Class 12 Chemistry Ch 5 Coordination Compounds - Download PDF

Strategic angle. Frame Werner's theory as the answer to one experimental fact: the conductivity / silver-nitrate test for the four cobalt-ammine chlorides. Once we explain the 1:3, 1:2, 1:1 and 0 ion ratios, the whole theory follows. The four postulates are not arbitrary axioms; they are the simplest set of rules that reproduces every laboratory observation.

• CoCl3.6NH3 gives 3 Cl-, 1 cation: 4 ions total (luteo, yellow).
• CoCl3.5NH3 gives 2 Cl-, 1 cation: 3 ions (purpureo, purple).
• CoCl3.4NH3 gives 1 Cl-, 1 cation: 2 ions (praseo green / violeo violet).
• CoCl3.3NH3 gives a non-electrolyte: 0 ions.

1. Werner asked: how can three Cl- disappear in stages while the cobalt stays Co³⁺? Answer: chlorides can enter a second, non-ionising sphere around the metal. The question itself is a clue: conductivity in water drops in unit steps as ammonia is replaced by chloride one-for-one.
2. He proposed two valencies. The primary (ionising) valency is 3 in every case (Co³⁺). The secondary (directional) valency is always 6 for Co³⁺. The two numbers are independent and can be satisfied by different species, the single deepest insight in the theory.
3. In [Co(NH3)6]Cl3 all six secondary valencies are taken by ammonia; the three chlorides sit outside as primary valencies and ionise. Molar conductivity _m ≈ 430 S cm² mol^-1 (a 1:3 electrolyte).
4. In [Co(NH3)5Cl]Cl2 one chloride sneaks inside the sphere, satisfying one secondary valency and one primary valency; only two chlorides remain outside. _m ≈ 260, a 1:2 electrolyte.
5. Continuing this logic gives [Co(NH3)4Cl2]Cl (_m ≈ 100, 1:1) and the non-electrolyte [Co(NH3)3Cl3] (_m ≈ 0). The model matches experiment exactly.
6. Alternative approach (geometric). The fixed secondary valency of 6 on Co³⁺ predicts an octahedron. From an octahedron with two Cl- ligands we get exactly two arrangements (cis, trans); both [Co(NH3)4Cl2]Cl isomers (violet and green) were isolated, again confirming Werner's geometric prediction.

Why this matters. The same primary/secondary split is the seed of the modern coordination number, oxidation state and coordination sphere vocabulary used in every remaining question of this chapter. Werner won the 1913 Nobel Prize for what is essentially a counting argument applied to a handful of cobalt-ammine salts.

Concept linkage. Modern bonding theories (VBT, CFT, MOT, Q 5.15–5.18) all start from Werner's coordination sphere. Without the sphere idea, the very ligand-field calculations we use to predict _o, magnetic moment and colour have nothing to ``split''. The chelate effect (Q 5.26) is an entropy argument that runs entirely inside Werner's secondary sphere.

Werner's postulates: two valency types (primary = ionisable, secondary = directional and fixed); the secondary defines the geometry and is today's coordination number.

Quick reading. Two salts; one falls apart, the other does not. The first is a mixture of crystals (a double salt); the second hides the copper inside a complex ion with four NH3 ligands. The acid test is whether the metal-ligand bond survives dissolution.

1. For Mohr's salt, write the dissolution as four free ion types: Fe²⁺, NH4⁺, SO4^2- and water. Free Fe²⁺ reacts with potassium ferricyanide K3[Fe(CN)6] to give the deep blue Turnbull's precipitate. Equivalently, with KSCN a pale colour is seen (Fe^2+ gives no red, Fe³⁺ does).
2. For the copper-ammonia mixture, ammonia is a moderately strong field ligand. Four ligands push into the four coordination sites of Cu²⁺ to give a square-planar complex [Cu(NH3)4]²⁺ with log₄ ≈ 12. Visually: the dirty-blue precipitate of Cu(OH)2 that forms with a few drops of ammonia redissolves as more ammonia is added, giving the famous deep blue Schweizer's reagent.
3. Because ₄ is so large, [Cu²⁺]_free ≈ [Cu(NH3)4²⁺]/₄ · [NH3]^-4 is negligible at any practical NH3 concentration. None of the usual tests for Cu²⁺ work.
4. Sanity check: passing H2S through the deep blue solution gives no black CuS precipitate, confirming absence of free Cu²⁺. Compare with K_sp(CuS) ≈ 10^-36: an enormous solubility product, yet not crossed because [Cu²⁺]_free is well below 10^-14 M inside the complex.
5. Numerical cross-check. Take 0.1 M [Cu(NH3)4]^2+ with [NH3]_free ≈ 0.1 M. Then [Cu²⁺]_free = 0.110¹² · (0.1)⁴ ≈ 10^-9 M. Below the precipitation threshold for CuS (∼ 10^-18 M in saturated H2S at neutral pH). Hence no CuS.

Why this matters. The contrast between double salts and complexes is the cleanest entry point into coordination chemistry. The whole rest of the chapter (Werner spheres, isomerism, CFT, the chelate effect) is the elaboration of one idea: a metal ion that is not free behaves differently from one that is.

Concept linkage. The stability-constant comparison (₄ ≈ 10¹²) is the prototype of the chelate effect discussion of Q 5.26 - except there a polydentate ligand adds further entropic stabilisation. The same logic explains Q 5.14 ([Cu(CN)4]^3- even more stable) and Q 5.30 (oxalato beats monodentate).

Mohr's salt: double salt, free ions, Fe²⁺ test positive. [Cu(NH3)4]SO4: complex, copper hidden inside the cation, Cu²⁺ test negative.

Structural observation. Each term either names a piece of the complex (entity, ligand) or describes its geometry (C.N., polyhedron) or the diversity of its donor set (homo- vs heteroleptic). The six terms are not isolated trivia - they are the orthogonal axes along which any complex can be classified.

1. Entity vs ligand. The entity is the whole thing inside the brackets ([M(L)_n]^q). The ligands are the satellites donating lone pairs to the metal. Together they carry the overall charge of the entity. Counter-ions sit outside the brackets and dissociate in solution (Werner's primary valencies, Q 5.1).
2. Coordination number. Count the donor atoms (not the ligands) attached to the metal: a didentate ligand counts twice. So [Co(en)3]³⁺ has C.N. 6 even though there are only three en ligands. Another fooler: [Pt(en)Cl2] has C.N. 4 (en counts 2 plus two chlorides), although there are only three ``ligands''.
3. Polyhedron. The shape is determined by C.N. and by the metal's d-electron count. C.N. 6 is almost always octahedral; C.N. 4 splits into tetrahedral (Ni(CO)4, [NiCl4]^2-) and square planar ([PtCl4]^2-, [Ni(CN)4]^2-). C.N. 5 is rare but real: Fe(CO)5 trigonal bipyramidal. C.N. 2: linear, [Ag(NH3)2]+. Roughly, d⁰, d¹⁰ and d⁵ high-spin metals prefer tetrahedral; d⁸ strong-field prefers square planar.
4. Homo- vs heteroleptic. Identify the donor atom (or the ligand name, since each ligand has one donor type unless ambidentate). One name in the formula ⇒ homoleptic; more than one ⇒ heteroleptic. The distinction matters because heteroleptic complexes routinely show geometrical isomerism (cis/trans, fac/mer), homoleptic ones with symmetric ligands do not.
5. Alternative classification. Some texts split further: cationic ([Co(NH3)6]³⁺), anionic ([Fe(CN)6]^4-), neutral ([Ni(CO)4]) entities. The same complex can be cationic with one counter-ion set and anionic with another (in salts: cation first, anion last).

Why this matters. These six words are used in every question that follows; absorbing them now makes the rest of the chapter a single coherent picture. Q 5.5 (oxidation state) and Q 5.6/5.7 (IUPAC names) lean directly on entity and ligand vocabulary. Q 5.9–5.12 (isomerism) lean on the homo/heteroleptic distinction. Q 5.15–5.16 (VBT, CFT) lean on coordination number and polyhedron.

Definitions and examples as in Steps 1–5.

Picture-first. Visualise each ligand as a hand: a unidentate ligand has one finger touching the metal, a didentate ligand grips with two fingers in the same molecule, and an ambidentate ligand has two different fingers but uses only one at a time. The denticity is the count of fingers actually touching.

1. Count lone pairs that actually bond. Ammonia has only one lone pair on N; chloride has lone pairs but only one binds at a time: both are unidentate. Carbon monoxide has a lone pair on C; despite having multiple lone pairs in total, only one binds, so CO is unidentate.
2. In en, both nitrogens donate to the same metal, and the -CH2-CH2- backbone is the right length to produce a strain-free 5-membered chelate ring (M-N-C-C-N). In oxalate, the two terminal oxygens both donate, again producing a 5-membered ring (M-O-C-C-O). Each ligand brings two donor atoms: didentate.
3. In NO2-, N has one lone pair, O has two. Either can donate, but never both to the same metal. This is the defining trait of ambidentate ligands. Hard metals (oxophilic, e.g. Co³⁺, Cr³⁺) bind O preferentially → nitrito-O; soft metals (e.g. Co³⁺ with strong-field set) often switch to nitro-N (Q 5.6).
4. Linkage isomers (Q 5.7 (x) and Q 5.6 (v)) come directly out of this dual binding mode. Two different colours, two different absorption maxima, same formula.
5. Alternative classification (denticity scale). Hexadentate ligands like EDTA bind through six donors (two N, four O), wrapping the metal in a cage. Such ligands give the most stable chelates and saturate a 6-C.N. site with a single molecule (preview of Q 5.26).
6. Numerical bookkeeping. If denticity of a ligand is k and it appears m times in a complex of C.N. 6, then k m = 6. Examples: k = 1, m = 6 for [Co(NH3)6]³⁺; k = 2, m = 3 for [Co(en)3]³⁺; k = 6, m = 1 for [Co(EDTA)]^-.

