NCERT Solutions for Class 12 Chemistry Chapter 5: Coordination Compounds

Rules (IUPAC). Central metal first, then ligands in alphabetical order (irrespective of charge); coordination entity in square brackets; polyatomic ligands in parentheses; charge of the entity must balance the counter ions. The oxidation state tells the metal charge, from which the charges of cation/anion and the number of counter ions follow.

Conclusion: (i) [Co(NH₃)₄(H₂O)₂]Cl₃; (ii) K₂[Ni(CN)₄]; (iii) [Cr(en)₃]Cl₃; (iv) [Pt(NH₃)BrCl(NO₂)]⁻; (v) [PtCl₂(en)₂](NO₃)₂; (vi) Fe₄[Fe(CN)₆]₃.

Method. Name the cation first; in the complex, ligands alphabetically, then the metal (with –ate if the complex is an anion) and its oxidation state in Roman numerals (found from charge balance).

Conclusion: (i) hexaamminecobalt(III) chloride; (ii) pentaamminechloridocobalt(III) chloride; (iii) potassium hexacyanidoferrate(III); (iv) potassium trioxalatoferrate(III); (v) potassium tetrachloridopalladate(II); (vi) diamminechlorido(methanamine)platinum(II) chloride.

(i) K[Cr(H₂O)₂(C₂O₄)₂] — type [M(AA)₂b₂]⁻ with ox = AA: geometrical isomerism (cis and trans) and, for the cis form, optical isomerism (a pair of enantiomers). Total 3 stereoisomers: trans (achiral, the two H₂O opposite), cis-d and cis-l (the two H₂O adjacent).

(ii) [Co(en)₃]Cl₃ — optical isomerism only: the tris-chelate octahedral cation is chiral; two enantiomers (d and l). No geometrical isomers (all ligands identical).

(iii) [Co(NH₃)₅(NO₂)](NO₃)₂ — linkage isomerism because NO₂⁻ is ambidentate: [Co(NH₃)₅(NO₂)](NO₃)₂ (yellow, N-bonded nitro) and [Co(NH₃)₅(ONO)](NO₃)₂ (red, O-bonded nitrito). It can also show ionisation isomerism (the counter ion NO₃⁻ can exchange with a ligand, e.g. [Co(NH₃)₅(NO₃)](NO₂)(NO₃)). The NCERT answer key mentions ten possible isomers in all (geometrical, ionisation and linkage types combined).

(iv) [Pt(NH₃)(H₂O)Cl₂] — square planar of the type MABX₂: geometrical isomerism (cis and trans with respect to the two Cl).

Conclusion: (i) geometrical + optical; (ii) optical; (iii) linkage (and ionisation); (iv) geometrical (cis/trans).

Concept. Ionisation isomers have the same formula but give different ions in solution because the ligand and the counter ion are interchanged. Chemical tests with reagents specific for the free ions distinguish them.

[Co(NH₃)₅Cl]SO₄ + Ba²⁺ → BaSO₄(s); [Co(NH₃)₅(SO₄)]Cl + Ag⁺ → AgCl(s)

Also, their conductivities and the colours of the solutions differ (the first is violet, the second red), confirming different structures although both are 1:1 electrolytes of the same composition.

Conclusion: The two compounds give different precipitates with BaCl₂ and AgNO₃, proving that they are ionisation isomers.

Ni²⁺ configuration. Ni (Z = 28) = [Ar]3d⁸4s²; Ni²⁺ = [Ar]3d⁸ (two unpaired electrons in the free ion).

[Ni(CN)₄]²⁻ (square planar). CN⁻ is a strong-field ligand: it forces the two unpaired 3d electrons to pair up, vacating one 3d orbital. One 3d, the 4s and two 4p orbitals undergo dsp² hybridisation giving four equivalent hybrid orbitals directed to the corners of a square. Four CN⁻ donate lone pairs into them. All electrons are paired (3d⁸ → eight electrons in four 3d orbitals), so the complex is diamagnetic (low spin).

