Class 9 Chemistry Chapter 5: Coordination Compounds MCQ Quiz with Detailed Answers

The rationalized CBSE Class 9 syllabus places heavy emphasis on competency-based assessment. Coordination Compounds—despite being a complex topic—are tested via MCQs because they demand: (1) recall of IUPAC nomenclature rules, (2) visual-spatial reasoning for isomerism, and (3) understanding of bonding models. Each MCQ in CBSE papers is designed to eliminate guessing: wrong options (distractors) are crafted from common student errors. For example, confusing coordination number with oxidation state, or misidentifying geometry (square planar vs. tetrahedral) can trap unprepared students. The new pattern rewards speed + accuracy: you have ~90 seconds per question in board exams. By solving 30 graded MCQs, you'll internalize the decision-making process, recognize trick questions instantly, and build the confidence needed to score 9–10 out of 10 on this section. MCQs also force you to eliminate ambiguity—something descriptive answers cannot do.

**Q1.** The coordination number of cobalt in [Co(NH₃)₆]³⁺ is: (A) 3 (B) 6 (C) 9 (D) 18 **Answer: (B) 6** *Reason: Coordination number = number of ligands directly bonded to the central metal atom. NH₃ is a monodentate ligand; 6 × 1 = 6.* **Q2.** In the complex [CuCl₄]²⁻, the oxidation state of copper is: (A) +1 (B) +2 (C) +3 (D) +4 **Answer: (B) +2** *Reason: 4 Cl⁻ ligands contribute 4 × (−1) = −4 charge. Total charge = −2; so Cu oxidation state = −2 − (−4) = +2.* **Q3.** Which of the following is a neutral ligand? (A) NH₃ (B) Cl⁻ (C) OH⁻ (D) NO₃⁻ **Answer: (A) NH₃** *Reason: Ammonia has no charge and donates electron pairs without contributing ionic charge.* **Q4.** The IUPAC name of [Pt(en)₂]²⁺ is: (A) Diethylenediamine platinum(II) ion (B) Bis(ethylenediamine)platinum(II) ion (C) Platinumethylenediamine (D) Ethylenediamine platinum complex **Answer: (B) Bis(ethylenediamine)platinum(II) ion** *Reason: 'Bis' prefix indicates 2 identical ligands (en = ethylenediamine). Always state oxidation state in Roman numerals.* **Q5.** The complex [Fe(CN)₆]³⁻ contains: (A) 3 Fe and 6 CN (B) 1 Fe and 6 CN (C) 3 Fe and 2 CN (D) 1 Fe and 3 CN **Answer: (B) 1 Fe and 6 CN** *Reason: Complex notation [M(L)ₙ]^charge shows 1 central metal ion and n ligands per formula unit.* **Q6.** A monodentate ligand: (A) Has 2 donor atoms (B) Forms 2 bonds with metal (C) Has 1 donor atom (D) Forms no bonds **Answer: (C) Has 1 donor atom** *Reason: Monodentate = 'one-toothed'; donates one electron pair to one metal atom (e.g., Cl⁻, NH₃, H₂O).* **Q7.** In the complex [Zn(NH₃)₄]²⁺, ammonia acts as: (A) Anion (B) Cation (C) Neutral ligand (D) Metal ion **Answer: (C) Neutral ligand** *Reason: NH₃ is uncharged and acts as a Lewis base, donating its lone pair to Zn²⁺.* **Q8.** Which is a bidentate ligand? (A) NH₃ (B) Cl⁻ (C) Ethylenediamine (en) (D) H₂O **Answer: (C) Ethylenediamine (en)** *Reason: Ethylenediamine has 2 nitrogen donor atoms, allowing it to bind to one metal atom at two points simultaneously.* **Q9.** The metal ion in [CrCl₃(NH₃)₃] is: (A) Cr⁺ (B) Cr²⁺ (C) Cr³⁺ (D) Cr⁴⁺ **Answer: (C) Cr³⁺** *Reason: 3 Cl⁻ ligands contribute −3 charge; neutral complex means Cr oxidation state = +3.* **Q10.** Ligands that contain sulfur (e.g., S²⁻) are called: (A) Thiols (B) Thioligands (C) Sulfides (D) Alkenes **Answer: (B) Thioligands** *Reason: Sulfur-containing ligands coordinate via S atoms; common example is thiosulfate (S₂O₃²⁻).*

**Q11.