NCERT Exercises and Solutions: Coordination Compounds
🎓 Class 12ChemistryCBSETheoryCh 5 – Coordination Compounds⏱ ~8 min
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NCERT Exercises and Solutions: Coordination Compounds
Chapter Summary
Coordination compounds form a vast and important branch of inorganic chemistry. The first systematic theory was proposed by Alfred Werner, who introduced the ideas of primary (ionisable) and secondary (non-ionisable) valences. The secondary valence corresponds to the modern coordination number; the primary to the oxidation state. Geometry is mostly octahedral, tetrahedral or square planar.
Definitions
Coordination entity, central atom (Lewis acid), ligand (Lewis base), donor atom, denticity (uni-, di-, poly-, ambidentate), chelate, coordination sphere, polyhedron, oxidation number, homoleptic vs heteroleptic.
Nomenclature (IUPAC)
Cation first, anion second. Inside a complex: ligands alphabetical, then metal. Anionic ligands end in -o/-ido; oxidation state in Roman numerals; -ate suffix for anionic complex.
Hybridisation: sp³ tetrahedral, dsp² square planar, sp³d² (outer-orbital, high spin) or d²sp³ (inner-orbital, low spin) octahedral. Magnetic moment μ = √n(n+2) BM.
CFT
d-orbital splitting: octahedral (t₂g lower, eg upper, gap Δₒ); tetrahedral inverted (e lower, t₂ upper, Δₜ = 4/9 Δₒ). Spectrochemical series ranks ligand field strength. Strong field → low spin; weak field → high spin. Colour from d–d transitions.
Stability & chelate effect
Formation constants β; chelates are more stable than equivalent monodentate complexes (entropic origin).
Metal carbonyls
Synergic σ (CO → M) and π* back-donation (M → CO).
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NCERT Exercises — Solutions
Exercise 5.1 — Bonding in coordination compounds (Werner)
Werner: a metal centre uses two valences. The primary valence equals today's oxidation state — ionisable, satisfied by negative ions. The secondary valence equals the coordination number — non-ionisable, satisfied by neutral molecules or anions held inside the coordination sphere. Secondary valences are fixed for a given metal and the ligands occupy fixed positions in space (octahedral/tetrahedral/square planar).
Exercise 5.2 — Why does Mohr's salt give Fe²⁺ test but [Cu(NH₃)₄]²⁺ does not give Cu²⁺ test?
FeSO4·(NH4)2SO4·6H2O is a double salt; in water it dissociates fully → free Fe2+ ions test positive (e.g. with K3[Fe(CN)6]).
CuSO4 + 4 NH3 forms the complex [Cu(NH3)4]2+ with very high formation constant. The free Cu2+ concentration is too small to give the usual tests.
Coordination entity: [Co(NH3)6]3+, [Fe(CN)6]4−. Ligand: H2O, NH3 (also Cl−, CN−, en, EDTA). Coordination number: 6 in [PtCl6]2−; 4 in [Ni(CO)4]. Coordination polyhedron: octahedron in [Co(NH3)6]3+; tetrahedron in [Ni(CO)4]. Homoleptic: [Co(NH3)6]3+, [Ni(CO)4] — only one type of ligand. Heteroleptic: [Co(NH3)4Cl2]+, [Pt(NH3)2Cl2] — more than one type.
Exercise 5.4 — Unidentate, didentate and ambidentate ligands
Unidentate: bonds via one donor atom — Cl−, NH3, H2O, CN−, CO. Didentate: bonds via two donor atoms simultaneously — en (NH2CH2CH2NH2) [N,N]; oxalate C2O42− [O,O]. Ambidentate: can bond via either of two different donor atoms — NO2− (N or O) and SCN− (S or N).
Exercise 5.5 — Oxidation numbers
(i) [Co(H2O)(CN)(en)2]2+: 0 + (−1) + 0 + Co = +2 → Co = +3.
(ii) [CoBr2(en)2]+: 2(−1) + 0 + Co = +1 → Co = +3.
(iii) [PtCl4]2−: 4(−1) + Pt = −2 → Pt = +2.
(iv) K3[Fe(CN)6]: 3(+1) + Fe + 6(−1) = 0 → Fe = +3.
(v) [Cr(NH3)3Cl3] (neutral): 3(0) + 3(−1) + Cr = 0 → Cr = +3.
Exercise 5.8 — Types of isomerism (one example each)
Geometrical: cis & trans-[Pt(NH3)2Cl2]. Optical: Δ & Λ-[Co(en)3]3+. Linkage: [Co(NH3)5(NO2)]Cl2 (yellow, –NO2) vs [Co(NH3)5(ONO)]Cl2 (red, –ONO). Coordination: [Co(NH3)6][Cr(CN)6] vs [Cr(NH3)6][Co(CN)6]. Ionisation: [Co(NH3)5SO4]Br vs [Co(NH3)5Br]SO4. Solvate (hydrate): [Cr(H2O)6]Cl3 vs [Cr(H2O)5Cl]Cl2·H2O.