Why this matters. Denticity directly fixes the C.N. of the metal: pick three en ligands and you have C.N. 6 without needing six separate molecules. It also fixes the entropy of complex formation (more donor atoms in one ligand ⇒ fewer displaced waters ⇒ larger chelate effect, Q 5.26).

Concept linkage. The ambidentate concept connects directly to Q 5.6 (formulas for nitrito-O vs nitrito-N) and Q 5.8 (linkage isomerism). The didentate concept underlies almost every chiral coordination complex (Q 5.10, 5.11).

Unidentate, didentate, ambidentate distinguished by donor-atom count and binding mode; examples in Steps 1–3.

Strategic angle. Write every charge balance as one line and solve. Don't memorise individual answers: the method is a single linear equation. The structure of every step is the same: ``metal + sum of ligand charges = overall entity charge''.

1. Bring all known charges to one side, the unknown x to the other. The unknown is always the metal oxidation state. Ligand-charge cheat-sheet to keep handy: neutral NH3, H2O, en, py, CO; -1 for halides, CN-, OH-, SCN-, NO2-; -2 for oxalate C2O4^2-, O^2-, peroxide.
2. Apply to each case in turn:
   (i) ammine and en are neutral, one CN- (-1); aqua neutral. Sum of ligand charges = -1. So x - 1 = +2 ⇒ x = +3. (d⁶.)
   (ii) en neutral, two Br- (-2). x - 2 = +1 ⇒ x = +3. (d⁶.)
   (iii) four Cl- (-4). x - 4 = -2 ⇒ x = +2. (d⁸.)
   (iv) charge on complex anion = -(+3) = -3 (from three K+); six CN- (-6). x - 6 = -3 ⇒ x = +3. (d⁵.)
   (v) neutral entity; three Cl- (-3): x - 3 = 0 ⇒ x = +3. (d³.)
3. Confirm with periodic trends: Co³⁺, Pt²⁺, Fe³⁺, Cr³⁺ are all common, stable oxidation states. Co²⁺ also exists, but the ligand-set of (i) and (ii) (CN-, en, Br-) stabilises +3 by ligand-field arguments.
4. Alternative approach (group number). For first-row metals: dⁿ = (group number) - x. Co in group 9 with x = +3 gives d⁶ - consistent. Pt is a 5d⁸ metal; Pt²⁺ is 5d⁸ (square planar, diamagnetic). Fe in group 8, Fe³⁺ is 3d⁵ (half-filled, very stable).
5. Numerical cross-check via μ (peek ahead to Q 5.19). The dⁿ count predicts the spin-only μ = √n(n+2). If a published μ is available it immediately confirms whether your oxidation-state choice is right.

Why this matters. Every magnetic-moment, CFSE and colour calculation in this chapter starts with the metal oxidation state. Get this step wrong and everything downstream is wrong. The same formula handles JEE numerical questions (``find μ for [Mn(CN)6]^3-'' etc.) which always begin with oxidation-state determination.

Concept linkage. Q 5.6 / 5.7 (IUPAC names ↔ formulas) need exactly the same charge-balance arithmetic. Q 5.15 (VBT bonding) takes x and dⁿ as inputs.

(i) Co(III); (ii) Co(III); (iii) Pt(II); (iv) Fe(III); (v) Cr(III).

Strategic angle. The trick is to compute the charge on the complex ion first; then everything else (counter ions, formula brackets) falls into place. Treat each name as a charge-balance exercise rather than a vocabulary test. Four checks per name: oxidation state, ligands and their charges, overall complex charge, counter ion ratio.

1. For neutral complexes (e.g. (iii) cisplatin and the carbonyl-class), the sum of metal + ligand charges must equal 0. No counter ion required outside the brackets.
2. For ionic complexes with K as cation, the number of potassium ions is fixed by the negative charge on the complex. Examples: (ii) K2[PdCl4] (-2 complex, 2 K's); (iv) K2[Ni(CN)4]; (vii) K3[Cr(C2O4)3].
3. For ionic complexes with SO4^2- as anion, you may need a ratio: two cations per three sulphates in (vi), because the cation charge is +3 and the anion charge is -2; LCM gives [Co(NH3)6]2(SO4)3.
4. Linkage isomers (v) vs (x): write ONO for O-donor nitrito and NO2 for N-donor nitrito. Same formula otherwise. The same Co, same five ammines, same overall +2 charge - only the donor atom (and the IR N-O stretching frequency) tells them apart.
5. Alternative bracket convention. Some recent IUPAC recommendations always parenthesise the ambidentate donor atom inside the ligand name (nitrito-κN, nitrito-κO). The Class 12 NCERT uses nitrito-N and nitrito-O; both are acceptable in the board answer.
6. Special case for (i): zincate(II). Zn(II) is the prototypical case where the metal sits inside an anion because the ligand-set is anionic (OH-). The naming suffix -ate on zinc signals that the entity is anionic; without the suffix the same formula could mislead.

Why this matters. The reverse mapping (formula → name) in Q 5.7 uses exactly the same rules read backwards. The charge-balance arithmetic is identical to what powers dⁿ determination (Q 5.5) and downstream all of VBT, CFT, magnetic moment.

Concept linkage. Linkage isomers (v) vs (x) are the prototypical pair used in Q 5.8 to illustrate linkage isomerism quantitatively. Cisplatin in (iii) is the medicinally relevant example revisited in Q 5.27.

Formulas: (i) [Zn(OH)4]^2-; (ii) K2[PdCl4]; (iii) [Pt(NH3)2Cl2]; (iv) K2[Ni(CN)4]; (v) [Co(ONO)(NH3)5]²⁺; (vi) [Co(NH3)6]2(SO4)3; (vii) K3[Cr(C2O4)3]; (viii) [Pt(NH3)6]⁴⁺; (ix) [CuBr4]^2-; (x) [Co(NO2)(NH3)5]²⁺.

Quick reading. The recipe is mechanical: compute the oxidation state, alphabetise the ligands, slap on prefixes, finish with the metal and Roman numeral. Cation first if salt; -ate on the metal if the complex is anionic. Ligand-name suffix decides one letter; metal-name suffix decides the next.

1. For salts (i, ii, iv, vii), the outer counter ions fix the complex's charge. Use that to deduce the oxidation state. E.g. in (vii) [Ni(NH3)6]Cl2, the outer two Cl- give complex +2 ⇒ Ni(II).
2. For free complex ions (iii, v, vi, viii), the listed charge is the complex's charge. The suffix ion closes the name.
3. For neutral homoleptic carbonyl (ix), the metal must be in zero oxidation state since carbonyl is neutral. Roman ``0'' is written in parentheses after the metal name: ``nickel(0)''. Same idea applies to Fe(CO)5 (iron(0)) and Cr(CO)6 (chromium(0)).
4. Alphabetising by ligand name: ``ammine'' before ``chlorido'' before ``nitrito-N''; ``aqua'' is its own letter A; en (ethane-1,2-diamine) starts with e. ``Methyl'' in (ii) is part of the parent ligand name (methanamine) which starts with m, so it follows chlorido alphabetically.
5. Alternative classical names (in passing). ``Cyanido'' was ``cyano'' in the 1970 IUPAC rules; ``chlorido'' was ``chloro''. Both forms appear in older textbooks. The 2005 IUPAC recommendation is the -ido ending; that is what NCERT uses.
6. Cross-check via charge. After writing a name, run the reverse charge balance: ligand charges + metal oxidation-state Roman numeral should equal the complex charge (zero for neutral). Catches arithmetic slips.

Why this matters. Solid IUPAC fluency is the price of entry to research papers and reference tables: every compound in a catalogue is named this way. The reverse direction (name → formula, Q 5.6) and isomer-naming (cis-, trans-, fac-, mer-, Δ-, Λ-) build on the same vocabulary.

Concept linkage. The IUPAC name of (ii) diamminechlorido(methanamine)platinum(II) chloride foreshadows the parametric form [Mabcd] of Q 5.12 (three geometric isomers, no optical activity).

Nine names as in Steps 1–9.

Picture-first. Structural isomers are like different jigsaw fits (different atoms in different sockets). Stereoisomers are like left/right hands: same pieces, mirrored arrangement. Six types in total, falling neatly into two families.

1. Structural isomers: change which atoms are bonded. Four sub-types correspond to four ways atoms can swap places: donor atom of an ambidentate ligand (linkage), between cation and anion (coordination), between sphere and counter ion (ionisation), between sphere and lattice water (solvate). All four give two compounds with the same molecular formula but different connectivities.
2. Stereoisomers: same bonds, different geometry. Cis/trans (or fac/mer) for non-chiral, d/l (Δ/Λ) for chiral pairs. Geometric isomers are not mirror images of each other; optical isomers are.
3. Test for optical isomers: look for a plane of symmetry. If you can superimpose the molecule on its mirror image, it is achiral (not optically active). All-square-planar complexes have a σ-plane (the molecular plane) and therefore never show optical isomerism.
4. Alternative classification (modern). Some coordination chemists group ionisation and solvate isomers together as ``ionisation isomerism'' (since both interchange species across the sphere boundary). The Class 12 NCERT keeps them separate; the JEE syllabus follows NCERT.
5. Counting trick. A typical exam question asks ``How many possible isomers?'' Run two passes: (a) enumerate geometric arrangements, (b) for each, check chirality (mirror-plane test). The total is _g (1 + c_g) where c_g is 1 if geometry g is chiral, 0 otherwise. This single formula handles Q 5.10–5.12.
6. Sub-types with two-line definitions.
   • Linkage: same ligand binds via two different donor atoms. Ambidentate NO2-, SCN-.
   • Coordination: in a salt where both ions are complexes, the metals swap their ligand-sets.
   • Ionisation: counter ion exchanges with one ligand in the sphere; releases different anions in water.
   • Solvate: water moves into/out of the sphere; gives different colours.
   • Geometrical: cis/trans, fac/mer.
   • Optical: non-superimposable mirror images.

