NCERT Solutions for Class 12 Chemistry Chapter 5: Coordination Compounds
These Class 12 Chemistry Chapter 5 solutions cover the complete NCERT chapter Coordination Compounds for session 2026–27. Every Intext Question and every numbered Exercise (5.1–5.31) is reproduced verbatim from the NCERT textbook and solved fully — with IUPAC names, oxidation-state working, VBT and CFT explanations, isomer structures described in words, and clear answers for all the objective questions. Answers are exam-ready and original.
Coordination compounds are the backbone of modern inorganic, bio-inorganic and industrial chemistry — chlorophyll (Mg), haemoglobin (Fe) and vitamin B12 (Co) are all coordination compounds. The chapter begins with Werner’s theory (primary and secondary valences) and the basic vocabulary: central atom/ion, ligand, coordination number, coordination sphere and counter ion. It then covers IUPAC nomenclature, the many kinds of isomerism (geometrical, optical, linkage, coordination, ionisation, solvate), and two models of bonding — Valence Bond Theory (VBT) with inner/outer-orbital (low-spin/high-spin) complexes, and Crystal Field Theory (CFT) with d-orbital splitting (Δo), the spectrochemical series and colour. It closes with bonding in metal carbonyls (synergic σ–π bonding) and the real-life importance of coordination compounds.
Key Concepts & Definitions
Coordination entity: a central metal atom/ion bonded to a fixed number of ions or molecules, e.g. [Co(NH3)6]3+.
Ligand: the ion/molecule bound to the central atom. Unidentate (one donor atom, e.g. NH3, Cl−), didentate (two donor atoms, e.g. en, C2O42−), polydentate/hexadentate (EDTA4−) and ambidentate (binds through either of two atoms, e.g. NO2−, SCN−).
Coordination number (CN): number of ligand donor atoms σ-bonded to the metal.
Coordination sphere / counter ion: the part inside [ ] is the coordination sphere; ionisable groups outside are counter ions.
Homoleptic vs heteroleptic: bound to one kind of donor group vs more than one kind.
Chelate: a ring formed when a di-/polydentate ligand binds one metal through two or more donor atoms; chelates are extra stable.
5.2 Write the IUPAC names of the following coordination compounds:
(i) [Co(NH3)6]Cl3 (ii) [Co(NH3)5Cl]Cl2 (iii) K3[Fe(CN)6](iv) K3[Fe(C2O4)3] (v) K2[PdCl4] (vi) [Pt(NH3)2Cl(NH2CH3)]Cl
5.3 Indicate the types of isomerism exhibited by the following complexes and draw the structures for these isomers:
(i) K[Cr(H2O)2(C2O4)2] (ii) [Co(en)3]Cl3(iii) [Co(NH3)5(NO2)](NO3)2 (iv) [Pt(NH3)(H2O)Cl2]
ANSWER(i) Geometrical (cis/trans) and optical isomerism. This octahedral complex has two H2O and two didentate oxalate ligands. The two aqua ligands can lie adjacent (cis) or opposite (trans). The cis form lacks a plane of symmetry, so it is chiral and exists as a pair of non-superimposable mirror images (d- and l-enantiomers); the trans form is optically inactive.(ii) Optical isomerism. [Co(en)3]3+ is an octahedral tris-chelate that has no plane of symmetry, so it exists as two non-superimposable mirror images (d- and l-forms), like a left and right hand.(iii) Linkage isomerism (the ambidentate NO2− binds through N to give –NO2 nitrito-N, or through O to give –ONO nitrito-O). Ionisation isomerism is also possible because NO2− inside and NO3− outside can exchange, e.g. [Co(NH3)5(NO3)](NO3)(NO2).(iv) Geometrical (cis/trans) isomerism. In this square-planar [Pt(NH3)(H2O)Cl2], the two Cl ligands may be adjacent (cis) or opposite (trans).
5.4 Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ionisation isomers.
ANSWERThese ionisation isomers release different counter ions in solution and so react differently to test reagents:[Co(NH3)5Cl]SO4 gives free SO42−, so adding BaCl2 gives a white precipitate of BaSO4; adding AgNO3 gives no precipitate (Cl− is coordinated).[Co(NH3)5(SO4)]Cl gives free Cl−, so adding AgNO3 gives a white precipitate of AgCl; adding BaCl2 gives no precipitate (SO42− is coordinated). The opposite responses confirm they are ionisation isomers.
