Coordination Compounds

Given: Names of coordination compounds. We apply IUPAC rules: write the coordination entity in square brackets with metal last, then counter-ions outside.

(i) tetraamminediaquacobalt(III) chloride
- Central metal: Co(III), i.e., Co³⁺
- Ligands: 4 NH₃ (tetraammine) + 2 H₂O (diaqua)
- Charge on complex ion: +3, so 3 Cl⁻ outside
$$[\text{Co}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_3$$

(ii) potassium tetracyanidonickelate(II)
- Central metal: Ni(II), i.e., Ni²⁺ (anionic complex → potassium outside)
- Ligands: 4 CN⁻ (tetracyanido)
- Charge on complex ion: 2−4 = −2, so K₂ outside
$$K_2[\text{Ni}(\text{CN})_4]$$

(iii) tris(ethane-1,2-diamine)chromium(III) chloride
- Central metal: Cr(III), i.e., Cr³⁺
- Ligands: 3 en (tris(ethane-1,2-diamine))
- Charge on complex ion: +3, so 3 Cl⁻ outside
$$[\text{Cr}(\text{en})_3]\text{Cl}_3$$

(iv) amminebromidochloridonitrito-N-platinate(II)
- Central metal: Pt(II), anionic complex
- Ligands: NH₃ (ammine), Br⁻ (bromido), Cl⁻ (chlorido), NO₂⁻ bonded through N (nitrito-N)
- Charge: 2 − 1 − 1 − 1 = −1, so no outer cation shown (the name implies it is an anion; written as the anion)
$$[\text{Pt}(\text{NH}_3)\text{BrCl}(\text{NO}_2)]^-$$

(v) dichloridobis(ethane-1,2-diamine)platinum(IV) nitrate
- Central metal: Pt(IV), i.e., Pt⁴⁺
- Ligands: 2 Cl⁻ (dichloro) + 2 en (bis(ethane-1,2-diamine))
- Charge on complex ion: 4 − 2 = +2, so 2 NO₃⁻ outside
$$[\text{PtCl}_2(\text{en})_2](\text{NO}_3)_2$$

(vi) iron(III) hexacyanidoferrate(II)
- Two metal centres: Fe³⁺ (cation) and Fe²⁺ (in anionic complex)
- Complex anion: [Fe(CN)₆]⁴⁻
- Charge balance: 3 Fe³⁺ balanced by 4 [Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃
$$\text{Fe}_4[\text{Fe}(\text{CN})_6]_3$$

Rules used: Name ligands alphabetically before the metal; use prefixes di-, tri-, etc. for simple ligands and bis-, tris- for complex ones; state oxidation state of metal in Roman numerals; for anionic complexes add suffix '-ate'.

(i) [Co(NH₃)₆]Cl₃
- Ligands: 6 NH₃ → hexaammine
- Metal: Co; charge = +3 → cobalt(III)
- Anion: Cl⁻ → chloride
Name: Hexaamminecobalt(III) chloride

(ii) [Co(NH₃)₅Cl]Cl₂
- Ligands: 5 NH₃ (pentaammine) + 1 Cl⁻ (chlorido)
- Metal: Co; charge: let x + 0×5 − 1 = +2 (overall cation charge) → x = +3 → cobalt(III)
- Anion: 2 Cl⁻ → chloride
Name: Pentaamminechloridocobalt(III) chloride

(iii) K₃[Fe(CN)₆]
- Cation: K⁺ (potassium)
- Ligands: 6 CN⁻ → hexacyanido
- Metal: Fe; charge: 3K⁺ balances −3 on complex; x − 6 = −3 → x = +3 → ferrate(III) (anionic complex)
Name: Potassium hexacyanidoferrate(III)

(iv) K₃[Fe(C₂O₄)₃]
- Cation: K⁺ (potassium)
- Ligands: 3 C₂O₄²⁻ (oxalato) → trioxalato (or tris(oxalato))
- Metal: Fe; x − 6 = −3 → x = +3 → ferrate(III)
Name: Potassium trioxalatoferrate(III)

(v) K₂[PdCl₄]
- Cation: K⁺ (potassium)
- Ligands: 4 Cl⁻ → tetrachloro
- Metal: Pd; x − 4 = −2 → x = +2 → palladate(II)
Name: Potassium tetrachloridopalladate(II)

(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
- Ligands (alphabetical): Cl⁻ (chlorido), NH₂CH₃ (methanamine), 2 NH₃ (diammine)
- Metal: Pt; charge: x − 1 = +1 → x = +2 → platinum(II)
- Anion: Cl⁻ → chloride
Name: Diamminechloridomethanamiineplatinum(II) chloride
*(More precisely: Diamminechlorido(methanamine)platinum(II) chloride)*

(i) K[Cr(H₂O)₂(C₂O₄)₂]

Types of isomerism: Geometrical isomerism (cis and trans) and Optical isomerism (for the cis form).

- The complex ion [Cr(H₂O)₂(C₂O₄)₂]⁻ is octahedral with two water molecules and two bidentate oxalate ligands.
- Cis isomer: The two H₂O ligands are adjacent (90° apart). This cis form is non-superimposable on its mirror image → optical isomers (Δ and Λ).
- Trans isomer: The two H₂O ligands are opposite (180° apart). This form has a plane of symmetry → optically inactive.

Structures (described):
- *trans*: H₂O–Cr–OH₂ along one axis; two oxalates in the equatorial plane.
- *cis*: H₂O and H₂O adjacent; two oxalates spanning adjacent positions → gives d and l (Δ and Λ) optical isomers.