[NiCl₄]²⁻ (tetrahedral). Cl⁻ is a weak-field ligand and cannot pair the 3d electrons. One 4s and three 4p orbitals undergo sp³ hybridisation; the four Cl⁻ donate pairs; the 3d⁸ configuration with two unpaired electrons remains — so the complex is paramagnetic (μ = √8 = 2.83 BM).

Conclusion: [Ni(CN)₄]²⁻ is dsp² (no unpaired electrons, diamagnetic); [NiCl₄]²⁻ is sp³ (two unpaired electrons, paramagnetic).

Oxidation states. In [NiCl₄]²⁻ nickel is Ni(II): 3d⁸ with two unpaired electrons; Cl⁻ is a weak-field ligand and cannot pair them, so the complex (sp³) is paramagnetic.

[Ni(CO)₄]. Nickel is in the zero oxidation state, Ni(0) = 3d⁸4s². CO is a very strong-field ligand: the two 4s electrons shift into 3d (3d¹⁰), the ten 3d electrons are all paired, and the vacant 4s and three 4p orbitals undergo sp³ hybridisation — tetrahedral geometry. With no unpaired electron the complex is diamagnetic.

Conclusion: In the presence of the strong-field CO the unpaired d electrons of Ni(0) pair up (diamagnetic), whereas the weak Cl⁻ cannot pair those of Ni²⁺ (paramagnetic).

Fe³⁺ = [Ar]3d⁵ (five unpaired electrons in the free ion).

[Fe(H₂O)₆]³⁺. H₂O is a weak-field ligand: it cannot pair the 3d electrons, so the hybridisation uses the outer 4d orbitals — sp³d² (outer orbital, high-spin complex) — leaving five unpaired electrons (μ = √35 = 5.92 BM): strongly paramagnetic.

[Fe(CN)₆]³⁻. CN⁻ is a strong-field ligand: the five 3d electrons pair up in three orbitals (t₂g⁵), vacating two 3d orbitals for d²sp³ hybridisation (inner orbital, low-spin complex). Only one unpaired electron remains (μ = √3 = 1.73 BM): weakly paramagnetic.

Conclusion: H₂O gives a high-spin sp³d² complex with five unpaired electrons; CN⁻ gives a low-spin d²sp³ complex with one.

[Co(NH₃)₆]³⁺. Co³⁺ = 3d⁶. NH₃ is a strong-field ligand, so the six 3d electrons pair up in three 3d orbitals, leaving two 3d orbitals vacant. These two 3d, one 4s and three 4p orbitals form six d²sp³ hybrid orbitals: an inner orbital (low spin) complex, diamagnetic.

[Ni(NH₃)₆]²⁺. Ni²⁺ = 3d⁸. Even though NH₃ is strong, eight electrons occupy all five 3d orbitals (three pairs and two unpaired); pairing cannot free two 3d orbitals for d²sp³. Hence the outer 4d orbitals are used: one 4s, three 4p and two 4d give sp³d² hybridisation — an outer orbital (high-spin) complex with two unpaired electrons (paramagnetic).

Conclusion: Co³⁺ (d⁶) frees two 3d orbitals for d²sp³ (inner); Ni²⁺ (d⁸) cannot, so it uses 4d orbitals — sp³d² (outer).

Pt²⁺ = [Xe]4f¹⁴5d⁸. A square planar complex requires dsp² hybridisation, i.e. one vacant (n−1)d orbital. The eight 5d electrons pair up in four of the five 5d orbitals (CN⁻ is a strong-field ligand), leaving one 5d orbital vacant; with the 6s and two 6p orbitals it forms four dsp² hybrids that receive the lone pairs of the four CN⁻.

Number of unpaired electrons in [Pt(CN)₄]²⁻ = 0 (diamagnetic)

Conclusion: [Pt(CN)₄]²⁻ has no unpaired electron (dsp², diamagnetic).

Mn²⁺ = 3d⁵. In an octahedral field the d orbitals split into t₂g (lower) and e_g (higher) with separation Δ_o. The electron distribution depends on Δ_o compared with the pairing energy P.

[Mn(H₂O)₆]²⁺. H₂O is a weak-field ligand: Δ_o < P, so electrons occupy all five orbitals singly before pairing — t₂g³ e_g² (high spin): five unpaired electrons.