5.5 Explain on the basis of valence bond theory that [Ni(CN)4]2− ion with square planar structure is diamagnetic and the [NiCl4]2− ion with tetrahedral geometry is paramagnetic.
ANSWERIn both ions Ni is in the +2 state with a 3d8 configuration (two unpaired electrons in the free ion).In [Ni(CN)4]2−, CN− is a strong-field ligand; it pairs up the two unpaired 3d electrons, vacating one 3d orbital. Hybridisation is dsp2 (square planar). With no unpaired electron, the ion is diamagnetic.In [NiCl4]2−, Cl− is a weak-field ligand and cannot pair the electrons. Hybridisation is sp3 (tetrahedral) and the two unpaired electrons remain, so the ion is paramagnetic.
5.6 [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
ANSWERIn [Ni(CO)4] nickel is in the zero oxidation state (3d84s2). CO is a strong-field ligand: the electrons rearrange so that all become paired (3d10), and sp3 hybridisation gives a tetrahedral, diamagnetic molecule.In [NiCl4]2− nickel is in the +2 state (3d8). Cl− is a weak-field ligand and cannot pair the two unpaired d electrons, so the sp3 tetrahedral ion is paramagnetic.
5.7 [Fe(H2O)6]3+ is strongly paramagnetic whereas [Fe(CN)6]3− is weakly paramagnetic. Explain.
ANSWERIn both, Fe is +3 (3d5). In [Fe(CN)6]3−, CN− is a strong-field ligand; the 3d electrons pair up to give t2g5 with only one unpaired electron. Hybridisation is d2sp3 (inner-orbital, low-spin), so it is weakly paramagnetic.In [Fe(H2O)6]3+, H2O is a weak-field ligand; the 3d electrons stay unpaired giving five unpaired electrons. Hybridisation is sp3d2 (outer-orbital, high-spin), so it is strongly paramagnetic.
5.8 Explain [Co(NH3)6]3+ is an inner orbital complex whereas [Ni(NH3)6]2+ is an outer orbital complex.
ANSWERIn [Co(NH3)6]3+, Co is +3 (3d6). NH3 pairs up the 3d electrons (t2g6), freeing two inner 3d orbitals. Hybridisation is d2sp3 using inner (3d) orbitals, so it is an inner-orbital (low-spin) complex.In [Ni(NH3)6]2+, Ni is +2 (3d8). The 3d orbitals contain pairs and the two unpaired electrons cannot be moved out to free two inner d orbitals. The metal therefore uses outer 4d orbitals, giving sp3d2 hybridisation — an outer-orbital (high-spin) complex.
5.9 Predict the number of unpaired electrons in the square planar [Pt(CN)4]2− ion.
ANSWERPt is +2 (5d8). For a square-planar geometry the hybridisation is dsp2, so the 5d electrons pair up to leave one d orbital empty. Hence there are zero unpaired electrons, and the ion is diamagnetic.
5.10 The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory.
ANSWERMn2+ is 3d5. In [Mn(H2O)6]2+, H2O is a weak-field ligand, so Δo < P (high spin): electrons spread out as t2g3eg2, giving five unpaired electrons.In [Mn(CN)6]4−, CN− is a strong-field ligand, so Δo > P (low spin): electrons pair up in the lower set as t2g5eg0, leaving only one unpaired electron.
NCERT Exercises 5.1–5.31 — Solutions
5.1 Explain the bonding in coordination compounds in terms of Werner’s postulates.
ANSWERWerner’s postulates: (1) A metal shows two kinds of valences — primary (ionisable) and secondary (non-ionisable). (2) Primary valences are satisfied by negative ions and equal the oxidation number. (3) Secondary valences are satisfied by neutral molecules or negative ions and equal the coordination number, which is fixed for a metal. (4) The groups bound by secondary valences have fixed spatial arrangements (coordination polyhedra) such as octahedral, tetrahedral or square planar.For example, in CoCl3·6NH3, written [Co(NH3)6]Cl3, the six NH3 satisfy the six secondary valences (octahedral, non-ionisable) and the three Cl− satisfy the primary valences (ionisable, precipitated as AgCl).