Summary: Geometrical isomers (cis and trans) + optical isomers of the cis form.

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(ii) [Co(en)₃]Cl₃

Type of isomerism: Optical isomerism only.

- The complex [Co(en)₃]³⁺ is octahedral with three bidentate en ligands. It has no plane of symmetry.
- Two non-superimposable mirror images exist: Δ (d) and Λ (l) forms.
- No geometrical isomerism is possible since all three ligands are identical bidentate ligands.

Summary: Two optical isomers (enantiomers).

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(iii) [Co(NH₃)₅(NO₂)](NO₃)₂

Types of isomerism:
1. Linkage isomerism: NO₂⁻ can coordinate through N (nitrito-N, –NO₂) or through O (nitrito-O, –ONO).
- Isomer 1: [Co(NH₃)₅(NO₂)]²⁺ — N-bonded (nitro)
- Isomer 2: [Co(NH₃)₅(ONO)]²⁺ — O-bonded (nitrito)
2. Ionisation isomerism: The NO₃⁻ outside can exchange with NO₂⁻ inside:
- [Co(NH₃)₅(NO₂)](NO₃)₂ ↔ [Co(NH₃)₅(NO₃)](NO₃)(NO₂) ↔ [Co(NH₃)₅(NO₃)₂]NO₂
3. Geometrical isomerism: Not applicable here (all NH₃ are equivalent in a MA₅B type complex — only one geometric form possible).

Summary: Linkage isomerism and ionisation isomerism (total ~10 possible isomers when all types are considered).

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(iv) [Pt(NH₃)(H₂O)Cl₂]

Type of isomerism: Geometrical isomerism (cis and trans).

- Square planar complex of type MA₂BC (two different unidentate ligands Cl₂, plus NH₃ and H₂O).
- Cis isomer: Two Cl ligands are adjacent to each other.
- Trans isomer: Two Cl ligands are opposite to each other.

Structures:
- *cis*-[Pt(NH₃)(H₂O)Cl₂]: Cl–Pt–Cl at 90°; NH₃ and H₂O adjacent.
- *trans*-[Pt(NH₃)(H₂O)Cl₂]: Cl–Pt–Cl at 180°; NH₃ and H₂O opposite each other.

Summary: Two geometrical isomers (cis and trans).

Concept: Ionisation isomers give different ions in solution and therefore react differently with specific reagents.

Evidence from chemical reactions:

Test with BaCl₂/Ba²⁺ solution (test for SO₄²⁻):
$$[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 \xrightarrow{\text{water}} [\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} + \text{SO}_4^{2-}$$
Adding Ba²⁺: $\text{Ba}^{2+} + \text{SO}_4^{2-} \rightarrow \text{BaSO}_4 \downarrow$ (white precipitate formed ✓)

$$[\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl} \xrightarrow{\text{water}} [\text{Co}(\text{NH}_3)_5(\text{SO}_4)]^{+} + \text{Cl}^-$$
Adding Ba²⁺: No free SO₄²⁻ in solution → No precipitate ✗

Test with AgNO₃/Ag⁺ solution (test for Cl⁻):
$$[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{SO}_4 + \text{Ag}^+ \rightarrow \text{No reaction}$$ (Cl⁻ is inside the coordination sphere, not free)

$$[\text{Co}(\text{NH}_3)_5(\text{SO}_4)]\text{Cl} + \text{Ag}^+ \rightarrow \text{AgCl} \downarrow$$ (white precipitate formed ✓)

Conclusion: The two compounds produce different ions in solution and respond differently to the same reagents, confirming they are ionisation isomers.

Given: [Ni(CN)₄]²⁻ (square planar, diamagnetic) and [NiCl₄]²⁻ (tetrahedral, paramagnetic).

Electronic configuration of Ni²⁺:
Ni: [Ar] 3d⁸ 4s²; Ni²⁺: [Ar] 3d⁸ (8 electrons in 3d)

$$\text{Ni}^{2+}: \underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow}_{3d^8}$$

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[Ni(CN)₄]²⁻ — Square Planar, Diamagnetic:

CN⁻ is a strong field ligand. It forces the two unpaired 3d electrons to pair up:
$$\text{Ni}^{2+} \text{ (in CN⁻ field)}: \underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{\phantom{x}}}_{3d}$$
One 3d orbital becomes empty. Hybridisation: dsp² (one 3d + one 4s + two 4p orbitals).

$$\underbrace{[\text{3d}]}_{\text{one empty}} \underbrace{[\text{4s}][\text{4p}][\text{4p}]}_{} \rightarrow dsp^2 \text{ (square planar)}$$

All electrons are paired → Diamagnetic ✓

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[NiCl₄]²⁻ — Tetrahedral, Paramagnetic:

Cl⁻ is a weak field ligand. It does not cause pairing of 3d electrons.
$$\text{Ni}^{2+} \text{ (in Cl⁻ field)}: \underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow}_{3d^8}$$
Hybridisation uses 4s and three 4p orbitals: sp³ (tetrahedral).

Two unpaired electrons remain in 3d orbitals → Paramagnetic ✓

Magnetic moment: $\mu = \sqrt{n(n+2)} = \sqrt{2(4)} = 2.83$ BM

Given: Both [NiCl₄]²⁻ and [Ni(CO)₄] are tetrahedral, yet one is paramagnetic and the other diamagnetic.

Key difference: Oxidation state of Ni and nature of ligand.

