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CBSE Class 12 Chemistry Chapter 5 Coordination Compounds — 20 MCQs with Answers

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Coordination Compounds is a high-scoring chapter in CBSE Class 12 Chemistry, contributing around 5 marks in the objective section and 3-5 marks in the subjective paper. Students often lose marks in nomenclature conventions and Crystal Field Theory application. This page offers 20 MCQs that mirror board exam standards, covering nomenclature rules, geometrical and optical isomerism, Crystal Field Theory splitting diagrams, colour origin, magnetic moment calculations and ligand strength. Each question is paired with a clear answer and rationale drawn from NCERT Chapter 5.

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Key takeaways

✓20 board-style MCQs cover nomenclature, isomerism, CFT, colour, magnetism and ligand field strength across difficulty levels.
✓Each question carries four options with the correct answer marked and a concise NCERT-aligned explanation.
✓Assertion-Reason format and application-based questions reflect the latest CBSE Class 12 Chemistry blueprint.
✓Crystal Field Theory questions test understanding of splitting patterns in octahedral and tetrahedral fields.
✓Isomerism MCQs distinguish geometrical, optical, linkage and ionisation isomers with real complex examples.
✓Nomenclature questions apply IUPAC rules for naming coordination compounds systematically.
✓Strategy section explains how to maximise MCQ scores in the 5-mark objective section of the CBSE board paper.

Nomenclature and Formula Writing MCQs

IUPAC naming of coordination compounds follows a strict protocol: ligands in alphabetical order (ignoring prefixes di-, tri-), metal centre with oxidation state in Roman numerals, and anionic complexes ending in '-ate'. Students often confuse ligand prefixes or forget to alphabetise. The 2024 CBSE board paper carried 1 MCQ on naming a platinum complex, and errors in oxidation-state counting cost many candidates a mark. These four questions drill ligand ordering, cation versus anion complex naming, and oxidation-state determination from neutral and charged ligands.

Q1. The IUPAC name of [Co(NH₃)₅Cl]Cl₂ is: (a) Pentaamminechloridocobalt(III) chloride (b) Chloropentaamminecobalt(III) chloride (c) Pentaamminechlorocobaltate(III) chloride (d) Pentaamminechloridocobalt(II) chloride | Answer: (a). Ligands in alphabetical order: 'ammine' before 'chlorido'; cobalt is +3 (5×0 + 1×(−1) + x = +2, x = +3).
Q2. Which complex has the formula [Pt(NH₃)₂Cl₂]? (a) Diamminedichloroplatinum(II) (b) Dichlorodiammineplatinum(II) (c) Diamminedichloridoplatinate(II) (d) Diamminedichloroplatinum(IV) | Answer: (a). Neutral complex, Pt is +2 (2×0 + 2×(−1) + x = 0, x = +2); alphabetically 'ammine' precedes 'chloro'.
Q3. Oxidation state of iron in [Fe(CN)₆]³⁻ is: (a) +2 (b) +3 (c) +4 (d) +6 | Answer: (b). 6×(−1) + x = −3, x = +3.
Q4. The correct IUPAC name for K₃[Fe(CN)₆] is: (a) Potassium hexacyanoferrate(III) (b) Tripotassium hexacyanidoferrate(III) (c) Potassium hexacyanoiron(III) (d) Potassium ferricyanide | Answer: (b). Anionic complex uses '-ate' suffix; 'cyanido' is modern IUPAC; oxidation state +3.

Isomerism in Coordination Compounds

Isomerism is a favourite topic for CBSE examiners. Geometrical isomerism occurs in square planar (cis/trans in MA₂B₂) and octahedral (cis/trans, fac/mer in MA₃B₃) complexes. Optical isomerism arises when a complex lacks a plane of symmetry, common in octahedral chelate complexes like [Co(en)₃]³⁺. Linkage isomerism involves ambidentate ligands (NO₂⁻/ONO⁻, SCN⁻/NCS⁻), and ionisation isomerism differs in the counter-ion. The 2023 board paper tested geometrical isomerism in [Pt(NH₃)₂Cl₂]. These MCQs cover all four structural isomer types with real NCERT examples to sharpen recognition skills.