5.2 FeSO4 solution mixed with (NH4)2SO4 solution in 1:1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ ion. Explain why?
ANSWERFeSO4·(NH4)2SO4·6H2O (Mohr’s salt) is a double salt. It dissociates completely in water into its simple ions (Fe2+, NH4+, SO42−), so free Fe2+ is available and gives the usual Fe2+ test.CuSO4 with excess NH3 forms the complex [Cu(NH3)4]SO4. The complex ion [Cu(NH3)4]2+ is stable and does not dissociate to give appreciable free Cu2+, so the ordinary Cu2+ test fails.
5.3 Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
ANSWERCoordination entity: a central metal bonded to a fixed number of ions/molecules. Examples: [Ni(CO)4], [Co(NH3)6]3+.Ligand: the ion/molecule bound to the central atom. Examples: Cl−, NH3 (also H2O, en).Coordination number: number of donor atoms σ-bonded to the metal. Examples: 6 in [PtCl6]2−; 4 in [Ni(NH3)4]2+.Coordination polyhedron: the spatial arrangement of donor atoms. Examples: octahedral [Co(NH3)6]3+; square planar [PtCl4]2− (or tetrahedral [Ni(CO)4]).Homoleptic: metal bound to one kind of donor group. Examples: [Co(NH3)6]3+, [Ni(CO)4].Heteroleptic: metal bound to more than one kind of donor group. Examples: [Co(NH3)4Cl2]+, [Co(NH3)5Cl]2+.
5.4 What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
ANSWERUnidentate: binds through a single donor atom. Examples: NH3, Cl− (also H2O).Didentate: binds through two donor atoms. Examples: ethane-1,2-diamine (en, NH2CH2CH2NH2), oxalate (C2O42−).Ambidentate: has two different donor atoms but binds through only one at a time. Examples: NO2− (through N or O), SCN− (through S or N).
5.5 Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H2O)(CN)(en)2]2+ (ii) [CoBr2(en)2]+ (iii) [PtCl4]2−(iv) K3[Fe(CN)6] (v) [Cr(NH3)3Cl3]
ANSWER(i) en and H2O neutral, CN = −1: x − 1 = +2 ⇒ x = +3.(ii) en neutral, 2 Br = −2: x − 2 = +1 ⇒ x = +3.(iii) 4 Cl = −4: x − 4 = −2 ⇒ x = +2.(iv) Complex charge = −3 (to balance 3 K+); 6 CN = −6: x − 6 = −3 ⇒ x = +3.(v) Neutral; 3 NH3 neutral, 3 Cl = −3: x − 3 = 0 ⇒ x = +3.
5.6 Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxidozincate(II) (ii) Potassium tetrachloridopalladate(II) (iii) Diamminedichloridoplatinum(II)(iv) Potassium tetracyanidonickelate(II) (v) Pentaamminenitrito-O-cobalt(III) (vi) Hexaamminecobalt(III) sulphate(vii) Potassium tri(oxalato)chromate(III) (viii) Hexaammineplatinum(IV) (ix) Tetrabromidocuprate(II) (x) Pentaamminenitrito-N-cobalt(III)
5.8 List various types of isomerism possible for coordination compounds, giving an example of each.
ANSWER(A) Stereoisomerism — same bonds, different spatial arrangement:• Geometrical (cis/trans): [Pt(NH3)2Cl2].• Optical (d/l): [Co(en)3]3+.(B) Structural isomerism — different bonds:• Linkage: [Co(NH3)5(NO2)]Cl2 vs [Co(NH3)5(ONO)]Cl2.• Coordination: [Co(NH3)6][Cr(CN)6] vs [Cr(NH3)6][Co(CN)6].• Ionisation: [Co(NH3)5(SO4)]Br vs [Co(NH3)5Br]SO4.• Solvate (hydrate): [Cr(H2O)6]Cl3 vs [Cr(H2O)5Cl]Cl2·H2O.