** [PtCl₂(NH₃)₂] exhibits: (A) Only cis isomerism (B) Only trans isomerism (C) Both cis and trans isomerism (D) No isomerism **Answer: (C) Both cis and trans isomerism** *Reason: Square planar complexes with 2 bidentate or 2 identical monodentate ligands can exist in both cis (same side) and trans (opposite side) configurations.* **Q12.** The geometry of [CoCl₄]²⁻ is: (A) Octahedral (B) Tetrahedral (C) Linear (D) Trigonal planar **Answer: (B) Tetrahedral** *Reason: Coordination number 4 with weak-field chloride ligands → tetrahedral geometry. Cl⁻ does not cause pairing in d-orbitals.* **Q13.** Coordination isomers differ in: (A) The spatial arrangement of ligands (B) The distribution of ligands between metal ions (C) Their oxidation states (D) Their molecular formulas **Answer: (B) The distribution of ligands between metal ions** *Reason: Example: [Co(NH₃)₆]³⁺[Cr(CN)₆]³⁻ vs. [Co(CN)₆]³⁻[Cr(NH₃)₆]³⁺ are coordination isomers.* **Q14.** Linkage isomerism occurs when: (A) A ligand can coordinate via different atoms (B) Ligands occupy different positions (C) Complexes have different colors (D) Metal ions have different charges **Answer: (A) A ligand can coordinate via different atoms** *Reason: NO₂⁻ can bind via N (nitro) or O (nitrito); SCN⁻ can bind via S or N.* **Q15.** The complex [Fe(H₂O)₆]²⁺ is octahedral. How many isomers are possible if 2 H₂O are replaced by 2 NH₃? (A) 1 (B) 2 (C) 3 (D) 4 **Answer: (C) 3** *Reason: For octahedral complexes [ML₄L'₂], three positional isomers exist: ortho (adjacent), meta, and para arrangements (geometric isomerism).* **Q16.** In [Cr(H₂O)₆]³⁺, the ligand is: (A) An anion (B) A neutral molecule (C) A radical (D) A cation **Answer: (B) A neutral molecule** *Reason: H₂O is neutral; oxidation state of Cr = +3 comes from the charge balance: 6(0) + Cr = 3+.* **Q17.** Optical isomerism is exhibited by: (A) [PtCl₂(NH₃)₂] trans (B) [PtCl₂(NH₃)₂] cis (C) [Zn(NH₃)₄]²⁺ (D) [FeCl₄]⁻ **Answer: (B) [PtCl₂(NH₃)₂] cis** *Reason: Square planar cis-isomer lacks a mirror plane and is chiral; trans-isomer is planar and achiral. Tetrahedral [Zn] is not chiral.* **Q18.** Which complex is diamagnetic? (A) [Fe(CN)₆]⁴⁻ (B) [FeCl₆]⁴⁻ (C) [CoF₆]³⁻ (D) [MnCl₆]⁴⁻ **Answer: (A) [Fe(CN)₆]⁴⁻** *Reason: CN⁻ is a strong-field ligand → pairing of d electrons in Fe²⁺ (d⁶ → all paired) → diamagnetic.* **Q19.** The coordination number of Al in [Al(H₂O)₆]³⁺ is: (A) 3 (B) 6 (C) 9 (D) 12 **Answer: (B) 6** *Reason: Six water molecules surround the Al³⁺ ion, each as a monodentate ligand.* **Q20.** Hydrate isomerism in [CrCl₃·6H₂O] produces isomers where: (A) Water and chloride swap coordination spheres (B) Chloride ions are exchanged (C) Color changes (D) Metal oxidation state changes **Answer: (A) Water and chloride swap coordination spheres** *Reason: [Cr(H₂O)₆]Cl₃, [Cr(H₂O)₅Cl]Cl₂·H₂O, and [Cr(H₂O)₄Cl₂]Cl·2H₂O are hydrate isomers.*

**Q21.** **Assertion:** [Fe(CN)₆]⁴⁻ is colorless while [Fe(H₂O)₆]²⁺ is pale green. **Reason:** The ligand field strength determines the magnitude of d-orbital splitting; CN⁻ causes larger Δ than H₂O. (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion true; reason false (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: Strong-field CN⁻ splits d-orbitals significantly (Δ > pairing energy); no visible light absorbed if all d electrons are paired. Weak-field H₂O allows transition between unpaired d electrons, absorbing green light.* **Q22.** **Assertion:** [CoF₆]³⁻ is paramagnetic with 4 unpaired electrons. **Reason:** Fluoride is a weak-field ligand; d electrons remain unpaired in Co³⁺ (d⁶). (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion false; reason true (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: F⁻ is weak-field → low Δ → electrons do not pair in d-orbitals. Co³⁺ (d⁶) configuration: t₂g⁴ eg² → 4 unpaired electrons.