Why this matters. Q 5.9–5.12 work through specific examples of geometric and optical isomerism in detail. Beyond NCERT, the cisplatin / transplatin distinction (medicinal chemistry) and ferrocene / non-ferrocene chiral catalysts (asymmetric synthesis) are real-world applications of the same vocabulary.

Concept linkage. Linkage isomerism is the chemistry of ambidentate ligands (Q 5.4). Coordination, ionisation and solvate isomerism all rely on Werner's ``two-sphere'' description (Q 5.1). Optical isomerism connects to the chelate-ring discussion of Q 5.10/Q 5.26.

Six types of isomerism with one example each, as above.

Strategic angle. Decide whether the ligand set can be ``re-arranged'' to give a different connectivity-class. Three chelates of one kind: no, just one way. Three monodentates of one kind plus three of another: yes, two ways (fac vs mer). The classification is purely topological - it depends only on the ligand-set and the polyhedron, not on what the ligands are.

1. For (i): three identical didentate ligands occupy the six sites symmetrically. The geometry is unique up to a mirror reflection ⇒ optical isomers exist but geometrical isomers do not. The two enantiomers are labelled Δ (right-handed helix) and Λ (left-handed).
2. For (ii): with three A and three B ligands, count independent arrangements on an octahedron. Either the three A's share a face (fac) or they lie around an equator on a meridian (mer). Both isomers have a σ-plane, so no optical activity.
3. Note: in fac-isomer the three M-A bonds are all cis to one another (90^∘); in mer-isomer two pairs are cis and one pair is trans (180^∘). The angular difference is detectable spectroscopically (different IR/visible bands).
4. Alternative approach via Burnside / orbit counting. Formally, the number of geometric isomers of [MA₃B₃] equals the number of orbits of S₃-permutations of the three A's modulo the octahedral symmetry group. Counting gives exactly 2 (fac, mer), matching our intuition. Similarly, [M(AA)₃] has 1 geometry (and 2 enantiomers).
5. Numerical check. The C.N. of (i) is 3 × 2 = 6 (oxalate is didentate). The C.N. of (ii) is 3 + 3 = 6 (six monodentates). Both octahedral, both 6-coordinate - but their isomer counts differ entirely because of how the ligand-set sits on the octahedron.

Why this matters. fac/mer designation is the standard way of distinguishing isomers of [MA₃B₃] compounds in lab manuals and reference works. The optical Δ/Λ pair of [Cr(C2O4)3]^3- was the basis of Werner's 1914 paper proving octahedral geometry by resolving optical isomers.

Concept linkage. The same logic, applied to [M(AA)₂X₂] in Q 5.10/5.11, gives cis (chiral) and trans (achiral) geometries. The σ-plane test recurs in every optical-isomerism question of this chapter.

(i) 0 (geometrical); (ii) 2 (fac, mer).

Picture-first. Imagine a right-handed and a left-handed helix wrapped around the metal. If the mirror image can't be turned to coincide with the original, you have a chiral pair. The Δ/Λ labels capture the screw sense viewed down the C₃ rotation axis of the chelating rings.

1. (i) Three oxalates wrap as a triple-helix; mirror image is the opposite-sense helix. No internal symmetry plane. Both enantiomers have identical absorption spectra and magnetic moments - only the sign of their optical rotation differs. In a polarimeter, Δ-form rotates plane-polarised light clockwise (dextrorotatory) and Λ-form anti-clockwise.
2. (ii) Trans-Cl₂ has a mirror plane containing both Cl and bisecting the two en ligands; achiral. Cis-Cl₂ does not have such a plane; chiral. So [PtCl2(en)2]²⁺ has a total of three isomers: 1 trans and 2 cis enantiomers.
3. (iii) Of the possible geometric isomers of [M(NH3)2Cl2(en)]+, only the cis-cis form (both pairs of monodentates mutually cis) is chiral. The all-trans and trans-cis forms have a mirror plane. Total: 4 isomers (2 achiral + 2 chiral).
4. Alternative test for chirality (improper rotation). A molecule is achiral iff it has any improper rotation axis S_n (mirror σ = S₁, inversion i = S₂, etc.). For trans-Cl2(en)2, the molecular plane containing the two Cl's is a _h (S₁). For cis-Cl2(en)2, no S_n axis exists; hence chiral.
5. Numerical perspective. The optical rotation [α] is large for Δ/Λ helical complexes - often ± 1000^∘ to ± 4000^∘ at the sodium D-line for chiral cobalt-en complexes. This is what makes [Co(en)3]³⁺ such a clean teaching example.

Why this matters. Chiral metal complexes are widely used in asymmetric catalysis (BINAP-based catalysts in industrial drug synthesis). The 2001 Nobel Prize (Knowles, Noyori, Sharpless) went to the development of chiral metal complexes for enantioselective hydrogenation and oxidation - sitting directly on top of the same chirality concept used in this question.

Concept linkage. The Δ/Λ labels reappear in the resolution of [Co(en)3]³⁺ (Q 5.26) and in the prochiral logic of enzymatic catalysis. Q 5.11 systematises the counting algorithm for all possible isomers.

Three chiral pairs as in Steps 1–3.

Strategic angle. For each complex, list geometric forms, then test each for a mirror plane. Each chiral form contributes two enantiomers; each achiral form contributes one isomer. The recipe is algorithmic and works for any octahedral mixed-ligand complex.

1. (i) [CoCl2(en)2]+: two arrangements of Cl₂: trans (achiral, 1) and cis (chiral, 2). Total 1+2 = 3. The chiral cis pair is labelled cis-Δ and cis-Λ.
2. (ii) [Co(NH3)Cl(en)2]²⁺: two arrangements of unique NH3/Cl: trans (achiral, 1) and cis (chiral, 2). Total 1+2 = 3. The trans isomer has a C₂ axis through the unique pair, which also serves as a σ-plane (containing both en ligands).
3. (iii) [Co(NH3)2Cl2(en)]+: the en occupies cis sites by force (the chelate ring would otherwise stretch impossibly), leaving the four remaining vertices as two trans pairs. The two NH3 and two Cl can therefore only sit as (a) one trans pair each (trans-NH3, trans-Cl; achiral) or (b) all-cis with each NH3 trans to one Cl (chiral, Δ/Λ). Total 1 + 2 = 3 isomers.
4. Alternative perspective (group theory). The possible geometries of [M(AA)₂ X Y] form orbits under the octahedral group O_h. Orbit-counting agrees with the case-by-case analysis above. For (i), [M(AA)₂ X₂], the orbit calculation gives 1 (trans) + 1 (cis with two enantiomers) = 3.
5. Numerical check via permutations. For (i) with two identical en's and two identical Cl's on six octahedral vertices, naive count is 62/2 = 7.5, which is not integer - the symmetry must be quotiented out carefully. Doing so correctly yields 3 isomers. For (iii), 3 isomers; for (ii) 3 isomers. These small integers are signature of high octahedral symmetry.

Why this matters. Counting isomers correctly is a recurring exam task; the recipe is always (a) list geometric forms, (b) test each for chirality. This recipe applies in any octahedral case, and a variant works for square planar.

Concept linkage. The same enantiomer-counting algorithm re-appears in Q 5.12 for square-planar [Mabcd], where the answer is 3 geometric, 0 optical (the molecular plane is always a σ-plane).

(i) 3; (ii) 3; (iii) 3.

Picture-first. Place Pt at the centre of a square. Decide who sits opposite to whom; only the trans pairs matter. With four different ligands there are three ways to pair them up across the diagonals. The geometry of the complex (square planar) automatically contains a σ-plane (the molecular plane), so no isomer can be chiral.

1. Pair Br with Cl across one diagonal: NH3 and py occupy the other. Equivalent to writing the trans-pair as ``Br/Cl''.
2. Pair Br with NH3: Cl and py opposite (trans-Br/NH3).
3. Pair Br with py: NH3 and Cl opposite (trans-Br/py).
4. For optical activity, the complex must be chiral. Square plane has a horizontal mirror plane (the plane of the four atoms); no chirality. This is a general fact: every square-planar complex is achiral.
5. Alternative approach: count permutations modulo symmetry. For [Mabcd] on a square, the symmetry group is D_4h. Distinct ligand arrangements are orbits under this group; there are exactly 4! / 8 = 3 such orbits. Same answer, via group theory.
6. Numerical accent. The IR fingerprint distinguishes the three isomers cleanly: each gives a different _Pt-X pattern depending on whose neighbour each Pt-ligand bond has. Mass spectroscopy shows the same parent ion for all three.

Why this matters. Pt(II) drug design (cisplatin family) heavily relies on cis vs trans geometric isomerism but never on optical isomerism. The cisplatin (cis-[Pt(NH3)2Cl2], a cis-arrangement with C.N. 4 and the two amines on the same side) is anticancer; trans is inert. The same logic - cis groups attack DNA simultaneously, trans cannot - extends to carboplatin, oxaliplatin and almost all modern Pt drugs.