5.9 How many geometrical isomers are possible in the following coordination entities?
(i) [Cr(C2O4)3]3− (ii) [Co(NH3)3Cl3]
ANSWER(i) Nil (zero). With three identical symmetrical didentate oxalate ligands, no cis/trans arrangement is possible, so there are no geometrical isomers (it does show optical isomerism, though).(ii) Two geometrical isomers — the facial (fac) isomer (three like ligands on one triangular face) and the meridional (mer) isomer (three like ligands around a meridian).
5.10 Draw the structures of optical isomers of:
(i) [Cr(C2O4)3]3− (ii) [PtCl2(en)2]2+ (iii) [Cr(NH3)2Cl2(en)]+
ANSWER(i) [Cr(C2O4)3]3− is an octahedral tris-chelate with no plane of symmetry; it exists as two non-superimposable mirror images — the d and l enantiomers (a left-handed and a right-handed three-bladed propeller).(ii) Only the cis-[PtCl2(en)2]2+ is optically active and exists as d and l mirror images; the trans form is achiral. (Drawn as the two cis enantiomers with the two Cl adjacent and the two en chelate rings spanning the remaining positions.)(iii) The cis-cis form of [Cr(NH3)2Cl2(en)]+ lacks a symmetry plane and exists as d and l enantiomers (the two NH3 cis, the two Cl cis, with the en chelate ring completing the octahedron).
5.11 Draw all the isomers (geometrical and optical) of:
(i) [CoCl2(en)2]+ (ii) [Co(NH3)Cl(en)2]2+ (iii) [Co(NH3)2Cl2(en)]+
ANSWER(i) [CoCl2(en)2]+ — 3 isomers. A trans form (two Cl opposite) which is optically inactive, and a cis form which is chiral and splits into d and l enantiomers.(ii) [Co(NH3)Cl(en)2]2+ — 3 isomers. A trans form (NH3 and Cl opposite) which is optically inactive, and a cis form (NH3 and Cl adjacent) which is chiral, giving d and l enantiomers.(iii) [Co(NH3)2Cl2(en)]+. Geometrical forms: NH3 can be cis or trans and Cl can be cis or trans. The form with both NH3 and both Cl cis is chiral and gives d and l enantiomers; the form with the two Cl trans (and NH3 cis) and the form with the two NH3trans (and Cl cis) are achiral — giving three geometrical isomers, one of which is optically active.
5.12 Write all the geometrical isomers of [Pt(NH3)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
ANSWERFor a square-planar complex of the type [Mabcd] there are three geometrical isomers, depending on which ligand sits trans to NH3: (1) NH3 trans to Br, (2) NH3 trans to Cl, (3) NH3 trans to py.Square-planar complexes are achiral (they have a plane of symmetry — the molecular plane), so none of these three exhibits optical isomerism.
5.13 Aqueous copper sulphate solution (blue in colour) gives:
(i) a green precipitate with aqueous potassium fluoride and(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
ANSWERIn water, CuSO4 exists as the complex [Cu(H2O)4]2+, whose blue colour gives the solution its colour.(i) On adding KF, the weak-field H2O ligands are replaced by F− to give [CuF4]2−, seen as a green precipitate: [Cu(H2O)4]2+ + 4F− → [CuF4]2− + 4H2O.(ii) On adding KCl, Cl− replaces H2O to give [CuCl4]2−, a bright green solution: [Cu(H2O)4]2+ + 4Cl− → [CuCl4]2− + 4H2O.
5.14 What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S(g) is passed through this solution?
ANSWERExcess KCN converts the copper to the very stable complex ion [Cu(CN)4]2− (also written [Cu(CN)4]3− after reduction): [Cu(H2O)4]2+ + 4CN− → [Cu(CN)4]2− + 4H2O.Because CN− is a strong-field ligand it forms a highly stable complex with very low dissociation, so the concentration of free Cu2+ in solution is extremely small. On passing H2S, the ionic product of Cu2+ and S2− cannot exceed the solubility product of CuS, so no CuS precipitate forms.