In [NiCl₄]²⁻:
- Ni is in +2 oxidation state → Ni²⁺: [Ar] 3d⁸
- Cl⁻ is a weak field ligand → does not cause pairing of electrons
- 3d⁸ configuration retains 2 unpaired electrons
- Hybridisation: sp³ (tetrahedral)
- Result: Paramagnetic ($\mu = 2.83$ BM)

In [Ni(CO)₄]:
- Ni is in zero oxidation state → Ni⁰: [Ar] 3d¹⁰ 4s⁰ (in the complex, 4s electrons shift to 3d)
- Actually Ni⁰: [Ar] 3d¹⁰ — all 3d orbitals are completely filled
- CO is a strong field ligand and also causes back-bonding; in Ni(0), the configuration is 3d¹⁰
- Hybridisation: sp³ (tetrahedral)
- All electrons are paired → Diamagnetic

Conclusion: The difference arises because in [NiCl₄]²⁻, Ni is Ni²⁺ (3d⁸, 2 unpaired electrons) with weak-field Cl⁻, while in [Ni(CO)₄], Ni is Ni⁰ (3d¹⁰, no unpaired electrons) with strong-field CO.

Given: Fe³⁺ is the central metal in both complexes. Fe³⁺: [Ar] 3d⁵ (5 unpaired electrons in free ion).

[Fe(H₂O)₆]³⁺ — Strongly Paramagnetic:
- H₂O is a weak field ligand → small crystal field splitting (Δ₀ < P, pairing energy)
- 3d electrons do not pair up → high spin complex
- Electronic configuration: $t_{2g}^3\ e_g^2$ → 5 unpaired electrons
- Hybridisation: sp³d² (outer orbital complex)
- Magnetic moment: $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM → strongly paramagnetic

[Fe(CN)₆]³⁻ — Weakly Paramagnetic:
- CN⁻ is a strong field ligand → large crystal field splitting (Δ₀ > P)
- 3d electrons pair up as much as possible → low spin complex
- Electronic configuration: $t_{2g}^5\ e_g^0$ → 1 unpaired electron
- Hybridisation: d²sp³ (inner orbital complex)
- Magnetic moment: $\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM → weakly paramagnetic

Conclusion: The difference is due to the field strength of the ligands. H₂O (weak field) gives high-spin Fe³⁺ with 5 unpaired electrons; CN⁻ (strong field) gives low-spin Fe³⁺ with only 1 unpaired electron.

[Co(NH₃)₆]³⁺ — Inner Orbital Complex:

- Co³⁺: [Ar] 3d⁶ → 6 electrons in 3d
- Free Co³⁺: $\underbrace{\uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow}_{3d^6}$ (4 unpaired)
- NH₃ is a strong field ligand → forces pairing of 3d electrons:
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{}\ \boxed{}}_{3d} \rightarrow \text{two 3d orbitals empty}$$
- Two empty 3d orbitals + one 4s + two 4p → d²sp³ hybridisation
- Uses inner (n−1)d = 3d orbitals → inner orbital complex
- All electrons paired → diamagnetic

[Ni(NH₃)₆]²⁺ — Outer Orbital Complex:

- Ni²⁺: [Ar] 3d⁸ → 8 electrons in 3d
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow}_{3d^8}$$
- NH₃, though a moderately strong ligand, cannot pair up the 3d electrons in Ni²⁺ (3d⁸ — no room to pair further without using 4d)
- Hybridisation uses 4s, three 4p, and two 4d orbitals → sp³d² hybridisation
- Uses outer nd = 4d orbitals → outer orbital complex
- 2 unpaired electrons → paramagnetic

Conclusion: Co³⁺ (3d⁶) with strong-field NH₃ undergoes d²sp³ hybridisation (inner orbital), while Ni²⁺ (3d⁸) cannot free inner 3d orbitals, so sp³d² hybridisation (outer orbital) occurs.

Given: [Pt(CN)₄]²⁻ is square planar.

Step 1: Find oxidation state of Pt.
Let oxidation state of Pt = x.
$$x + 4(-1) = -2 \Rightarrow x = +2$$
So Pt is in +2 oxidation state.

Step 2: Electronic configuration of Pt²⁺.
Pt: [Xe] 4f¹⁴ 5d⁹ 6s¹; Pt²⁺: [Xe] 4f¹⁴ 5d⁸
$$5d^8: \underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ \uparrow}_{5d^8}$$

Step 3: Effect of CN⁻ (strong field ligand).
For square planar geometry, hybridisation is dsp². One 5d orbital must be emptied.
CN⁻ forces pairing:
$$5d^8 \rightarrow \underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{}}_{\text{one empty 5d}}$$

Step 4: Hybridisation.
One empty 5d + 6s + two 6p → dsp² (square planar)

Result: All electrons are paired.

$$\boxed{\text{Number of unpaired electrons} = 0}$$

The complex is diamagnetic.

Given: Mn²⁺ is the central metal ion in both complexes.