Concept linkage. The lack of optical isomerism for any square planar complex is a recurring board fact; it follows from the existence of the molecular plane as a σ-plane. Q 5.19 relies on this (square-planar d⁸ Ni(CN)4^2- is diamagnetic and achiral).

3 geometrical isomers; 0 optical isomers.

Quick reading. Both experiments are simple ligand substitution reactions on the aqua-copper cation. The colour change tracks the change in ligand-field strength. Cu(II) is d⁹; one unpaired electron, one d-d band; the band position is set by _o which is set by the ligand.

1. In [Cu(H2O)4]^2+, the d-d transition absorbs in the red region (around λ ≈ 700 nm), transmitting blue. Aqua is a moderate-field ligand, and _o for tetragonal Cu²⁺ (Jahn-Teller distorted) gives this red absorption.
2. Add 4 F- ⇒ a slightly weaker (longer-wavelength) field ⇒ green colour. The fluoride product happens to be sparingly soluble, giving a precipitate. [Cu(H2O)4]²⁺ + 4 F- -> [CuF4]^2- (s) + 4 H2O.
3. Add 4 Cl- to the same blue solution: chloride is even weaker than fluoride, the d-d band shifts further to longer λ, and we see bright green. The chloride complex stays in solution. [Cu(H2O)4]²⁺ + 4 Cl- -> [CuCl4]^2- + 4 H2O.
4. Sanity check with the spectrochemical series: Cl- < F- < H2O; the observed colour gradient (blue → green) matches. Decreasing _o shifts the d-d absorption to longer λ (red-shift); the complementary colour shifts blue → green → yellow.
5. Alternative explanation (Jahn-Teller). Cu²⁺ (d⁹) suffers a strong Jahn-Teller distortion: two trans ligands move out, four equatorial ligands move in. The d-d transition then has a fine structure of 3 closely spaced bands. The dominant red/green absorption survives the substitution analysis.
6. Numerical check. Estimated _equiv: Δ(Cl) ≈ 12,000 cm^-1, Δ(F) ≈ 13,500 cm^-1, Δ(H2O) ≈ 14,500 cm^-1 for Cu(II). Converted to λ = 10⁷/ν nm: ∼ 830, 740, 690 nm respectively, matching the visible observations.

Why this matters. Substitution reactions on Cu²⁺ are textbook demonstrations of how _o controls colour. The same logic applies to chromium and cobalt complexes (Q 5.20, 5.21, 5.25, 5.31). Industrially, the colour of [CuCl4]^2- is exploited in semiconductor etchant baths.

Concept linkage. This question is the operational version of Q 5.20 (Ni complexes), Q 5.21 (Fe complexes) and Q 5.25 ([Ti(H2O)6]³⁺): replace a ligand, watch _o change, watch the colour change.

F- and Cl- replace H2O in [Cu(H2O)4]²⁺; new ligand-field shifts the d-d band, producing green colours.

Strategic angle. Two steps in one: a redox plus a complexation. After both, the copper is in +1 state and is hidden inside a tightly bound cyanido complex. With the metal sequestered, no sulphide test can detect it. Cyanide is rare among ligands in that it can both reduce and coordinate; this pair of roles is what makes the chemistry surprising.

1. Add a few moles of KCN: white CuCN precipitates and a brown cloud of (CN)2 (cyanogen) briefly forms. The copper has been reduced from +2 to +1; cyanide has been oxidised from -1 (CN-) to zero ((CN)2). 2 Cu²⁺ + 4 CN- -> 2 CuCN(s) + (CN)2(g) .
2. Continue adding KCN: the precipitate dissolves to give a colourless solution of [Cu(CN)4]^3- ions. Cyanide bonds via its C atom (the donor); the C atom has sp hybridisation with a lone pair pointing along the C≡ N axis. CuCN(s) + 3 CN- -> [Cu(CN)4]^3-.
3. Bubble H2S: nothing happens. Free Cu+ would be needed; its concentration is ∼ 10^-30 M, far below the precipitation threshold for Cu2S.
4. Alternative path: do a redox first, complex later. Cu²⁺ is reduced by CN- because the product [Cu(CN)4]^3- is exceptionally stable (log₄ ≈ 30). Energetically the +1 state becomes accessible only because the cyanide complex stabilises it by ∼ 170 kJ/mol; without that stabilisation Cu+ would disproportionate to Cu(0) and Cu²⁺ (as it does in water).
5. Numerical check. If [Cu(CN)4^3-] ≈ 0.1 M and [CN-]_free ≈ 0.1 M, then [Cu+]_free = 0.110³⁰· (0.1)⁴ ≈ 10^-27 M. With [S^2-] ≈ 10^-15 M (saturated H2S at pH 7), the ionic product [Cu+]²[S^2-] ≈ 10^-69, vastly below K_sp(Cu2S) ≈ 10^-47. No precipitate.

Why this matters. The same chemistry is exploited in electroplating (cyanide baths give smooth Cu deposits because the free Cu+ is dosed gradually by the complex) and in cyanide leaching of gold/silver ores: dissolved as [Au(CN)2]^- or [Ag(CN)2]^-, transported, then recovered by zinc displacement. The toxicity of cyanide is the reason these processes need careful waste management.

Concept linkage. The same masking principle (large β suppresses free metal) underlies Q 5.2 (Cu(NH3)4²⁺ masks Cu²⁺ from H2S test), Q 5.27 (EDTA chelation therapy traps lead), and Q 5.30 (oxalate gives the most stable Fe³⁺ complex).

[Cu(CN)4]^3- forms; its stability suppresses free Cu+ so H2S cannot precipitate copper sulphide.

Strategic angle. Three pieces of data fix the VBT picture: metal oxidation state (→ dⁿ), ligand-field strength (strong → inner; weak → outer) and coordination geometry (here all octahedral). The hybridisation and magnetic moment follow mechanically. Memorise: inner = d²sp³, outer = sp³d².

1. Compute the metal oxidation state and dⁿ for each complex (see Q 5.5 method). (i) Fe(II), d⁶; (ii) Fe(III), d⁵; (iii) Co(III), d⁶; (iv) Co(III), d⁶.
2. Read off the ligand-field strength from the spectrochemical series. (i) CN- strong; (ii) F- weak; (iii) oxalate moderate-to-strong (here strong enough to pair); (iv) F- weak.
3. Decide whether (n-1)d electrons pair (strong-field) or stay unpaired (weak-field) before counting unpaired electrons. Strong → inner d²sp³ → low-spin; weak → outer sp³d² → high-spin.
4. Apply μ = √n(n+2) BM to get the magnetic moment.
   • (i) d⁶ low-spin: n = 0 ⇒ μ = 0.
   • (ii) d⁵ high-spin: n = 5 ⇒ μ = √35 ≈ 5.92 BM.
   • (iii) d⁶ low-spin: n = 0 ⇒ μ = 0.
   • (iv) d⁶ high-spin: t_2g⁴ e_g², n = 4 ⇒ μ = √24 ≈ 4.90 BM.
5. Alternative approach (CFT & MOT angle). VBT is the orbital-counting picture. CFT instead views the same complex as an electrostatic perturbation: the same number of unpaired electrons emerges, but the interpretation differs. MOT goes further and assigns ligand σ and π orbitals to a molecular orbital diagram; the spin state is then read off the HOMO/LUMO gap. All three theories predict the same n and μ for octahedral complexes; VBT is just the simplest.
6. Cross-check via observed colours/magnetic moments. [FeF6]^3- is colourless-to-pale (spin-forbidden d-d), high μ ≈ 5.9 BM (consistent with d⁵ high-spin). [Fe(CN)6]^4- is yellow (low _o in the visible), μ = 0 (diamagnetic). Matches VBT.

Why this matters. Magnetic moment measurements are routine characterisation tools; VBT predictions are how we link them to electronic structure. The same VBT framework, with dsp² hybridisation for square planar (Q 5.19), sp³ for tetrahedral (Q 5.24 (iv)) and sp for linear ([Ag(NH3)2]⁺), covers every geometry in the chapter.

Concept linkage. The high-spin/low-spin distinction is quantified in Q 5.18 via _o vs P; the explicit CFT picture is in Q 5.16. The colour difference of Q 5.20 / 5.21 between [Fe(CN)6]^4- and [Fe(H2O)6]²⁺ uses exactly this analysis.

(i) inner, μ = 0; (ii) outer, μ ≈ 5.92 BM; (iii) inner, μ = 0; (iv) outer, μ ≈ 4.90 BM.

Picture-first. Place the metal at the origin and six point charges along the Cartesian axes. The two orbitals whose lobes lie along these axes (d_z² and d_x²-y², the e_g pair) feel the largest repulsion. The other three (d_xy, d_yz, d_zx, the t_2g set) lie between the axes and are stabilised. Two sets, one gap, one number: _o.

1. Five orbitals start degenerate. In a uniformly negative spherical field they all rise by the same amount. This ``spherical field'' is a mathematical fiction useful as a zero-of-energy reference.
2. Concentrate the spherical charge into six point charges on the axes: degeneracy is lifted. The two on-axis orbitals rise by +0.6 _o = +35_o, the three off-axis ones fall by -0.4 _o = -25_o.
3. Barycentre check: 2· 0.6 - 3· 0.4 = 0. Total d-electron energy is conserved relative to spherical field. This conservation is a direct consequence of trace of the perturbation matrix being zero.
4. Alternative geometries.
   • Tetrahedral: invert the labels - e down, t₂ up; _t ≈ 49_o (smaller, because only 4 ligands and they sit off-axis, never directly along the orbital lobes).
   • Square planar: derive from octahedral by removing two trans ligands (along z); d_z² drops a lot, d_x²-y² stays high; results in 4 distinct levels and the largest gap of any 4-coordinate geometry.
5. Numerical placeholders. For aqua ligands on a first-row M³⁺ ion, _o ≈ 17,000 to 20,000 cm^-1 ≈ 200 kJ/mol. For strong-field ligands _o can be ∼ 30,000 cm^-1; for weak-field halides it can drop below 10,000 cm^-1.