5.15 Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)6]4− (ii) [FeF6]3− (iii) [Co(C2O4)3]3− (iv) [CoF6]3−
ANSWER(i) [Fe(CN)6]4−: Fe is +2 (3d6). CN− is strong-field, so the 3d electrons pair up to t2g6. Hybridisation d2sp3 → octahedral, inner-orbital, diamagnetic (0 unpaired e−).(ii) [FeF6]3−: Fe is +3 (3d5). F− is weak-field, electrons stay unpaired. Hybridisation sp3d2 → octahedral, outer-orbital, paramagnetic (5 unpaired e−).(iii) [Co(C2O4)3]3−: Co is +3 (3d6). Oxalate behaves as a strong-field ligand here, so electrons pair to t2g6. Hybridisation d2sp3 → octahedral, inner-orbital, diamagnetic (0 unpaired e−).(iv) [CoF6]3−: Co is +3 (3d6). F− is weak-field, electrons stay unpaired. Hybridisation sp3d2 → octahedral, outer-orbital, paramagnetic (4 unpaired e−).
5.16 Draw figure to show the splitting of d orbitals in an octahedral crystal field.
ANSWERIn an octahedral field the five degenerate d orbitals split into two sets. The lower-energy t2g set (dxy, dyz, dxz) points between the axes and is lowered by −(2/5)Δo; the higher-energy eg set (dx2−y2, dz2) points along the axes towards the ligands and is raised by +(3/5)Δo. The energy gap between the two sets is the crystal field splitting energy Δo.Schematically: eg (dx2−y2, dz2) —— (higher); ↑↓Δo; t2g (dxy, dyz, dxz) ——— (lower); both measured from the barycentre (average) energy.
5.17 What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
ANSWERThe spectrochemical series is an experimentally determined arrangement of ligands in order of increasing crystal-field strength (based on light absorbed by complexes): I− < Br− < SCN− < Cl− < F− < OH− < C2O42− < H2O < NH3 < en < CN− < CO.Weak-field ligands (e.g. I−, Cl−, F−, H2O) cause small splitting (Δo < P), so electrons stay unpaired → high-spin complexes.Strong-field ligands (e.g. CN−, CO, NH3, en) cause large splitting (Δo > P), so electrons pair up in the lower set → low-spin complexes.
5.18 What is crystal field splitting energy? How does the magnitude of Δo decide the actual configuration of d orbitals in a coordination entity?
ANSWERCrystal field splitting energy (Δo) is the energy separation between the t2g and eg sets of d orbitals produced by the ligand field in an octahedral complex.The configuration is decided by comparing Δo with the pairing energy P (for d4–d7 ions):If Δo < P (weak field), the fourth electron enters the higher eg set rather than pairing → t2g3eg1 → high spin.If Δo > P (strong field), it is cheaper to pair in the lower set → t2g4eg0 → low spin.
5.19 [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2− is diamagnetic. Explain why?
ANSWER[Cr(NH3)6]3+: Cr is +3 (3d3). The three 3d electrons occupy three t2g orbitals singly. With d2sp3 hybridisation it is octahedral, and the three unpaired electrons make it paramagnetic.[Ni(CN)4]2−: Ni is +2 (3d8). The strong-field CN− pairs the unpaired electrons; dsp2 hybridisation gives a square-planar ion with no unpaired electrons, so it is diamagnetic.
5.20 A solution of [Ni(H2O)6]2+ is green but a solution of [Ni(CN)4]2− is colourless. Explain.
ANSWERIn [Ni(H2O)6]2+, the weak-field H2O ligands give a small Δo; an unpaired d electron can undergo a d–d transition that absorbs in the visible region, so the complex is coloured (green).In [Ni(CN)4]2−, the strong-field CN− pairs all the d electrons, so there is no unpaired electron and the d–d transition would need very high energy (absorption shifts into the UV, outside the visible region). Hence the complex appears colourless.
5.21 [Fe(CN)6]4− and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?
ANSWERIn both, Fe is +2 (3d6), but the ligands differ in field strength. CN− is a much stronger field ligand than H2O, so Δo is larger for [Fe(CN)6]4−.A larger Δo means light of higher energy (shorter wavelength) is absorbed for the d–d transition. Since the two complexes absorb different wavelengths, they transmit different complementary colours — hence they appear different in colour.
5.22 Discuss the nature of bonding in metal carbonyls.
ANSWERThe metal–carbon bond in metal carbonyls has both σ and π character. The M–C σ bond forms by donation of the carbon lone pair of CO into a vacant metal orbital.The M–C π bond forms by back-donation of electrons from a filled metal d orbital into the vacant antibonding π* orbital of CO. This synergic effect (the two donations reinforce each other) strengthens the metal–carbon bond and weakens the C–O bond.
