Using Valence Bond Theory, explain why [Ni(CN) 4 ] 2- is square planar and diamagnetic while [NiCl 4 ] 2- is tetrahedral and paramagnetic.

In both ions, Ni is in +2 oxidation state (3d 8 ) with 2 unpaired electrons in free Ni 2+ . [Ni(CN) 4 ] 2- : Strong-field CN - forces pairing of the two unpaired 3d electrons, vacating one 3d orbital for dsp 2 hybridisation ( Square Planar ). Since n = 0, it is diamagnetic . [NiCl 4 ] 2- : Weak-field Cl - does not pair 3d electrons, so one 4s and three 4p orbitals undergo sp 3 hybridisation ( Tetrahedral ). Two unpaired electrons remain (n = 2), making it paramagnetic (μ = 2.83 BM).

Write the IUPAC names of: (a) [CoCl 2 (en) 2 ]Cl (b) K 3 [Fe(C 2 O 4 ) 3 ] (c) [Pt(NH 3 ) 2 Cl(NO 2 )] (d) [Ni(CO) 4 ]

(a) dichloridobis(ethane-1,2-diamine)cobalt(III) chloride (b) potassium trioxalatoferrate(III) (c) diamminechloridonitrito-N-platinum(II) (d) tetracarbonylnickel(0)

How many moles of AgCl are precipitated when 1 mole of CoCl 3 ·6NH 3 and 1 mole of CoCl 3 ·4NH 3 are separately treated with excess AgNO 3 ? Write their modern Werner formulas.

CoCl 3 ·6NH 3 → [Co(NH 3 ) 6 ]Cl 3 → gives 3 mol AgCl (3 ionisable Cl - counter-ions). CoCl 3 ·4NH 3 → [CoCl 2 (NH 3 ) 4 ]Cl → gives 1 mol AgCl (2 Cl - satisfy secondary valence inside brackets; only 1 Cl - is ionisable).

Draw the geometrical isomers of [Pt(NH 3 ) 2 Cl 2 ] and [Co(NH 3 ) 3 (NO 2 ) 3 ]. Which isomer of [Pt(NH 3 ) 2 Cl 2 ] is used in cancer treatment?

[Pt(NH 3 ) 2 Cl 2 ] (Square planar): Exists as cis (two Cl - adjacent at 90°) and trans (two Cl - opposite at 180°). The cis-isomer (cis-platin) inhibits tumour growth in chemotherapy. [Co(NH 3 ) 3 (NO 2 ) 3 ] (Octahedral Ma 3 b 3 ): Exists as facial (fac) (three identical ligands at corners of one octahedral triangular face) and meridional (mer) (around a meridian plane of the octahedron).

Give chemical test evidence to prove that [Co(NH 3 ) 5 Br]SO 4 and [Co(NH 3 ) 5 (SO 4 )]Br are ionisation isomers.

They give different ions in aqueous solution: [Co(NH 3 ) 5 Br]SO 4 yields SO 4 2- → gives a white precipitate of BaSO 4 with BaCl 2 , and no precipitate with AgNO 3 . [Co(NH 3 ) 5 (SO 4 )]Br yields Br - → gives a pale-yellow precipitate of AgBr with AgNO 3 , and no precipitate with BaCl 2 .

NCERT Plus Two (Class 12) Chemistry • Chapter 5

Coordination Compounds

Q: Draw a neat energy-level diagram showing the splitting of d orbitals in an octahedral crystal field. Give the energies of the split sets relative to the barycentre.

In an octahedral field, the 5 degenerate d orbitals split into two sets: e g set (d x 2 -y 2 , d z 2 ): Point directly along the axes toward ligands → raised in energy by +(3/5) Δ o (+0.6 Δ o ). t 2g set (d xy , d yz , d xz ): Lie between the axes → lowered in energy by -(2/5) Δ o (-0.4 Δ o ). The total energy separation between t 2g and e g is Δ o .

Q: Out of cis -[CrCl 2 (ox) 2 ] 3- and trans -[CrCl 2 (ox) 2 ] 3- , which one is optically active (chiral)? Give reason.

cis-[CrCl 2 (ox) 2 ] 3- is optically active (chiral) because it lacks a plane of symmetry and forms a non-superimposable mirror-image pair ( d and l enantiomers). The trans isomer possesses a plane of symmetry and is superimposable on its mirror image (achiral).

Q: Explain why [Ti(H 2 O) 6 ] 3+ is violet in colour while it becomes colourless upon heating.

In [Ti(H 2 O) 6 ] 3+ , Ti 3+ has a 3d 1 configuration ( t 2g 1 e g 0 ). When blue-green light (498 nm) is absorbed, the electron undergoes a d–d transition to e g 1 , causing the solution to transmit complementary violet colour. Upon heating, H 2 O ligands are lost; crystal field splitting disappears (Δ o = 0), so no d–d transition occurs and the substance turns colourless.