[Mn(CN)₆]⁴⁻. CN⁻ is a strong-field ligand: Δ_o > P, so the electrons pair in the lower t₂g level — t₂g⁵ e_g⁰ (low spin): one unpaired electron.

Conclusion: H₂O (weak field) gives t₂g³e_g² with 5 unpaired electrons; CN⁻ (strong field) gives t₂g⁵ with 1 unpaired electron.

Werner’s postulates (1898).

1. In coordination compounds the metal shows two types of linkages (valences) — primary and secondary.
2. The primary valences are normally ionisable and are satisfied by negative ions (they correspond to the oxidation state of the metal).
3. The secondary valences are non-ionisable; they are satisfied by neutral molecules or negative ions; the secondary valence equals the coordination number and is fixed for a metal.
4. The groups bound by the secondary linkages have characteristic spatial arrangements corresponding to the coordination number (octahedral for six, tetrahedral or square planar for four).

Application — the cobalt(III) chloride–ammonia complexes (Table 5.1). Excess AgNO₃ precipitates 3, 2, 1, 1 mol of AgCl per mole of the yellow, purple, green and violet compounds respectively, showing that only the primary-valence Cl⁻ (outside the square bracket) are free; the secondary valence of Co(III) is six in all.

Thus a chloride ion may satisfy both primary and secondary valences. The octahedral shape of the six secondary bonds explains the existence of the two isomers (green = trans, violet = cis) of [CoCl₂(NH₃)₄]⁺. Werner’s theory is the modern distinction between ionic (primary) and covalent/coordinate (secondary) bonds.

Conclusion: Primary valence = oxidation state (ionisable); secondary valence = coordination number (non-ionisable, directional) — explaining the Co(III)–NH₃–Cl⁻ complexes.

FeSO₄ + (NH₄)₂SO₄ (1:1). The product, FeSO₄·(NH₄)₂SO₄·6H₂O (Mohr’s salt), is a double salt. Double salts dissociate completely into their simple ions in water: Fe²⁺, NH₄⁺ and SO₄²⁻. Hence free Fe²⁺ ions are present and give their characteristic tests (e.g. blue precipitate with K₃[Fe(CN)₆]).

FeSO₄·(NH₄)₂SO₄·6H₂O → Fe²⁺ + 2NH₄⁺ + 2SO₄²⁻ + 6H₂O

CuSO₄ + 4NH₃. Ammonia is a strong ligand and forms the stable complex tetraamminecopper(II) sulphate:

CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄ (deep blue)

The complex ion [Cu(NH₃)₄]²⁺ does not dissociate appreciably to give Cu²⁺ ions (very high stability constant); the Cu²⁺ is bound by four coordinate bonds in the coordination sphere. The concentration of free Cu²⁺ is too small to give the usual Cu²⁺ tests (e.g. with H₂S).

Conclusion: Mohr’s salt is a double salt giving free Fe²⁺; [Cu(NH₃)₄]SO₄ is a complex whose [Cu(NH₃)₄]²⁺ does not release Cu²⁺.

Conclusion: Definitions with examples are tabulated above.

Ambidentate ligands give rise to linkage isomerism, e.g. [Co(NH₃)₅(NO₂)]²⁺ (yellow) and [Co(NH₃)₅(ONO)]²⁺ (red). Polydentate ligands (e.g. N(CH₂CH₂NH₂)₃, hexadentate EDTA⁴⁻) have more than two donor atoms.

Conclusion: Unidentate: Cl⁻, NH₃; didentate: en, C₂O₄²⁻; ambidentate: NO₂⁻, SCN⁻.

Method. Oxidation number of the metal = charge of the entity − sum of the charges of the ligands (neutral ligands: H₂O, en, NH₃ = 0; Cl⁻, Br⁻, CN⁻ = −1).

Conclusion: (i) +3; (ii) +3; (iii) +2; (iv) +3; (v) +3.

Conclusion: See the formulas in the table above.

Conclusion: See the names in the table above.

Conclusion: Geometrical, optical, linkage, coordination, ionisation and solvate isomerism, with the examples above.

