NCERT Exemplar Solutions Class 12 Chemistry Ch 5 - Download PDF

Quick reading. Stability constant K has no upper limit, but the spectrochemical series fixes the relative ordering of donor strengths to Cu²⁺. Reading the four ligands in series order directly predicts the order of log K.

1. Place the four ligands on the spectrochemical series from weak to strong field: H2O < NH3 < en < CN^-. Stronger field ⇒ stronger Cu²⁺-L σ-bond (plus π-back-bonding for CN^-) ⇒ larger K.
2. Compare the listed log K against this order: H2O (8.9) < NH3 (11.6) < en (15.4) < CN^- (27.3). The data match the prediction term-for-term.
3. Note that the chelate effect bumps en above NH3 by ∼ 4 orders in K, but the cyanide complex still wins because CN^- is both a stronger σ-donor and a π-acceptor, and four cyanides simply trump two chelating en.

Why this matters. Sodium cyanide is used in industrial gold leaching, 4 Au + 8 CN^- + O2 + 2 H2O -> 4 [Au(CN)2]^- + 4 OH^-, for exactly the same reason this question turns on: CN^- forms outrageously stable complexes with d-block metals.

[Cu(CN)4]^2-, option (ii).

Strategic angle. Two simple proportions take you from spectrochemical position straight to absorbed wavelength: ligand strength ∝ _o and λ ∝ 1/_o. So ligand-strength order reverses to give wavelength order.

1. Spectrochemical series snippet (left = weak, right = strong): H2O < NH3 < CN^-.
2. Reverse this to read off λ-absorbed order (weakest field absorbs the longest wavelength): [Co(H2O)6]³⁺ (∼ 600 nm) > [Co(NH3)6]³⁺ (∼ 470 nm) > [Co(CN)6]^3- (∼ 310 nm).
3. Numerical cross-check with _o([Co(NH3)6]³⁺) ≈ 22,900 cm^-1 and λ = 10⁷/22900 = 437 nm — in the violet, so the complex looks yellow-orange. Pattern confirmed.

Why this matters. The same logic predicts why [Ti(H2O)6]³⁺ is purple (_o = 20,300 cm^-1, absorbs ∼ 493 nm) and why exchanging H2O for NH3 on the same metal usually shifts a complex's colour visibly toward shorter λ.

Option (iii): [Co(H2O)6]³⁺ > [Co(NH3)6]³⁺ > [Co(CN)6]^3-.

Strategic angle. Drop one chloride at a time into the coordination sphere and stop the moment the AgCl count matches the data.

1. Possible isomers of CoCl3(NH3)5 and their AgCl output per mole of complex:
   • [Co(NH3)5]Cl33 Cl outside → 3 mol AgCl (but this isomer needs CN =5 for Co, which is wrong — included only for completeness).
   • [Co(NH3)5 Cl]Cl22 Cl outside, 1 inside → 2 mol AgCl.
   • [Co(NH3)5 Cl2]Cl1 Cl outside, 2 inside → 1 mol AgCl.
   • [Co(NH3)5 Cl3]0 Cl outside → 0 mol AgCl.
2. Observed: 0.2 mol AgCl per 0.1 mol complex = 2 per formula unit. Only [Co(NH3)5 Cl]Cl2 fits.
3. Conductivity link: one +2 cation and two -1 anions means the molar conductivity sits around 245–260 S cm² mol^-1 at infinite dilution in water, the signature of a 1:2 electrolyte (compare a 1:1 NaCl-like complex at ∼ 100 S cm² mol^-1).

Why this matters. Werner deduced the 1:1, 1:2, 1:3 pattern of cobalt-amine chlorides purely from AgCl titration and conductivity, decades before any X-ray crystal structure. The same logic is fair game in board derivations.

[Co(NH3)5 Cl]Cl2, a 1:2 electrolyte.

Picture-first. Draw Cr³⁺ at the centre of an octahedron. Whichever six donors sit at the vertices are inside the coordination sphere; anything beyond is a counter-ion (or lattice water). Count the chlorides outside the vertices.

1. For each option, write the inner-sphere occupancy: arrayl|l|l Option & Inner sphere & Outer Cl
   (i) & 3 Cl + 3 H2O & 0
   (ii) & 2 Cl + 4 H2O & 1
   (iii)& 1 Cl + 5 H2O & 2
   (iv) & 6 H2O & 3
   array
2. Match outer Cl to the observed 3 mol AgCl: only (iv) gives 3.
3. Conductivity confirmation: [Cr(H2O)6]Cl3 dissociates into 4 ions per formula unit, the highest of the candidates, and the experimental molar conductance is ∼ 430 S cm² mol^-1 — consistent with a 1:3 electrolyte at 25^∘C.

Why this matters. Identifying solvate isomers from precipitation data was Werner's experimental signature, and this exact problem shows up almost every alternate year on JEE.

[Cr(H2O)6]Cl3.

Quick reading. Three independent IUPAC lookups stack up: oxidation state, alphabetical ligand order, ligand spelling (ammine vs chlorido). Each lookup is a filter that eliminates wrong options, so even if one filter is forgotten the other two still pin (i).

1. Filter 1 — oxidation state. The complex is neutral overall, Cl ligands are -1 each, NH3 is neutral: x + 0 + 2(-1) = 0 ⇒ x = +2. So Pt is in +2, named platinum(II). Options (ii) and (iv) list +4 (would need four Cl or oxide ligands); option (iii) lists 0 (needs zero Cl). Three options die at this filter, only (i) survives.
2. Filter 2 — alphabetical ligand order. Compare first letters of the bare ligand names (not their counting prefixes): ammine starts with `a', chlorido starts with `c'. So a before c: ``diammine'' is written before ``dichlorido''. Option (iv) reverses this order, doubly failing.
3. Filter 3 — 2005 IUPAC spelling. Coordinated NH3 is ammine (two m's, distinct from the organic amine); the 2005 recommendation renamed chloro to chlorido (ending -ido signals a singly charged anionic ligand). NCERT uses the post-2005 forms throughout this chapter — answer in the same convention to avoid losing the half-mark.

Why this matters. The compound is the parent skeleton of cisplatin, cis-[PtCl2(NH3)2] — one of the most successful anticancer drugs ever synthesised (FDA-approved in 1978). Naming it correctly is the gateway to discussing its biology: it is the cis geometrical isomer that binds DNA's guanine bases and disrupts replication in cancer cells, while the trans isomer is inactive. Same molecular formula, dramatically different drug.

Diamminedichloridoplatinum(II).

Picture-first. Sketch each complex; only the one whose ligand bites the metal at two points (a chelate ring) wins.

1. Of the four ligands, only C2O4^2- has two donor atoms (the two carboxylate O^-). Three such ligands wrap three 5-membered rings around Fe³⁺.
2. Chelate ring count for the four options: 0, 0, 3, 0. Highest entropy gain ⇒ highest stability.
3. Quantitatively, going from [Fe(H2O)6]³⁺ to [Fe(C2O4)3]^3- releases six water molecules and captures three oxalates: Δ n_gas-like = +3, so TΔ S^∘ is large and positive.

Why this matters. The chelate effect is exploited in EDTA (EDTA^4-, hexadentate) used in food preservation and metal sequestering in water treatment.

[Fe(C2O4)3]^3-, the only chelated complex.

Strategic angle — pattern matching. Geometrical isomerism needs at least two different ligand types AND the right ligand count pattern AND a coordination geometry that allows distinct cis/trans arrangements. Run the three filters as a 2-second sieve and only one option survives.