Werner's Theory, Ligands & Coordination Number, IUPAC Nomenclature, Isomerism, Valence Bond Theory & Crystal Field Splitting

📄 Official NCERT Textbook PDF

Topic 5.1: Werner's Theory of Coordination Compounds & Double Salts vs Complexes

Picture a crowded wedding hall in 1893. The groom sits inside the VIP stage surrounded closely by six bodyguards who never leave his side, while a few extra relatives stand outside near the exit gate. When a guest shakes hands at the gate, only the outside relatives greet them—the six VIP bodyguards stay locked around the groom! Back in 1898, Swiss chemist Alfred Werner realised cobalt chloride–ammonia compounds behave identically: only the outside Cl^- counter-ions drop out as white AgCl precipitate when excess AgNO₃ is added, while everything inside the square brackets [...] stays tightly bonded!

Coordination compounds are compounds in which a central metal atom or ion is bonded to a fixed number of anions or neutral molecules by sharing electron pairs. They are the backbone of biology—chlorophyll (Mg), haemoglobin (Fe), and vitamin B₁₂ (Co) are all coordination compounds.

Werner's Postulates (1898, Nobel Prize 1913)

Metals exhibit two types of linkages (valences): Primary valence and Secondary valence.
Primary valences are normally ionisable and satisfied by negative ions (correspond to oxidation state).
Secondary valences are non-ionisable, satisfied by neutral molecules or negative ions, fixed for a metal, and equal to its Coordination Number.
Groups bound by secondary linkages have characteristic spatial arrangements called coordination polyhedra (octahedral, tetrahedral, square planar).

Empirical Formula & Colour	Modern Werner Formula	AgCl ppt per mol	Electrolyte Type
CoCl₃·6NH₃ (Yellow)	[Co(NH₃)₆]³⁺ 3Cl^-	3 mol AgCl	1 : 3 electrolyte (4 ions)
CoCl₃·5NH₃ (Purple)	[CoCl(NH₃)₅]²⁺ 2Cl^-	2 mol AgCl	1 : 2 electrolyte (3 ions)
CoCl₃·4NH₃ (Green)	[CoCl₂(NH₃)₄]⁺ Cl^-	1 mol AgCl	1 : 1 electrolyte (2 ions)
CoCl₃·4NH₃ (Violet)	[CoCl₂(NH₃)₄]⁺ Cl^-	1 mol AgCl	1 : 1 electrolyte (2 ions)

NCERT Example 5.1: Assign secondary valences from AgCl precipitated per mole with excess AgNO₃:
• PdCl₂·4NH₃ → 2 mol AgCl ⇒ [Pd(NH₃)₄]Cl₂ ⇒ Secondary valence = 4
• NiCl₂·6H₂O → 2 mol AgCl ⇒ [Ni(H₂O)₆]Cl₂ ⇒ Secondary valence = 6
• PtCl₄·2HCl → 0 mol AgCl ⇒ H₂[PtCl₆] ⇒ Secondary valence = 6
• PtCl₂·2NH₃ → 0 mol AgCl ⇒ [PtCl₂(NH₃)₂] ⇒ Secondary valence = 4

Double Salt vs Complex Compound

Both form by combining two or more stable compounds in stoichiometric ratio, but double salts like Carnallite (KCl·MgCl₂·6H₂O), Mohr's salt (FeSO₄·(NH₄)₂SO₄·6H₂O), and Potash alum (KAl(SO₄)₂·12H₂O) dissociate completely into simple ions in water. In contrast, complexes like K₄[Fe(CN)₆] keep the complex ion [Fe(CN)₆]^4- intact and do not give tests for Fe²⁺ or CN^-!

In CoCl₃·5NH₃, total Cl is 3, yet ONLY 2 mol AgCl precipitate because one Cl^- sits inside the coordination bracket satisfying BOTH primary and secondary valence! Also remember: Mohr's salt tests positive for Fe²⁺, whereas K₄[Fe(CN)₆] does NOT.

Topic 5.2: Key Definitions: Ligands, Denticity, Chelate Effect, Coordination Number & Polyhedra

Think of a central metal ion as a person holding hands in a circle. A unidentate ligand like Cl^- or NH₃ holds on with one hand. A didentate ligand like ethylenediamine (en) grabs the metal with both hands like a crab claw—literally chele in Greek! That two-handed crab grip makes chelate complexes dramatically harder to break apart. And EDTA^4-? It's an octopus gripping the metal with six donor arms at once!