* **Q23.** In Crystal Field Theory (CFT), the d-orbitals of a metal ion split into: (A) 2 groups in octahedral; 3 groups in tetrahedral geometry (B) 3 groups in octahedral; 2 groups in tetrahedral geometry (C) Same groups in both geometries (D) 5 equal-energy levels in both **Answer: (B) 3 groups in octahedral; 2 groups in tetrahedral geometry** *Reason: Octahedral: t₂g (3 orbitals) and eg (2 orbitals). Tetrahedral: e (2 orbitals, lower) and t₂ (3 orbitals, higher). Opposite order due to different geometry.* **Q24.** **Assertion:** Nickel complexes often display square planar geometry. **Reason:** Ni²⁺ (d⁸) with strong-field ligands undergoes d-orbital splitting favoring square planar structure. (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion false; reason true (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: Ni²⁺ (d⁸) with strong-field ligands (e.g., CN⁻, PR₃) can undergo electronic reorganization to form square planar [Ni(CN)₄]²⁻, which is more stable than tetrahedral or octahedral.* **Q25.** Which complex is expected to show the maximum crystal field stabilization energy (CFSE)? (A) [Fe(H₂O)₆]³⁺ (B) [Fe(CN)₆]³⁻ (C) [FeCl₆]³⁻ (D) [Fe(OH)₆]³⁻ **Answer: (B) [Fe(CN)₆]³⁻** *Reason: CN⁻ is the strongest-field ligand; largest Δ → maximum CFSE. Fe³⁺ (d⁵) in [Fe(CN)₆]³⁻: t₂g⁵ eg⁰ configuration maximizes splitting benefit.* **Q26.** **Assertion:** In tetrahedral complexes, the d-orbital splitting energy (Δ_t) is less than in octahedral complexes (Δ_o) for the same metal and ligand. **Reason:** Lower coordination number in tetrahedral → weaker crystal field effect. (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion false; reason true (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: Δ_t ≈ (4/9)Δ_o. Tetrahedral has 4 ligands vs. 6 in octahedral; weaker electrostatic field → smaller orbital splitting.* **Q27.** Which statement about the spectrochemical series is correct? (A) It ranks ligands by their coordination ability only (B) It ranks ligands by crystal field strength (weak to strong) (C) It is independent of metal ion (D) All ligands in the series have identical CFT effects **Answer: (B) It ranks ligands by crystal field strength (weak to strong)** *Reason: Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO. Applies to most metal ions.* **Q28.** **Assertion:** Coordination compounds are used as catalysts in industrial processes like the hydroformylation reaction. **Reason:** Metal ions in complexes can facilitate electron transfer and stabilize transition states more effectively than free metal ions. (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion true; reason partially true (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: Ligands modulate the reactivity of metal centers. Complexes like Co-based hydroformylation catalysts allow precise control of orbital energies and reaction selectivity.* **Q29.** The color of [Cr(H₂O)₆]³⁺ (violet) arises because: (A) Water is a strong absorber of blue light (B) An electron transitions from lower t₂g to higher eg d-orbital level, absorbing blue light (C) Cr³⁺ is inherently violet (D) Water emits violet light **Answer: (B) An electron transitions from lower t₂g to higher eg d-orbital level, absorbing blue light** *Reason: Cr³⁺ (d³, octahedral): t₂g³ eg⁰. Visible light (λ ≈ 460 nm, blue) provides energy for t₂g → eg transition. Complementary color (violet) is observed.* **Q30.