Q5. [Co(NH₃)₅(NO₂)]Cl₂ and [Co(NH₃)₅(ONO)]Cl₂ are an example of: (a) Geometrical isomerism (b) Linkage isomerism (c) Ionisation isomerism (d) Optical isomerism | Answer: (b). NO₂⁻ is ambidentate; binding through N or O gives linkage isomers.
Q6. Which complex exhibits geometrical isomerism? (a) [Co(NH₃)₆]³⁺ (b) [Pt(NH₃)₂Cl₂] (c) [Ni(CO)₄] (d) [Co(en)₃]³⁺ | Answer: (b). Square planar MA₂B₂ shows cis/trans isomers; others are octahedral MA₆ or tetrahedral, no geometrical isomerism.
Q7. [Cr(H₂O)₆]Cl₃ and [Cr(H₂O)₅Cl]Cl₂·H₂O are: (a) Linkage isomers (b) Ionisation isomers (c) Hydrate isomers (d) Coordination isomers | Answer: (b). Different ions inside and outside coordination sphere; ionisation isomerism.
Q8. Which is optically active? (a) cis-[PtCl₂(en)] (b) trans-[Co(en)₂Cl₂]⁺ (c) [Co(en)₃]³⁺ (d) [NiCl₄]²⁻ | Answer: (c). Octahedral tris-chelate lacks plane of symmetry; exhibits optical isomerism (d and l forms).

Crystal Field Theory — Octahedral Splitting

Crystal Field Theory (CFT) explains colour and magnetism by describing how d-orbitals split in a ligand field. In an octahedral complex, five degenerate d-orbitals split into t₂g (lower, dxy, dyz, dzx) and eg (higher, dx²−y², dz²) sets with energy gap Δₒ. Strong-field ligands (CN⁻, CO) cause large Δₒ leading to low-spin (maximum pairing), while weak-field ligands (I⁻, Br⁻) yield small Δₒ and high-spin. The 2024 board paper asked about the number of unpaired electrons in [Fe(CN)₆]⁴⁻ versus [FeF₆]³⁻. These questions test splitting diagrams, electron filling order and spin state prediction using the spectrochemical series from NCERT.

Q9. In octahedral [CoF₆]³⁻, Co³⁺ is d⁶. Number of unpaired electrons (F⁻ is weak field): (a) 0 (b) 1 (c) 2 (d) 4 | Answer: (d). Weak field → high spin: t₂g⁴ eg² gives 4 unpaired.
Q10. [Fe(CN)₆]⁴⁻ is diamagnetic because: (a) CN⁻ is strong field, pairing occurs (b) Fe²⁺ is d⁸ (c) Octahedral splitting is zero (d) Fe is in +3 state | Answer: (a). Fe²⁺ is d⁶; CN⁻ strong field → t₂g⁶ eg⁰, all paired, diamagnetic.
Q11. Crystal field stabilisation energy is highest for: (a) d³ octahedral high spin (b) d⁵ octahedral low spin (c) d⁸ octahedral (d) d¹⁰ octahedral | Answer: (b). d⁵ low spin t₂g⁵ eg⁰ maximises occupancy of lower t₂g orbitals.
Q12. Δₒ (octahedral splitting) is largest for ligand: (a) I⁻ (b) Br⁻ (c) H₂O (d) CN⁻ | Answer: (d). Spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO.

Tetrahedral and Square Planar Complexes

Tetrahedral complexes show smaller splitting (Δₜ ≈ 4/9 Δₒ) and are almost always high-spin because the energy gap is insufficient for pairing. Common examples include [NiCl₄]²⁻ and [CoCl₄]²⁻. Square planar geometry is favoured by d⁸ metal ions (Ni²⁺, Pd²⁺, Pt²⁺) with strong-field ligands. [Ni(CN)₄]²⁻ is square planar and diamagnetic (dsp² hybridisation), whereas [NiCl₄]²⁻ is tetrahedral and paramagnetic (sp³). The 2023 sample paper included an MCQ distinguishing these geometries. Understanding hybridisation and magnetic behaviour is key to scoring in this sub-topic, which students often confuse with octahedral splitting rules.