Why this matters. The pattern is the workhorse for the rest of the chapter; tetrahedral splitting just inverts it with _t ≈ 49_o. Once you know _o and the dⁿ count, you can predict colour (Q 5.20, 5.21, 5.25), magnetic moment (Q 5.19), spin state (Q 5.18), and rough lattice stabilisation (CFSE).

Concept linkage. The _o formula and the splitting pattern enter every downstream CFT question. The spectrochemical series (Q 5.17) is essentially a ranking of ligands by the _o they produce on a given metal.

Octahedral CFT diagram: e_g up 0.6 _o, t_2g down 0.4 _o.

Quick reading. Memorise the rough order I- < Cl- < F- < H2O < NH3 < en < CN- < CO and treat all the rest as interpolations. The series is empirical (established from optical absorption spectra) but rationalised by MOT.

1. Strong-field ligands carry empty π^* orbitals (CN-, CO) or use chelation (en, oxalate) to push _o up. The empty π^* accepts electrons from filled metal t_2g, stabilising the bonding (lowering t_2g) and effectively raising e_g relative to it.
2. Weak-field ligands are mostly heavy halides and water; their lone pairs interact weakly with the metal e_g. The π-donation from F-, Cl- into the empty metal t_2g^* (in the MO picture) destabilises t_2g, shrinking _o.
3. Pairing-energy comparison: if _o > P, low-spin wins. If _o < P, high-spin wins. The series tells you which side of P you're on. Boundary cases are the d⁴–d⁷ first-row metals.
4. Alternative classification (MOT, π-acceptor vs π-donor). Classify ligands by their π character:
   • π-donor (e.g. F-, Cl-, OH-, O^2-): weak-field, small _o.
   • Pure σ-donor (e.g. NH3, H2O): intermediate.
   • π-acceptor (e.g. CN-, CO, PR3): strong-field, large _o.
   This MOT viewpoint explains why the empirical spectrochemical series has the order it does.
5. Numerical examples. For Co³⁺: _o([CoF6]^3-) ≈ 13,100 cm^-1; _o([Co(NH3)6]³⁺) ≈ 22,900 cm^-1; _o([Co(CN)6]^3-) ≈ 34,800 cm^-1. Almost a factor of 3 across the series - enough to flip Co(III) from high-spin (fluoride) to low-spin (cyanide).

Why this matters. The series is the bridge between an empirically measured colour or magnetic moment and an electronic structure prediction. Memorise the short version (``I, Br, Cl, F, OH, H2O, py, NH₃, en, NO₂^-, phen, CN, CO''), and almost every problem of this chapter follows mechanically.

Concept linkage. The series is invoked again in Q 5.20 (Ni complexes), Q 5.21 (Fe complexes), Q 5.25 (Ti), Q 5.30 (stability) and Q 5.31 (wavelength order). It is the single most reused list in the chapter.

Spectrochemical series ranks ligands by _o; weak-field → high-spin; strong-field → low-spin.

Strategic angle. It's a tug-of-war: orbital energy vs pairing energy. The winner sets the configuration. The pairing energy P has two contributions: the Coulombic cost of two electrons sharing one orbital, and the loss of exchange energy when two parallel spins become antiparallel.

1. In d¹, d², d³: only t_2g fills singly; the magnitude of _o doesn't matter for the configuration. All such ions (Ti³⁺, V³⁺, Cr³⁺) are always high-spin trivially.
2. In d⁴: high-spin gives t_2g³ e_g¹ (4 unpaired); low-spin gives t_2g⁴ e_g⁰ (2 unpaired). The _o vs P comparison decides. For [Mn(H2O)6]³⁺, _o ≈ 21,000 cm^-1, P ≈ 28,000 cm^-1, so high-spin. For [Mn(CN)6]^3-, _o ≈ 33,500 cm^-1 > P, so low-spin.
3. In d⁸, d⁹, d¹⁰: e_g must be partly filled, so the spin state is forced by the count. d⁸: t_2g⁶ e_g², 2 unpaired (octahedral). d¹⁰: all paired. Hence colourless aqua-zinc and aqua-copper(I) are different from coloured aqua-copper(II) (d⁹).
4. Alternative approach (LFSE in Dq units). Substitute _o = 10 Dq: t_2g at -4 Dq, e_g at +6 Dq. For d⁶ low-spin LFSE = 6 · (-4Dq) + 0 = -24 Dq; for d⁶ high-spin LFSE = 4 · (-4Dq) + 2 · (+6Dq) = -4 Dq. Difference is 20 Dq = 2_o, the largest LFSE difference of any dⁿ.
5. Numerical check (the Co(III) example). Co³⁺ (d⁶) with H2O: _o ≈ 18,200 cm^-1, P ≈ 21,000 cm^-1 - borderline, mostly low-spin in [Co(H2O)6]³⁺ because of the large LFSE gain. Excellent fit between theory and experiment for almost all Co(III) complexes (low-spin).

Why this matters. _o is the central knob of CFT: turn it and the colour, magnetism and stability of the complex all change. The same logic explains why Fe³⁺ (d⁵) with weak-field ligands is high-spin and colourless/very pale (spin- forbidden d-d), but with strong-field CN- becomes low-spin and yellow.

Concept linkage. The whole spectrochemical-series chapter (Q 5.17) and the colour/magnetic moment questions (Q 5.19–5.21, 5.29) depend on this single _o vs P comparison.

If _o > P, low-spin; if _o < P, high-spin. Only d⁴ to d⁷ ions show this dichotomy.

Strategic angle. Two facts settle the question: dⁿ of the metal and the geometry forced by the ligand. Then the spin-only formula gives the magnetic moment without needing CFT calculations.

1. Cr³⁺ is d³: always three unpaired electrons in t_2g, regardless of the ligand. So [Cr(NH3)6]³⁺ is paramagnetic. Charge balance: x + 6(0) = +3 ⇒ x = +3, d³. Ammine is moderate-field, octahedral geometry, t_2g³ e_g⁰. _calc = √15 ≈ 3.87 BM. Observed: _exp = 3.85 BM - excellent agreement.
2. Ni²⁺ is d⁸. With strong-field CN- in a 4-coordinate complex, the geometry is square planar. In a square-planar d⁸ complex the d_x²-y² orbital is far above the other four; the eight electrons fill the lower four orbitals in pairs: (d_xy)²(d_yz)²(d_zx)²(d_z²)²(d_x²-y²)⁰. VBT: dsp² hybridisation. n = 0, μ = 0 BM (diamagnetic).
3. Apply μ = √n(n+2) BM: n=3 gives √15 ≈ 3.87; n=0 gives 0.
4. Alternative approach (geometry from _o). For Ni²⁺, the choice between tetrahedral and square planar is settled by the ligand: weak-field Cl- gives [NiCl4]^2- tetrahedral (paramagnetic, n=2); strong-field CN- gives [Ni(CN)4]^2- square planar (diamagnetic). The same Ni²⁺ d⁸, completely different magnetic behaviour.
5. Numerical cross-check (hybridisation vs MOT). VBT predicts μ = 0 for square-planar d⁸ because dsp² uses an empty d_x²-y². MOT also predicts μ = 0: the four-coordinate σ donors lift only d_x²-y² very high; the other four remain low and are all doubly occupied. Same answer, different language.

Why this matters. Square-planar d⁸ complexes (Ni, Pd, Pt) are central to organometallic catalysis and all happen to be diamagnetic. The Wacker process, hydroformylation, and many cross-coupling reactions cycle through square-planar d⁸ Pd or Pt intermediates.

Concept linkage. The same machinery answers Q 5.20 (d⁸ green vs colourless), Q 5.21 (same metal, different ligand, different colour), Q 5.29 (highest μ in a list).

Cr(III): d³, 3 unpaired, paramagnetic. Square-planar Ni(II) with CN-: all paired, diamagnetic.

Quick reading. Same d⁸ metal; very different Δ: aqua gives a small split (visible absorption → green); cyanide gives a huge split (UV absorption → colourless). One question, two complexes, one explanation: the spectrochemical series.

1. Aqua complex: octahedral [Ni(H2O)6]²⁺, _o ≈ 8,500 cm^-1. Configuration t_2g⁶ e_g². d-d transitions absorb around λ ≈ 700–1200 nm (red and near-IR); transmits green. The solution is green.
2. Cyanide complex: square planar [Ni(CN)4]^2-, with a very large splitting (much larger than octahedral _o). Absorption shifts well into UV (the d_x²-y² ← d_z² transition is at ∼ 285 nm); nothing in the visible is absorbed; the solution is colourless. The d⁸ electron count fills the four lower-energy d orbitals; d_x²-y² remains empty.
3. Numerical cross-check via E = hc/λ. For green absorption (λ ≈ 500 nm), E ≈ 240 kJ/mol = 20,000 cm^-1. The [Ni(H2O)6]²⁺ absorption maximum near 720 nm corresponds to _o ≈ 14,000 cm^-1 (close to value above). The strong-field cyanide complex gives Δ ≈ 35,000 cm^-1, in the UV.
4. Alternative explanation (charge transfer). Some intense colours of transition-metal complexes come from ligand-to-metal or metal-to-ligand charge transfer (LMCT, MLCT), not d-d. For Ni(II) the absorptions are pure d-d (Laporte forbidden, weak), so the colour-to-no-colour difference here is genuinely a Δ effect.
5. Concept linkage. The same logic - strong field ⇒ large Δ ⇒ UV absorption ⇒ colourless or pale - explains why [Co(CN)6]^3- is pale yellow (smaller-band shift) and [Co(en)3]³⁺ is orange (intermediate).