1. Filter 1 — two or more ligand types? Option (iii) [Co(NH3)6]³⁺ has only one ligand type (MA6). All six octahedral vertices are equivalent, so only one structure exists — zero GI. Option (iii) dies immediately.
2. Filter 2 — coordination geometry compatible with cis/trans? Option (ii) [Pt(NH3)3 Cl] has only four ligands. In a tetrahedral MA3 B all positions are equivalent (the unique B has no ``cis vs trans'' partner); in square planar MA3 B the unique ligand again has no cis/trans choice (its three identical neighbours fix it uniquely). Either way, no GI.
3. Filter 3 — ligand count pattern. The GI-active octahedral patterns are MA4 B2, MA3 B3, MA2 B2 C2 etc. — patterns where at least two identical ligands can sit either adjacent (90^∘, cis) or opposite (180^∘, trans). Option (i) [Cr(H2O)4 Cl2]⁺ is exactly MA4 B2 — passes. Option (iv) [Co(CN)5 (NC)]^3- is MA5 B: the single different ligand has no cis/trans partner. Option (iv) shows linkage isomerism (because CN^-/NC^- is ambidentate) but not geometrical isomerism — a textbook trap.
4. Survivor: option (i) only. The two isomers: cis-[Cr(H2O)4 Cl2]⁺ (two Cl at 90^∘; lacks a mirror plane through both chlorines, only C_2v symmetry, no chirality but distinct connectivity) and trans-[Cr(H2O)4 Cl2]⁺ (two Cl at 180^∘; possesses D_4h symmetry, including a centre of inversion, hence achiral and visibly different in colour and dipole moment).

Why this matters. Werner's 1913 Nobel Prize hinged on exactly this kind of count: by enumerating the number of distinct isomers of [Co(NH3)4 Cl2]⁺ and [Co(NH3)3 Cl3], he proved the octahedral coordination geometry of Co(III) decades before X-ray crystallography existed. The cis/trans split is the most powerful inference tool inorganic chemistry has without a diffractometer.

[Cr(H2O)4 Cl2]⁺ (cis/trans), option (i).

Picture-first. Draw the two ligand fields and notice why _t < _o: in a tetrahedron the ligands sit on the alternate corners of a cube, none pointing along the x, y or z axes; in an octahedron they sit exactly on the axes.

1. Energy book-keeping. Octahedral d splits as t_2g (-0.4_o, 3 orbitals) below e_g (+0.6_o, 2 orbitals). Tetrahedral inverts: e below t₂, separation _t = (4/9)_o.
2. Plug in: _t = 49(18,000) = 8,000 cm^-1.
3. Wavelength check (for fun): octahedral absorption near _o = 10⁷ / 18,000 ≈ 556 nm (green/yellow, explaining the pink colour of [Co(H2O)6]²⁺ solutions); tetrahedral absorption near _t = 10⁷ / 8,000 ≈ 1250 nm — that's in the near-IR, so the tetrahedral cobalt(II) chloride absorbs additional bands in the visible too, producing the deep blue colour.

Why this matters. The pink-to-blue colour change when CoCl2 · 6H2O is heated (or when chloride is added in excess) is precisely the octahedral → tetrahedral conversion governed by this _t/_o ratio.

_t = 8,000 cm^-1.

Structural observation. Read the ligand-atom order: SCN versus NCS. The atom written first after the opening parenthesis is the donor. So SCN binds through S and NCS binds through N.

1. Identify the ambidentate ligand: SCN^- (thiocyanate). Its two donor atoms: S (soft) and N (harder).
2. Match HSAB: Pd(II) is soft, prefers S-bonding → Pd-SCN predominates; Cr(III) (hard) would prefer N-bonding. So the Pd(C6H5)2(SCN)2 form is usually more stable than Pd(C6H5)2(NCS)2, but both exist and are isolable.
3. The pair fits the definition of linkage (or ``ambidentate-ligand'') isomerism exactly.

Why this matters. Werner cited the [Co(NH3)5 NO2]²⁺ vs [Co(NH3)5 ONO]²⁺ pair as the first known case of linkage isomerism. Same conceptual setup, same HSAB analysis.

Linkage isomers, option (i).

Strategic angle. The first filter for every ``what kind of isomerism?'' MCQ is: do the two formulas have identical atom counts?

1. Count atoms in each compound: [Co(SO4)(NH3)5]BrCo₁ S₁ O₄ N₅ H₁₅ Br₁, [Co(SO4)(NH3)5]ClCo₁ S₁ O₄ N₅ H₁₅ Cl₁. Identical in every element except the halide. Different molecular formulas ⇒ not isomers.
2. Compare with a true ionisation isomer pair: [Co(NH3)5 Br]SO4 vs [Co(NH3)5 SO4]Br both have Co N5 H15 Br S O4 — identical formulas — and do differ only by swap of inner/outer-sphere anion.
3. Answer: no isomerism between the two compounds in this question.

Why this matters. A board examiner often plants this distractor (swap one halide for another) to test whether the student remembers the formula-conservation requirement. Mark the trap once, never miss it again.

No isomerism — option (iv).

Strategic angle. For each ligand, sketch the donor atoms and the ring it would close. If you cannot close a 5- or 6-membered ring, it cannot chelate.

1. Oxalato: ^-O-C(=O)-C(=O)-O^-, donors O ⋯ O separated by two carbons → 5-membered ring with M.
2. Glycinato: H2N-CH2-COO^-, donors N⋯ O separated by two atoms → 5-membered ring.
3. Ethane-1,2-diamine: H2N-CH2-CH2-NH2, donors N⋯ N separated by two carbons → 5-membered ring.
4. Thiosulphato: S=S(=O)(=O)-O^-. The four oxygens and the two sulfurs are bunched on a single tetrahedral S atom, too close together to allow simultaneous coordination to one metal — it sits as a monodentate ligand.

Why this matters. ``Number of donor atoms'' alone is not the criterion; geometry of the donor positions is. Thiosulphate has many potential donors but cannot chelate because of its compact geometry.

Thiosulphato, option (i).

Quick reading. Draw each Lewis structure mentally and look for a lone pair on the donor atom.

1. NO: :N=O with an extra electron in π^*. Lone pair on N → ligand.
2. NH4+: four single bonds to four hydrogens. All four electron pairs are committed. No lone pair → not a ligand.
3. H2N-CH2-CH2-NH2: each N has one lone pair → bidentate ligand.
4. :C≡O:: lone pair on C and one on O. Almost always binds through C → ligand.

Why this matters. Distinguishing ``NH3 is a ligand'' from ``NH4+ is not a ligand'' tests whether the student understands lone pairs and not just chemical names.

NH4+, option (ii).

Strategic angle. Same molecular formula + difference confined to solvent-vs-halide partitioning between spheres = solvate isomerism. Verify the formula match first.

1. Both formulas reduce to CrCl3 · 6 H2O. Without this match, ``isomerism'' is the wrong word.
2. Find what is moving: an H2O from outer → inner is balanced by a Cl^- from inner → outer (or vice versa). This is the solvate-isomerism fingerprint.
3. Distinguish from ionisation isomerism: the latter requires the anion (not the solvent) to swap; the moving partner here is H2O, hence ``solvate''/``hydrate''.

Why this matters. Solvate isomers are also called ``hydrate'' isomers in older NCERT editions. Both names appear in JEE answer keys; they refer to the same phenomenon.

Solvate (hydrate) isomerism, option (ii).

Quick reading. Each option is wrong for a specific reason. Knock them out one by one.

1. Option (i): ``platinum diaminechloronitrite'' — wrong word order (metal at the front), wrong spelling (``diamine'' with one m would mean an organic -NH2 group), no oxidation state. Eliminated.
2. Option (ii): ``chloronitrito-N-ammineplatinum(II)'' — wrong alphabetical order (chloro before ammine), missing the ``di'' multiplicity for NH3. Eliminated.
3. Option (iv): `` platinate(II)'' — the complex is neutral, so the metal name should be ``platinum'', not the anionic suffix ``platinate''. Eliminated.
4. Option (iii): correct word order (alphabetical: ammine, chlorido, nitrito-N), correct multiplicity (di-ammine), correct neutral metal name (platinum), correct oxidation state in roman numerals. Survives all four checks.

Why this matters. A clean nomenclature drill: spelling (ammine), 2005 update (chlorido), ambidentate specification (-N vs -O), alphabetical order, and -ate/-um distinction — all in one question.

Diamminechloridonitrito-N-platinum(II), option (iii).