5.23 Give the oxidation state, d orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K3[Co(C2O4)3] (ii) cis-[CrCl2(en)2]Cl (iii) (NH4)2[CoF4] (iv) [Mn(H2O)6]SO4
ANSWER(i) K3[Co(C2O4)3]: Co OS = +3, CN = 6, d6, low-spin t2g6eg0.(ii) cis-[CrCl2(en)2]Cl: Cr OS = +3, CN = 6, d3, t2g3.(iii) (NH4)2[CoF4]: Co OS = +2, CN = 4, d7, t2g5eg2 (high spin).(iv) [Mn(H2O)6]SO4: Mn OS = +2, CN = 6, d5, t2g3eg2 (high spin).
5.24 Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H2O)2(C2O4)2]·3H2O (ii) [Co(NH3)5Cl]Cl2 (iii) [CrCl3(py)3](iv) Cs[FeCl4] (v) K4[Mn(CN)6]
5.25 Explain the violet colour of the complex [Ti(H2O)6]3+ on the basis of crystal field theory.
ANSWERTi3+ is a 3d1 system. In the octahedral field of the six H2O ligands, the single electron occupies the lower t2g set in the ground state (t2g1eg0).When the complex absorbs light in the blue-green region (about 498–500 nm), the electron is excited from t2g to the empty eg level (t2g1eg0 → t2g0eg1) — a d–d transition. The colour transmitted is complementary to the blue-green absorbed, so the complex appears violet.
5.26 What is meant by the chelate effect? Give an example.
ANSWERThe chelate effect is the extra stability that a complex gains when a di- or polydentate ligand binds the same metal ion through two or more donor atoms to form a ring (a chelate), compared with a similar complex of unidentate ligands.Example: [Ni(en)3]2+ (en is a didentate chelating ligand) is much more stable than [Ni(NH3)6]2+, even though both have six N-donor atoms around nickel.
5.27 Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems (ii) medicinal chemistry (iii) analytical chemistry (iv) extraction/metallurgy of metals.
ANSWER(i) Biological systems: chlorophyll (a Mg complex) drives photosynthesis; haemoglobin (an Fe complex) carries oxygen; vitamin B12 (a Co complex) prevents pernicious anaemia.(ii) Medicinal chemistry: cis-platin [Pt(NH3)2Cl2] inhibits tumour growth; EDTA chelate therapy treats lead poisoning; D-penicillamine removes excess copper.(iii) Analytical chemistry: coloured complexes are used to detect and estimate metal ions — e.g. Ni2+ with dimethylglyoxime (DMG); water hardness is estimated by EDTA titration of Ca2+/Mg2+.(iv) Extraction/metallurgy: gold and silver are extracted as cyanide complexes, e.g. gold as [Au(CN)2]−, then displaced by zinc; impure nickel is purified via [Ni(CO)4] (Mond process).
5.28 How many ions are produced from the complex Co(NH3)6Cl2 in solution?
(i) 6 (ii) 4 (iii) 3 (iv) 2
ANSWER(iii) 3. The compound is [Co(NH3)6]Cl2, which dissociates into one [Co(NH3)6]2+ cation and two Cl− ions, giving 3 ions in total.
5.29 Amongst the following ions which one has the highest magnetic moment value?
(i) [Cr(H2O)6]3+ (ii) [Fe(H2O)6]2+ (iii) [Zn(H2O)6]2+
ANSWER(ii) [Fe(H2O)6]2+. Unpaired electrons: Cr3+ (d3) = 3; Fe2+ (d6, high spin) = 4; Zn2+ (d10) = 0. With the most unpaired electrons (4), Fe2+ has the highest μ = √[4(6)] = √24 ≈ 4.90 BM.
5.30 Amongst the following, the most stable complex is
(i) [Fe(H2O)6]3+ (ii) [Fe(NH3)6]3+ (iii) [Fe(C2O4)3]3− (iv) [FeCl6]3−
ANSWER(iii) [Fe(C2O4)3]3−. Oxalate (C2O42−) is a didentate chelating ligand; the chelate effect makes its complex the most stable of the four.