Coordination Entity & Central Atom/Ion: Central metal atom/ion (a Lewis acid) bonded to a fixed number of ligands (Lewis bases), e.g. [CoCl₃(NH₃)₃], [Fe(CN)₆]^4-.
Coordination Sphere & Counter Ions: Central atom + ligands inside square brackets [...] form the coordination sphere; ionisable species outside are counter ions (in K₄[Fe(CN)₆], counter ion is K⁺).
Classification of Ligands by Denticity (number of ligating donor atoms):
- Unidentate: 1 donor atom — Cl^-, H₂O, NH₃, CN^-, CO.
- Didentate: 2 donor atoms — en (ethane-1,2-diamine, H₂NCH₂CH₂NH₂), C₂O₄^2- (oxalate).
- Polydentate: Multiple donor atoms — N(CH₂CH₂NH₂)₃; hexadentate EDTA^4- binds via 2 N and 4 O atoms simultaneously.
Chelate Ligand & Chelate Effect: When a di- or polydentate ligand binds through two or more donor atoms to a single metal ion forming a ring, it is a chelate ligand. Chelate complexes are far more stable than analogous complexes with unidentate ligands.
Ambidentate Ligand: Has two different donor atoms, only one of which ligates at a time: NO₂^- (via N or O) and SCN^- (via S or N).

Coordination Number (CN) = Total number of σ (sigma) bonds between ligand donor atoms and the central metal (π bonds are NEVER counted!).
Examples: [PtCl₆]^2- → CN = 6 | [Ni(NH₃)₄]²⁺ → CN = 4 | [Fe(C₂O₄)₃]^3- → CN = 3 × 2 = 6 | [Co(en)₃]³⁺ → CN = 3 × 2 = 6

Coordination Polyhedron: Spatial arrangement of directly attached donor atoms — Octahedral ([Co(NH₃)₆]³⁺), Tetrahedral ([Ni(CO)₄]), Square Planar ([PtCl₄]^2-).
Oxidation Number: Charge the central atom would carry if all ligands were removed with their shared electron pairs, written in Roman numerals: e.g. Cu(I) in [Cu(CN)₄]^3-.
Homoleptic vs Heteroleptic: Only one kind of donor group = homoleptic ([Co(NH₃)₆]³⁺); more than one kind = heteroleptic ([Co(NH₃)₄Cl₂]⁺).

NEVER equate the number of ligands with the Coordination Number! In [Co(en)₃]³⁺ or [Fe(C₂O₄)₃]^3-, there are only 3 ligands, yet CN is 6 because each ligand is didentate (2 donor atoms). And remember: in EDTA^4-, denticity = 6 (2 N + 4 O).

Topic 5.3: IUPAC Nomenclature & Writing Formulas of Coordination Compounds

Think of naming a complex like introducing a royal visitor and then writing their passport formula—the two follow opposite outer steps, yet both sort guests alphabetically! When writing a formula, the metal king steps out first inside [...], followed by ligands alphabetically. When speaking the IUPAC name, the ligands introduce themselves alphabetically first, and if the complex bracket carries a negative charge, the metal takes a formal Latin-style -ate title like ferrate or argentate!

A. Rules for Writing Formulas (Mononuclear Entities)

List the central atom first, followed by ligands in alphabetical order (regardless of charge; abbreviations use their first letter).
Enclose polyatomic ligands and abbreviations in parentheses (...); enclose the whole entity in square brackets [...] with no spaces inside.
If written without counter ion, show net charge as right superscript with number before sign: [Co(CN)₆]^3-.

B. Rules for IUPAC Naming

Name the cation first, then the anion (never add di/tri for counter ions outside brackets!).
Inside the complex, name ligands in alphabetical order (ignoring prefixes di/tri/bis) BEFORE the central metal.
Anionic ligands end in -ido / -o (chlorido, cyanido, hydroxido, oxalato, carbonato, nitrito-N, thiocyanato-S). Neutral ligands keep their name EXCEPT: H₂O = aqua, NH₃ = ammine (double 'm'!), CO = carbonyl, NO = nitrosyl.
Use di, tri, tetra for simple ligands; use bis, tris, tetrakis (with ligand in parentheses) if the ligand name already contains a number (e.g. bis(ethane-1,2-diamine), bis(triphenylphosphine)).
Write oxidation state as Roman numeral in parentheses immediately after the metal: (0), (I), (II), (III).
If the complex ion is an anion, the metal ends in -ate (e.g. Fe → ferrate, Ag → argentate, Cu → cuprate, Zn → zincate, Co → cobaltate). Cationic & neutral complexes keep the regular English metal name!

