Coordination Compounds

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Exercise 5.1Nomenclature, Isomerism, VBT and CFT

Write the formulas for the following coordination compounds: (i) Tetraamminediaquacobalt(III) chloride (ii) Potassium tetracyanonickelate(II) (iii) Tris(ethane-1,2-diamine)chromium(III) chloride (iv) Amminebromidochloridonitrito-N-platinate(II) (v) Dichloridobis(ethane-1,2-diamine)platinum(IV) bromide (vi) Iron(III) hexacyanidoferrate(II)

Solution

(i) Tetraamminediaquacobalt(III) chloride: [Co(NH₃)₄(H₂O)₂]Cl₃ (ii) Potassium tetracyanonickelate(II): K₂[Ni(CN)₄] (iii) Tris(ethane-1,2-diamine)chromium(III) chloride: [Cr(en)₃]Cl₃ (iv) Amminebromidochloridonitrito-N-platinate(II): [Pt(NH₃)(Br)(Cl)(NO₂)]⁻ (as an anion) or [Pt(NH₃)(Br)(Cl)(NO₂)] (v) Dichloridobis(ethane-1,2-diamine)platinum(IV) bromide: [Pt(en)₂Cl₂]Br₂ (vi) Iron(III) hexacyanidoferrate(II): Fe₄[Fe(CN)₆]₃ (Prussian blue)

Write the IUPAC names for the following coordination compounds: (i) [Co(NH₃)₆]Cl₃ (ii) [Co(NH₃)₅Cl]Cl₂ (iii) K₃[Fe(CN)₆] (iv) K₃[Fe(C₂O₄)₃] (v) K₂[PdCl₄] (vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl

Solution

(i) [Co(NH₃)₆]Cl₃: Hexaamminecobalt(III) chloride (ii) [Co(NH₃)₅Cl]Cl₂: Pentaamminechloridocobalt(III) chloride (iii) K₃[Fe(CN)₆]: Potassium hexacyanidoferrate(III) (iv) K₃[Fe(C₂O₄)₃]: Potassium tris(oxalato)ferrate(III) (v) K₂[PdCl₄]: Potassium tetrachloridopalladate(II) (vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl: Amminechiloridomethanaminplatinum(II) chloride (or: Amminechlorido(methanamine)platinum(II) chloride)

Indicate the types of isomerism exhibited by the following complexes and draw the structures of isomers: (i) K[Cr(H₂O)₂(C₂O₄)₂] (ii) [Co(en)₃]Cl₃ (iii) [Co(NH₃)₅(NO₂)]Cl₂

Solution

(i) K[Cr(H₂O)₂(C₂O₄)₂]: Geometric isomerism (cis/trans): • cis: both H₂O on same side, both oxalates on same side • trans: the two H₂O ligands trans to each other (ii) [Co(en)₃]Cl₃: Optical isomerism (enantiomers): • Δ (delta) isomer — right-handed propeller arrangement • Λ (lambda) isomer — left-handed propeller arrangement Both are non-superimposable mirror images (optical isomers) (iii) [Co(NH₃)₅(NO₂)]Cl₂: Linkage isomerism: • Nitro complex: [Co(NH₃)₅(NO₂)]Cl₂ — N is the donor atom (N-nitro) • Nitrito complex: [Co(NH₃)₅(ONO)]Cl₂ — O is the donor atom (O-nitrito) Same molecular formula, different bonding mode of NO₂⁻ ligand

What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S is passed through this solution?

Solution

When excess KCN is added to CuSO₄ solution: First: Cu²⁺ + 2CN⁻ → Cu(CN)₂ (unstable, immediately reduces) Then: 2Cu(CN)₂ → 2CuCN + (CN)₂ (Cu²⁺ oxidises CN⁻) Excess CN⁻: CuCN + 3CN⁻ → [Cu(CN)₄]³⁻ The coordination entity formed is tetracyanidocuprate(I): [Cu(CN)₄]³⁻ Why no CuS precipitate with H₂S: [Cu(CN)₄]³⁻ is an extremely stable complex with very low dissociation constant (Kd very small). The concentration of free Cu²⁺ (or Cu⁺) ions in solution is so low that the ionic product [Cu²⁺][S²⁻] does not exceed the Ksp of CuS. Therefore, CuS does not precipitate — the copper is effectively locked in the stable cyanide complex.

In a solution of acetic acid treated with ferric chloride, why does a blood-red colour appear?

Solution

When FeCl₃ is added to acetic acid (CH₃COOH) solution: Fe³⁺ ions react with acetate ions (CH₃COO⁻) to form an iron(III) acetate complex: Fe³⁺ + 3CH₃COO⁻ → [Fe(CH₃COO)₃] or related complex The blood-red (or dark reddish-brown) colour is due to the formation of iron(III) acetate complex (specifically [Fe(CH₃COO)]²⁺ or polynuclear acetate complexes). This is used as a qualitative test for acetic acid: addition of neutral FeCl₃ to acetate solutions gives a blood-red coloration which disappears on acidification (since H⁺ protonates CH₃COO⁻ to form undissociated acid, breaking the complex).

Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory: (i) [Fe(CN)₆]⁴⁻ (ii) [FeF₆]³⁻ (iii) [Co(C₂O₄)₃]³⁻ (iv) [CoF₆]³⁻

Solution

(i) [Fe(CN)₆]⁴⁻ — Fe²⁺ (d⁶): CN⁻ is a strong field ligand → causes pairing of electrons Fe²⁺ d⁶: rearranges to t₂g⁶ (all 6d electrons paired in 3 d orbitals) Hybridisation: d²sp³ (inner orbital complex) Geometry: Octahedral Magnetic: Diamagnetic (no unpaired electrons) (ii) [FeF₆]³⁻ — Fe³⁺ (d⁵): F⁻ is weak field ligand → no pairing Fe³⁺ d⁵: 5 unpaired electrons remain in d orbitals Hybridisation: sp³d² (outer orbital complex) Geometry: Octahedral Magnetic: Paramagnetic (5 unpaired electrons) (iii) [Co(C₂O₄)₃]³⁻ — Co³⁺ (d⁶): Oxalate is moderate field ligand → causes pairing (Co³⁺ is low spin) Hybridisation: d²sp³ Geometry: Octahedral Magnetic: Diamagnetic (iv) [CoF₆]³⁻ — Co³⁺ (d⁶): F⁻ is weak field → no pairing, d⁶ high spin Hybridisation: sp³d² Geometry: Octahedral Magnetic: Paramagnetic (4 unpaired electrons)

Draw the structures of optical isomers of: (i) [Cr(C₂O₄)₃]³⁻ (ii) [PtCl₂(en)₂]²⁺ (iii) [Cr(NH₃)₂Cl₂(en)]⁺

Solution

(i) [Cr(C₂O₄)₃]³⁻: Tris(oxalato)chromate(III) — octahedral complex with 3 bidentate oxalate ligands. Exhibits optical isomerism: • Δ (delta) form — right-handed propeller (Δ enantiomer) • Λ (lambda) form — left-handed propeller (Λ enantiomer) Both are non-superimposable mirror images. (ii) [PtCl₂(en)₂]²⁺: Octahedral — trans isomer has a plane of symmetry (optically inactive). Cis isomer: the two en ligands on same side — optically active. The cis form has Δ and Λ optical isomers (non-superimposable mirror images). (iii) [Cr(NH₃)₂Cl₂(en)]⁺: Octahedral — en occupies two cis positions. Depending on arrangement of NH₃ and Cl: multiple geometric isomers possible. The form where all asymmetric arrangement exists shows optical isomerism (Δ and Λ).

Calculate the magnetic moment (spin only) of [NiCl₄]²⁻.

Solution

Ni²⁺: electron configuration [Ar]3d⁸ Cl⁻ is weak field ligand → does not cause pairing [NiCl₄]²⁻ is tetrahedral (sp³ hybridisation) In tetrahedral field, crystal field splitting is small → high spin Ni²⁺ (d⁸) in tetrahedral field: e⁴ t₂⁴ arrangement — 2 unpaired electrons Number of unpaired electrons n = 2 Spin-only magnetic moment: μ = √(n(n+2)) BM = √(2(2+2)) = √8 = 2.83 BM

Calculate the magnetic moment of [Fe(CN)₆]³⁻.

Solution

Fe³⁺: electron configuration [Ar]3d⁵ CN⁻ is a strong field ligand → causes maximum pairing In octahedral field with strong ligand: t₂g⁵ eg⁰ arrangement Unpaired electrons = 1 (one electron in t₂g unpaired) Spin-only magnetic moment: μ = √(n(n+2)) = √(1(1+2)) = √3 = 1.73 BM The complex is low spin with 1 unpaired electron, μ = 1.73 BM.

Q10

Using crystal field theory, draw energy level diagrams for d orbitals, write electronic configurations of the central metal ion and determine the magnetic moments of the following complexes: (i) [CoF₆]³⁻ (ii) [Co(C₂O₄)₃]³⁻ (iii) [CrCl₆]³⁻

Solution

(i) [CoF₆]³⁻ — Co³⁺ (d⁶), F⁻ weak field: High spin d⁶: t₂g⁴ eg² Unpaired electrons: 4 (t₂g has 3 pairs + 1 single; eg has 2 singles) Actually: t₂g: ↑↓ ↑ ↑, eg: ↑ ↑ → 4 unpaired μ = √(4×6) = √24 = 4.90 BM (ii) [Co(C₂O₄)₃]³⁻ — Co³⁺ (d⁶), oxalate strong/moderate: Low spin d⁶: t₂g⁶ eg⁰ Unpaired electrons: 0 μ = 0 (diamagnetic) (iii) [CrCl₆]³⁻ — Cr³⁺ (d³), Cl⁻ weak field: d³: t₂g³ eg⁰ (same for both high and low spin with 3d electrons) Unpaired electrons: 3 μ = √(3×5) = √15 = 3.87 BM

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Coordination Compounds

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