Electronic configuration of Mn²⁺: [Ar] 3d⁵ (5 electrons)

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[Mn(H₂O)₆]²⁺ — 5 unpaired electrons (High Spin):

- H₂O is a weak field ligand → small crystal field splitting energy $\Delta_o$
- Since \Delta_o &lt; P (pairing energy), electrons occupy all five 3d orbitals singly before pairing
- Crystal field configuration: $t_{2g}^3\ e_g^2$
$$\underbrace{\uparrow\ \uparrow\ \uparrow}_{t_{2g}} \underbrace{\uparrow\ \uparrow}_{e_g}$$
- 5 unpaired electrons → strongly paramagnetic
- $\mu = \sqrt{5 \times 7} = \sqrt{35} \approx 5.92$ BM

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[Mn(CN)₆]⁴⁻ — 1 unpaired electron (Low Spin):

- CN⁻ is a strong field ligand → large crystal field splitting energy $\Delta_o$
- Since \Delta_o &gt; P, electrons pair up in the lower energy $t_{2g}$ orbitals
- Crystal field configuration: $t_{2g}^4\ e_g^0$
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\ }_{t_{2g}^4 \text{ (wait: 5 electrons)}}$$
Actually for 3d⁵ in strong field: $t_{2g}^5\ e_g^0$
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow}_{t_{2g}^5} \underbrace{\phantom{x}}_{e_g^0}$$
- 1 unpaired electron → weakly paramagnetic
- $\mu = \sqrt{1 \times 3} = \sqrt{3} \approx 1.73$ BM

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Conclusion: The difference is due to the magnitude of $\Delta_o$. Weak-field H₂O gives high-spin (5 unpaired e⁻); strong-field CN⁻ gives low-spin (1 unpaired e⁻) for Mn²⁺ (3d⁵).

Werner's Postulates for Bonding in Coordination Compounds:

Alfred Werner (1893) proposed the following postulates to explain bonding in coordination compounds:

Postulate 1 — Primary (Ionisable) Valence:
- Every metal exhibits a primary valence (Hauptvalenz), which is its normal oxidation state (electrovalence).
- Primary valences are satisfied by negative ions and are ionisable.
- Example: In [Co(NH₃)₆]Cl₃, the primary valence of Co is 3, satisfied by 3 Cl⁻ ions.

Postulate 2 — Secondary (Non-ionisable) Valence:
- Every metal also has a secondary valence (Nebenvalenz), which is the coordination number.
- Secondary valences are satisfied by neutral molecules or negative ions (ligands) and are non-ionisable.
- Secondary valences are directed in space, giving the complex a definite geometry.
- Example: In [Co(NH₃)₆]Cl₃, the secondary valence (coordination number) of Co is 6, satisfied by 6 NH₃ molecules.

Postulate 3 — Geometry:
- The secondary valences are directed towards fixed positions in space around the central metal, giving rise to definite geometrical shapes (octahedral, tetrahedral, square planar, etc.).

Illustration:
$$[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$$
- Co has primary valence = 3 (satisfied by 3 Cl⁻ outside the bracket)
- Co has secondary valence = 6 (satisfied by 6 NH₃ inside the bracket)
- Geometry: Octahedral

Werner's theory successfully explained the existence of isomers and the number of ions produced in solution by coordination compounds.

Given:
- FeSO₄ + (NH₄)₂SO₄ (1:1) → gives Fe²⁺ test
- CuSO₄ + NH₃(aq) (1:4) → does not give Cu²⁺ test

Explanation:

Case 1: FeSO₄ + (NH₄)₂SO₄

When FeSO₄ and (NH₄)₂SO₄ are mixed in 1:1 ratio, they form a double salt:
$$\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O} \quad (\text{Mohr's salt})$$

Double salts completely dissociate in aqueous solution to give all their constituent ions:
$$\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \rightarrow \text{Fe}^{2+} + 2\text{NH}_4^+ + 2\text{SO}_4^{2-}$$

Free Fe²⁺ ions are present in solution → Fe²⁺ test is positive.

Case 2: CuSO₄ + NH₃(aq) (1:4)

When CuSO₄ reacts with excess aqueous ammonia, it forms a complex compound (coordination compound):
$$\text{CuSO}_4 + 4\text{NH}_3 \rightarrow [\text{Cu}(\text{NH}_3)_4]\text{SO}_4$$

This is tetraamminecopper(II) sulphate, a coordination compound. In solution, it ionises as:
$$[\text{Cu}(\text{NH}_3)_4]\text{SO}_4 \rightarrow [\text{Cu}(\text{NH}_3)_4]^{2+} + \text{SO}_4^{2-}$$

The Cu²⁺ ions are not free — they are firmly held within the coordination sphere as [Cu(NH₃)₄]²⁺. The complex ion does not dissociate appreciably to give free Cu²⁺ ions → Cu²⁺ test is negative.

Conclusion: Double salts dissociate completely in solution releasing all ions, while coordination compounds retain the metal ion within the complex entity, so the metal ion test fails.

(i) Coordination Entity:
A coordination entity is the central metal atom/ion bonded to a fixed number of ions or molecules (ligands). It is enclosed in square brackets.

*Examples:*
1. $[\text{Fe}(\text{CN})_6]^{4-}$ — hexacyanidoferrate(II) ion
2. $[\text{Co}(\text{NH}_3)_6]^{3+}$ — hexaamminecobalt(III) ion

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(ii) Ligand:
Ligands are ions or molecules that donate electron pairs to the central metal atom/ion to form coordinate bonds. They must have at least one lone pair of electrons.

*Examples:*
1. $\text{NH}_3$ (ammonia) — neutral unidentate ligand
2. $\text{C}_2\text{O}_4^{2-}$ (oxalate ion) — anionic bidentate ligand

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(iii) Coordination Number:
The coordination number of a metal in a complex is the total number of ligand donor atoms directly bonded to the central metal atom/ion.