Quick reading. Inner-orbital (low-spin) means electrons pair in t_2g. Only d⁶ ions can be fully paired in t_2g alone. Look for d⁶ and you have the diamagnetic ones.

1. Map each option to dⁿ: d⁶, d⁴, d⁶, d⁵ for (i), (ii), (iii), (iv).
2. Low-spin d⁶: t_2g⁶ e_g⁰, all paired → diamagnetic. Low-spin d⁴: one unpaired pair, so n=2. Low-spin d⁵: n=1.
3. Both (i) and (iii) are d⁶, hence diamagnetic; (ii) and (iv) carry unpaired electrons even in the low-spin limit.

Why this matters. [Fe(CN)6]^4- is the diamagnetic ``ferrocyanide'' anion; [Fe(CN)6]^3- is paramagnetic ferricyanide. One electron's difference, totally different magnetic behaviour — the cleanest demonstration that dⁿ matters.

(i) [Co(NH3)6]³⁺ and (iii) [Fe(CN)6]^4-.

Strategic angle. Outer-orbital means high-spin. Read off n_unpaired from the standard table; find two complexes that match.

1. Mn³⁺ (d⁴, high-spin): 4 unpaired. Fe³⁺ (d⁵, high-spin): 5 unpaired. Co³⁺ (d⁶, high-spin): 4 unpaired. Ni²⁺ (d⁸, high-spin): 2 unpaired.
2. Match: Mn³⁺ and Co³⁺ both have 4. They also have the same spin-only μ = √4(4+2) = √24 ≈ 4.90 BM.
3. Sanity check on (ii): [FeF6]^3- is famously μ = 5.92 BM (the highest possible for first-row transition metals).

Why this matters. ``Same number of unpaired electrons'' is the same as ``same spin-only magnetic moment'', which is what an experimentalist actually measures with a Gouy balance.

(i) and (iii), both with 4 unpaired electrons.

Strategic angle. Strong-field CN^- → low-spin → inner-orbital d² sp³. The geometry is therefore octahedral and the magnetic moment is small but nonzero (d⁵ low-spin keeps one unpaired electron).

1. Place CN^- on the spectrochemical series (very high) ⇒ pairing energy P < _o ⇒ low-spin.
2. Low-spin d⁵ in octahedral field: t_2g⁵ e_g⁰, so n_unpaired = 1. Two 3d orbitals are now empty and join the 4s, 4p orbitals in d² sp³ hybridisation (the two d's come from the (n-1) shell, hence ``inner orbital'').
3. Therefore: paramagnetic (one unpaired), d² sp³ inner orbital, μ ≈ 1.73 BM.

Why this matters. Compare with [FeF6]^3- (outer orbital, sp³ d², 5 unpaired, μ ≈ 5.92 BM). Same metal oxidation state, very different magnetic moments — the textbook demonstration of how ligand field strength rules.

d² sp³ hybridisation, paramagnetic — options (i) and (iii).

Strategic angle — two simultaneous truths. The colour flip pink → blue is driven by two facts that must hold at once: (a) a geometry change (oct → tet, drop in coordination number from 6 to 4) and (b) a splitting change (_o → _t = 49_o, much smaller). The correct options are exactly the ones that name these two facts.

1. Identify the new species. Excess Cl^- peels off the six aqua ligands and locks the cobalt into the cheap, low-coordination [CoCl4]^2-. Coordination number drops 6 → 4; geometry switches octahedral → tetrahedral. Option (ii) names this species — correct.
2. Quote the CFT inequality. Tetrahedral splitting is always smaller than octahedral for the same ligand set: _t = 49_o (≈ 0.45 _o). Because _abs = hc/Δ, smaller Δ means longer λ absorbed — and the observed colour (complement) shifts toward blue. Option (iii) states this — correct.
3. Rule out the distractors. Option (iv) flips the sign: it claims _t > _o, contradicting the geometric factor (only 4 ligands instead of 6, none on a d-orbital axis). Option (i) names [CoCl6]^4-, which would still be octahedral and would not explain the colour change — and it isn't even the species formed.

Why this matters. The pink ↔ blue colour flip is a single observation that probes three chapter ideas simultaneously: ligand substitution kinetics (H2O → Cl^-), coordination-number change (6 → 4), and _t/_o ratio — which is why it recurs in board, JEE (2019) and NEET papers year after year.

Options (ii) and (iii).

Quick reading. ``Homo'' = ``same''. Look at the coordination sphere; if you see only one symbol for ligands, it's homoleptic.

1. (i) and (iii) each list a single ligand species (six NH3 and four CN^- respectively).
2. (ii) and (iv) each list two different ligand species (NH3 + Cl^-).
3. Homoleptic: (i), (iii). Heteroleptic: (ii), (iv).

Why this matters. Many quick-fire NEET MCQs are decided by this single distinction. ``Homoleptic'' vs ``heteroleptic'' is also the key to predicting whether a complex can show geometrical isomerism.

(i) [Co(NH3)6]³⁺ and (iii) [Ni(CN)4]^2-.

Quick reading. Mirror of Q19. Two different ligand species in the brackets ⇒ heteroleptic.

1. (ii) and (iv) each have NH3 and Cl ligands on the same metal — heteroleptic.
2. (i) is six NH3 and (iii) is six CN^- — both homoleptic.
3. Heteroleptic complexes are necessary for geometrical isomerism (in suitable patterns) and for many catalytic complexes because mixed-ligand environments allow tunability.

Why this matters. Most useful catalysts (Wilkinson's, Grubbs, cisplatin) are heteroleptic — the differing ligands tune the metal centre's reactivity in opposite directions on the same molecule.

(ii) [Fe(NH3)4 Cl2]⁺ and (iv) [Co(NH3)4 Cl2].

Picture-first. Build the three-blade propeller in your head and look for a mirror plane.

1. [Co(en)3]³⁺ — three en chelates form a propeller; mirror image has the opposite handedness. No improper symmetry. Chiral.
2. cis-[Co(en)2 Cl2]⁺ — the two Cl adjacent; the two en chelates wrap the remaining four vertices in two possible helical orientations → two enantiomers. Chiral.
3. trans-[Co(en)2 Cl2]⁺ — the two Cl on the z-axis; the two en in the xy plane. Mirror plane is the xy plane itself → achiral.
4. [Cr(NH3)5 Cl]²⁺ — C_4v symmetry, multiple mirror planes → achiral.

Why this matters. Werner's resolution of [Co(en)2(NH3) Cl]²⁺ into two enantiomers (1911) clinched his octahedral geometry over the prism alternative — one of the foundational experiments in inorganic chemistry.

(i) [Co(en)3]³⁺ and (iii) cis-[Co(en)2 Cl2]⁺.

Strategic angle. Apply three quick tests: charge, number of donor atoms, and ring formation.

1. Charge: count formal charges on H2N-CH2-CH2-NH2 — all zero. Neutral → (i).
2. Donor count: two N atoms, each with a lone pair directed at the same metal → bidentate → (ii). Hence (iv) is wrong.
3. Ring test: M-N-C-C-N closes a 5-membered ring (the ``magic-number'' ring size) → chelating → (iii).

Why this matters. en is the most-used bidentate ligand in NCERT problems. Knowing its three signature properties (neutral, bidentate, chelating) handles half the chapter's nomenclature and isomerism questions.

Options (i), (ii), (iii).

Strategic angle — ambidentate scan. Three ambidentate ligand families show up in NCERT-level problems: NO2^- (nitrito-N vs nitrito-O), SCN^- (thiocyanato-S vs thiocyanato-N) and CN^- (cyano-C vs isocyano-N). Spot any of these in the formula and linkage isomerism follows automatically; spot none and it doesn't.