NCERT Solved Examples (5.2 & 5.3):
1. [Cr(NH₃)₃(H₂O)₃]Cl₃ → triamminetriaquachromium(III) chloride
2. [Co(H₂NCH₂CH₂NH₂)₃]₂(SO₄)₃ → tris(ethane-1,2-diamine)cobalt(III) sulphate
3. [Ag(NH₃)₂][Ag(CN)₂] → diamminesilver(I) dicyanidoargentate(I) (Notice silver vs argentate!)
4. K₃[Cr(C₂O₄)₃] → potassium trioxalatochromate(III)
5. Hg[Co(SCN)₄] → mercury(I) tetrathiocyanato-S-cobaltate(III)
6. K₂[Zn(OH)₄] ← potassium tetrahydroxidozincate(II)

Watch three classic exam traps! (1) Spell ammonia ligand as ammine (two m's), NOT amine! (2) Alphabetical sorting ignores multiplying prefixes: triammine comes before trichlorido (a before c, ignoring tri). (3) Never write "trichloride" for 3 counter Cl^- ions outside brackets—just write chloride!

Topic 5.4: Isomerism: Structural (Linkage, Coordination, Ionisation, Solvate) & Stereoisomerism

Look at your left and right hands: identical thumb and four fingers, identical wrist joints—yet try sliding your right glove onto your left hand. It will never fit! Chemists call non-superimposable mirror-image pairs optical enantiomers. Meanwhile, sitting side-by-side at a square dining table versus sitting directly across each other gives cis and trans isomers—even tiny spatial swaps turn an ordinary platinum salt into the life-saving cancer drug cisplatin!

Isomers share the same chemical formula yet differ in physical or chemical properties.

A. Structural Isomerism (Different Bonds)

Type	Origin	Classic NCERT Example Pair
1. Linkage	Ambidentate ligand binds via two different donor atoms	`[Co(NH₃)₅(NO₂)]Cl₂` (Yellow, N-bound) vs `[Co(NH₃)₅(ONO)]Cl₂` (Red, O-bound)
2. Coordination	Ligand interchange between cationic & anionic metal spheres	`[Co(NH₃)₆][Cr(CN)₆]` vs `[Cr(NH₃)₆][Co(CN)₆]`
3. Ionisation	Counter ion swaps with a ligand inside the sphere	`[Co(NH₃)₅(SO₄)]Br` (gives AgBr ppt with Ag⁺) vs `[Co(NH₃)₅Br]SO₄` (gives BaSO₄ ppt with Ba²⁺)
4. Solvate (Hydrate)	H₂O bonded inside sphere vs free lattice crystal water	`[Cr(H₂O)₆]Cl₃` (Violet) vs `[Cr(H₂O)₅Cl]Cl₂·H₂O` (Grey-green)

B. Stereoisomerism (Same Bonds, Different 3D Spatial Arrangement)

1. Geometrical Isomerism (CN 4 & 6):
- Square planar [MX₂L₂]: Adjacent (90°) = cis, opposite (180°) = trans, e.g. [Pt(NH₃)₂Cl₂]. Square planar [MABXL] gives 3 geometrical isomers (2 cis, 1 trans).
- Tetrahedral [MX₂L₂]: Does NOT show geometrical isomerism because all 4 positions are symmetrically equivalent relative to each other!
- Octahedral [MX₂L₄] & [MX₂(L-L)₂]: cis (90°) vs trans (180°), e.g. [Co(NH₃)₄Cl₂]⁺, [CoCl₂(en)₂]⁺.
- Octahedral [Ma₃b₃]: e.g. [Co(NH₃)₃(NO₂)₃] shows facial (fac) (3 identical donors on one triangular face) and meridional (mer) (around a central meridian).
2. Optical Isomerism: Non-superimposable mirror images are called enantiomers (dextro, d rotates plane-polarised light right; laevo, l rotates left). Common in octahedral didentate chelates like [Co(en)₃]³⁺ and [Cr(C₂O₄)₃]^3-.

NCERT Example 5.5: Between cis-[CrCl₂(ox)₂]^3- and trans-[CrCl₂(ox)₂]^3-, ONLY the cis-isomer is chiral (optically active)! Similarly, in [PtCl₂(en)₂]²⁺ or [CoCl₂(en)₂]⁺, only the cis form gives d and l optical isomers; the trans form has a plane of symmetry and is optically inactive.

Remember two huge exam favorites: (1) Tetrahedral complexes NEVER show geometrical isomerism! (2) For [MX₂(en)₂], the cis form gives 2 optical isomers (d and l), whereas the trans form is optically INACTIVE—so total stereoisomers = 3 (cis-d, cis-l, trans), NOT 4!

Topic 5.5.1 – 5.5.3: Valence Bond Theory (VBT): Inner vs Outer Orbital Hybridisation & Magnetic Behaviour

Imagine a hotel with five inner VIP suites (3d orbitals) and empty outer rooms (4s, 4p, 4d). When six aggressive guests (strong-field ligands like NH₃ or CN^-) arrive, they push the existing metal electrons to double up inside three 3d suites, freeing two inner 3d suites to build a tight d²sp³ inner-orbital complex! Mild guests (like F^-) refuse to crowd anyone, so the metal spills out into the outer 4d annex to form a high-spin sp³d² outer-orbital complex!