Why this matters. Strong-field ligands often whiten a coloured complex by pushing absorption out of the visible window. Industrially, this is exploited to create colour-stable Pt(II) and Au(I) cyanide solutions for plating. Conversely, for sensors and dyes one chooses ligands with _o tuned into the visible.

Concept linkage. Q 5.21 is the same question phrased slightly differently for Fe(II); Q 5.25 for Ti(III); Q 5.31 orders three Ni(II) complexes by wavelength.

Different Δ values: aqua → green; cyanide → colourless (UV absorption).

Strategic angle. Hold the metal and oxidation state fixed, then read off colour straight from the spectrochemical position of the ligand. The colour of a given dⁿ ion changes continuously as the ligand is replaced - a direct visualisation of the CFT prediction.

1. Fe²⁺: d⁶ in both complexes. Charge balance for [Fe(CN)6]^4-: x + 6(-1) = -4 ⇒ x = +2. For [Fe(H2O)6]²⁺: x + 6(0) = +2 ⇒ x = +2.
2. CN- ≫ H2O in the spectrochemical series. _o(CN-) ≈ 33,000 cm^-1 for Fe(II); _o(H2O) ≈ 10,400 cm^-1. A factor of ∼ 3 difference in _o, which is enough to also flip the spin state of d⁶ Fe(II) from high-spin (4 unpaired, with H2O) to low-spin (0 unpaired, with CN-).
3. Bigger _o ⇒ blue-shifted absorption ⇒ different transmitted colour. Aqua complex: t_2g⁴ e_g², absorbs in orange/red, transmits pale green / bluish-green. Cyano complex: t_2g⁶ e_g⁰, absorbs near violet, transmits yellow.
4. Numerical cross-check. For [Fe(CN)6]^4-, _o = 33,000 cm^-1 corresponds to λ = 10⁷/33,000 ≈ 303 nm (UV), but spin- allowed transitions occur at longer wavelength because of coupling - observed _max ≈ 420 nm (violet), so transmits yellow. Matches observation.
5. Alternative angle: pairing-energy crossover. At _o ≈ 17,000 cm^-1 for Fe(II), the spin state crosses over from high-spin to low-spin. Aqua is below (∼ 10,000) so high-spin; cyanide is far above (∼ 33,000) so low-spin. The magnetic moment differs: aqua μ ≈ 5.2 BM, cyanide μ = 0.

Why this matters. Most analytical colorimetric methods exploit precisely this: change the ligand and the colour changes. The same metal ion, in two different complexes, can be detected with two different absorption bands.

Concept linkage. Q 5.20 is the same comparison for Ni(II); Q 5.25 generalises to Ti(III) (d¹, single d-d transition); Q 5.31 ranks Ni complexes by wavelength.

Different _o for CN- vs H2O shifts the d-d band, giving different colours.

Picture-first. Imagine two arrows on the same axis: one from CO to metal (σ), one from metal back to CO (π^*). Each strengthens the other. The C-O multiple bond stays mostly intact, but it weakens slightly because antibonding π^* gets populated. The metal-CO bond order ends up around 1.5 to 2 depending on the metal's electron density.

1. Carbon's sp lone pair (in CO) donates to an empty hybrid on the metal: that's the dative σ bond. The HOMO of CO is a weakly antibonding 3σ orbital on carbon, well-aligned for end-on σ donation.
2. A filled metal d_π overlaps with CO's empty π^* (a node between C and O), so electron density goes back. This is the π back-bond and gives M-C partial double-bond character. The LUMO of CO is a low-lying π^* orbital - perfectly placed to accept electrons from filled metal d_xy, d_yz, d_zx.
3. The two flows are coupled: more σ donation means more metal electron density and so more π^* back-donation. This is the synergic effect. The result is a much stronger M-C bond than a simple dative σ bond would suggest.
4. Numerical evidence. Free CO: _CO = 2143 cm^-1, bond order ≈ 3. Cationic [Mn(CO)6]⁺: _CO ≈ 2090 cm^-1 (small drop, less back-donation because metal is electron-poor). Neutral Cr(CO)6: ν ≈ 2000 cm^-1. Anionic [V(CO)6]^-: ν ≈ 1860 cm^-1 (largest drop, biggest back-donation). The systematic ∼ 300 cm^-1 decrease confirms the synergic π back-bond.
5. Alternative perspective (MO picture). Build a MO diagram for the M-CO unit: the 5σ HOMO of CO and the 1π^* LUMO of CO interact with metal hybrid and d_π orbitals respectively. Bonding MOs delocalise charge from C to M and from M to π^*. The simultaneous operation of both interactions is the synergic effect in MO language.
6. Cross-check via 18-electron rule. Ni(CO)4: Ni(0) gives 10, four CO give 4× 2 = 8, total 18. Fe(CO)5: 8 + 10 = 18. Cr(CO)6: 6 + 12 = 18. The 18-electron count works because σ donation places exactly two electrons per CO ligand into metal-based MOs.

Why this matters. Variants of the same idea explain bonding in alkene complexes (Dewar-Chatt-Duncanson model), dinitrogen complexes (N2 has a π^*), and many catalytic intermediates. Industrial catalysis (hydroformylation, the Monsanto acetic-acid process, Reppe synthesis) sits on top of metal carbonyl chemistry.

Concept linkage. The same σ/π^* synergy explains why CN- is a strong-field ligand (Q 5.17): it has both an excellent lone-pair σ donor and a low-lying π^* accept set, just like CO.

Synergic σ (CO → M) and π^* (M → CO) bonding gives a strong M-C bond and a slightly weakened C-O bond.

Quick reading. Three numbers per complex; each from a one-line calculation. Charge balance for oxidation state, dⁿ = (group number) - (ox. state) for the first row, and sum of donor atoms for C.N. No memorisation required - just the periodic table and the ligand-charge table.

1. For oxalate-, en-bearing complexes don't forget to multiply the chelate by 2 when counting C.N. In (i) three oxalates ⇒ 3 × 2 = 6 donor atoms.
2. For Co: group 9, so Co²⁺ is d⁷ and Co³⁺ is d⁶. For (i) the complex charge is -3, ligand-set -6, so x = +3, d⁶. For (iii) ammonium ions outside give complex charge -2, ligand-set -4 (four fluorides), so x = +2, d⁷.
3. For Cr: group 6, so Cr³⁺ is d³. In (ii) one outer Cl- gives complex charge +1, ligand-set -2 (two chlorides inside), so x = +3.
4. For Mn: group 7, so Mn²⁺ is d⁵. In (iv) outer SO4^2- gives complex charge +2, ligand-set 0, so x = +2.
5. Alternative approach (electron-count check). Use the 18-electron rule to verify your dⁿ: for [Co(C2O4)3]^3- - Co(III) is d⁶ = 6; six donor atoms × 2 = 12 from ligands; total 18. Tidy. Same for [CrCl2(en)2]Cl at 15 electrons (the Cr³⁺ d³ ion is short of 18, as expected for early transition metals).
6. Cross-check via predicted μ. Once dⁿ is known, _spin-only = √n(n+2). For (iv) μ = √35 ≈ 5.92 BM, the typical Mn²⁺ value - confirms x = +2 and d⁵.

Why this matters. These three numbers feed directly into the magnetic-moment and CFSE calculations of Q 5.24. They are the ``inputs'' to every electronic-structure prediction in this chapter.

Concept linkage. The same charge-balance arithmetic appears in Q 5.5; the dⁿ count appears in Q 5.15, 5.19; the C.N. feeds into the geometric-isomer count of Q 5.9, 5.11.

(i) +3, d⁶, 6; (ii) +3, d³, 6; (iii) +2, d⁷, 4; (iv) +2, d⁵, 6.

Strategic angle. Same recipe applied five times: charge balance → x; group number → dⁿ; ligand-field strength → spin state; n unpaired → μ. After two or three complexes the routine becomes automatic.

1. For (i) K[Cr(H2O)2(C2O4)2].3H2O: two anionic oxalates contribute -4, two waters contribute 0; the complex anion must be -1 (one K+ outside). So x = +3. Cr(III) is d³, C.N. = 6 (oxalate is didentate, so 2 + 2× 2 = 6). Three unpaired in t_2g, μ ≈ 3.87 BM. Stereochemistry: cis (chiral) / trans (achiral) of two oxalates and two waters.
2. For (ii) [Co(NH3)5Cl]Cl2, the ammine ligand on Co³⁺ is strong enough to give a low-spin d⁶ closed shell (diamagnetic). Charge balance: x + 5(0) - 1 = +2 ⇒ x = +3. μ = 0. Only one geometry ([MA5B], no isomers).
3. For (iii) [CrCl3(py)3], pyridine is comparable to NH3, weakly strong-field. The d³ count of Cr³⁺ doesn't permit a choice of spin: three unpaired regardless. μ ≈ 3.87 BM. fac and mer isomers possible (Q 5.9 logic).
4. For (iv) Cs[FeCl4], tetrahedral fields are inherently small (_t ≈ 49_o), so almost all tetrahedral complexes are high-spin. Fe(III) is d⁵ high-spin, n = 5, μ = √35 ≈ 5.92 BM. No isomers (single tetrahedral arrangement).
5. For (v) K4[Mn(CN)6], CN- is the strongest common ligand and forces low-spin even on Mn²⁺ (d⁵). Configuration t_2g⁵ e_g⁰, n = 1, μ = √3 ≈ 1.73 BM. Octahedral, no isomers.
6. Alternative approach (CFSE comparison). CFSE gives a cross-check for the spin assignment. For (ii) d⁶ low-spin: CFSE = -2.4_o + 2P, a substantial stabilisation that explains why Co(III) is almost always low-spin. For (v) d⁵ low-spin: CFSE = -2.0_o + 2P, also substantial, hence cyanide does force low-spin.
7. Numerical cross-check. Experimental μ values: [Co(NH3)5Cl]²⁺ shows _exp = 0 (diamagnetic, confirming d⁶ low-spin); [Cr(NH3)6]³⁺ shows _exp = 3.85 BM (d³); [FeCl4]^- shows _exp = 5.9 BM (d⁵ high-spin). All consistent with the table above.