*Examples:*
1. In $[\text{Co}(\text{NH}_3)_6]^{3+}$: Co is bonded to 6 N atoms → coordination number = 6
2. In $[\text{Pt}(\text{Cl})_4]^{2-}$: Pt is bonded to 4 Cl atoms → coordination number = 4

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(iv) Coordination Polyhedron:
The spatial arrangement of the ligand atoms directly attached to the central metal defines the coordination polyhedron.

*Examples:*
1. $[\text{Co}(\text{NH}_3)_6]^{3+}$: 6 ligands arranged at corners of an octahedron
2. $[\text{Ni}(\text{CO})_4]$: 4 ligands arranged at corners of a tetrahedron

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(v) Homoleptic Complexes:
Complexes in which the metal is bonded to only one type of donor groups/ligands.

*Examples:*
1. $[\text{Co}(\text{NH}_3)_6]^{3+}$ — only NH₃ ligands
2. $[\text{Fe}(\text{CN})_6]^{4-}$ — only CN⁻ ligands

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(vi) Heteroleptic Complexes:
Complexes in which the metal is bonded to more than one type of donor groups/ligands.

*Examples:*
1. $[\text{Co}(\text{NH}_3)_4\text{Cl}_2]^+$ — NH₃ and Cl⁻ ligands
2. $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ — NH₃ and Cl⁻ ligands

(i) Unidentate Ligands:
Ligands that coordinate to the central metal through only one donor atom (one lone pair donated).

*Examples:*
1. $\text{Cl}^-$ (chlorido) — donates through Cl
2. $\text{NH}_3$ (ammine) — donates through N

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(ii) Didentate (Bidentate) Ligands:
Ligands that coordinate to the central metal through two donor atoms simultaneously, forming a ring (chelate).

*Examples:*
1. Ethane-1,2-diamine (en), $\text{H}_2\text{N}$–$\text{CH}_2$–$\text{CH}_2$–$\text{NH}_2$ — donates through two N atoms
2. Oxalate ion ($\text{C}_2\text{O}_4^{2-}$) — donates through two O atoms

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(iii) Ambidentate Ligands:
Ligands that can coordinate to the central metal through two different donor atoms (but only one at a time), giving rise to linkage isomerism.

*Examples:*
1. $\text{NO}_2^-$ (nitrite ion) — can bond through N (nitro, –NO₂) or through O (nitrito, –ONO)
2. $\text{SCN}^-$ (thiocyanate ion) — can bond through S (thiocyanato-S) or through N (thiocyanato-N)

Method: Sum of oxidation states of all species = overall charge on the complex ion.

(i) [Co(H₂O)(CN)(en)₂]²⁺
- H₂O: neutral (0); CN⁻: −1; en: neutral (0); overall charge: +2
$$x + 0 + (-1) + 0 = +2$$
$$x = +3$$
Oxidation state of Co = +3

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(ii) [CoBr₂(en)₂]⁺
- Br⁻: −1 each (×2 = −2); en: neutral (0); overall charge: +1
$$x + (-2) + 0 = +1$$
$$x = +3$$
Oxidation state of Co = +3

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(iii) [PtCl₄]³⁻ (as written; note: unusual, but solving as given)
- Cl⁻: −1 each (×4 = −4); overall charge: −3
$$x + (-4) = -3$$
$$x = +1$$
Oxidation state of Pt = +1
*(Note: If the formula is [PtCl₆]³⁻, then x − 6 = −3, giving Pt = +3)*

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(iv) K₃[Fe(CN)₆]
- 3 K⁺ outside; complex ion [Fe(CN)₆]³⁻
- CN⁻: −1 each (×6 = −6); overall charge on complex: −3
$$x + (-6) = -3$$
$$x = +3$$
Oxidation state of Fe = +3

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(v) [Cr(NH₃)₃Cl₃]
- NH₃: neutral (0); Cl⁻: −1 each (×3 = −3); overall charge: 0
$$x + 0 + (-3) = 0$$
$$x = +3$$
Oxidation state of Cr = +3

(i) Tetrahydroxidozincate(II)
- Zn²⁺, 4 OH⁻ ligands, anionic complex
- Charge: 2 − 4 = −2
$$[\text{Zn}(\text{OH})_4]^{2-}$$

(ii) Potassium tetrachloridopalladate(II)
- K⁺ outside; Pd²⁺, 4 Cl⁻ ligands
- Charge on complex: 2 − 4 = −2 → K₂ outside
$$K_2[\text{PdCl}_4]$$

(iii) Diamminedichloridoplatinum(II)
- Pt²⁺, 2 NH₃ + 2 Cl⁻; neutral complex
$$[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$$

(iv) Potassium tetracyanidonickelate(II)
- K⁺ outside; Ni²⁺, 4 CN⁻
- Charge: 2 − 4 = −2 → K₂ outside
$$K_2[\text{Ni}(\text{CN})_4]$$

(v) Pentaamminenitrito-O-cobalt(III)
- Co³⁺, 5 NH₃ + NO₂⁻ bonded through O (–ONO)
- Charge: 3 − 1 = +2 → no counter ion specified (cation)
$$[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$$

(vi) Hexaamminecobalt(III) sulphate
- Co³⁺, 6 NH₃; charge = +3; SO₄²⁻ outside
- 2 formula units of complex needed for 3 SO₄²⁻: $[\text{Co}(\text{NH}_3)_6]_2(\text{SO}_4)_3$
$$[\text{Co}(\text{NH}_3)_6]_2(\text{SO}_4)_3$$