** **Assertion:** [Cu(NH₃)₄]²⁺ is square planar, while [Cu(H₂O)₄]²⁺ is tetrahedral. **Reason:** NH₃ is a strong-field ligand causing d-orbital splitting that stabilizes square planar geometry for d⁹ Cu²⁺; H₂O is weak-field. (A) Both true; reason explains assertion (B) Both true; reason does not explain (C) Assertion false; reason true (D) Both false **Answer: (A) Both true; reason explains assertion** *Reason: Cu²⁺ (d⁹): With strong-field NH₃, the dx²-y² orbital (eg) is stabilized → square planar. With weak H₂O, tetrahedral is favored. Jahn-Teller distortion also plays a role.*

**Trap 1: Confusing Coordination Number with Oxidation State** Students often mix these up. [Co(NH₃)₆]³⁺ has coordination number 6 (number of ligands) but Co oxidation state +3 (determined by charges). The MCQ distractors will offer both numbers as options—choose only the one the question asks. **Trap 2: Misidentifying Ligand Charge** Neutral ligands (NH₃, H₂O, CO) are frequently confused with anionic ligands (Cl⁻, OH⁻, CN⁻). A wrong assumption here cascades into incorrect oxidation state calculations. Always check: Does the ligand have a formal charge symbol? **Trap 3: Geometry & Coordination Number Mix-up** Coordination number 4 → tetrahedral OR square planar (context matters). Don't assume all 4-coordinate complexes are tetrahedral. Check the metal ion and ligand field strength using the spectrochemical series. **Trap 4: Naming Errors in IUPAC Nomenclature** Common mistakes: forgetting to add 'bis,' 'tris,' 'tetrakis' prefixes for multiple identical ligands; reversing the order (anions named before neutral ligands, but both before cations); omitting Roman numerals for oxidation state. Practice the correct order: (1) Cation, (2) Anion or neutral ligands (alphabetical), (3) Metal name + oxidation state. **Trap 5: Isomerism Type Misidentification** Cis-trans (geometric) isomerism is often conflated with optical isomerism. Only chiral complexes (lacking mirror planes) show optical isomerism. For example, trans-[PtCl₂(NH₃)₂] is planar and achiral; cis-[PtCl₂(NH₃)₂] is chiral. **Trap 6: Crystal Field Stabilization Energy (CFSE) Confusion** CFSE depends on both Δ (field strength) and electron configuration. [Fe(CN)₆]⁴⁻ (d⁶, all paired) and [Fe(CN)₆]³⁻ (d⁵) have different CFSE values. Always state the d-orbital configuration first. **Trap 7: Paramagnetic vs. Diamagnetic Decision** Unpaired electrons = paramagnetic. Paired electrons = diamagnetic. Use CFT splitting diagrams to fill electrons: high-spin (weak field, max unpaired) vs. low-spin (strong field, min unpaired). MCQs asking about magnetism often hide the answer in ligand strength. **Trap 8: Ignoring Spectrochemical Series Order** Weak-field ligands (I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O) vs. strong-field (CN⁻ ≈ CO). Reversing this order will give a completely wrong answer about geometry and magnetism. Memorize the sequence or recognize it in MCQ options. **Trap 9: Color & Absorption Wavelength Inversion** If a complex absorbs blue light (λ ≈ 450 nm), it appears yellow (complementary). MCQs test whether you know: smaller Δ → longer wavelength absorbed → red/infrared shift → complex appears more green/blue. Higher Δ → shorter wavelength absorbed → violet/blue shift → complex appears more yellow/orange. **Trap 10: Forgetting Counter-Ions in Charge Calculations** In [Co(NH₃)₆]Cl₃, the Cl⁻ ions are counter-ions outside the coordination sphere. They do NOT affect the coordination number or geometry of the complex—only the overall charge balance. Distractors may list these as part of the complex, which is wrong.