According to Valence Bond Theory (VBT), under the influence of ligands, a metal uses its (n-1)d, ns, np OR ns, np, nd orbitals for hybridisation, forming equivalent directional orbitals that overlap with filled ligand orbitals.

Complex	Ion & Config	Hybridisation	Geometry	Unpaired e^- & Magnetic
[Co(NH₃)₆]³⁺	Co³⁺ (3d⁶)	d²sp³ (Inner / Low spin)	Octahedral	n = 0 ⇒ Diamagnetic
[CoF₆]^3-	Co³⁺ (3d⁶)	sp³d² (Outer / High spin)	Octahedral	n = 4 ⇒ Paramagnetic (4.90 BM)
[Ni(CN)₄]^2-	Ni²⁺ (3d⁸)	dsp² (Inner / Low spin)	Square Planar	n = 0 ⇒ Diamagnetic
[NiCl₄]^2-	Ni²⁺ (3d⁸)	sp³ (High spin)	Tetrahedral	n = 2 ⇒ Paramagnetic (2.83 BM)
[Ni(CO)₄]	Ni⁰ (3d⁸4s² → 3d¹⁰)	sp³	Tetrahedral	n = 0 ⇒ Diamagnetic
[Fe(CN)₆]^3-	Fe³⁺ (3d⁵)	d²sp³ (Inner)	Octahedral	n = 1 ⇒ Weakly Paramagnetic (1.73 BM)
[FeF₆]^3-	Fe³⁺ (3d⁵)	sp³d² (Outer)	Octahedral	n = 5 ⇒ Strongly Paramagnetic (5.92 BM)

Spin-only Magnetic Moment: μ = √(n(n + 2)) B.M. | (where n = number of unpaired electrons)
NCERT Example 5.7: [MnBr₄]^2- has μ = 5.9 BM ⇒ n = 5 unpaired e^- ⇒ must be Tetrahedral (sp³) rather than square planar!

Limitations of Valence Bond Theory (5.5.3)

Involves several assumptions; cannot give quantitative magnetic values.
Does NOT explain the colour of coordination compounds.
Cannot explain thermodynamic/kinetic stability or distinguish cleanly between weak and strong ligands.
Cannot predict without magnetic data whether a 4-coordinate complex is tetrahedral (sp³) or square planar (dsp²).

Always check both oxidation state AND d-electron count! [NiCl₄]^2- and [Ni(CO)₄] are BOTH tetrahedral (sp³), yet [NiCl₄]^2- is paramagnetic (Ni²⁺ has 3d⁸ → 2 unpaired e^-) whereas [Ni(CO)₄] is diamagnetic (Ni⁰ pairs all 10 electrons into 3d¹⁰)! Also remember [Ni(NH₃)₆]²⁺ is outer orbital (sp³d²) because Ni²⁺ (d⁸) cannot vacate two 3d orbitals even if paired!

Topic 5.5.4: Crystal Field Theory (CFT): Octahedral & Tetrahedral Splitting and Spectrochemical Series

Picture five identical d orbitals as five sleepy commuters sitting at the same energy level. Suddenly, six negatively charged ligands march straight down the x, y, and z axes to form an octahedron! Two axial orbitals (d_x²-y² and d_z²) sit right in the path of oncoming ligands and suffer fierce electrostatic repulsion—shooting upward to become the e_g set! Meanwhile, the three between-axis orbitals (d_xy, d_yz, d_xz) dodge the traffic and drop safely to the lower t_2g set!

Crystal Field Theory (CFT) is an electrostatic model treating ligands as point charges (anions) or point dipoles (neutral molecules). In an asymmetrical crystal field, the 5 degenerate d orbitals split!

A. Octahedral Crystal Field Splitting (Δ_o)

Higher e_g set (d_x²-y², d_z², pointing along axes): energy rises by +(3/5) Δ_o (or +0.6 Δ_o).
Lower t_2g set (d_xy, d_yz, d_xz, lying between axes): energy drops by -(2/5) Δ_o (or -0.4 Δ_o).

Spectrochemical Series (Increasing Field Strength / Δ_o):
I^- < Br^- < SCN^- < Cl^- < S^2- < F^- < OH^- < C₂O₄^2- < H₂O < NCS^- < edta^4- < NH₃ < en < CN^- < CO

B. High-Spin vs Low-Spin (Δ_o vs Pairing Energy P)

For d¹, d², d³ ions, electrons enter t_2g singly by Hund's rule. Starting at d⁴ (up to d⁷), two competing paths appear:

Δ_o < P (Weak Field Ligand): Cheaper to jump to e_g than pay pairing energy P ⇒ t_2g³ e_g¹ (for d⁴) → High-spin complex.
Δ_o > P (Strong Field Ligand): Cheaper to pair in t_2g ⇒ t_2g⁴ e_g⁰ (for d⁴) → Low-spin complex.