Exercise 5.9 — How many geometrical isomers?
(i) [Cr(C2O4)3]3−: 0 geometrical isomers — only one geometric arrangement (it has 2 optical isomers though).
(ii) [Co(NH3)3Cl3]: 2 geometrical isomers — facial (fac) and meridional (mer).
Exercise 5.10 — Optical isomers
(i) [Cr(C2O4)3]3−: two optical isomers (Δ and Λ enantiomers — three didentate ligands wrap helically).
(ii) [PtCl2(en)2]2+: only the cis isomer is chiral; gives Δ and Λ enantiomers (2 optical isomers). The trans isomer is achiral.
(iii) [Cr(NH3)2Cl2(en)]+: cis isomers are chiral and exist as Δ and Λ enantiomers (2 optical isomers).
Exercise 5.11 — Geometrical and optical isomers
(i) [CoCl2(en)2]+: cis & trans (2 geometrical); the cis form is chiral → 2 optical isomers. Total 3 stereoisomers.
(ii) [Co(NH3)Cl(en)2]2+: cis (chiral, gives 2 enantiomers) and trans (achiral) → 3 stereoisomers (cis-Δ, cis-Λ, trans).
(iii) [Co(NH3)2Cl2(en)]+: 2 geometrical (cis, trans). The cis form has 2 optical isomers. Total 3 stereoisomers.
A square-planar MABCD shows 3 geometrical isomers (each pair of trans ligands defines one isomer). None is optically active — the molecular plane itself is a mirror plane, so the complex is achiral.
Exercise 5.13 — CuSO₄ + KF vs CuSO₄ + KCl
With KF, the Cu2+ aqua complex is replaced by a green CuF2·2H2O precipitate (or [CuF4]2− in excess F−) — the d–d transition energy is changed by the new ligand field.
With KCl, soluble [CuCl4]2− forms (bright green). Cl− sits low in the spectrochemical series; the absorption shifts so the complementary colour is bright green.
Exercise 5.14 — CuSO₄ + excess KCN; why no CuS with H₂S?
Excess KCN forms [Cu(CN)4]3− (Cu(I)) — the very high formation constant (β ≈ 1030) makes the free [Cu+] concentration extremely small. The ionic product [Cu+]²[S2−] never reaches Ksp(Cu2S), so no precipitate forms when H2S is bubbled in.
t2g (3 orbitals) drops by (2/5)Δₒ; eg (2 orbitals) rises by (3/5)Δₒ.
Exercise 5.17 — Spectrochemical series; weak vs strong field ligand
Spectrochemical series: I− < Br− < SCN− < Cl− < F− < OH− < C2O42− < H2O < NH3 < en < CN− < CO. Weak-field ligands produce a small Δₒ (less than the pairing energy P) → high-spin complexes; strong-field ligands produce large Δₒ (> P) → pairing of electrons → low-spin complexes.
Exercise 5.18 — Crystal field splitting energy and configuration
Crystal-field splitting energy Δₒ is the energy gap between t2g and eg orbitals in an octahedral complex. For d4–d7 ions: if Δₒ < pairing energy P, the next electron enters eg giving t2g³eg¹ (high-spin). If Δₒ > P, the electron pairs in t2g giving t2g⁴eg⁰ (low-spin). So Δₒ vs P decides the configuration and hence the magnetic moment.
Cr3+ = 3d3: the three d-electrons sit singly in the t2g set regardless of field strength → 3 unpaired → paramagnetic.
Ni2+ = 3d8: with the strong-field CN−, the geometry is square planar (dsp²); all electrons are paired → diamagnetic.
Exercise 5.20 — [Ni(H₂O)₆]²⁺ green vs [Ni(CN)₄]²⁻ colourless
[Ni(H2O)6]2+ is octahedral with a small Δₒ (H2O weak field). It absorbs in the red region; complementary green colour observed.
[Ni(CN)4]2− is square-planar with a very large dx²-y²–dxy gap (CN− very strong field). Absorption shifts to the UV region — no visible light is absorbed → solution appears colourless.
Exercise 5.21 — [Fe(CN)₆]⁴⁻ vs [Fe(H₂O)₆]²⁺ colour
Both have Fe2+ (3d6). With strong-field CN−, the complex is low-spin; Δₒ is large → absorbs higher-energy (shorter λ) light → complementary colour appears yellow.