1. (i) Has NO2^-. Two attachment isomers: [Co(NH3)5 -NO2]²⁺ (nitrito-N, N-bound, yellow) and [Co(NH3)5 -ONO]²⁺ (nitrito-O, O-bound, red). Linkage isomerism — yes.
2. (ii) Has CO (only). Although CO has a lone pair on O as well, the O-bound ``isocarbonyl'' form is observed only in matrix-isolated exotic species. NCERT treats CO as C-bound only. Linkage isomerism — no.
3. (iii) Has SCN^-. Two isomers: [Cr(NH3)5 -SCN]²⁺ (thiocyanato-S) and [Cr(NH3)5 -NCS]²⁺ (thiocyanato-N). Iron and chromium typically bind N-end; soft metals like Pd²⁺, Pt²⁺ prefer S-end. Linkage isomerism — yes.
4. (iv) Has en (chelating N-only) and Cl^-. Neither offers a second donor atom. Linkage isomerism — no.

Why this matters. The Werner [Co(NH3)5 NO2]Cl2 vs [Co(NH3)5 ONO]Cl2 pair was the very first documented case of linkage isomerism (yellow → red on gentle warming) and helped cement the idea that the metal–ligand bond involves a specific donor atom, not just a generic anion.

(i) and (iii).

Strategic angle. Convert each compound into a ratio of the form 1:n by reading off the outer-sphere charge balance, then order by n.

1. For each complex, the outer-sphere anion count equals the positive charge of the inner complex:
   • [Co(NH3)3 Cl3]: inner sphere neutral, no outer ion, 1:0 ratio.
   • [Co(NH3)4 Cl2]Cl: inner +1, outer Cl^-, 1:1.
   • [Cr(NH3)5 Cl]Cl2: inner +2, outer 2 Cl^-, 1:2.
   • [Co(NH3)6]Cl3: inner +3, outer 3 Cl^-, 1:3.
2. Apply _m^∞(1:n) ≈ (n+1) × 100 S cm² mol^-1 (very rough), giving roughly 0, 100, 300, 400 — same monotonic ranking.
3. Hence increasing conductivity is exactly the increasing n order.

Why this matters. Werner used precisely this conductivity ladder to assign the inner/outer sphere of his series of Co(NH3)_x Cl_y compounds — long before X-ray work confirmed the geometry.

[Co(NH3)3 Cl3] < [Co(NH3)4 Cl2]Cl < [Cr(NH3)5 Cl]Cl2 < [Co(NH3)6]Cl3.

Strategic angle — constraints first. ``Total ions = 2'' is the strongest single constraint: it forces a 1:1 electrolyte (one cation + one anion). With that fixed, the AgNO3 test fixes which ion sits outside, and inner-sphere charge balance fixes the rest.

1. Conductance constraint. Two ions per formula unit means a 1:1 electrolyte. So the dissociation is [complex]⁺ + Cl^- (or the reverse). The only ionisable anion in CrCl3 · 4 H2O is chloride, so the outer-sphere ion is Cl^- (one chloride) and the complex cation must have charge +1.
2. Cl accounting. Total Cl in the formula = 3. Outer Cl = 1. Therefore 2 Cl sit inside the coordination sphere. Place all four water molecules inside as well: inner sphere occupancy is 4 H2O + 2 Cl^-. Coordination number = 4 + 2 = 6, matching octahedral Cr(III).
3. Charge check. Inner-sphere charge = (+3) + 4(0) + 2(-1) = +1, balanced by the outer Cl^-. Reacting with AgNO3 would liberate only the outer Cl^- as AgCl — exactly 1 mol of AgCl per mol of complex, a directly testable prediction.
4. IUPAC name. Inner sphere written in alphabetical order: aqua before chlorido → ``tetraaquadichlorido''. Metal: chromium(III) (oxidation state confirmed by inner charge balance). Outer: chloride. Full name: tetraaquadichloridochromium(III) chloride.

Why this matters. This compound is the greyish-green hydrate isomer of CrCl3 · 6H2O (see Q13). The three hydrate isomers of CrCl3 · 6H2O produce 3, 2 and 1 ions in solution respectively — a direct conductance test distinguishes them without any spectroscopy.

[Cr(H2O)4 Cl2]Cl, tetraaquadichloridochromium(III) chloride.

Strategic angle — symmetry first. The keyword ``optically active'' means lacks any improper symmetry element (no σ plane, no S_n axis, no centre of inversion). For an octahedral [M(AA)2 X2]ⁿ⁺ the only arrangement that satisfies that constraint is the cis isomer.

1. Geometry. Coordination number = 2 × 2 + 2 = 6 ⇒ octahedral. Two bidentate AA ligands occupy four cis vertices and lock the geometry into one of two possibilities: cis-X₂ or trans-X₂.
2. trans isomer: symmetry kills chirality. The two X sit along one axis; both AA chelates lie in the perpendicular plane. That equatorial plane is a σ plane of the molecule, hence the molecule is superimposable on its mirror image — achiral, optically inactive.
3. cis isomer: no improper symmetry. The two X sit adjacent (90^∘); the two AA chelates wrap the remaining four cis vertices into a helical pattern. There is no mirror plane, no S_n, no inversion centre. The molecule is chiral and exists as two non-superimposable enantiomers, conventionally labelled Δ (right-handed helix) and Λ (left-handed helix).
4. Resolution. Werner resolved cis-[Co(en)2 Cl2]⁺ into (+) and (-) enantiomers by fractional crystallisation of the bromocamphor sulfonate salt — the first non-carbon optical resolution ever performed (1911, Nobel 1913). Each enantiomer rotates plane-polarised light by roughly ± 90^∘ in opposite senses.

Why this matters. A polarimeter measurement is the cleanest pre-X-ray test for ``cis or trans?'' in a [M(AA)2 X2]ⁿ⁺ complex — observe optical activity, conclude cis. The same logic generalises to [M(AA)3]ⁿ⁺ tris-chelates (always chiral) and is the basis for understanding helical metal complexes that mimic DNA's double-helix chirality.

Cis-octahedral; e.g. cis-[Co(en)2 Cl2]⁺.

Strategic angle — two equations, two unknowns. The spin-only formula maps μ → n, and n + dⁿ together pin the hybridisation. Run the inversion and only one geometry is left consistent with the measured moment.

1. Invert the spin-only formula. μ = √n(n+2) ⇒ (5.92)² = 35.05 = n(n+2). Testing integer n: n=4 ⇒ 4 · 6 = 24 (gives μ = 4.9, too small); n=5 ⇒ 5 · 7 = 35 (gives μ = 5.92, match); n=6 ⇒ 42 (too large). So n = 5.
2. Map to dⁿ and oxidation state. The complex [MnCl4]^2- has overall charge -2, four Cl^- at -1 each, giving Mn = +2, configuration [Ar] 3d⁵. Five unpaired electrons in d⁵ means all five d electrons are unpaired — no pairing anywhere.
3. Eliminate square planar. A four-coordinate complex can be tetrahedral (sp³) or square planar (dsp²). Square planar on d⁵ would require pairing electrons into only four orbitals — at most 1 unpaired electron in the dsp² scheme, giving μ ≤ √3 = 1.73 BM. Incompatible with 5.92 BM. So square planar is killed.
4. Conclude: tetrahedral, sp³, high-spin. The tetrahedral sp³ scheme leaves the five d electrons in a e² t₂³ pattern (small _t keeps Hund's rule in charge). Plug n = 5 back: μ = √5 · 7 = √35 = 5.92 BM, matching observation.

Why this matters. Mn²⁺ (d⁵) is the textbook ``high-spin always'' example: pairing five electrons costs five times the pairing energy P, almost always larger than Δ (especially the smaller _t). Even strong-field ligands rarely beat it.

Tetrahedral, sp³, high-spin, 5 unpaired electrons, μ = 5.92 BM.

Picture-first. Sketch the t_2g/e_g splitting and feed 6 electrons in by either of two rules: ``Hund first'' (weak field) or ``pair first'' (strong field).

1. Weak-field (_o < P): the 6 electrons fan out across t_2g and e_g following Hund's rule. End state: t_2g⁴ e_g², n=4 unpaired, paramagnetic.
2. Strong-field (_o > P): the same 6 electrons pair up in t_2g because promotion to e_g costs more than pairing. End state: t_2g⁶ e_g⁰, n=0, diamagnetic.
3. Magnetic moments: 4.9 BM (weak) vs 0 BM (strong) — easy to distinguish in a Gouy balance.