C. Tetrahedral Crystal Field Splitting (Δ_t)

In a tetrahedral complex, splitting is inverted (e set is lower by -(3/5) Δ_t, t₂ set is higher by +(2/5) Δ_t; note NO g subscript because a tetrahedron lacks a centre of symmetry!).

Δ_t = (4/9) Δ_o ⇒ Splitting is too small to exceed pairing energy P, so tetrahedral low-spin configurations are rarely observed!

Watch two classic CFT traps: (1) Tetrahedral orbitals are written simply as e and t₂ (WITHOUT the 'g' subscript — gerade applies only when a centre of symmetry exists, as in octahedral/square planar!). (2) In tetrahedral fields, t₂ is HIGHER in energy and e is LOWER—exact opposite of octahedral!

Topic 5.5.5 – 5.5.6: Colour in Coordination Compounds, d–d Transitions & Limitations of CFT

Why is a sapphire or copper sulphate crystal deep blue, while anhydrous CuSO₄ powder turns boring white when heated? Picture white sunlight—all rainbow colours mixed together—shining into a vial of purple [Ti(H₂O)₆]³⁺. Its lone d electron snatches 498 nm blue-green photons to leap from t_2g up to e_g. What passes out to your eyes? Everything left over—bright, glowing complementary violet! Heat away the water ligands, the d–d jump disappears, and the crystal turns completely colourless!

When white light passes through a transition metal complex, a specific visible wavelength is absorbed to excite an electron from a lower d level to a higher d level (d–d transition). The observed colour is the complementary colour of the absorbed wavelength.

Classic Example: [Ti(H₂O)₆]³⁺ (Violet)

Ti³⁺ is a 3d¹ system with ground state t_2g¹ e_g⁰. Absorption of 498 nm (blue-green) light excites the electron: t_2g¹ e_g⁰ → t_2g⁰ e_g¹, making the solution look violet!

Coordination Entity (Table 5.3)	λ Absorbed (nm)	Colour Absorbed	Observed Colour
[CoCl(NH₃)₅]²⁺	535	Yellow	Violet
[Co(NH₃)₅(H₂O)]³⁺	500	Blue-Green	Red
[Co(NH₃)₆]³⁺	475	Blue	Yellow-Orange
[Co(CN)₆]^3-	310 (UV)	Ultraviolet	Pale Yellow
[Cu(H₂O)₄]²⁺	600	Red	Blue
[Ti(H₂O)₆]³⁺	498	Blue-Green	Violet

Ligand Control of Colour (NCERT Fig. 5.11): Progressively replacing H₂O in green [Ni(H₂O)₆]²⁺ with stronger en ligands increases Δ_o, shifting absorption to shorter wavelengths and shifting colour smoothly:
[Ni(H₂O)₆]²⁺ (Green) → [Ni(H₂O)₄(en)]²⁺ (Pale Blue) → [Ni(H₂O)₂(en)₂]²⁺ (Blue/Purple) → [Ni(en)₃]²⁺ (Violet).
Gemstones: Both Ruby (Cr³⁺ in Al₂O₃) and Emerald (Cr³⁺ in beryl Be₃Al₂Si₆O₁₈) owe their vivid colours to d–d transitions of Cr³⁺ (d³) in different host crystal fields!

Limitations of Crystal Field Theory (5.5.6)

Treating ligands strictly as negative point charges wrongly predicts that anionic ligands should exert the largest Δ_o—yet in reality anions sit at the weak end of the spectrochemical series!
It ignores the covalent character of metal–ligand bonding (addressed later by Ligand Field Theory & MOT).

Two classic board questions: (1) Heating [Ti(H₂O)₆]Cl₃ or blue CuSO₄·5H₂O turns them colourless/white because removing ligands removes crystal field splitting! (2) Wavelength absorbed is inversely proportional to Δ_o (E = hc/λ), so stronger ligand ⇒ larger Δ_o ⇒ SHORTER λ absorbed: thus λ_abs order is [Ni(NO₂)₆]^4- < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺!

Topic 5.6 – 5.7: Metal Carbonyls (Synergic Bonding) & Applications in Biology, Medicine and Industry

Picture a two-way friendship handshake between a metal atom and carbon monoxide (CO). First, CO hands a lone pair of electrons from carbon into a vacant metal orbital to form a strong σ bond. Immediately, the electron-rich metal returns the favour by pushing electrons from its filled d orbitals back into the empty π* antibonding orbital of CO! That mutual give-and-take—synergic bonding—locks CO onto zero-valent metals so tightly that Mond's nickel refinery and haemoglobin oxygen transport rely on similar metal–ligand chemistry!