5.31 What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(NO2)6]4−, [Ni(NH3)6]2+, [Ni(H2O)6]2+
ANSWERField strength (spectrochemical series): H2O < NH3 < NO2−. So the splitting energy Δo increases as [Ni(H2O)6]2+ < [Ni(NH3)6]2+ < [Ni(NO2)6]4−.Since energy absorbed E = hc/λ, a larger Δo means a shorter wavelength absorbed. Therefore the wavelength of absorption is in the opposite order: [Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4−.
Extra Practice Questions
Short Answer Type Questions
Q1. Differentiate between a double salt and a complex with one example each.
ANSWERA double salt (e.g. Mohr’s salt FeSO4·(NH4)2SO4·6H2O) dissociates completely into all its simple ions in water. A complex (e.g. K4[Fe(CN)6]) keeps the complex ion [Fe(CN)6]4− intact and does not give free Fe2+ and CN−.
Q2. Why does the ‘g’ subscript appear in t2g/eg for octahedral but not for tetrahedral complexes?
ANSWERThe subscript g (gerade) denotes a centre of symmetry. Octahedral (and square-planar) complexes have a centre of symmetry, so g is used; tetrahedral complexes lack a centre of symmetry, so the labels (e and t2) carry no g.
Q3. Calculate the spin-only magnetic moment of [MnBr4]2− and predict its geometry.
ANSWERMn is +2 (d5); Br− is weak field, so 5 unpaired electrons. μ = √[5(7)] = √35 ≈ 5.92 BM. A magnetic moment near 5.9 BM corresponds to five unpaired electrons, which is consistent only with a tetrahedral (sp3) shape, not square planar.
Q4. Why is CO a stronger field ligand than Cl−?
ANSWERCO is a neutral π-acceptor ligand that can both donate a σ lone pair and accept metal d-electron density into its π* orbital (synergic bonding). This strong metal–ligand interaction gives a large Δo. Cl− is a π-donor anion with weak interaction, giving a small Δo; hence CO lies far above Cl− in the spectrochemical series.
Q5. Name the type of isomerism shown by [Co(NH3)6][Cr(CN)6] and [Cr(NH3)6][Co(CN)6].
ANSWERCoordination isomerism — the ligands are interchanged between the cationic and anionic complex ions of the two different metals.
Long Answer Type Questions
Q1. Using VBT, explain the geometry and magnetic behaviour of [Co(NH3)6]3+ and [CoF6]3−.
ANSWERBoth have Co3+ (3d6). In [Co(NH3)6]3+, NH3 is a strong-field ligand: the six 3d electrons pair into the t2g set (t2g6), freeing two inner 3d orbitals. These with one 4s and three 4p orbitals give d2sp3 hybridisation — an octahedral, inner-orbital, low-spin complex with zero unpaired electrons, so it is diamagnetic. In [CoF6]3−, F− is weak-field and cannot pair the electrons; the metal uses outer 4d orbitals giving sp3d2 hybridisation — an octahedral, outer-orbital, high-spin complex with four unpaired electrons, so it is paramagnetic (μ ≈ 4.90 BM). The difference in magnetic behaviour for the same metal and oxidation state is explained by the difference in ligand field strength.
Q2. Explain crystal field splitting in octahedral and tetrahedral fields, and why low-spin tetrahedral complexes are rare.
ANSWERIn an octahedral field the six ligands approach along the axes. The eg orbitals (dx2−y2, dz2) point at the ligands and are raised by +(3/5)Δo; the t2g orbitals (dxy, dyz, dxz) point between the ligands and are lowered by −(2/5)Δo. In a tetrahedral field the splitting is inverted (the e set lies lower and the t2 set higher) and much smaller, Δt = (4/9)Δo. Because Δt is so small, it is almost always less than the pairing energy P, so electrons prefer to stay unpaired rather than pair; hence low-spin tetrahedral complexes are very rarely observed.
Q3. Describe the postulates of Werner’s theory and how they explain the behaviour of the CoCl3·nNH3 series with AgNO3.