Q13. [NiCl₄]²⁻ is paramagnetic and tetrahedral. Hybridisation is: (a) dsp² (b) sp³ (c) d²sp³ (d) sp³d² | Answer: (b). Tetrahedral uses sp³; Ni²⁺ d⁸ has 2 unpaired (high spin).
Q14. [Ni(CN)₄]²⁻ is diamagnetic. Its geometry and hybridisation: (a) Tetrahedral, sp³ (b) Square planar, dsp² (c) Octahedral, d²sp³ (d) Square planar, sp³d | Answer: (b). CN⁻ strong field forces pairing; d⁸ → dsp² square planar, zero unpaired.
Q15. Assertion: [CoCl₄]²⁻ is tetrahedral. Reason: Cl⁻ is a weak field ligand. (a) Both true, Reason explains Assertion (b) Both true, Reason does not explain (c) Assertion true, Reason false (d) Both false | Answer: (a). Weak field + smaller size favour tetrahedral geometry over octahedral.
Q16. Splitting in tetrahedral field (Δₜ) compared to octahedral (Δₒ) for same metal and ligand: (a) Δₜ = Δₒ (b) Δₜ ≈ 4/9 Δₒ (c) Δₜ = 2 Δₒ (d) Δₜ = 1/2 Δₒ | Answer: (b). Δₜ ≈ 4/9 Δₒ; smaller splitting means tetrahedral complexes are always high-spin.

Colour and Magnetic Properties Applications

Colour in coordination compounds arises from d-d electronic transitions when Δ falls in the visible range. A complex absorbs one colour and transmits the complementary colour (e.g. [Cu(H₂O)₄]²⁺ absorbs red, appears blue). Magnetic moment μ = √[n(n+2)] BM, where n is the number of unpaired electrons. The 2024 CBSE board carried a 1-mark MCQ asking for the magnetic moment of [Fe(H₂O)₆]²⁺. Students must count unpaired electrons from CFT splitting, then apply the spin-only formula. These questions integrate CFT, spectrochemical series and magnetic measurement, testing deeper conceptual linkages beyond rote memorisation of ligand order.

Q17. [Ti(H₂O)₆]³⁺ is violet. The d-d transition involves: (a) t₂g → eg (b) eg → t₂g (c) No d-d transition (d) Charge transfer | Answer: (a). Ti³⁺ is d¹; electron excites from lower t₂g to higher eg level in octahedral field.
Q18. Magnetic moment of [Mn(H₂O)₆]²⁺ (Mn²⁺ d⁵ high spin): (a) 0 BM (b) 1.73 BM (c) 5.92 BM (d) 3.87 BM | Answer: (c). High spin d⁵ has 5 unpaired; μ = √[5×7] = √35 ≈ 5.92 BM.
Q19. Which is colourless? (a) [Cu(NH₃)₄]²⁺ (b) [Zn(NH₃)₄]²⁺ (c) [Ni(H₂O)₆]²⁺ (d) [Fe(H₂O)₆]²⁺ | Answer: (b). Zn²⁺ is d¹⁰, no d-d transition possible; complex is colourless.
Q20. Assertion: [CoF₆]³⁻ is paramagnetic. Reason: F⁻ is a weak field ligand causing high spin d⁶. (a) Both true, Reason explains (b) Both true, Reason does not explain (c) Assertion true, Reason false (d) Both false | Answer: (a). Weak field → high spin t₂g⁴ eg², 4 unpaired, paramagnetic.