A. Homoleptic Metal Carbonyls & Synergic Bonding (5.6)

Transition metals in zero oxidation state form well-defined homoleptic carbonyls: [Ni(CO)₄] (Tetrahedral), [Fe(CO)₅] (Trigonal bipyramidal), [Cr(CO)₆] (Octahedral), [Mn₂(CO)₁₀] (two square-pyramidal Mn(CO)₅ units joined by a Mn–Mn bond), and [Co₂(CO)₈] (Co–Co bond bridged by two CO groups).

Synergic Bonding (M–C has both σ and π character):
• M–C σ bond: Donation of lone pair from carbonyl carbon into a vacant metal orbital.
• M–C π bond (Back-bonding): Donation of an electron pair from a filled metal d orbital into the vacant antibonding π* orbital of CO.
Result: Mutual reinforcement (synergic effect) strengthens the M–C bond!

B. High-Yield Applications of Coordination Compounds (5.7)

Field	Coordination Compound / Ligand & Specific Role
Biological Systems	Chlorophyll (Mg, photosynthesis); Haemoglobin (Fe, oxygen carrier); Vitamin B₁₂ / cyanocobalamin (Co, anti-pernicious anaemia); enzymes carboxypeptidase A & carbonic anhydrase.
Medicinal Chemistry	cis-platin (cis-[Pt(NH₃)₂Cl₂]) inhibits tumour growth; EDTA treats lead poisoning; D-penicillamine & desferrioxime B remove excess toxic Cu/Fe via chelate therapy.
Analytical Chemistry	Na₂EDTA titration estimates hardness of water (Ca²⁺ & Mg²⁺); DMG (for Ni²⁺), α-nitroso-β-naphthol, cupron detect metal ions.
Metallurgy & Refining	Gold & Silver extraction via cyanide complex `[Au(CN)₂]^-` displaced by Zn; Nickel purification via volatile `[Ni(CO)₄]` (Mond process).
Catalysis & Photography	Wilkinson catalyst `[(Ph₃P)₃RhCl]` hydrogenates alkenes; Black & white film fixing uses hypo (Na₂S₂O₃) dissolving AgBr as `[Ag(S₂O₃)₂]^3-`; smooth Ag/Au electroplating from `[Ag(CN)₂]^-` and `[Au(CN)₂]^-`.

Do NOT confuse back-bonding target orbitals! In metal carbonyls, the metal donates d electrons into the vacant antibonding π* orbital of CO—NOT into a π bonding orbital! Also memorize the exact central metals: Chlorophyll = Mg, Haemoglobin = Fe, Vitamin B₁₂ = Co, Wilkinson catalyst = Rh.

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March (Repeated Favorite) • 3 Marks): Using Valence Bond Theory, explain why [Ni(CN)₄]^2- is square planar and diamagnetic while [NiCl₄]^2- is tetrahedral and paramagnetic.

In both ions, Ni is in +2 oxidation state (3d⁸) with 2 unpaired electrons in free Ni²⁺.

[Ni(CN)₄]^2-: Strong-field CN^- forces pairing of the two unpaired 3d electrons, vacating one 3d orbital for dsp² hybridisation (Square Planar). Since n = 0, it is diamagnetic.
[NiCl₄]^2-: Weak-field Cl^- does not pair 3d electrons, so one 4s and three 4p orbitals undergo sp³ hybridisation (Tetrahedral). Two unpaired electrons remain (n = 2), making it paramagnetic (μ = 2.83 BM).

Q2 (Kerala HSE SAY / NCERT • 2 Marks): Draw a neat energy-level diagram showing the splitting of d orbitals in an octahedral crystal field. Give the energies of the split sets relative to the barycentre.

In an octahedral field, the 5 degenerate d orbitals split into two sets:

e_g set (d_x²-y², d_z²): Point directly along the axes toward ligands → raised in energy by +(3/5) Δ_o (+0.6 Δ_o).
t_2g set (d_xy, d_yz, d_xz): Lie between the axes → lowered in energy by -(2/5) Δ_o (-0.4 Δ_o).

The total energy separation between t_2g and e_g is Δ_o.

Q3 (Kerala HSE March • 2 Marks): Write the IUPAC names of: (a) [CoCl₂(en)₂]Cl (b) K₃[Fe(C₂O₄)₃] (c) [Pt(NH₃)₂Cl(NO₂)] (d) [Ni(CO)₄]

(a) dichloridobis(ethane-1,2-diamine)cobalt(III) chloride
(b) potassium trioxalatoferrate(III)
(c) diamminechloridonitrito-N-platinum(II)
(d) tetracarbonylnickel(0)

Q4 (Kerala HSE Model / NCERT • 2 Marks): How many moles of AgCl are precipitated when 1 mole of CoCl₃·6NH₃ and 1 mole of CoCl₃·4NH₃ are separately treated with excess AgNO₃? Write their modern Werner formulas.