(i) [Cr(C₂O₄)₃]³⁻. Three identical didentate ligands: M(AA)₃ — no geometrical isomers (only one arrangement of three symmetric chelate rings). It does show optical isomerism (two enantiomers, d and l).

(ii) [Co(NH₃)₃Cl₃]. Octahedral type [Ma₃b₃]: two geometrical isomers —

• fac (facial): the three NH₃ (and three Cl) occupy the corners of one triangular face of the octahedron;
• mer (meridional): the three NH₃ lie in a plane through the metal (a meridian of the octahedron), two of them trans to each other.

Conclusion: (i) None (only optical isomers); (ii) two (fac and mer).

Concept. Optical isomers are non-superimposable mirror images (enantiomers); they occur when the complex has no plane or centre of symmetry. Red arcs below represent the chelating ligands (ox or en) spanning two cis positions.

(i) [Cr(C₂O₄)₃]³⁻ — three oxalate chelate rings arranged like a three-bladed propeller; the two enantiomers are mirror images (left- and right-handed propellers).

(ii) cis-[PtCl₂(en)₂]²⁺ — two en ligands and two Cl at cis positions; the cis form is chiral (the trans form has a mirror plane and is optically inactive).

(iii) [Cr(NH₃)₂Cl₂(en)]⁺ — of the three geometrical isomers [Cr(AA)a₂b₂] only the isomer in which each of the monodentate ligands is trans to a different kind of group (NH₃ trans to Cl, NH₃ trans to N of en, Cl trans to N of en) is chiral; it exists as a pair of enantiomers:

Conclusion: All three entities form enantiomeric pairs; for (ii) it is the cis isomer and for (iii) the isomer with all three kinds of trans pairs different.

(i) [CoCl₂(en)₂]⁺ (type M(AA)₂b₂): 3 stereoisomers — trans (achiral) and cis (a pair of enantiomers).

(ii) [Co(NH₃)Cl(en)₂]²⁺ (type M(AA)₂ab): 3 stereoisomers — trans (achiral; NH₃ and Cl opposite) and cis (pair of enantiomers).

(iii) [Co(NH₃)₂Cl₂(en)]⁺ (type M(AA)a₂b₂): 3 geometrical isomers (A: NH₃ trans; B: Cl trans; C: all cis), of which C is chiral → 4 stereoisomers in total.

Conclusion: (i) 3 stereoisomers (trans + cis d/l); (ii) 3 (trans + cis d/l); (iii) 4 (two achiral geometrical isomers + one chiral pair).

Type. Square planar [MABCD] (all four ligands different) has three geometrical isomers. Fix NH₃ at one position and place each of the other three ligands trans to it in turn:

Optical isomerism. A square planar complex lies in a plane that is itself a mirror plane (the plane of the molecule), so its mirror image is superimposable on it. None of the three isomers is optically active.

Conclusion: Three geometrical isomers; none shows optical isomerism.

Species in solution. Aqueous CuSO₄ contains the blue complex ion [Cu(H₂O)₄]²⁺ (Cu²⁺ is d⁹; the colour is due to a d–d transition, which absorbs red light).

(i) With KF. The ligand F⁻ replaces the water ligands to give a fluoro-complex of Cu(II). The new ligand set changes the crystal-field splitting Δ, so a different wavelength is absorbed and the compound is green (and sparingly soluble, so it separates as a green precipitate):

[Cu(H₂O)₄]²⁺ + 4F⁻ → [CuF₄]²⁻ + 4H₂O (green)

(ii) With KCl. Cl⁻ similarly substitutes H₂O to give the chloro-complex, which is bright green and remains in solution:

[Cu(H₂O)₄]²⁺ + 4Cl⁻ → [CuCl₄]²⁻ + 4H₂O (bright green)

Both F⁻ and Cl⁻ are weak-field ligands but different from H₂O in the spectrochemical series, so the absorption shifts and the transmitted (observed) colour changes from blue to green.

Conclusion: Substitution of H₂O by F⁻ or Cl⁻ changes the ligand field (Δ) and hence the colour: green [CuF₄]²⁻ and bright green [CuCl₄]²⁻.