**Step 1: Read the Question Stem First (10–15 seconds)** Underline the exact question being asked. Is it asking for: (A) coordination number, (B) geometry, (C) oxidation state, (D) isomerism type? Misreading costs marks. For assertion-reason MCQs, separately evaluate if assertion is true, if reason is true, and if reason explains assertion. **Step 2: Identify Question Type (5 seconds)** - Nomenclature → Check IUPAC rules (ligand name, oxidation state Roman numeral). - Geometry/Coordination → Use coordination number + ligand field strength. - Isomerism → Recall definitions (cis-trans, optical, coordination, linkage). - CFT/Color → Reference spectrochemical series and d-orbital splitting. **Step 3: Eliminate Impossible Options (10–20 seconds)** Remove 1–2 distractors that are obviously wrong based on basic facts. For example, if a question asks for coordination number in [Cu(NH₃)₄]²⁺, eliminate any option > 4 immediately. **Step 4: Work Through Remaining Options (20–30 seconds)** For each remaining option, ask: "Does this fit the rule/definition?" Use quick calculations or mental checks. Write down oxidation states or electron configurations on scrap paper if needed. **Step 5: Double-Check Answer (5 seconds)** Re-read the question and confirm your answer matches what was asked. For tough questions, use the process of elimination to narrow down to 2 choices, then choose the most chemically reasonable one. **Timing Breakdown for a 90-Minute CBSE Exam:** - Easy MCQs (10 questions): 10 × 1.5 min = 15 minutes (1 min per Q + buffer). - Medium MCQs (10 questions): 10 × 2 min = 20 minutes (1.5 min per Q + buffer). - Hard/Assertion-Reason (10 questions): 10 × 2.5 min = 25 minutes (2 min per Q + buffer). - **Total MCQ time: 60 minutes.** Leaves 30 minutes for long-form questions or review. **Pro Tips for Speed Without Sacrificing Accuracy:** 1. **Memorize key facts**: Spectrochemical series, common complex geometries, oxidation states of common metals. 2. **Use acronyms**: IUPAC (Indicate, Name ligands, Use Roman numerals for oxidation state, prefix for count, Add charge). CFT (Crystal Field Theory: Count electrons, Fill d-orbitals, Transitions cause color). 3. **Skip-and-Return strategy**: If a question takes > 90 seconds, mark it and return after finishing easy questions. Don't get stuck. 4. **Visualize geometry**: For cis-trans and optical isomerism MCQs, quickly sketch the complex (2D square or octahedral) on scrap paper. 30 seconds of drawing saves 2 minutes of confusion. 5. **Verify oxidation states mentally**: Always check that (ligand charges) + (metal oxidation state) = (complex charge). This catches ~40% of errors. 6. **For color MCQs**: Small Δ (weak field) → red light absorbed → green complex. Large Δ (strong field) → blue light absorbed → yellow/orange complex. Use this inverse logic. With practice on these 30 MCQs, you'll internalize the decision trees and solve new questions in <90 seconds, freeing time for detailed problem-solving. Start a 3-day free trial at cbsetutor.ai to access video explanations of each MCQ and adaptive quizzes that scale to your level.