CoCl₃·6NH₃ → [Co(NH₃)₆]Cl₃ → gives 3 mol AgCl (3 ionisable Cl^- counter-ions).
CoCl₃·4NH₃ → [CoCl₂(NH₃)₄]Cl → gives 1 mol AgCl (2 Cl^- satisfy secondary valence inside brackets; only 1 Cl^- is ionisable).

Q5 (Kerala HSE March (High Yield) • 3 Marks): Draw the geometrical isomers of [Pt(NH₃)₂Cl₂] and [Co(NH₃)₃(NO₂)₃]. Which isomer of [Pt(NH₃)₂Cl₂] is used in cancer treatment?

[Pt(NH₃)₂Cl₂] (Square planar): Exists as cis (two Cl^- adjacent at 90°) and trans (two Cl^- opposite at 180°). The cis-isomer (cis-platin) inhibits tumour growth in chemotherapy.

[Co(NH₃)₃(NO₂)₃] (Octahedral Ma₃b₃): Exists as facial (fac) (three identical ligands at corners of one octahedral triangular face) and meridional (mer) (around a meridian plane of the octahedron).

Q6 (Kerala HSE SAY • 2 Marks): Out of cis-[CrCl₂(ox)₂]^3- and trans-[CrCl₂(ox)₂]^3-, which one is optically active (chiral)? Give reason.

cis-[CrCl₂(ox)₂]^3- is optically active (chiral) because it lacks a plane of symmetry and forms a non-superimposable mirror-image pair (d and l enantiomers). The trans isomer possesses a plane of symmetry and is superimposable on its mirror image (achiral).

Q7 (Kerala HSE March • 2 Marks): Explain why [Ti(H₂O)₆]³⁺ is violet in colour while it becomes colourless upon heating.

In [Ti(H₂O)₆]³⁺, Ti³⁺ has a 3d¹ configuration (t_2g¹ e_g⁰). When blue-green light (498 nm) is absorbed, the electron undergoes a d–d transition to e_g¹, causing the solution to transmit complementary violet colour.

Upon heating, H₂O ligands are lost; crystal field splitting disappears (Δ_o = 0), so no d–d transition occurs and the substance turns colourless.

Q8 (Kerala HSE / NCERT Intext 5.4 • 2 Marks): Give chemical test evidence to prove that [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅(SO₄)]Br are ionisation isomers.

They give different ions in aqueous solution:

[Co(NH₃)₅Br]SO₄ yields SO₄^2- → gives a white precipitate of BaSO₄ with BaCl₂, and no precipitate with AgNO₃.
[Co(NH₃)₅(SO₄)]Br yields Br^- → gives a pale-yellow precipitate of AgBr with AgNO₃, and no precipitate with BaCl₂.

Q9 (Kerala HSE March • 2 Marks): Explain the nature of bonding in metal carbonyls with a labelled interaction scheme.

The metal–carbon bond in metal carbonyls has both σ and π character (synergic bonding):

M–C σ bond: Formed by donation of the lone pair on the carbonyl carbon into a vacant orbital of the metal.
M–C π bond: Formed by donation of an electron pair from a filled d orbital of the metal into the vacant antibonding π* orbital of CO.

This mutual metal–ligand interaction produces a synergic effect that strengthens the M–C bond.

Q10 (Kerala HSE Improvement • 2 Marks): What is meant by (a) Ambidentate ligand and (b) Chelate effect? Give one example of each.

(a) Ambidentate ligand: A unidentate ligand having two different donor atoms, either of which can coordinate to the metal at a time. Examples: NO₂^- (via N or O) and SCN^- (via S or N).
(b) Chelate effect: Enhanced stability of complexes formed by di- or polydentate chelating ligands forming closed rings compared to analogous complexes with unidentate ligands. Example: [Fe(C₂O₄)₃]^3- or [Co(en)₃]³⁺.

Q11 (Kerala HSE March • 1 Marks): Name the central metal ions present in: (a) Chlorophyll (b) Haemoglobin (c) Vitamin B₁₂.

(a) Chlorophyll: Magnesium (Mg) | (b) Haemoglobin: Iron (Fe) | (c) Vitamin B₁₂ (cyanocobalamin): Cobalt (Co).

Q12 (Kerala HSE SAY / NCERT Ex 5.1 • 1 Marks): FeSO₄ mixed with (NH₄)₂SO₄ (1:1) gives the test for Fe²⁺ ion, whereas CuSO₄ mixed with aqueous NH₃ (1:4) does not give the test for Cu²⁺ ion. Why?

FeSO₄·(NH₄)₂SO₄·6H₂O (Mohr's salt) is a double salt that dissociates completely into free Fe²⁺, NH₄⁺, and SO₄^2- ions in water. In contrast, CuSO₄ + 4NH₃ forms the coordination complex [Cu(NH₃)₄]SO₄; its complex ion [Cu(NH₃)₄]²⁺ does not dissociate into free Cu²⁺ ions.