Coordination entity. CN⁻ is a very strong ligand. With excess KCN, copper(II) sulphate gives the highly stable cyanido complex; in the NCERT answer it is written as [Cu(CN)₄]²⁻:

Cu²⁺(aq) + 4CN⁻(aq) → [Cu(CN)₄]²⁻(aq)

(In practice CN⁻ also reduces Cu(II) to Cu(I), and the stable species formed in excess KCN is [Cu(CN)₄]³⁻ with cyanogen, (CN)₂, evolved; the argument below is the same for either species.)

Why no CuS. The cyanido complex is extremely stable (very large formation constant), so it releases hardly any free Cu²⁺ ions into solution. When H₂S is passed, the ionic product [Cu²⁺][S²⁻] does not reach the very small solubility product of CuS (CuS is extremely insoluble — the textbook uses K_sp = 6 × 10⁻¹⁶ in Ex. 1.27 of Unit 1), so no black precipitate of CuS forms.

Conclusion: The entity is the stable cyanido complex of copper ([Cu(CN)₄]²⁻ in the NCERT key); its negligible dissociation gives too little Cu²⁺ to precipitate CuS.

Explanation. (i) The six 3d electrons of Fe²⁺ are paired in three orbitals by the strong CN⁻, vacating two 3d orbitals which, with 4s and 4p, accommodate six lone pairs (d²sp³). (ii) Weak F⁻ cannot pair the five unpaired 3d electrons of Fe³⁺, so the outer 4d orbitals are used (sp³d²). (iii) Oxalate pairs the six d electrons of Co³⁺ (d²sp³): diamagnetic. (iv) Weak F⁻ leaves four unpaired electrons in Co³⁺ and uses 4d orbitals (sp³d²).

Conclusion: (i) d²sp³ diamagnetic; (ii) sp³d², 5 unpaired; (iii) d²sp³ diamagnetic; (iv) sp³d², 4 unpaired.

Explanation. In an isolated gaseous metal ion the five d orbitals are degenerate. In a spherically symmetrical negative field they rise together (barycentre). In an octahedral field the six ligands approach along the ±x, ±y, ±z axes: the orbitals d_x²−y² and d_z², directed towards the ligands, experience greater repulsion and are raised (e_g, +3/5 Δ_o); the orbitals d_xy, d_yz and d_zx, pointing between the axes, are lowered (t₂g, −2/5 Δ_o). The energy separation is Δ_o, the crystal field splitting energy.

Energy of e_g = +(3/5)Δ_o ; energy of t₂g = −(2/5)Δ_o ; 2 × (3/5) = 3 × (2/5)

Conclusion: In an octahedral field the d orbitals split into a lower t₂g set (three) and a higher e_g set (two) separated by Δ_o.

Spectrochemical series. The arrangement of ligands in order of increasing crystal-field splitting (field strength), established experimentally from the absorption of light by their complexes:

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO

Conclusion: The spectrochemical series ranks ligands by Δ_o; weak-field ligands (Δ_o < P) give high-spin and strong-field ligands (Δ_o > P) give low-spin complexes.

Crystal field splitting energy (Δ_o). The energy difference between the e_g and t₂g sets of d orbitals produced when ligands approach the metal ion in an octahedral arrangement (e_g raised by 3/5 Δ_o, t₂g lowered by 2/5 Δ_o). It depends on the ligand (spectrochemical series), the charge on the metal ion and the metal.

Role of Δ_o. For d¹–d³ the electrons enter t₂g singly (Hund’s rule). For d⁴–d⁷ the electron configuration depends on the competition between Δ_o and the pairing energy P:

• If Δ_o < P (weak-field ligand): the fourth electron enters e_g — t₂g³e_g¹ (high spin).
• If Δ_o > P (strong-field ligand): the fourth electron pairs in t₂g — t₂g⁴e_g⁰ (low spin).

For d⁸–d¹⁰ only one arrangement is possible. The number of unpaired electrons, and hence the magnetic moment, is thus decided by Δ_o.

Conclusion: Δ_o is the t₂g–e_g gap; Δ_o < P gives high spin, Δ_o > P gives low spin for d⁴–d⁷ ions.

