₃) + copper + (II).
3. Which of the following is a bidentate ligand? (a) Cl⁻ (b) NH₃ (c) Ethylenediamine (en) (d) CO
Answer(c) Ethylenediamine (en) en = H₂N−CH₂−CH₂−NH₂ has two −NH₂ groups, both of which can donate lone pairs → bidentate.
4. The complex [Fe(CN)₆]⁴⁻ is: (a) Paramagnetic with 4 unpaired e⁻ (b) Diamagnetic (c) Paramagnetic with 2 unpaired e⁻ (d) Paramagnetic with 6 unpaired e⁻
Answer(b) Diamagnetic CN⁻ is a strong field ligand → causes pairing of electrons. Fe²⁺ = d⁶; with strong field → all 6 electrons paired in lower t₂g → 0 unpaired → diamagnetic.
5. Geometrical isomerism is possible in: (a) [Co(NH₃)₆]³⁺ (b) [Pt(NH₃)₂Cl₂] (c) [Cu(NH₃)₄]²⁺ (d) [Ag(NH₃)₂]⁺
Answer(b) [Pt(NH₃)₂Cl₂] Square planar complexes of the type MA₂B₂ show cis-trans isomerism. The other complexes are either octahedral with same ligands or linear.
6. The hybridisation of [Ni(CN)₄]²⁻ is: (a) sp³ (b) dsp² (c) sp³d² (d) d²sp³
Answer(b) dsp² CN⁻ is a strong field ligand. Ni²⁺ = d⁸. Strong field causes pairing → one d orbital freed for bonding → dsp² hybridisation → square planar complex, diamagnetic.
7. Which of the following ligands shows the chelate effect? (a) Cl⁻ (b) NH₃ (c) EDTA (d) H₂O
Answer(c) EDTA EDTA is hexadentate — it wraps around the metal forming 5 five-membered rings. The chelate effect gives exceptional thermodynamic stability (high Kf).
II. Short Answer Questions
Q1. What is the chelate effect? Why are chelate complexes more stable?
Answer
When a polydentate ligand binds to a metal through multiple donor atoms, it forms ring structures called chelate rings. This increases stability compared to the same number of monodentate ligands.
Reasons for stability: 1. Entropic advantage: replacing monodentate ligands with a polydentate one releases more solvent molecules → ΔS increases → ΔG more negative. 2. Pre-organisation: polydentate ligands are geometrically suited to wrap around the metal.
Example: [Cu(en)₂]²⁺ is more stable than [Cu(NH₃)₄]²⁺ even though both have CN = 4.
Q2. What is Crystal Field Theory (CFT)? How does it explain the colour of complexes?
Answer
CFT treats ligands as point charges that create an electrostatic field around the central metal. The five degenerate d orbitals of the metal split into two sets in the presence of this field:
- In an octahedral field: lower energy t₂g (3 orbitals) and higher energy eₘ (2 orbitals).
- The energy gap between them is called crystal field splitting energy (Δₒ).
When visible light falls on the complex, electrons absorb photons of energy exactly equal to Δₒ and jump from t₂g to eₘ. The complementary colour of the absorbed light is what we observe.
Example: [Ti(H₂O)₆]³⁺ absorbs green light (500 nm) → appears purple.
Q3. Distinguish between strong-field and weak-field ligands.
Answer
| Property | Strong-field Ligands | Weak-field Ligands |
|---|
| CFSE (Δₒ) | High | Low |
| Electron pairing | Forces pairing | Does not force pairing |
| Spin type | Low spin | High spin |
| Magnetic nature | Diamagnetic or less paramagnetic | More paramagnetic |
| Examples | CN⁻, CO, NO₂⁻ | Cl⁻, Br⁻, I⁻, F⁻, H₂O |
Spectrochemical series: CO > CN⁻ > NO₂⁻ > en > NH₃ > H₂O > OH⁻ > F⁻ > Cl⁻ > Br⁻ > I⁻
Q4. Explain optical isomerism in coordination compounds with an example.