ANSWERWerner proposed primary (ionisable) and secondary (non-ionisable, fixed = coordination number, with definite geometry) valences. For cobalt the secondary valence is 6 (octahedral). In CoCl3·6NH3 = [Co(NH3)6]Cl3, all three Cl− are outside the sphere (ionisable), so all 3 are precipitated as AgCl (1:3 electrolyte). In CoCl3·5NH3 = [CoCl(NH3)5]Cl2, one Cl− is coordinated, so only 2 AgCl precipitate (1:2 electrolyte). In CoCl3·4NH3 = [CoCl2(NH3)4]Cl, two Cl− are coordinated, so only 1 AgCl precipitates (1:1 electrolyte). The amount of AgCl precipitated thus matches the number of ionisable (primary-valence) chloride ions, exactly as Werner predicted; the two CoCl3·4NH3 forms (green and violet) being geometrical isomers further supported his idea of fixed spatial geometry.
MCQs & Assertion–Reason
1. The coordination number of cobalt in [Co(en)3]3+ is:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: [Fe(CN)6]4− is diamagnetic.
Reason: CN− is a strong-field ligand that pairs all the d electrons of Fe2+.
A-R 2. Assertion: [Ni(CO)4] and [NiCl4]2− are both tetrahedral and both paramagnetic.
Reason: In both, nickel is in the +2 oxidation state.
A-R 3. Assertion: Chelate complexes are more stable than complexes of unidentate ligands.
Reason: Ring formation by a didentate/polydentate ligand gives extra (chelate-effect) stability.
A-R 4. Assertion: Square-planar complexes of the type [Mabcd] do not show optical isomerism.
Reason: A square-planar complex has a plane of symmetry (the molecular plane).
A-R 5. Assertion: [Ti(H2O)6]3+ is coloured.
Reason: A d–d transition of its single d electron absorbs light in the visible region.
Answer key: 1-(A), 2-(C), 3-(A), 4-(A), 5-(A).
Common Mistakes & Exam Tips
Common mistakes to avoid
Forgetting that ligands are named in alphabetical order (ignoring multiplying prefixes like di-, tri-) but written after the metal in the formula.
Adding the suffix -ate to the metal only when the complex ion is an anion (e.g. ferrate, cuprate) — not for cationic/neutral complexes.
Counting π bonds towards coordination number — only σ bonds (donor atoms) count.
Treating C2O42− or en as one donor — each is didentate, so it contributes 2 to the coordination number.
Mixing up Δo with wavelength: larger Δo ⇒ shorter wavelength absorbed (higher energy).
Assuming tetrahedral complexes show geometrical isomerism — they do not.
How to score full marks in this chapter
Always show the oxidation-state calculation explicitly (charge of complex = OS of metal + sum of ligand charges). For magnetic-moment questions, write the metal’s dn, decide high-/low-spin from the ligand’s place in the spectrochemical series, count unpaired electrons, then apply μ = √[n(n + 2)] BM. Learn the spectrochemical series and the hybridisation–geometry table by heart, and remember the standard examples (chlorophyll, haemoglobin, vitamin B12, cis-platin, EDTA) for the “importance” question.
Frequently Asked Questions
What is Class 12 Chemistry Chapter 5 Coordination Compounds about?
It covers Werner’s theory, the basic terms (ligand, coordination number, coordination sphere), IUPAC nomenclature, the types of isomerism, and the bonding theories VBT and CFT, ending with metal carbonyls and the importance of coordination compounds in biology, medicine, analysis and metallurgy.
How do I decide whether a complex is high-spin or low-spin?
Compare the ligand’s field strength using the spectrochemical series. Strong-field ligands (CN−, CO, NH3, en) give Δo > P and form low-spin (inner-orbital, d2sp3) complexes; weak-field ligands (F−, Cl−, H2O) give Δo < P and form high-spin (outer-orbital, sp3d2) complexes.
How is the magnetic moment of a complex calculated?
Use the spin-only formula μ = √[n(n + 2)] BM, where n is the number of unpaired electrons. For example, n = 4 gives μ = √24 ≈ 4.90 BM and n = 1 gives μ = √3 ≈ 1.73 BM.
Are these Class 12 Chemistry Chapter 5 solutions free?
Yes. All ClearStudy NCERT Solutions for Class 12 Chemistry are free and follow the official NCERT textbook for session 2026–27.