**Oxidation State Calculation:** Oxidation state of metal = (Complex charge) − (Sum of ligand charges) Example: In [PtCl₄]²⁻, Pt = −2 − (4 × −1) = −2 + 4 = +2. **Coordination Number (CN):** CN = Total number of ligand donor atoms bonded to metal ion. - CN = 4 → Tetrahedral (most transition metals, Zn, Pt) or Square Planar (d⁸ metals like Ni, Pd, Pt with strong-field ligands). - CN = 6 → Octahedral (most common, Fe, Co, Cr, Al). - CN = 2 → Linear (Cu(I), Au(I), Ag(I) complexes). **IUPAC Nomenclature Order:** 1. Cation (metal + oxidation state). 2. Anions (alphabetical by element symbol, not ligand name). 3. Neutral ligands (alphabetical by ligand name). 4. Prefix (mono-, di-, tri-, tetra-, penta-, hexa-; use bis-, tris-, tetrakis- for complex names). Example: [Co(NH₃)₆]Cl₃ → Hexaamminecobalt(III) chloride. **Spectrochemical Series (Ligand Field Strength):** I⁻ < Br⁻ < SCN⁻ < Cl⁻ < NO₃⁻ < F⁻ < OH⁻ < H₂O < NCS⁻ < NH₃ < en < NO₂⁻ < CN⁻ ≈ CO (Weak field) ←—————————————————→ (Strong field) **d-Orbital Splitting in Octahedral Geometry:** Lower energy: t₂g (dxy, dyz, dzx) — 3 orbitals Higher energy: eg (dx²-y², dz²) — 2 orbitals Δ_o = Energy difference between eg and t₂g **d-Orbital Splitting in Tetrahedral Geometry:** Lower energy: e (dz², dx²-y²) — 2 orbitals Higher energy: t₂ (dxy, dyz, dzx) — 3 orbitals Δ_t ≈ (4/9)Δ_o (Note: reverse order, smaller magnitude) **Crystal Field Stabilization Energy (CFSE) for Octahedral d-Configurations:** - d¹: −0.4Δ_o - d⁶ (low-spin, all paired): −2.4Δ_o (maximum CFSE) - d⁵ (high-spin, all unpaired): 0 (no CFSE benefit) **Jahn-Teller Effect:** Octahedral complexes with d⁷, d⁸, d⁹, or d⁴ configurations (unequal electron distribution in eg level) undergo distortion. Example: [Cu(H₂O)₆]²⁺ (d⁹) distorts to square planar. **Color & Light Absorption:** Visible light (400–700 nm) is absorbed when ΔE (d-orbital splitting) matches photon energy. - Small Δ → Infrared or red light absorbed → Complex appears green/cyan. - Large Δ → Blue/UV light absorbed → Complex appears yellow/orange/red. - Complementary color rule: Absorbed + Observed = White light. **Isomerism Types:** - **Geometric (Cis-Trans):** Same formula, different spatial arrangement of ligands (octahedral [ML₄L'₂], square planar [ML₂L'₂]). - **Optical (Enantiomers):** Non-superimposable mirror images; complex lacks mirror plane or center of symmetry. - **Coordination:** Ligands exchanged between coordination spheres of different metal ions in a multi-metal complex. - **Linkage:** Ambidentate ligands (NO₂⁻, SCN⁻) bind via different donor atoms. - **Hydrate (Solvate):** Water molecules in/out of coordination sphere; affects complex charge and properties. **Magnetism Prediction:** No. of unpaired electrons → Magnetic susceptibility (χ_m). - 0 unpaired electrons → Diamagnetic (χ_m < 0). - ≥ 1 unpaired electron → Paramagnetic (χ_m > 0). Use d-orbital filling (respecting Hund's rule for high-spin, pairing for low-spin) based on ligand field strength.