Ligand Types and Coordination Number

Ligands are classified by the number of donor atoms: monodentate (Cl⁻, NH₃, H₂O), bidentate (en, ox²⁻), and polydentate (EDTA⁴⁻ is hexadentate). Coordination number is the total number of ligand donor atoms bonded to the central metal. Common coordination numbers are 4 (tetrahedral or square planar) and 6 (octahedral). Ambidentate ligands like NO₂⁻/ONO⁻ and SCN⁻/NCS⁻ can bind through different atoms, leading to linkage isomerism. The 2023 board paper asked students to identify a bidentate ligand from a list. These MCQs test recognition of denticity, calculation of coordination number from formula, and understanding of chelate ring stability, all core NCERT concepts.

Q21. EDTA⁴⁻ is: (a) Monodentate (b) Bidentate (c) Tridentate (d) Hexadentate | Answer: (d). Ethylenediaminetetraacetate has 2 N and 4 O donors, total 6 coordination sites.
Q22. Coordination number of Pt in [Pt(en)₂Cl₂] is: (a) 2 (b) 4 (c) 6 (d) 8 | Answer: (c). Two bidentate en (4 sites) + 2 Cl (2 sites) = 6.
Q23. Which is an ambidentate ligand? (a) en (b) EDTA (c) NO₂⁻ (d) NH₃ | Answer: (c). NO₂⁻ binds via N (nitro) or O (nitrito); ambidentate means two possible donor atoms.
Q24. [Co(NH₃)₆]³⁺ has coordination number: (a) 3 (b) 6 (c) 12 (d) 9 | Answer: (b). Six monodentate NH₃ ligands give coordination number 6 (octahedral).

Werner's Theory and Bonding Basics

Alfred Werner proposed primary valence (oxidation state, satisfied by anions) and secondary valence (coordination number, satisfied by ligands). For [Co(NH₃)₆]Cl₃, primary valence is +3 (three Cl⁻ counter-ions), secondary valence is 6 (six NH₃). Valence Bond Theory describes bonding via hybridisation: octahedral d²sp³ or sp³d², square planar dsp², tetrahedral sp³. Limitations include inability to explain colour and magnetic variations; CFT fills that gap. The 2024 sample paper featured a 1-mark question on Werner's postulates. These MCQs ensure foundational clarity on historical theory before moving to modern CFT applications, aligning with the NCERT sequence of Chapter 5.

How to Attempt MCQs in the CBSE Chemistry Paper

The CBSE Class 12 Chemistry paper allocates 5 marks (5×1) to Section A multiple-choice questions, including 1-2 from Coordination Compounds. Speed and accuracy are critical. Read the question stem carefully—many errors arise from misreading oxidation state or ligand type. Eliminate obviously wrong options first. For assertion-reason questions, verify both statements independently, then check logical connection. In numerical magnetic moment or CFSE questions, quickly write the d-electron configuration and apply the formula rather than guessing. If unsure, mark your best guess and move on—no negative marking. Reserve final 5 minutes to review marked answers. CBSETUTOR.ai offers unlimited MCQ mock tests with instant photo-upload doubt solving at ₹999/month for all classes 6-12, complete with a 3-day free trial, helping students build the exam tempo needed to finish Section A in under 8 minutes.

Underline keywords: 'high spin', 'diamagnetic', 'IUPAC name', 'oxidation state' to avoid careless misreads.
Use the spectrochemical series printed in NCERT Table 5.1 as a mental checklist for ligand field strength.
For isomerism MCQs, sketch a quick 2D structure to confirm cis/trans or optical activity rather than relying on memory.
In assertion-reason format, treat assertion and reason as independent true/false, then test causality.
Practice 50+ MCQs from NCERT Exemplar and past papers to recognise question patterns and common distractors.
Time budget: aim for 45-60 seconds per MCQ; don't spend more than 90 seconds on any single question.

Frequently asked questions

How many MCQs from Coordination Compounds appear in the CBSE Class 12 Chemistry board exam?+

Typically 1-2 MCQs out of the 20-question Section A carry content from Chapter 5 Coordination Compounds. Expect questions on nomenclature, isomerism types, or magnetic properties, each worth 1 mark with no negative marking.