Why this matters. Vitamin B₁₂ has a Co(III) centre surrounded by strong-field ligands and is therefore diamagnetic — directly the right-hand side of this question.

Weak field: t_2g⁴ e_g², 4 unpaired, paramagnetic. Strong field: t_2g⁶ e_g⁰, 0 unpaired, diamagnetic.

Quick reading. The single fact you need: _t ≈ 0.45 _o. Since octahedral d-block complexes only barely cross into low-spin territory in the strong-field limit, anything less than half their splitting will not.

1. For low spin, need Δ > P.
2. Even with the strongest field ligands, _t ≈ 49_o ≈ 0.45 _o < P for first-row transition metals.
3. Therefore low-spin tetrahedral complexes are essentially never observed in standard chemistry (a handful of exceptions with second/third row metals exist but are exotic).

Why this matters. Tetrahedral Co²⁺, Ni²⁺ and Mn²⁺ are always high-spin in practice — saves you from ever having to do the ``what if _t > P ?'' calculation.

_t is too small (49_o) to overcome P, so tetrahedral complexes are always high spin.

Strategic angle — three-step CFT algorithm. For each octahedral complex, run the same three-step pipeline: (a) oxidation state → dⁿ; (b) place the ligand on the spectrochemical series to decide high vs low spin; (c) fill t_2g/e_g using Hund (high spin) or pairing (low spin). The magnetic moment follows from μ = √n(n+2) BM.

1. [CoF6]^3-. Co oxidation: x + 6(-1) = -3 ⇒ x = +3, so Co³⁺ is [Ar] 3d⁶. Ligand F^- sits at the weak-field end of the spectrochemical series, so _o < P ⇒ high-spin filling: t_2g⁴ e_g², n = 4, μ = √24 = 4.9 BM (paramagnetic).
2. [Fe(CN)6]^4-. Fe oxidation: x + 6(-1) = -4 ⇒ x = +2, so Fe²⁺ is 3d⁶. Ligand CN^- is the strongest practical σ-donor plus π-acceptor, so _o > P ⇒ low-spin: t_2g⁶ e_g⁰, n = 0, μ = 0 BM (diamagnetic).
3. [Cu(NH3)6]²⁺. Cu oxidation: +2, configuration 3d⁹. With 9 electrons in 5 orbitals only one filling is possible — t_2g⁶ e_g³, n = 1 regardless of field strength. μ = √3 = 1.73 BM. Note: the asymmetric e_g³ occupancy triggers a Jahn–Teller distortion (axial elongation), but the spin count is unaffected.

Why this matters. This same trio of complexes (CoF6, Fe(CN)6, Cu(NH3)6) recurs in nearly every NCERT and CBSE board CFT question — memorise their configurations and you have a one-look answer for half the chapter's magnetic-moment problems.

t_2g⁴ e_g², t_2g⁶ e_g⁰, t_2g⁶ e_g³ respectively; μ = 4.9, 0, 1.73 BM.

Strategic angle — controlled experiment. Both complexes share the same metal centre (Fe³⁺, d⁵) and the same coordination number (6). The only variable is ligand field strength — and that variable alone controls n_unpaired, hence μ.

1. [Fe(H2O)6]³⁺, weak field. H2O sits in the middle of the spectrochemical series but is far below the strong-field threshold for d⁵ on Fe³⁺. So _o < P ⇒ high-spin d⁵: t_2g³ e_g² (all five orbitals singly occupied, Hund's rule). n = 5, μ = √5(5+2) = √35 = 5.92 BM. VBT picture: outer-orbital sp³ d² hybridisation using 4d orbitals.
2. [Fe(CN)6]^3-, strong field. CN^- is a powerful σ-donor and π-acceptor, pushing _o above P. So low-spin: t_2g⁵ e_g⁰ (four electrons pair up in the t_2g set; one stays unpaired). n = 1, μ = √1(1+2) = √3 = 1.73 BM. VBT: inner-orbital d² sp³ using 3d orbitals.
3. Pattern check. The ratio _H2O/ _CN = 5.92/1.73 ≈ 3.4 — a single ligand swap drops the moment by a factor of ∼ 3.4. The ordering matches the spectrochemical series H2O ≪ CN^-.

Why this matters. It is the cleanest experimental demonstration that _o can flip the magnetic class of a complex: same dⁿ, same coordination number, two utterly different magnetic moments and (because the colour also tracks _o) two different colours.

Weak-field H2O keeps 5 unpaired (μ = 5.92 BM); strong-field CN^- leaves only 1 (μ = 1.73 BM).

Quick reading. Same metal, same oxidation state → rank by ligand. Read the spectrochemical series and you're done.

1. Ligand position: Cl^- (weak), NH3 (moderate), CN^- (strong).
2. _o tracks ligand strength linearly.
3. Order: [CrCl6]^3- < [Cr(NH3)6]³⁺ < [Cr(CN)6]^3-.

Why this matters. The same logic explains the visible-light absorption shift from yellow ([CrCl6]^3-) to violet ([Cr(NH3)6]³⁺) to nearly colourless ([Cr(CN)6]^3-, where absorption has moved into the near-UV).

[CrCl6]^3- < [Cr(NH3)6]³⁺ < [Cr(CN)6]^3-.

Strategic angle — two independent variables. Geometry and ligand strength play different roles. Geometry pins the splitting pattern (t_2g/e_g for octahedral, e/t₂ for tetrahedral); ligand strength pins the splitting magnitude Δ. Whether Δ is bigger or smaller than the pairing energy P then decides high vs low spin, hence n_unpaired, hence μ. Confusing these two variables is the most common student mistake.

1. Same geometry ⇒ same Δ. Octahedral [CoF6]^3- and octahedral [Co(NH3)6]³⁺ share the same orbital labels (t_2g^? e_g^?) but _o(F^-) ≪ _o(NH3), by a factor of nearly two. So the same d⁶ count fills differently: t_2g⁴ e_g² (high spin, n = 4, μ = 4.9 BM, paramagnetic) versus t_2g⁶ e_g⁰ (low spin, n = 0, μ = 0 BM, diamagnetic).
2. Ligand strength flips the inequality. Weak field (Δ < P) ⇒ Hund spreading ⇒ high spin ⇒ many unpaired electrons. Strong field (Δ > P) ⇒ pairing in t_2g first ⇒ low spin ⇒ few unpaired.
3. Conclusion. Identical geometry plus identical dⁿ plus different ligand ⇒ different n_unpaired ⇒ different μ. Geometry alone is never enough to predict the magnetic moment — the ligand identity is co-essential.

Why this matters. Predicting magnetic behaviour from a complex's formula is impossible without knowing the spectrochemical position of every ligand. This is why every exam problem on μ deliberately mentions both the metal and the ligand — they are both essential inputs.

Same geometry can hold different ligand-field strengths, giving different n_unpaired and therefore different μ.

Picture-first. The cleanest mental model is two d-level diagrams side by side: one with a measurable t_2g/e_g splitting (the hydrate), and one with all five d orbitals degenerate (the anhydrous salt). The presence or absence of that gap controls whether a visible-light d–d transition can happen, and therefore whether the solid has a colour.

1. Hydrate, CuSO4 · 5 H2O. Four H2O molecules coordinate equatorially to Cu²⁺ (the fifth water is hydrogen-bonded to SO4^2-). The aqua ligand field splits d into t_2g and e_g with _o ≈ 12,500 cm^-1 — corresponding to λ ≈ 800 nm (red). The d⁹ Cu²⁺ has one empty e_g slot, so a d–d transition t_2g⁶ e_g³ → t_2g⁵ e_g⁴ absorbs that red band. Complementary colour = blue, the characteristic colour of blue vitriol.
2. Anhydrous CuSO4. Removing the aqua ligands removes the source of crystal-field splitting. The Cu²⁺ sits in a sulfate lattice where the next-nearest oxygens contribute only a very weak, nearly spherical field. The d orbitals are effectively degenerate, so no visible d–d band exists. The solid appears nearly white (very pale greenish-grey to the eye).
3. Reversibility — the school-lab demo. CuSO4 · 5 H2O CuSO4 + 5 H2O (Δ forward, +H2O reverse). Heating drives the equilibrium right (blue → white); adding water drives it left (white → blue). This is the standard CBSE chemistry-lab test for the presence of water.