(vii) Potassium tri(oxalato)chromate(III)
- K⁺ outside; Cr³⁺, 3 C₂O₄²⁻
- Charge: 3 − 6 = −3 → K₃ outside
$$K_3[\text{Cr}(\text{C}_2\text{O}_4)_3]$$

(viii) Hexaammineplatinum(IV)
- Pt⁴⁺, 6 NH₃; charge = +4 (cation only, no anion specified)
$$[\text{Pt}(\text{NH}_3)_6]^{4+}$$

(ix) Tetrabromidocuprate(II)
- Cu²⁺, 4 Br⁻; anionic complex
- Charge: 2 − 4 = −2
$$[\text{CuBr}_4]^{2-}$$

(x) Pentaamminenitrito-N-cobalt(III)
- Co³⁺, 5 NH₃ + NO₂⁻ bonded through N (–NO₂)
- Charge: 3 − 1 = +2
$$[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$$

Rules: Name ligands alphabetically before metal; use oxidation state in Roman numerals; anionic complexes end in '-ate'.

(i) [Co(NH₃)₆]Cl₃
- 6 NH₃ (hexaammine); Co: x = +3; Cl⁻ outside
Hexaamminecobalt(III) chloride

(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
- Ligands: 2 NH₃ (diammine), Cl⁻ (chlorido), NH₂CH₃ (methanamine) — alphabetical: chlorido, diammine, methanamine
- Pt: x − 1 = +1 → x = +2; Cl⁻ outside
Diamminechloridomethanamiineplatinum(II) chloride
*(Diamminechlorido(methanamine)platinum(II) chloride)*

(iii) [Ti(H₂O)₆]³⁺
- 6 H₂O (hexaaqua); Ti: x = +3
Hexaaquatitanium(III) ion

(iv) [Co(NH₃)₄Cl(NO₂)]Cl
- Ligands: 4 NH₃ (tetraammine), Cl⁻ (chlorido), NO₂⁻ (nitrito-N or nitro)
- Co: x − 1 − 1 = +1 → x = +3; Cl⁻ outside
Tetraamminechloridonitrito-N-cobalt(III) chloride
*(or Tetraamminechloridonitrocobalt(III) chloride)*

(v) [Mn(H₂O)₆]²⁺
- 6 H₂O (hexaaqua); Mn: x = +2
Hexaaquamanganese(II) ion

(vi) [NiCl₄]²⁻
- 4 Cl⁻ (tetrachloro); Ni: x − 4 = −2 → x = +2; anionic → nickelate
Tetrachloronickelate(II) ion
*(Tetrachloridodickelate(II) ion)*

(vii) [Ni(NH₃)₆]Cl₂
- 6 NH₃ (hexaammine); Ni: x = +2; Cl⁻ outside
Hexaamminenickel(II) chloride

(viii) [Co(en)₃]³⁺
- 3 en (tris(ethane-1,2-diamine)); Co: x = +3
Tris(ethane-1,2-diamine)cobalt(III) ion

(ix) [Ni(CO)₄]
- 4 CO (tetracarbonyl); Ni: x = 0 (neutral complex, CO is neutral)
Tetracarbonylnickel(0)

Types of Isomerism in Coordination Compounds:

## A. Stereoisomerism

(i) Geometrical Isomerism:
Arises due to different spatial arrangements of ligands around the central metal. Common in square planar and octahedral complexes.

*Example:* cis- and trans-$[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$
- *cis*: two Cl adjacent (90°)
- *trans*: two Cl opposite (180°)

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(ii) Optical Isomerism:
Arises when a complex and its mirror image are non-superimposable (chiral). The two forms are called enantiomers (d and l, or Δ and Λ).

*Example:* $[\text{Co}(\text{en})_3]^{3+}$ — exists as Δ (right-handed) and Λ (left-handed) forms.

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## B. Structural Isomerism

(iii) Linkage Isomerism:
Arises when an ambidentate ligand coordinates through different donor atoms.

*Example:* $[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$ (nitro, N-bonded) and $[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$ (nitrito, O-bonded)

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(iv) Coordination Isomerism:
Arises in compounds containing both cationic and anionic complex ions, where the distribution of ligands between the two metal centres differs.

*Example:* $[\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6]$ and $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$

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(v) Ionisation Isomerism:
Arises when the counter ion and a ligand exchange positions (inside and outside the coordination sphere), giving different ions in solution.

*Example:* $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ and $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$

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(vi) Solvate (Hydrate) Isomerism:
Arises when solvent molecules (e.g., water) are either inside the coordination sphere as ligands or outside as free solvent molecules.

*Example:* $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (violet) and $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$ (grey-green)

(i) [Cr(C₂O₄)₃]³⁻

- This is an octahedral complex with three identical bidentate ligands (oxalate, C₂O₄²⁻).
- For a complex of type [M(AA)₃] where AA is a symmetric bidentate ligand, all three ligands are equivalent.
- There is only one geometric arrangement possible — no geometric isomers.

$$\boxed{\text{Number of geometrical isomers} = 1}$$

*(However, it does show optical isomerism — Δ and Λ forms)*

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(ii) [Co(NH₃)₃Cl₃]

- This is an octahedral complex of type [MA₃B₃].
- Two geometric isomers are possible:
1. fac (facial) isomer: Three NH₃ (and three Cl) occupy the three positions of one triangular face of the octahedron. Each NH₃ is adjacent to two other NH₃ molecules.
2. mer (meridional) isomer: Three NH₃ (and three Cl) occupy positions along a meridian (one axis and two equatorial positions). The three NH₃ are not all equivalent.

$$\boxed{\text{Number of geometrical isomers} = 2 \text{ (fac and mer)}}$$

*(Note: Structural drawings are described in detail as they cannot be rendered as images in text format.)*

(i) [Cr(C₂O₄)₃]³⁻

- Octahedral complex with three bidentate oxalate ligands: [M(AA)₃] type.
- This complex is chiral — it has no plane of symmetry.
- Two optical isomers exist:
- Δ (delta) form: Right-handed propeller arrangement of the three oxalate ligands around Cr.
- Λ (lambda) form: Left-handed propeller arrangement (mirror image of Δ).
- The two forms are non-superimposable mirror images (enantiomers).