With weak-field H2O the complex is high-spin; Δₒ is small → absorbs longer λ → pale green colour.
Exercise 5.22 — Bonding in metal carbonyls
The M–C bond in carbonyls has dual character: a σ-bond (CO carbon's lone pair → empty metal hybrid orbital) and a π-bond (filled metal d-orbital → empty CO π* antibonding orbital). The two donations reinforce each other — the synergic effect — making the M–C bond strong and the C–O bond weakened (longer than free CO).
Exercise 5.25 — Violet colour of [Ti(H₂O)₆]³⁺ by CFT
Ti3+ is a 3d1 ion. The single d-electron sits in the t2g level. Light corresponding to Δₒ (≈ blue-green, λ ≈ 498 nm) excites the electron to the eg level (t2g¹eg⁰ → t2g⁰eg¹ — a d–d transition). The transmitted/reflected light is the complement of blue-green, namely violet, so the complex appears violet.
Exercise 5.26 — The chelate effect
The chelate effect is the extra thermodynamic stability that a complex gains when monodentate ligands are replaced by an equivalent number of polydentate (chelating) ligands. The driving force is largely entropic — replacing several small ligands by one polydentate ligand releases more particles into solution. Example: log β for [Ni(en)3]2+ ≈ 18 vs log β for [Ni(NH3)6]2+ ≈ 8 — about 1010-fold more stable.
Exercise 5.27 — Role of coordination compounds in (i) biology (ii) medicine (iii) analysis (iv) extraction
(i) Biology: haemoglobin (Fe in haem) carries O2 in blood; chlorophyll (Mg in porphyrin) traps light for photosynthesis; vitamin B12 (Co in corrin) prevents pernicious anaemia.
(ii) Medicine: cisplatin (cis-[Pt(NH3)2Cl2]) treats testicular and ovarian cancers; EDTA chelation removes lead and other heavy metals from the body.
(iii) Analytical chemistry: EDTA titration of Ca2+/Mg2+ (water hardness); DMG forms a brick-red Ni complex used for gravimetric Ni determination.
(iv) Extraction: 4 Au + 8 CN− + O2 + 2 H2O → 4 [Au(CN)2]− + 4 OH−; gold is then displaced by zinc. Mond's process refines Ni via Ni(CO)4.
Exercise 5.28 — How many ions from [Co(NH₃)₆]Cl₂ in solution? (Note: NCERT typo — should be Cl₃)
Reading as [Co(NH3)6]Cl3, the complex dissociates into [Co(NH3)6]3+ and 3 Cl− → 4 ions in total. Answer: (ii) 4.
Exercise 5.29 — Highest magnetic moment
(i) [Cr(H2O)6]3+: Cr3+ = 3d3; n = 3; μ = √15 ≈ 3.87 BM.
(ii) [Fe(H2O)6]2+: Fe2+ = 3d6; H2O weak-field, so high spin; n = 4; μ = √24 ≈ 4.90 BM.
(iii) [Zn(H2O)6]2+: Zn2+ = 3d10; n = 0; μ = 0 BM. Highest: (ii) [Fe(H2O)6]2+.
Exercise 5.30 — Most stable complex
[Fe(C2O4)3]3− is most stable: oxalate is a chelating (didentate) ligand, so this complex enjoys the chelate effect, giving the largest β. Answer: (iii).
Exercise 5.31 — Order of absorption wavelengths
Field strength: NO2− > NH3 > H2O. Larger Δₒ means absorption at higher energy (shorter wavelength). Therefore wavelength of absorption decreases as Δₒ increases: λ([Ni(NO2)6]4−) < λ([Ni(NH3)6]2+) < λ([Ni(H2O)6]2+).
🎯 Mixed Competency-Based Questions
Q1. The IUPAC name of [Co(en)3](SO4)3/2 (or written as [Co(en)3]2(SO4)3) is: L1 Remember
tris(ethane-1,2-diamine)cobalt(III) sulphate.
Q2. CN− sits very high in the spectrochemical series. Predict the magnetic moment of [Mn(CN)6]3−. L3 Apply
Mn3+ = 3d4. Strong-field CN− → low spin t2g4eg0; n = 2 unpaired electrons → μ = √8 ≈ 2.83 BM (in practice ≈ 2 unpaired electrons; a single value of 1 unpaired is also seen due to Jahn-Teller).
Q3. Why does [Ni(en)3]2+ exhibit optical isomerism but [Ni(NH3)6]2+ does not? L4 Analyse
Three didentate en ligands wrap helically around Ni — left- or right-handed; the two helices are non-superimposable mirror images, so the complex exists as Δ and Λ enantiomers. [Ni(NH3)6]2+ has six identical monodentate ligands and possesses several mirror planes — it is achiral.
Q4. Predict the geometry, magnetic behaviour and hybridisation of [Pt(CN)4]2−. L3 Apply
Pt2+ = 5d8. Strong-field CN− → square-planar dsp² hybridisation; all 8 d electrons pair up; diamagnetic; n = 0; μ = 0.
Q5. HOT (Create): Invent a coordination compound that simultaneously demonstrates (i) the chelate effect, (ii) optical isomerism, and (iii) high paramagnetism. Justify each. L6 Create
Try [Fe(ox)3]3− (tris-oxalatoferrate(III)). (i) Three didentate oxalates → chelate effect → high stability. (ii) Three didentate ligands wrap helically → Δ and Λ enantiomers → optical isomerism. (iii) Fe3+ = 3d5; oxalate is at the weak-to-moderate end of the spectrochemical series → high spin → 5 unpaired electrons → μ ≈ 5.92 BM (highly paramagnetic).
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: [Co(NH3)6]3+ is diamagnetic but [CoF6]3− is paramagnetic.
R: NH3 is a strong-field ligand causing pairing of 3d electrons; F− is weak-field.
Answer: (A). Both true; R correctly explains A. Inner-orbital d²sp³ vs outer-orbital sp³d² hybridisation respectively.
A: [Cr(C2O4)3]3− is more stable than [CrCl6]3−.
R: Oxalate is a chelating ligand and forms five-membered rings.
Answer: (A). Both true; R correctly explains A — the chelate effect.
A: Tetrahedral [NiCl4]2− shows no geometrical isomerism but octahedral [Co(NH3)4Cl2]+ shows cis-trans isomerism.
R: All vertices of a tetrahedron are mutually equivalent, but in an octahedron a pair of ligands can be either adjacent (cis) or opposite (trans).
Answer: (A). Both true; R correctly explains A.
Frequently Asked Questions - NCERT Exercises and Solutions: Coordination Compounds
What are the key NCERT exercise types in Chapter 5 Coordination Compounds?
NCERT Class 12 Chemistry Chapter 5 Coordination Compounds exercises cover definitions, structure-property relationships, reaction mechanisms, numerical problems, and predict-the-product questions. The MyAiSchool solution set provides step-by-step worked solutions for every NCERT question, aligned with the CBSE board exam pattern. Students should focus on reaction mechanisms, IUPAC nomenclature, and stoichiometric reasoning to score full marks.
How should students approach reaction mechanism questions in Coordination Compounds?
For reaction mechanism questions in NCERT Class 12 Chemistry Chapter 5 Coordination Compounds: (1) identify the substrate, reagent, solvent, and conditions, (2) draw curly-arrow electron movement at each step, (3) label intermediates (carbocation, carbanion, free radical, transition state), (4) state stereochemistry where relevant. The MyAiSchool solutions show full curly-arrow mechanisms for every multi-step reaction.
What are the most-asked CBSE board questions from Chapter 5?
From NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds), the most-asked CBSE board questions test conceptual understanding, structure-property logic, distinguishing tests, IUPAC naming, and short numerical problems. 5-mark questions usually combine structural reasoning + mechanism + application. The MyAiSchool exercise set tags each question by mark weight and Bloom level for prioritized prep.
How do I balance chemical equations in NCERT exercises?
For balancing equations in NCERT Class 12 Chemistry Chapter 5: (1) write skeletal equation with correct formulas, (2) balance atoms other than H, O first, (3) balance O, then H (using H2O for organic reactions; or in acidic/basic medium for redox), (4) balance charge using e- in redox, (5) cross-check atom count and charge on both sides. The MyAiSchool solutions include balanced redox half-reactions where applicable.
What are common mistakes students make in Chapter 5 exercises?
Common mistakes in NCERT Class 12 Chemistry Chapter 5 (Coordination Compounds) include: (1) incorrect IUPAC names (wrong locants or suffix), (2) skipping reaction conditions (catalyst, temperature, solvent), (3) wrong mechanism arrows, (4) sign errors in numerical (Ecell, Kc, ΔG), (5) forgetting stereochemistry (retention/inversion/racemisation). The MyAiSchool solutions flag these traps for each question.
How does the MyAiSchool solution differ from other NCERT solution sets?
MyAiSchool Class 12 Chemistry Chapter 5 Coordination Compounds solutions use NEP 2024-aligned pedagogy with Bloom Taxonomy tagged questions (L1 Remember to L6 Create), step-by-step working with chemical reasoning, curly-arrow mechanisms, IUPAC naming verifications, interactive simulations, and Competency-Based Questions (CBQs) for board exam practice. Verified against NCERT and CBSE marking schemes.
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