Why this matters. A practising inorganic chemist runs this recipe almost unconsciously every time a new complex appears. For board-paper style synthesis: 5 complexes, 5 sets of (name, x, dⁿ, C.N., stereochemistry, μ) is a 12-marker that rewards systematic application of the same formula.

Concept linkage. Builds on Q 5.5 (oxidation state), Q 5.7 (IUPAC name), Q 5.9–5.12 (isomerism), Q 5.15 (VBT), Q 5.17 (spectrochemical series), Q 5.19 (magnetic moment). It is the chapter's integrative question.

Five complexes, five sets of (name, x, dⁿ, C.N., stereochemistry, μ) as above.

Picture-first. One d electron, one t_2g → e_g transition, one absorbed wavelength. The transmitted light is the complementary colour: violet. [Ti(H2O)6]³⁺ is the textbook prototype of a d¹ ion because its absorption spectrum is the simplest possible - a single broad band centred near 500 nm.

1. Find dⁿ: Ti³⁺ is d¹. Charge balance: x + 6(0) = +3 ⇒ x = +3. Then dⁿ = 4 - 3 = d¹ (Ti is group 4).
2. Place the electron in t_2g. Promotion to e_g needs _o of energy. There is only one possible d-d transition for a d¹ ion in an octahedral field.
3. For aqua at C.N. 6, this _o corresponds to absorption at ∼ 500 nm (green-yellow). Numerically, _o = hc/λ: _o = (6.626 × 10^-34)(3 × 10⁸)500 × 10^-9 J = 3.98 × 10^-19 J. Per mole: E_mol = 3.98 × 10^-19 × 6.022 × 10²³ ≈ 240 kJ/mol, equivalent to ∼ 20,000 cm^-1. The observed _o for [Ti(H2O)6]³⁺ is 20,300 cm^-1 - excellent match.
4. Colour wheel: green-yellow absorbed ⇒ violet transmitted. The observed solution colour is a pale, attractive violet.
5. Alternative approach (Jahn-Teller asymmetry). The absorption band of [Ti(H2O)6]³⁺ shows a slight shoulder, betraying a Jahn-Teller distortion (the t_2g¹ state is JT-active for an octahedral d¹ ion). The asymmetry of the band is one of the smallest clean experimental measurements of JT effects.
6. Why broad and not sharp? d-d transitions are Laporte forbidden in centrosymmetric (octahedral) complexes - the transition gains intensity through vibronic coupling with antisymmetric stretches. The absorption is therefore always broad (∼ 100 nm wide) and weak (ε < 100).

Why this matters. The same logic explains the colour of every transition-metal aqua-ion: Cu²⁺ blue, Ni²⁺ green, Fe³⁺ pale violet, Cr³⁺ violet. A single d electron, the simplest possible case, fixes the whole _o scale for aqua complexes.

Concept linkage. Ti³⁺ is the calibration point for the spectrochemical series for aqua ligands. Q 5.20, 5.21, 5.31 all extend this single-band analysis to multi-electron dⁿ cases.

d¹ Ti(III), single d-d transition t_2g → e_g, absorbs green-yellow, complex appears violet.

Strategic angle. It is entropy, not enthalpy. Six water molecules leave; three ethylenediamines arrive. Particle count rises, Δ S > 0, Δ G becomes very negative. The enthalpic contributions from M-N bond formation are very similar in [Ni(en)3]²⁺ and [Ni(NH3)6]²⁺, so the entropy term carries almost all of the stability difference.

1. Write the substitution: [M(H2O)6]ⁿ⁺ + 3 en -> [M(en)3]ⁿ⁺ + 6 H2O; Δ n = +3 on the right. Particles in =4, particles out =7. The entropy gain of the system is positive.
2. Compute Δ S^∘ ≈ R ln(ratio); for Δ n = +3, this is ∼ +25 J/(K mol) when considering translational entropy alone. Including rotational and configurational contributions can push it to +40 to +50 J/(K mol).
3. At T = 298 K, T Δ S ∼ 7.5 to 15 kJ/mol, which translates to a large factor in K_eq, on top of any enthalpic advantage. Numerically: Δ(logβ) = -ΔΔ G^∘ / (2.303 RT) ≈ -1.3 per kJ/mol at room temp, so 7–15 kJ/mol explains Δ(logβ) ∼ 9–20, fitting the experimental gap.
4. Larger ring sizes (5- or 6-membered) give the strongest chelate effect; very small or very large rings strain and weaken the effect. Three-membered rings (no chelation possible). Four-membered rings (rare; M-N-C-N geometry is unfavourable). Five-membered (en, oxalate): widely observed. Six-membered (acac, glycinato): also common. Seven-membered: rare.
5. Alternative perspective (energy-decomposition). Modern DFT analysis decomposes the chelate effect into electrostatic, exchange-repulsion, polarisation, charge- transfer and dispersion contributions. The entropy term dominates, but the slight extra enthalpic gain (geometric rigidity of the 5-membered ring) accounts for the residual few kJ/mol stabilisation.
6. Numerical example: EDTA chelates. EDTA (Y^4-) is hexadentate: it replaces 6 waters with one ligand. Δ n = +5 (1 + 1 in to 1 + 6 out). logβ([CaY]^2-) ≈ 10.7; logβ([FeY]^-) ≈ 25.1. Used for water hardness titrations.

Why this matters. EDTA and the [Ca(EDTA)]^2- complex are stabilised by exactly the same entropy argument scaled up to a hexadentate ligand. In pharmacology, chelation therapy (Na2[Ca(EDTA)] swallowed by a lead-poisoned patient) works because Pb²⁺ displaces Ca²⁺ from EDTA and is excreted as the soluble Pb-EDTA chelate (no chelate effect ⇒ no medicine).

Concept linkage. The same idea explains the stability of porphyrin-based chelates (haemoglobin, chlorophyll, vitamin B₁₂) in Q 5.27. The cyanide complex of Q 5.14 also benefits from a related effect (very high ₄).

Chelate effect: extra stability of chelate complexes due to favourable entropy. Example: logβ(Ni(en)3²⁺) - logβ(Ni(NH3)6²⁺) ≈ 10.

Strategic angle. Each application exploits one of three features: (a) selective metal binding, (b) reversible ligand-exchange, (c) thermodynamic stability of a chelate. Often two or three of these features work together - a single coordination compound can be both selective and reversible and thermodynamically stable.

1. Biology: porphyrin and corrin ligands tune the metal's reactivity. Iron in haemoglobin binds O2 reversibly; same iron in cytochromes does electron transfer. Magnesium in chlorophyll converts photons to electrochemical potential. Cobalt in B₁₂ catalyses rearrangements via a Co-C bond that is unusually weak. Zinc in carbonic anhydrase activates a water molecule for nucleophilic attack on CO₂. Each example combines a metal with a polydentate ligand that fine-tunes its reactivity.
2. Medicine: cisplatin's two cis-Cl- ligands are replaced inside the cell by N7 of guanine, locking onto DNA. The trans-isomer doesn't work (geometry mismatch). The IC₅₀ of cisplatin for testicular cancer cells is ∼ 1 ; transplatin is essentially inert. EDTA chelation therapy: Na2[Ca(EDTA)] swaps Ca for Pb/Hg/As, mobilising the toxic metal for renal excretion.
3. Analysis: chelates have intense, narrow absorption bands, useful for colorimetric / titrimetric detection. EDTA titrations measure Ca²⁺ and Mg²⁺ (water hardness). DMG gives a rose-red precipitate with Ni²⁺ at ∼ 1 ppm sensitivity. SCN gives a blood-red colour with Fe³⁺ at ∼ 0.1 ppm. Each test relies on the chelate effect to ensure that the analyte ion is fully captured by the indicating ligand.
4. Metallurgy: form a soluble metal complex, separate, then thermally decompose. Mond and cyanide leaching are textbook examples. Mond process: Ni + 4 CO <=> Ni(CO)4 at ≈ 330 K; the volatile carbonyl distils away from impurities; decomposition at ∼ 470 K gives 99.99% pure Ni. Cyanide leaching: gold dissolves as [Au(CN)2]^-; recovered by Zn cementation. Bayer process: bauxite dissolved as [Al(OH)4]^-; gibbsite separated, then converted to Al2O3 for Hall-Heroult.
5. Alternative classification. One can group applications by which property is exploited: (a) reactivity (catalysis, organic synthesis); (b) colour (analysis, dyes); (c) magnetism (single-molecule magnets); (d) solubility (extraction, hydrometallurgy).
6. Numerical examples. Average adult haemoglobin carries ∼ 4 × 10¹⁹ Fe(II) ions; chlorophyll in a typical maple leaf has ∼ 10¹⁸ Mg(II) ions per cm². Cisplatin's annual market is ∼ 100 tonnes worldwide. Mond process delivers ∼ 100,000 tonnes of pure nickel per year.