*Description of structure:* In the Δ isomer, looking down the C₃ axis, the three oxalate ligands spiral clockwise. In the Λ isomer, they spiral anticlockwise.

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(ii) [PtCl₂(en)₂]²⁺

- Octahedral complex with two bidentate en ligands and two Cl⁻.
- First, geometric isomers exist:
- *cis* isomer: two Cl adjacent → chiral → shows optical isomers (Δ and Λ)
- *trans* isomer: two Cl opposite → has a plane of symmetry → optically inactive
- So optical isomers exist only for the cis form:
- Δ-cis-[PtCl₂(en)₂]²⁺
- Λ-cis-[PtCl₂(en)₂]²⁺

---

(iii) [Cr(NH₃)₂Cl₂(en)]⁺

- Octahedral complex with one bidentate en, two NH₃, and two Cl⁻.
- The en ligand occupies two cis positions.
- Geometric isomers are possible depending on the relative positions of NH₃ and Cl.
- For the isomer where the two Cl are cis to each other and the en spans two adjacent positions, the complex is chiral → optical isomers (Δ and Λ) exist.
- Δ form and Λ form are non-superimposable mirror images.

*In all cases, the optical isomers rotate plane-polarised light in opposite directions (+/−) and are called dextrorotatory (d) and laevorotatory (l) forms.*

(i) [CoCl₂(en)₂]⁺

Octahedral complex with two bidentate en ligands and two Cl⁻ (type [MA₂B₂] with bidentate A).

Geometric Isomers:
1. trans isomer: Two Cl⁻ are trans (opposite) to each other. The two en ligands are in the equatorial plane. This isomer has a plane of symmetry → optically inactive.

2. cis isomer: Two Cl⁻ are cis (adjacent) to each other. This isomer is chiral (no plane of symmetry) → shows optical isomers:
- Δ-cis isomer (d-form)
- Λ-cis isomer (l-form)

Total isomers: 3 (1 trans + 2 optical isomers of cis)

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(ii) [Co(NH₃)Cl(en)₂]²⁺

Octahedral complex with two bidentate en, one NH₃, and one Cl⁻.

- The two en ligands can be arranged so that NH₃ and Cl are either *trans* or *cis* to each other.

Geometric Isomers:
1. trans isomer: NH₃ and Cl are trans to each other. This has a plane of symmetry → optically inactive.

2. cis isomer: NH₃ and Cl are cis to each other. This is chiral → shows optical isomers:
- Δ-cis isomer
- Λ-cis isomer

Total isomers: 3 (1 trans + 2 optical isomers of cis)

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(iii) [Co(NH₃)₂Cl₂(en)]⁺

Octahedral complex with one bidentate en, two NH₃, and two Cl⁻. The en occupies two cis positions.

Geometric Isomers (based on positions of NH₃ and Cl relative to en):

1. Isomer A: Both Cl trans to each other (trans-Cl,Cl); en and two NH₃ in remaining positions.
- Has a plane of symmetry → optically inactive.

2. Isomer B: Both NH₃ trans to each other; en and two Cl in remaining positions.
- Has a plane of symmetry → optically inactive.

3. Isomer C: One Cl trans to NH₃, one Cl trans to en nitrogen.
- Chiral → shows optical isomers (Δ and Λ).

Total isomers: 4 (2 optically inactive geometric isomers + 2 optical isomers of the chiral form)

Given: [Pt(NH₃)(Br)(Cl)(py)] — Square planar complex of type [MABCD] where A = NH₃, B = Br, C = Cl, D = py (pyridine), all unidentate.

Number of Geometrical Isomers:
For a square planar complex [MABCD], the number of geometrical isomers = 3.

The three isomers arise from the three different ways to place the four different ligands:

Isomer 1:
$$\begin{array}{ccc} &amp; \text{NH}_3 &amp; \\ \text{Cl} &amp; \text{Pt} &amp; \text{Br} \\ &amp; \text{py} &amp; \end{array}$$
(NH₃ trans to py; Cl trans to Br)

Isomer 2:
$$\begin{array}{ccc} &amp; \text{NH}_3 &amp; \\ \text{py} &amp; \text{Pt} &amp; \text{Br} \\ &amp; \text{Cl} &amp; \end{array}$$
(NH₃ trans to Cl; py trans to Br)

Isomer 3:
$$\begin{array}{ccc} &amp; \text{NH}_3 &amp; \\ \text{py} &amp; \text{Pt} &amp; \text{Cl} \\ &amp; \text{Br} &amp; \end{array}$$
(NH₃ trans to Br; py trans to Cl)

Total geometrical isomers = 3

Optical Isomers:
Square planar complexes generally have a plane of symmetry (the plane of the molecule) and are therefore optically inactive.