Which topics within Coordination Compounds are most frequently tested in MCQs?+

Nomenclature (IUPAC naming rules), geometrical and linkage isomerism, Crystal Field Theory (octahedral splitting, high-spin/low-spin), and magnetic moment calculations dominate MCQ selections. The 2024 board included one CFT and one nomenclature MCQ.

How do I quickly determine oxidation state in a coordination complex MCQ?+

Sum the charges of all ligands (neutral ligands contribute 0, anionic −1 each) and equate to the overall charge. For [Co(NH₃)₅Cl]Cl₂, inside sphere: 5×0 + 1×(−1) + x = +1 (complex cation charge), so x = +2? No—check counter-ions: two Cl⁻ outside means complex is +2, so x = +3.

What is the difference between high-spin and low-spin complexes?+

High-spin occurs with weak-field ligands (small Δ): electrons occupy higher eg orbitals before pairing in t₂g. Low-spin occurs with strong-field ligands (large Δ): electrons pair in t₂g first. Example: [FeF₆]³⁻ is high-spin (5 unpaired), [Fe(CN)₆]³⁻ is low-spin (1 unpaired).

How do I remember the spectrochemical series for ligand field strength?+

Use the mnemonic 'I Believe Cute Foxes Have Nibbled Every Cute Nose': I⁻ < Br⁻ < Cl⁻ < F⁻ < H₂O < NH₃ < en < CN⁻ < CO. Strong-field ligands (right end) cause low-spin; weak-field (left) cause high-spin.

Why is [Ni(CN)₄]²⁻ square planar but [NiCl₄]²⁻ tetrahedral?+

CN⁻ is a strong-field ligand causing electron pairing in Ni²⁺ d⁸, favouring dsp² square planar geometry and diamagnetism. Cl⁻ is weak-field, leaving electrons unpaired in sp³ tetrahedral geometry, making [NiCl₄]²⁻ paramagnetic.

How do I distinguish between ionisation and linkage isomerism in MCQs?+

Ionisation isomers differ in which ion is inside versus outside the coordination sphere (e.g. [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br). Linkage isomers have the same ions but ligand binds via different atoms (e.g. -NO₂ vs -ONO). Check the ligand donor atom to decide.

What is the spin-only formula for magnetic moment and when do I use it?+

μ = √[n(n+2)] Bohr Magneton (BM), where n is the number of unpaired electrons. Use it in MCQs asking for magnetic moment; first find n from CFT splitting (t₂g and eg filling), then substitute. Example: 3 unpaired gives μ = √[3×5] = √15 ≈ 3.87 BM.

Can a d¹⁰ complex show colour according to Crystal Field Theory?+

No. d¹⁰ configuration (e.g. Zn²⁺, Cu⁺) has completely filled d-orbitals, so d-d transitions are impossible. Such complexes are colourless unless charge-transfer or ligand-based transitions occur. [Zn(NH₃)₄]²⁺ is a classic colourless example.

How should I approach assertion-reason MCQs in Coordination Compounds?+

First verify if the assertion is true by recalling NCERT facts. Then check if the reason is independently true. Finally, decide if the reason correctly explains the assertion. For example, 'Assertion: [CoF₆]³⁻ is paramagnetic. Reason: F⁻ is weak field'—both true and reason explains high-spin paramagnetism.

Is practicing only NCERT in-text and end-chapter questions enough for MCQs?+

NCERT is essential for concept clarity, but board MCQs often twist phrasing or combine topics. Supplement with NCERT Exemplar MCQs, previous years' board questions, and sample papers. CBSETUTOR.ai provides chapter-wise unlimited MCQ drills with detailed explanations at ₹999/month for classes 6-12.

What are common mistakes students make in nomenclature MCQs?+

Forgetting to alphabetise ligands, miscounting oxidation state, using incorrect suffixes for anionic complexes ('-ate' vs '-ium'), and ignoring modern IUPAC terms like 'chlorido' instead of 'chloro'. Practice writing full IUPAC names for 10 complexes daily to avoid these errors.

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