[Cr(NH₃)₆]³⁺. Cr³⁺ = 3d³. With six NH₃ the complex is octahedral, d²sp³ hybridised, and the three 3d electrons occupy the three t₂g orbitals singly (t₂g³, Hund’s rule). Hence there are three unpaired electrons (μ = √15 = 3.87 BM): paramagnetic.

[Ni(CN)₄]²⁻. Ni²⁺ = 3d⁸. CN⁻ is a strong-field ligand and pairs the electrons, vacating one 3d orbital for dsp² hybridisation — square planar geometry. All the electrons are paired: no unpaired electron, so the complex is diamagnetic.

Conclusion: Cr³⁺ (d³) retains three unpaired electrons; Ni²⁺ (d⁸) with strong CN⁻ becomes dsp² with all electrons paired.

[Ni(H₂O)₆]²⁺. Ni²⁺ is d⁸ (t₂g⁶e_g²) with two unpaired electrons. H₂O is a weak-field ligand and Δ_o corresponds to visible light: an electron in a t₂g orbital is excited to an e_g orbital by absorbing light in the red region (d–d transition). The complementary colour, green, is transmitted.

[Ni(CN)₄]²⁻. CN⁻ is a strong-field ligand: the complex is square planar (dsp²) with all electrons paired and the d-orbital splitting is so large that the energy needed for a d–d transition falls in the ultraviolet region; no visible light is absorbed, so the solution is colourless.

Conclusion: [Ni(H₂O)₆]²⁺ absorbs visible (red) light by d–d transition and looks green; [Ni(CN)₄]²⁻ has no visible d–d absorption and is colourless.

Same metal ion, different ligands. In both complexes Fe²⁺ is d⁶, but the ligands differ in field strength: CN⁻ is a strong-field ligand and H₂O a weak-field ligand. The crystal-field splitting Δ_o is therefore much larger in [Fe(CN)₆]⁴⁻ than in [Fe(H₂O)₆]²⁺.

The d–d transition (t₂g → e_g) in each ion needs energy equal to Δ_o, so each complex absorbs light of a different wavelength — the larger Δ_o of the cyanide complex corresponds to shorter wavelength absorption. The complementary colours transmitted (observed) are therefore different (the aqua ion is pale green, the cyanido complex is pale yellow in dilute solution).

Conclusion: The different ligand field strengths (CN⁻ ≫ H₂O) give different Δ_o and hence different light absorption and colour.

Metal carbonyls are homoleptic complexes of CO, formed by most transition metals — [Ni(CO)₄] (tetrahedral), [Fe(CO)₅] (trigonal bipyramidal), [Cr(CO)₆] (octahedral), [Mn₂(CO)₁₀] (Mn–Mn bond) and [Co₂(CO)₈] (Co–Co bond with two bridging CO).

Bonding. The metal–carbon bond has both σ and π character:

1. M ← C σ bond: the lone pair of electrons on the carbon of CO is donated into a vacant orbital of the metal.
2. M → C π bond (back-bonding): a pair of electrons from a filled d orbital of the metal is donated into a vacant antibonding π* orbital of CO.

The two interactions reinforce each other: σ-donation increases the electron density on the metal, which favours π back-donation; back-donation in turn removes negative charge from the metal and makes the CO a better σ-donor. This synergic effect strengthens the M–CO bond (and slightly weakens the C≡O bond), giving the carbonyls their stability.

Conclusion: M–CO bonding is synergic: σ donation from CO to metal plus π back-donation from metal to CO π* strengthens the bond.

Conclusion: (i) +3, t₂g⁶, 6; (ii) +3, t₂g³, 6; (iii) +2, e⁴t₂³, 4; (iv) +2, t₂g³e_g², 6.

The magnetic moments are the spin-only values √[n(n + 2)] BM. For (v) the observed value is about 2.2 BM (see Ex. 4.38 of Unit 4) — close to one unpaired electron.

Conclusion: See the table for names, oxidation states, d-configurations, coordination numbers, geometries and magnetic moments.