Why this matters. The same ``need a d splitting AND a partially filled d shell'' logic predicts why ZnSO4 · 7H2O is colourless (Zn²⁺ is d¹⁰ — no empty d orbital to receive the excited electron) and why Sc2(SO4)3 is colourless (Sc³⁺ is d⁰ — no ground-state electron to excite). The colour of a compound is a fast diagnostic for the metal's d-electron count.

Aqua ligands in CuSO4 · 5H2O split the d orbitals and allow a d–d transition (absorbs red, appears blue). Anhydrous CuSO4 lacks ligand-induced splitting and is colourless.

Quick reading. ``Ambidentate ligand → linkage isomerism'' is a one-to-one tag. Memorise two ligand examples and you have full marks.

1. Isomerism name: linkage (a.k.a. ambidentate) isomerism.
2. Two clean ambidentate ligands: NO2^- (N- or O-bound) and SCN^- (S- or N-bound).
3. Hard/soft preference: hard metals prefer N donor (Cr³⁺, Co³⁺); soft metals prefer S donor (Pd²⁺, Pt²⁺, Hg²⁺).

Why this matters. The HSAB principle (Hard/Soft Acid Base) predicts which linkage isomer is the thermodynamically stable one for any specific metal–ligand pair.

Linkage isomerism; ambidentate ligands include NO2^- and SCN^-.

Strategic angle — _o to colour via the wheel. For each complex, estimate _o from metal+ligand, convert to _abs = hc/_o, then read the complementary colour off the colour wheel. The four complexes span the full range of _o from low (Ti-aqua) to high (Co-ammine).

1. A. [Co(NH3)6]³⁺. Co³⁺ d⁶ with strong-field NH3 gives the largest _o (∼ 23,000 cm^-1). Absorbed wavelength ∼ 470 nm (blue-violet); complement is yellow-orange. → A–4 (yellowish orange).
2. B. [Ti(H2O)6]³⁺. Ti³⁺ d¹ with aqua ligand; the textbook _o ≈ 20,300 cm^-1 absorbs near 493 nm (green). The transmitted colour is pale violet/blue. → B–3 (pale blue).
3. C. [Ni(H2O)6]²⁺. Ni²⁺ d⁸, aqua moderate field. Two visible bands (∼ 400 and ∼ 720 nm); the eye reads the survivor as green. → C–2 (green).
4. D. [Ni(H2O)4(en)]²⁺. Replacing two aqua ligands with one chelating en raises _o (en is stronger than aqua), shifts absorption toward shorter λ, and the complementary colour shifts to violet. → D–1 (violet).

Why this matters. Successive replacement of H2O by en in [Ni(H2O)6]²⁺ shifts the colour continuously: green → pale blue → violet — a visual, classroom-friendly demonstration of the spectrochemical series in real time.

A–4, B–3, C–2, D–1; option (ii).

Quick reading. Four iconic coordination compounds; commit their central metals to memory.

1. Chlorophyll = Mg-porphyrin (plants are green because Mg is bound in a porphyrin).
2. Haemoglobin = Fe-porphyrin (red colour from haem).
3. Wilkinson's catalyst = Rh(PPh3)3 Cl (Geoffrey Wilkinson, Nobel 1973).
4. Vitamin B₁₂ = Co-corrin (only known biological role of Co in mammals).

Why this matters. These four ``trophy'' compounds appear in every NCERT bioinorganic question; the metal–compound mapping is worth one mark per match.

A–5, B–4, C–1, D–2; option (i).

Strategic angle — VBT pipeline per complex. For each complex run the same three-step pipeline: (a) oxidation state → dⁿ; (b) coordination number + ligand field strength → hybridisation; (c) electron occupancy → n_unpaired. The matching column-II entries follow uniquely.

1. A. [Cr(H2O)6]³⁺. Cr³⁺ is d³ (three electrons in t_2g, all unpaired by Hund). With only three d electrons in an octahedral field, two inner 3d orbitals are automatically empty — d² sp³ inner-orbital regardless of ligand strength. n_unp = 3 ⇒ pair (3).
2. B. [Co(CN)4]^2-. Co²⁺ d⁷ with strong-field CN^- in a 4-coordinate setting — NCERT treats this as square planar dsp² with the 7 electrons packed into the lower four d orbitals plus one unpaired in the highest filled. n_unp = 1 ⇒ pair (1).
3. C. [Ni(NH3)6]²⁺. Ni²⁺ d⁸; NH3 is borderline but NCERT treats hexammine-nickel(II) as outer-orbital sp³ d² (high-spin) — t_2g⁶ e_g². n_unp = 2 ⇒ pair (5).
4. D. [MnF6]^4-. Mn²⁺ d⁵ with weak-field F^- — outer-orbital sp³ d² (high-spin), t_2g³ e_g² with all five unpaired. n_unp = 5 ⇒ pair (2).

Why this matters. A single match-the-column question compactly tests three independent skills at once — oxidation state arithmetic, ligand-field assignment (high vs low spin) and coordination-number → hybridisation mapping. Drilling all four together is the most efficient revision of VBT.

A–3, B–1, C–5, D–2.

Quick reading — pattern recognition. Four canonical formula patterns map one-to-one to four canonical isomerism types. Spot the pattern, name the isomerism.

1. A. [Co(NH3)4 Cl2]⁺. Formula pattern MA4 B2 — the classic octahedral pattern that supports cis/trans pairs → geometrical isomerism. Match (4).
2. B. cis-[Co(en)2 Cl2]⁺. The cis form of the M(AA)2 X2 pattern lacks any mirror plane — chiral → optical isomerism with Δ/Λ enantiomers. Match (1).
3. C. [Co(NH3)5 (NO2)]Cl2. The NO2^- is ambidentate (N- or O-bound), enabling [Co(NH3)5 (ONO)]Cl2 as the linkage partner → linkage isomerism. Match (5).
4. D. [Co(NH3)6][Cr(CN)6]. Both cation and anion are themselves coordination complexes. Swap the inner-sphere ligands between metals to get [Co(CN)6]^3- + [Cr(NH3)6]³⁺ — the coordination partner. → coordination isomerism. Match (3).

Why this matters. Pattern recognition is the entire isomerism game: four canonical formula patterns, four canonical isomerism types. Once these four mappings are internalised the question is reduced to a single glance at each formula.

A–4, B–1, C–5, D–3.

Strategic angle. Strip the outer ions to read the inner complex's charge, then solve for x given the ligand charges.

1. A: SO3^2- outside means cation +2. Inner: -1 from NCS, so x = +3.
2. B: SO4^2- outside means cation +2. Inner: -2 from two Cl, so x = +4.
3. C: 4 Na⁺ outside means anion -4. Inner: -6 from three S2O3^2-, so x = +2.
4. D: neutral molecule, all CO neutral, so each Co is 0.

Why this matters. Cobalt happens to be one of the few metals that exists in oxidation states ranging from -1 to +5 in coordination compounds. This question samples four of them in one go.

A: +3; B: +4; C: +2; D: 0. Match A–5, B–1, C–4, D–2.

Strategic angle — three-part A/R check. Every assertion/reason MCQ collapses to three sub-questions: (a) is the assertion true? (b) is the reason true? (c) does the reason actually explain the assertion (causal link)? Only when all three are yes does option (i) hold.

1. Assertion check. EDTA chelation therapy is the standard clinical intervention for Pb²⁺ poisoning; BAL (British anti-Lewisite, 2,3-dimercaptopropanol) is used for Hg²⁺ and As³⁺; deferoxamine is used for Fe³⁺ overload. Toxic metals are removed by chelating ligands. Assertion true.
2. Reason check. The chelate effect: a polydentate ligand replacing several monodentate ligands releases free water molecules, giving a large positive Δ S^∘ and therefore a much larger K than the corresponding monodentate complex (e.g. log K for [Ni(en)3]²⁺ is ∼ 18.3, vastly larger than log K ∼ 8.6 for [Ni(NH3)6]²⁺). Chelate complexes are more stable. Reason true.
3. Causal link. The very reason chelation works as a therapy is that the metal–chelator complex has a sufficiently large K to stay intact in blood and be excreted whole through the kidneys. If the chelate weren't stable, the metal would re-release into tissue. So the reason is the explanation ⇒ option (i).

Why this matters. Bioinorganic chemistry's most direct clinical application — chelation therapy literally saves lives in the emergency room and is part of the WHO essential medicines list.

Option (i): assertion true, reason true, reason explains assertion.

Strategic angle. Reducing tendency is governed by the stability of the product after losing one electron, not by the electronic structure of the starting complex. Test both parts independently and then test the logical link.

1. Assertion check. The standard reduction potential E^∘(Cr³⁺/Cr²⁺) = -0.41 V (highly negative, Cr²⁺ is a strong reductant) and E^∘(Fe³⁺/Fe²⁺) = +0.77 V (mildly positive, Fe²⁺ is a mild reductant). Both complexes are indeed reducing. Assertion true.
2. Reason check. Cr²⁺ is d⁴, n=4; Fe²⁺ is d⁶, n=4 (high-spin in H2O). Both have unpaired electrons. Reason true.
3. Logical link. Reducing power is set by the stability of the oxidised state:
   • Cr²⁺ → Cr³⁺ (loses one e^-) gives d³, t_2g³, a half-filled t_2g — extra stability, large driving force.
   • Fe²⁺ → Fe³⁺ gives d⁵, half-filled d shell, extra-stable.
   The cause is half-filled stability of the product, not unpaired electrons of the starting material.
4. Counter-example to the reason: [Mn(H2O)6]²⁺ (d⁵, n=5) is loaded with unpaired electrons yet is a poor reductant (E^∘(Mn³⁺/Mn²⁺) = +1.51 V) — exactly because oxidising it destroys a half-filled d⁵ configuration.
5. Therefore the reason is true but does not explain the assertion ⇒ option (ii).

Why this matters. The board examiner uses this question to test whether students confuse ``has unpaired electrons'' with ``is easily oxidised''. They are not the same; product stability is the real driver.

Option (ii): both true, reason not the correct explanation.

Quick reading — biconditional. Linkage isomerism and ambidentate ligand are biconditionally linked: an ambidentate ligand is exactly what enables linkage isomerism, and linkage isomerism exists only when an ambidentate ligand is present. The two-donor-atom property of the reason is precisely the structural feature that makes linkage isomerism possible.

1. No ambidentate ⇒ no linkage isomerism. With a single fixed donor atom (e.g. Cl^-, en, aqua), only one attachment is possible and no isomerism arises from donor-atom choice.
2. Ambidentate present ⇒ linkage isomers exist. Two attachment isomers form, e.g. M-NO2 (nitrito-N) vs M-ONO (nitrito-O), or M-SCN (thiocyanato-S) vs M-NCS (thiocyanato-N). They differ in colour, IR spectrum and stability.
3. Logical link. The definition of linkage isomerism rests on the dual-donor possibility — the reason is exactly the structural mechanism behind the assertion, so the reason is the correct explanation.

Why this matters. A clean ``definition + consequence'' pair — perfect option (i) pattern. Recognising this structure in A/R questions saves seconds on every exam.

Option (i): both true, reason explains assertion.

Strategic angle. Two separate sentences — validate each independently, then test logical implication.

1. Assertion analysis. MX6 has all six vertices identically occupied, only one octahedral arrangement is possible. MX5 L: the unique L can go to any of the six octahedral vertices; all six positions are equivalent under the O_h symmetry of the X5 framework, so the 5+1 arrangement collapses to a single structure. Both ligand sets indeed lack geometrical isomerism. Assertion true.
2. Reason analysis. The statement is the blanket claim ``coordination number 6 never shows GI''. Counter-examples from this very chapter:
   • [Cr(H2O)4 Cl2]⁺ (Q7): MA4 B2, two geometrical isomers (cis, trans).
   • [Co(NH3)4 Cl2]⁺ (Q39): same pattern, GI.
   • [Co(NH3)3 Cl3]: MA3 B3, two GIs (fac/mer).
   So octahedral complexes do show GI in many ligand-count patterns. Reason is too broad. Reason false.
3. Logical conclusion: assertion true + reason false ⇒ option (iii).

Why this matters. Assertion/reason questions often hide a distractor by stating an over-general reason that contradicts the detailed knowledge from elsewhere in the chapter. Recognising over-generalisation is half the skill of A/R MCQs.

Option (iii): assertion true, reason false.

Strategic angle — numbers first. Plug the observed magnetic moment into the spin-only formula. The integer n that falls out either confirms or falsifies the assertion's claim.

1. Solve for n from μ. The observed μ = 1.74 BM for [Fe(CN)6]^3- gives n(n+2) = (1.74)² ≈ 3. Integer trial: n = 1 ⇒ 1 · 3 = 3 (match, μ = √3 = 1.73); n = 2 ⇒ 2 · 4 = 8 (gives μ = 2.83, too large). So n = 1, not 2. Assertion claim of ``two unpaired electrons'' is false.
2. Verify the reason. Fe³⁺ is d⁵; strong-field CN^- produces low-spin t_2g⁵ e_g⁰, freeing two inner 3d orbitals for d² sp³ inner-orbital hybridisation. The reason ``d² sp³ hybridisation'' is therefore correct. Reason true.
3. Combine. Assertion false + reason true ⇒ option (iv).

Why this matters. The spin-only formula is the bridge between magnetic measurement and electronic structure — always calculate n from μ (or the reverse) before accepting any ``unpaired count'' claim in an A/R question.

Option (iv): assertion false (one unpaired electron, not two), reason true.

Strategic angle. Walk the same six complexes through one unified table — dⁿ, ligand strength, hybridisation, t_2g/e_g configuration, n_unp, μ — and verify the predictions against the spectrochemical series. This consolidates the six CFT computations into a single grid.

1. Compute dⁿ for each metal centre:
   • Co(III) in [CoF6]^3-, [Co(CN)6]^3-: Co³⁺, [Ar] 3d⁶.
   • Co(II) in [Co(H2O)6]²⁺: Co²⁺, [Ar] 3d⁷.
   • Fe(III) in [FeF6]^3-: Fe³⁺, [Ar] 3d⁵.
   • Fe(II) in [Fe(H2O)6]²⁺, [Fe(CN)6]^4-: Fe²⁺, [Ar] 3d⁶.
2. Place each ligand on the spectrochemical series: F^- < H2O < NH3 < CN^-. Hence F^- and H2O are weak-field (high-spin); CN^- is strong-field (low-spin).
3. Apply the high-spin or low-spin filling rule and read off n_unp. For d⁶: high-spin gives t_2g⁴ e_g², n=4; low-spin gives t_2g⁶ e_g⁰, n=0. For d⁷ high-spin: t_2g⁵ e_g², n=3. For d⁵ high-spin: t_2g³ e_g², n=5.
4. Plug into μ = √n(n+2) BM: arrayl|c|c|c Complex & n & n(n+2) & μ/BM
   [CoF6]^3- & 4 & 24 & 4.90
   [Co(H2O)6]²⁺ & 3 & 15 & 3.87
   [Co(CN)6]^3- & 0 & 0 & 0
   [FeF6]^3- & 5 & 35 & 5.92
   [Fe(H2O)6]²⁺ & 4 & 24 & 4.90
   [Fe(CN)6]^4- & 0 & 0 & 0
   array
5. Notice the pattern across each metal: replacing weak-field F^-/H2O with strong-field CN^- on the same d⁶ metal sends μ from 4.90 BM (paramagnetic) all the way to 0 (diamagnetic) — a 5-electron pairing event.
6. Hybridisation check: weak-field octahedral complexes use sp³ d² (outer-orbital, 4d orbitals involved); strong-field use d² sp³ (inner-orbital, 3d orbitals involved). So [CoF6]^3-, [Co(H2O)6]²⁺, [FeF6]^3-, [Fe(H2O)6]²⁺ are sp³ d²; [Co(CN)6]^3- and [Fe(CN)6]^4- are d² sp³.