Why this matters. Coordination chemistry is one of the broadest unifying ideas in chemistry: same principles in a hospital, a forensic lab and a copper smelter. The four examples in this question span biology, medicine, analytics and industrial chemistry - together they illustrate why the coordination compound is among the most useful molecular architectures known.

Concept linkage. The chelate effect (Q 5.26) underlies every biological example. The spectrochemical-series colour arguments (Q 5.13, 5.20, 5.25) explain why analytical reagents work. VBT/CFT (Q 5.15, 5.16) explains why Mond-process Ni(CO)₄ is tetrahedral and stable.

Four domains, four key examples (haemoglobin, cisplatin, EDTA, Mond / cyanide leaching).

Quick reading. The cobalt + ammines travel together as a single 2+ cation; the two chlorides break off as Cl- ions. Two species + one cation = three ions.

1. Identify the bracket: Co(NH3)6Cl2 is the same as [Co(NH3)6]Cl2. The six ammines are inside the coordination sphere; the two chlorides are outside as counter ions.
2. Charge balance: x + 6(0) = +2 ⇒ x = +2. So Co(II), and the cation is [Co(NH3)6]²⁺.
3. Write the dissociation [Co(NH3)6]Cl2 -> [Co(NH3)6]²⁺ + 2 Cl-.
4. Count: one complex cation, two chloride anions: 1 + 2 = 3.
5. Alternative check via Werner's model. Co(II) prefers C.N. 6 (octahedral) and primary valency 2 (since the cation is Co²⁺). All six secondary valencies are taken by ammonia; the two chlorides sit outside as primary valencies and ionise. The number of chlorides outside equals the cation charge, equals the number of Cl- ions released = 2.
6. Numerical conductivity check. Molar conductivity of [Co(NH3)6]Cl2 in dilute aqueous solution is ∼ 260 S cm² mol^-1, characteristic of a 1:2 electrolyte. Compare with the 1:3 electrolyte [Co(NH3)6]Cl3 (_m ≈ 430) - the higher value corresponds to the higher ion count.

Why this matters. Werner's original conductivity measurements (Q 5.1) relied on exactly this counting argument. Modern day quality-control assays for inorganic drug formulations still measure conductivity to confirm complex stoichiometry.

Concept linkage. Same counting recipe as Q 5.1 (Werner). Bonus: contrasts well with [Co(NH3)5Cl]Cl which gives two ions (one complex cation + one Cl-), illustrating how a chloride inside the sphere lowers the ion count.

3 ions. Option (iii).

Quick reading. Count n. The one with the most unpaired electrons wins. Aqua is weak field, so all three are high-spin octahedral cases.

1. Cr³⁺ (group 6, x = +3): d³ ⇒ t_2g³ e_g⁰ ⇒ n = 3. μ = √15 ≈ 3.87 BM.
2. Fe²⁺ (group 8, x = +2): d⁶, weak-field high-spin ⇒ t_2g⁴ e_g² ⇒ n = 4. μ = √24 ≈ 4.90 BM.
3. Zn²⁺ (group 12, x = +2): d¹⁰, all paired ⇒ n = 0. μ = 0.
4. Largest n is 4 (Fe²⁺); μ ≈ 4.90 BM. Option (ii).
5. Alternative cross-check. For high-spin first-row octahedral ions, n_max peaks at d⁵ (n = 5, μ = 5.92 BM) and decreases symmetrically on both sides: d⁴ ⇔ d⁶ both give n = 4; d³ ⇔ d⁷ both give n = 3. Fe(II) is at d⁶ = near the peak; Cr(III) at d³ = further away.
6. Numerical experimental data. _exp([Cr(H2O)6]³⁺) = 3.84 BM (vs spin-only 3.87, excellent agreement); _exp([Fe(H2O)6]²⁺) = 5.20 BM (slightly higher than spin-only 4.90 because of orbital contribution to T-state ground term - the orbital contribution that NCERT mentions in passing); _exp([Zn(H2O)6]²⁺) = 0 (diamagnetic).

Why this matters. Magnetic susceptibility is the simplest experimental window into d-electron count and spin state. For a board-paper MCQ, the recipe ``count n from dⁿ and compute μ'' is the fastest route.

Concept linkage. The spin-only formula μ = √n(n+2) is the most-used formula in the chapter; recurs in Q 5.15, 5.19, 5.24, here and in the d-block chapter (Q 4.x). Always know it cold.

Option (ii): [Fe(H2O)6]²⁺, μ ≈ 4.90 BM.

Structural observation. Chelation gives a thermodynamic advantage of ∼ 10¹⁰ over monodentate equivalents. Oxalate is the only chelate ligand in the list. Combine that with its moderately strong-field character and the answer is clear.

1. Compare ligand-field strength: oxalate is intermediate but beats Cl-, H2O and NH3 for Fe³⁺. Position in series: Cl- < H2O < NH3 < C2O4^2- (for hard, oxophilic Fe³⁺).
2. Add the chelate bonus: a 5-membered ring is the most favourable size. Three oxalates per iron give three 5-membered rings ⇒ very high entropy gain on ligand substitution.
3. Compare stability constants numerically:
   • log₃([Fe(C2O4)3]^3-) ≈ 20.
   • log₆([Fe(NH3)6]³⁺) ≈ 5 (unstable because NH3 is a poor donor for hard Fe³⁺ - actually decomposes in water).
   • log₆([Fe(H2O)6]³⁺) ≈ 0 (water is the reference for aqueous Fe³⁺).
   • log₆([FeCl6]^3-) ≈ 2.
   Oxalato wins by ∼ 10¹⁵ over the chloride and by ∼ 10²⁰ over the aqua reference.
4. Add up: by both metrics, [Fe(C2O4)3]^3- wins. Option (iii).
5. Alternative perspective (HSAB). Fe³⁺ is a hard Lewis acid (small, highly charged). It prefers hard Lewis bases: F-, O^2-, OH-, C2O4^2- (O-donor). Soft NH3 and Cl- are less compatible. HSAB plus chelate effect together explain the clear win for oxalate.
6. Numerical cross-check via free energy. Δ G^∘ = -2.303 RT log K. For oxalato: log K ≈ 20 ⇒ Δ G^∘ ≈ -114 kJ/mol. For ammonia: log K ≈ 5 ⇒ Δ G^∘ ≈ -29 kJ/mol. The 85 kJ/mol gap is almost entirely entropic - the chelate effect in thermodynamic units.

Why this matters. The same reasoning explains why EDTA (logβ ≈ 25 with Fe³⁺) sequesters Fe³⁺ even more effectively than oxalate. EDTA is hexadentate; the entropic advantage scales with denticity. In rust-removal and food-preservation, EDTA is the active sequestering agent for free Fe³⁺.

Concept linkage. Q 5.26 (chelate effect) is the theoretical justification; Q 5.27 (applications) is the practical use. The HSAB principle hinted at here was developed by Pearson in 1963 and is the rationale for ligand selection in catalysis, medicine and environmental chemistry.

Option (iii): [Fe(C2O4)3]^3-.

Quick reading. Larger _o ⇒ higher absorption energy ⇒ smaller λ. Order the ligands on the spectrochemical scale and flip. Same metal (Ni²⁺, d⁸), same geometry (octahedral); only the ligand differs.

1. Order _o: H2O < NH3 < NO2-. Aqua at ∼ 8,500 cm^-1 for Ni(II); ammonia at ∼ 11,000 cm^-1; nitrite at ∼ 22,000 cm^-1. A factor of ∼ 3 from end to end.
2. Invert to get λ: aqua complex has the largest λ (∼ 1200 nm and red region), nitrite complex the smallest (∼ 450 nm, violet).
3. Observed colours. [Ni(H2O)6]²⁺ green (transmits green because it absorbs red); [Ni(NH3)6]²⁺ blue/violet; [Ni(NO2)6]^4- yellow/orange (because it absorbs violet-blue).
4. Numerical sanity check. Using E = hc/λ, ν = 1/λ: [Ni(NO2)6]^4- at 22,000 cm^-1 ⇒ λ ≈ 455 nm (violet absorption, yellow transmission). [Ni(H2O)6]²⁺ at 8500 cm^-1 ⇒ λ ≈ 1176 nm (near-IR, but tail extends into visible red - hence green solution).
5. Alternative angle: π-donor vs π-acceptor. H2O is a weak σ-donor (and weak π-donor). NH3 is a stronger σ-donor only (no π capability). NO2- is both a strong σ-donor and a π-acceptor, so it has the largest _o. The order in the spectrochemical series follows directly from this MOT analysis.
6. Cross-check via colour wheel. Aqua green; NH₃ blue; nitrite yellow. The visible colour sequence (green → blue/violet → yellow) is exactly what one expects from the absorption sequence moving from red to violet (in the complementary direction).

Why this matters. This is the standard exam template for ``predict the colour'' style questions on coordination compounds. Variants of this question appear in every JEE and NEET paper on coordination chemistry. Mastering the recipe ``spectrochemical order → _o order → λ inverse order → transmitted-colour order'' is essential.

Concept linkage. Q 5.13 (Cu colour change), Q 5.20 (Ni complexes by colour), Q 5.21 (Fe complexes by colour), Q 5.25 (Ti(III) violet) all use the same logic.

λ: [Ni(H2O)6]²⁺ > [Ni(NH3)6]²⁺ > [Ni(NO2)6]^4-.