Answer
Optical isomers (enantiomers) are non-superimposable mirror images of each other. They rotate plane-polarised light in opposite directions.
Octahedral complexes of the type [M(en)₃]ⁿ⁺ or [M(AA)₂B₂]ⁿ⁺ show optical isomerism.
Example: [Co(en)₃]³⁺ — the d-form rotates light to the right, the l-form to the left. They have identical chemical properties but differ in their effect on polarised light.
Q5. Write the IUPAC names of the following complexes:
(i) [Fe(CN)₆]³⁻ → hexacyanoferrate(III) ion
(ii) [Co(NH₃)₅Cl]²⁺ → pentaamminechloridocobalt(III) ion
(iii) K₂[PtCl₄] → potassium tetrachloridoplatinate(II)
(iv) [Cr(en)₂Cl₂]⁺ → dichlorido(ethylenediamine)chromium(III) ion
→ or: bis(ethane-1,2-diamine)dichloridochromium(III) ion
Worked Examples
Example 1 — Finding oxidation state of central metal
Find the oxidation state of metal in [Fe(CN)₆]³⁻, [Co(NH₃)₆]³⁺, [PtCl₄]²⁻
[Fe(CN)₆]³⁻: CN⁻ charge = −1 each. x + 6(−1) = −3 → x = +3
[Co(NH₃)₆]³⁺: NH₃ is neutral. x + 0 = +3 → x = +3
[PtCl₄]²⁻: Cl⁻ charge = −1 each. x + 4(−1) = −2 → x = +2
Example 2 — VBT hybridisation and shape
[Co(NH₃)₆]³⁺: Co³⁺ = d⁶; NH₃ strong field → all 6 e⁻ paired in 3d
Free orbitals: two 3d + one 4s + three 4p = d²sp³
Hybridisation: d²sp³ → octahedral → diamagnetic (inner complex)
[CoF₆]³⁻: Co³⁺ = d⁶; F⁻ weak field → electrons not paired
Uses outer d orbitals: one 4s + three 4p + two 4d = sp³d²
Hybridisation: sp³d² → octahedral → paramagnetic (outer complex)
Example 3 — Cis-trans isomerism in [Pt(NH₃)₂Cl₂]
Square planar MA₂B₂ type → two geometrical isomers:
cis-isomer: Both Cl on same side of Pt
trans-isomer: Both Cl on opposite sides
Significance: cisplatin (cis-isomer) is an anticancer drug;
trans-isomer has no anticancer activity.
Example 4 — EAN rule
EAN = atomic number of metal − electrons lost + electrons donated by ligands
[Fe(CO)₅]: Fe (Z=26); no charge loss; CO donates 2e⁻ each; 5 CO ligands
EAN = 26 − 0 + (5 × 2) = 36 = Kr (noble gas) → stable complex ✓
[Ni(CO)₄]: Ni (Z=28); no charge loss; 4 CO × 2e⁻ each
EAN = 28 − 0 + 8 = 36 = Kr → stable ✓
IUPAC Naming Pattern
1. Name cation before anion. 2. Name ligands alphabetically (ignore multiplying prefixes di-, tri- for alphabetical order). 3. Name central metal. 4. Write oxidation state in Roman numerals in brackets. 5. For anionic complexes, add -ate suffix to metal name (e.g. ferrate, cobaltate, platinate). 6. Neutral complexes: name as written, no suffix.
Common Mistakes
- Naming ligands in wrong alphabetical order (e.g. chlorido before ammine is wrong — ammine comes first).
- Confusing coordination number with oxidation state.
- Forgetting Roman numeral for oxidation state.
- Not using -ate suffix for anionic complexes.
- Calling en a monodentate ligand — it is bidentate.
- Confusing inner orbital complex (d²sp³) and outer orbital complex (sp³d²).
Exam Checklist
- Ligand charges are correctly assigned.
- Oxidation state of metal is shown with working.
- IUPAC name follows the correct order.
- Isomer type is correctly identified (geometric/optical/ionisation).
- VBT hybridisation matches the number of available d electrons.