Why this matters. Knowing both the energy-level diagram and the numerical μ tells the examiner you can both visualise the orbital occupancy and convert it into the measured property. The full table is also a one-page reference for every CFT/VBT problem in this chapter. The pattern across the six complexes is also pedagogically clean: weak-field F^- and H2O give outer-orbital paramagnetic species, strong-field CN^- gives inner-orbital diamagnetic species, and the crossover happens cleanly in d⁶ where pairing energy and _o are most evenly matched.

Cross-check via Gouy balance. Each of the six predicted moments is directly measurable by a Gouy balance experiment: hang a sample in a magnetic field gradient and weigh the apparent mass change. Paramagnetic samples are pulled into the field, diamagnetic samples are pushed out. Predicted versus observed agreement is ± 0.1 BM for all six complexes — the spin-only formula is quantitative for first-row 3d ions because orbital contribution to the moment is largely quenched by the ligand field.

(i) μ = 4.90, 3.87, 0 BM. (ii) μ = 5.92, 4.90, 0 BM. Strong-field CN^- drops d⁶ moments from ∼ 4.9 BM to 0.

Strategic angle. VBT decides inner vs outer orbital from ligand field strength; hybridisation pins the geometry (here all octahedral) and orbital occupancy gives the magnetic moment. Run the algorithm on each complex and assemble the table.

1. [Mn(CN)6]^3-: Mn³⁺, d⁴. Strong-field CN^- pairs the d electrons in t_2g⁴ e_g⁰; two empty 3d orbitals are freed for hybridisation ⇒ d² sp³ inner orbital. n_unp = 2, μ = √2(2+2) = √8 = 2.83 BM (paramagnetic).
2. [Co(NH3)6]³⁺: Co³⁺, d⁶. NH3 behaves as a strong-field ligand here ⇒ t_2g⁶ e_g⁰, d² sp³ inner orbital. n_unp = 0, μ = 0 BM (diamagnetic).
3. [Cr(H2O)6]³⁺: Cr³⁺, d³. Only three d electrons occupy the t_2g³ set; two empty 3d orbitals are automatically available, so d² sp³ inner orbital regardless of field strength. n_unp = 3, μ = √3(3+2) = √15 = 3.87 BM (paramagnetic).
4. [FeCl6]^4-: Fe²⁺, d⁶. Weak-field Cl^- gives high-spin t_2g⁴ e_g². No empty 3d orbital is available for hybridisation; instead 4s, 4p and 4d are used ⇒ sp³ d² outer orbital. n_unp = 4, μ = √4(4+2) = √24 = 4.90 BM (paramagnetic).
5. Summary: d² sp³ (inner orbital) for the first three; sp³ d² (outer orbital) for the last. All four are octahedral (coordination number 6).

Why this matters. VBT's inner vs outer orbital distinction is the direct precursor to CFT's high-spin vs low-spin labelling. Either framework predicts identical μ values from the same ligand set; the difference is that CFT also explains why (energy level diagram), whereas VBT only states the hybridisation.

μ values 2.83, 0, 3.87, 4.90 BM; first three are d² sp³ inner orbital, last is sp³ d² outer orbital.

Strategic angle — two diagnostic precipitations. Two classic qualitative-analysis tests pinpoint which anion is free in solution (i.e. outside the coordination sphere) for each isomer. The remaining anion must be locked inside.

1. Decode the tests. ``Gives white precipitate with AgNO3'' identifies free Cl<sup>-</sup> (Cl<sup>-</sup> + Ag<sup>+</sup> -> AgCl ↓). ``Gives white precipitate with BaCl2'' identifies free SO4^2- (SO4^2- + Ba²⁺ -> BaSO4 ↓). ``Does not react'' means the corresponding anion is hidden in the inner sphere, where it isn't free to encounter the test reagent.
2. Build A. A reacts with AgNO3 ⇒ outer Cl^-. A does not react with BaCl2 ⇒ inner SO4^2-. Structure: [Co(NH3)5 SO4]Cl. Charge check: inner +3 + 5(0) + (-2) = +1, balanced by one outer Cl^-.
3. Build B. B reacts with BaCl2 ⇒ outer SO4^2-. B does not react with AgNO3 ⇒ inner Cl^-. Structure: [Co(NH3)5 Cl]SO4. Charge check: inner +3 + 5(0) + (-1) = +2, balanced by outer SO4^2-.
4. Classify the isomerism. Same molecular formula CoSO4 Cl · 5 NH3, but the inner and outer anions are swapped between A and B. This is the textbook definition of ionisation isomerism.

Why this matters. The ``two anions, swap inner/outer'' diagnostic via precipitation is one of Werner's most elegant inferences — same molecular formula, different chemistry, different colours (A is brick-red, B is purple-violet). It illustrates how qualitative tests can probe coordination structure long before any modern spectroscopy.

A = [Co(NH3)5 SO4]Cl; B = [Co(NH3)5 Cl]SO4; ionisation isomerism.

Quick reading — one formula, one rule. The observed colour of a coordination complex is determined by two simultaneous facts: the Planck–Einstein relation _abs = hc/_o, and the colour-wheel complementarity rule ``observed = complement of absorbed''. Together they form a direct map from _o to visible colour.

1. Energy → wavelength. A d–d transition promotes a t_2g electron to e_g, absorbing a photon of energy E = _o = hc/_abs. Hence larger _o ⇒ shorter _abs (absorbs bluer/UV light) ⇒ transmits the redder/yellower half of the spectrum.
2. Wavelength → observed colour. A complex that absorbs around 400 nm (violet) appears yellow; around 500 nm (green) appears red; around 700 nm (red) appears green or blue. The two colours are diametrically opposite on the colour wheel.
3. Chain summary. Ligand identity sets _o via the spectrochemical series. _o sets _abs. _abs sets the observed colour (its complement). Three steps from formula to visual colour.

Why this matters. ``Why are coordination compounds coloured?'' is one of the most common board questions in this chapter. The full chain _o → _abs → observed colour is worth memorising verbatim because the question recurs across CBSE and NCERT exemplar in nearly every paper cycle.

Observed colour = complement of absorbed colour; _abs = hc/_o.

Strategic angle — geometry as an independent variable. Even when the metal and the ligands stay the same, the geometry alone changes the splitting magnitude. That in turn changes the absorbed wavelength and therefore the observed colour. Two facts drive the entire argument: the _t/_o = 4/9 ratio and the _abs = hc/Δ relation.

1. Quantify the geometry effect. The tetrahedral splitting is smaller than the octahedral splitting for two independent reasons: only 4 ligands instead of 6, and none of them sits along a d-orbital axis. Group-theoretical calculation gives _t = (4/9)_o ≈ 0.45_o.
2. Map to wavelength. Smaller Δ means longer _abs: _abs,tet/_abs,oct = _o/_t = 9/4 = 2.25. So the tetrahedral complex absorbs at roughly 2.25× longer wavelength, often moving the absorption from green/yellow into the red.
3. Map to observed colour. Octahedral [Co(H2O)6]²⁺ absorbs near 510 nm (green) and appears pink. Tetrahedral [CoCl4]^2- absorbs near 670 nm (red) and appears deep blue. Same Co²⁺ centre, dramatically different observed colour purely because the coordination geometry changed.

Why this matters. The colour-shifting test for the oct ↔ tet equilibrium of cobalt is a one-second visual diagnostic, used industrially in cobalt-impregnated humidity-indicator papers, anhydrous-cobalt drying agents and the classic blue-to-pink moisture test taught in every school lab.

_t ≈ (4/9)_o; different geometry ⇒ different absorbed wavelength ⇒ different colour.