Configuration. Ti³⁺ is a 3d¹ system. In the octahedral complex [Ti(H₂O)₆]³⁺ the t₂g and e_g levels are split by Δ_o; the single d electron is in the lower t₂g level in the ground state (t₂g¹e_g⁰).

Excitation. The complex absorbs light of energy equal to Δ_o: the blue-green part of the visible spectrum (≈ 498 nm, Table 5.3), which excites the electron from t₂g to e_g (t₂g¹e_g⁰ → t₂g⁰e_g¹). The light that is transmitted is the complementary colour — violet (Fig 5.10).

Without ligands (e.g. dehydrated [Ti(H₂O)₆]Cl₃ on heating) there is no splitting and the compound is colourless.

Conclusion: Violet colour arises from the d–d transition (t₂g → e_g) of Ti³⁺ (3d¹) absorbing blue-green light.

Chelate effect. A complex containing a chelate ligand (a didentate or polydentate ligand that forms a ring by binding through two or more donor atoms) is more stable than a similar complex containing unidentate ligands. This extra stability is called the chelate effect.

Reason. Replacing several unidentate ligands (e.g. six NH₃) by chelate ligands (three en) releases more free ligand molecules, which increases the entropy of the system (ΔS more positive), and the ring structure reduces the chance that the ligand is completely lost (once one end is detached the other end keeps it nearby, so it re-attaches).

[Ni(H₂O)₆]²⁺ + 3en ⇌ [Ni(en)₃]²⁺ + 6H₂O (the equilibrium lies far to the right; the number of particles rises from 4 to 7)

Example. [Co(en)₃]³⁺ and [Ni(en)₃]²⁺ (with five-membered rings M–N–C–C–N) are more stable than [Co(NH₃)₆]³⁺ and [Ni(NH₃)₆]²⁺ respectively. The stability is greatest for five- and six-membered rings; EDTA⁴⁻, a hexadentate ligand, gives particularly stable chelates (used for estimating water hardness).

Conclusion: The chelate effect is the extra stability of complexes with ring-forming (chelating) ligands, e.g. [Ni(en)₃]²⁺ is more stable than [Ni(NH₃)₆]²⁺.

Other applications: catalysts such as Wilkinson’s catalyst [(Ph₃P)₃RhCl] (hydrogenation of alkenes); electroplating of Ag and Au from [Ag(CN)₂]⁻ and [Au(CN)₂]⁻; photography (hypo dissolves AgBr as [Ag(S₂O₃)₂]³⁻).

Conclusion: Coordination compounds are essential in biology, medicine, analysis and metallurgy, with the examples above.

Formulation. The compound is [Co(NH₃)₆]Cl₂: the coordination entity [Co(NH₃)₆]²⁺ stays intact in solution (secondary valence), and the two chloride counter ions ionise.

[Co(NH₃)₆]Cl₂ → [Co(NH₃)₆]²⁺ + 2Cl⁻ (total 3 ions; a 1:2 electrolyte)

Conclusion: Option (iii): 3 ions are produced.

Conclusion: Option (ii) [Fe(H₂O)₆]²⁺ has the highest magnetic moment (4.90 BM).

Reasoning. All four contain Fe³⁺ with six donor atoms, but [Fe(C₂O₄)₃]³⁻ contains three didentate (chelating) oxalate ligands forming three five-membered rings. By the chelate effect, chelate complexes are much more stable than complexes of unidentate ligands (H₂O, NH₃, Cl⁻).

Conclusion: Option (iii) [Fe(C₂O₄)₃]³⁻ is the most stable complex.

Concept. The energy of light absorbed corresponds to the splitting Δ_o (ΔE = hc/λ): the stronger the ligand field, the larger Δ_o and the shorter the wavelength absorbed.

Ligand field strength (spectrochemical series). NO₂⁻ (N-bonded nitro) > NH₃ > H₂O, hence Δ_o: [Ni(NO₂)₆]⁴⁻ > [Ni(NH₃)₆]²⁺ > [Ni(H₂O)₆]²⁺.

λ(absorbed): [Ni(NO₂)₆]⁴⁻ < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Conclusion: Increasing wavelength of absorption: [Ni(NO₂)₆]⁴⁻ < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺.