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Coordination Compounds
Chemistry · Class 12NCERT pp. 118–1372 sets × 25 questions
Key facts and formulas for Coordination Compounds, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Naming order | Name: ligands (alphabetical) then metal, oxidation state in Roman numerals; anionic complex metal ends in -ate |
| Coordination number | CN = number of donor atoms = ligands × denticity (en = 2, EDTA = 6, dien = 3) |
| Oxidation state | Sum of metal oxidation number and ligand charges = charge on complex |
| Spin-only moment | μ = √[n(n + 2)] BM; n = 1 gives 1.73, n = 2 gives 2.83, n = 3 gives 3.87, n = 4 gives 4.90, n = 5 gives 5.92 |
| CFSE (octahedral) | CFSE = −0.4 Δo × (t₂g electrons) + 0.6 Δo × (e_g electrons), pairing energy ignored |
| Spectrochemical series | I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO; Δo rises 3d < 4d < 5d and with oxidation state |
| Hybridisation | Strong field: d²sp³ inner orbital; weak field: sp³d² outer orbital; four-coordinate: dsp² square planar or sp³ tetrahedral |
| EAN | EAN = Z − oxidation state + 2 × (number of ligands for monodentate ligands) |
| Isomerism | Geometrical: [MA₄X₂] 2, square planar [Mabcd] 3; optical: [M(en)₃]ⁿ⁺ and cis-[MA₂(en)₂]; coordination, ionisation, linkage, solvate |
| Colour | Observed colour is complementary to absorbed colour; d⁰ and d¹⁰ ions are colourless; wavelength rises as Δo falls |
Three sample questions
Q1. An excess of AgNO₃ is added to 100 mL of 0.01 M [Cr(H₂O)₄Cl₂]Cl. Moles of AgCl precipitated are
- 0.002
- 0.001
- 0.003
- 0.01
Hint and solution
Hint: Only chloride outside the square bracket is free to react with silver ions. Count those first, then find moles of complex.
- n = V × M = 0.1 L × 0.01 mol/L = 0.001 mol — Convert 100 mL to 0.1 L and multiply by molarity to get moles of complex.
- [Cr(H₂O)₄Cl₂]Cl → complex⁺ + Cl⁻ — Only the chloride outside the bracket ionises in water. The two inside are ligands.
- Free Cl⁻ per complex = 1 — One mole of complex releases one mole of Cl⁻, which gives one mole of AgCl.
- AgCl = 0.001 × 1 = 0.001 mol — So 0.001 mole of AgCl precipitates. Option B.
Answer: B. Moles of complex = 0.1 L × 0.01 M = 0.001. One Cl⁻ is outside the bracket, so 0.001 mol AgCl.
Q2. Moles of AgCl from 1 mol each of CoCl₃·6NH₃, CoCl₃·5NH₃ and CoCl₃·4NH₃ (excess AgNO₃) are respectively
- 1, 2, 3
- 3, 3, 2
- 2, 2, 1
- 3, 2, 1
Hint and solution
Hint: Write the formula of each compound with the ammonia and any bound chloride inside the bracket. Count the chlorides left outside.
- Co CN = 6 in all three — Cobalt keeps six donor positions, filled by NH₃ first and then Cl⁻.
- 6NH₃ → [Co(NH₃)₆]Cl₃ — Six ammonia fill all positions, so all 3 chlorides are free.
- 5NH₃ → [CoCl(NH₃)₅]Cl₂; 4NH₃ → [CoCl₂(NH₃)₄]Cl — With fewer ammonia, one and then two chlorides enter the bracket, leaving 2 and 1 outside.
- Only free Cl⁻ gives AgCl: 3, 2, 1 — Moles of AgCl equal moles of free chloride, so the sequence is 3, 2, 1. Option D.
Answer: D. The formulas are [Co(NH₃)₆]Cl₃, [CoCl(NH₃)₅]Cl₂ and [CoCl₂(NH₃)₄]Cl, with 3, 2, 1 free chlorides.
Q3. Consider: I. In a formula the metal is written first. II. In a name the metal is named first. III. Ligands are in alphabetical order in both. The correct statements are
- I and III only
- II and III only
- I and II only
- I, II and III
Hint and solution
Hint: In a name, which comes first, the ligands or the metal? Test statement II against that, then check the other two.
- Formula: metal first, then ligands — In a formula the metal symbol is written first inside the bracket. So I is true.
- Name: ligands first, metal last — In a name the ligands are written before the metal, for example hexaamminecobalt(III). So II is false.
- Alphabetical order of ligands — Ligands are arranged alphabetically in both the formula and the name, ignoring prefixes like di and tri. So III is true.
- True statements: I and III — So the correct choice is I and III only. Option A.
Answer: A. In a name the ligands come first and the metal last. So II is false.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.