All three geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] lie in a plane → each has a plane of symmetry → none of them exhibit optical isomerism.

$$\boxed{\text{Geometrical isomers} = 3; \quad \text{Optical isomers} = 0}$$

Background: Aqueous CuSO₄ contains the complex ion $[\text{Cu}(\text{H}_2\text{O})_4]^{2+}$ which is blue in colour.

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(i) Green precipitate with aqueous KF:

When KF is added to CuSO₄ solution, F⁻ ions (being a hard base and small ion) displace water ligands and form copper(II) fluoride:
$$\text{CuSO}_4 + 2\text{KF} \rightarrow \text{CuF}_2 \downarrow + \text{K}_2\text{SO}_4$$

Copper(II) fluoride is insoluble in water and precipitates as a green precipitate. F⁻ is not a good ligand for forming stable soluble complexes with Cu²⁺ under these conditions, so a precipitate forms rather than a soluble complex.

---

(ii) Bright green solution with aqueous KCl:

When excess KCl is added to CuSO₄ solution, Cl⁻ ions displace water ligands and form a stable soluble complex:
$$[\text{Cu}(\text{H}_2\text{O})_4]^{2+} + 4\text{Cl}^- \rightarrow [\text{CuCl}_4]^{2-} + 4\text{H}_2\text{O}$$

The tetrachloridocuprate(II) ion $[\text{CuCl}_4]^{2-}$ is bright green in colour and remains in solution.

Conclusion: The difference in behaviour is due to the different abilities of F⁻ and Cl⁻ to form stable soluble complexes with Cu²⁺. F⁻ forms an insoluble precipitate (CuF₂), while Cl⁻ forms a soluble green complex [CuCl₄]²⁻.

Step 1: Coordination entity formed with excess KCN:

When excess KCN is added to CuSO₄ solution, CN⁻ ions first reduce Cu²⁺ to Cu⁺ and then form a stable complex:
$$2\text{Cu}^{2+} + 4\text{CN}^- \rightarrow 2\text{Cu}^+ + (\text{CN})_2$$
$$\text{Cu}^+ + 3\text{CN}^- \rightarrow [\text{Cu}(\text{CN})_3]^{2-}$$

With excess KCN:
$$\text{Cu}^+ + 4\text{CN}^- \rightarrow [\text{Cu}(\text{CN})_4]^{3-}$$

The coordination entity formed is tetracyanidocuprate(I): $[\text{Cu}(\text{CN})_4]^{3-}$

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Step 2: Why no CuS precipitate with H₂S?

In the complex $[\text{Cu}(\text{CN})_4]^{3-}$, CN⁻ is a very strong field ligand and forms an extremely stable complex with Cu⁺. The complex has a very low dissociation constant (very high stability constant).

As a result, the concentration of free Cu²⁺ (or Cu⁺) ions in solution is negligibly small — far below the concentration required to exceed the solubility product ($K_{sp}$) of CuS.

$$K_{sp}(\text{CuS}) = [\text{Cu}^{2+}][\text{S}^{2-}]$$

Since $[\text{Cu}^{2+}]$ is extremely low, the ionic product [\text{Cu}^{2+}][\text{S}^{2-}] &lt; K_{sp}, and therefore no precipitate of CuS is formed when H₂S is passed through the solution.

(i) [Fe(CN)₆]⁴⁻

- Fe in [Fe(CN)₆]⁴⁻: oxidation state = +2; Fe²⁺: [Ar] 3d⁶
- Free Fe²⁺: $\underbrace{\uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow}_{3d^6}$ (4 unpaired)
- CN⁻ is a strong field ligand → forces pairing of 3d electrons:
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{}\ \boxed{}}_{3d} \quad \underbrace{[\text{4s}][\text{4p}][\text{4p}][\text{4p}]}$$
- Two empty 3d orbitals available; hybridisation: d²sp³ (inner orbital)
- 6 CN⁻ donate electron pairs into these 6 hybrid orbitals
- Geometry: Octahedral; Diamagnetic (no unpaired electrons)

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(ii) [FeF₆]³⁻

- Fe in [FeF₆]³⁻: oxidation state = +3; Fe³⁺: [Ar] 3d⁵
- Free Fe³⁺: $\underbrace{\uparrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow}_{3d^5}$ (5 unpaired)
- F⁻ is a weak field ligand → no pairing of 3d electrons
- 3d orbitals remain half-filled; hybridisation uses outer orbitals: sp³d² (outer orbital)
- 6 F⁻ donate electron pairs into 6 sp³d² hybrid orbitals
- Geometry: Octahedral; Paramagnetic (5 unpaired electrons)
- $\mu = \sqrt{5 \times 7} \approx 5.92$ BM

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(iii) [Co(C₂O₄)₃]³⁻

- Co in [Co(C₂O₄)₃]³⁻: oxidation state = +3; Co³⁺: [Ar] 3d⁶
- Free Co³⁺: $\underbrace{\uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow}_{3d^6}$ (4 unpaired)
- Oxalate (C₂O₄²⁻) is a moderate to strong field ligand → causes pairing:
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{}\ \boxed{}}_{3d}$$
- Hybridisation: d²sp³ (inner orbital, octahedral)
- 3 bidentate oxalate ligands donate 6 electron pairs
- Geometry: Octahedral; Diamagnetic (no unpaired electrons)

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(iv) [CoF₆]³⁻

- Co in [CoF₆]³⁻: oxidation state = +3; Co³⁺: [Ar] 3d⁶
- Free Co³⁺: $\underbrace{\uparrow\downarrow\ \uparrow\ \uparrow\ \uparrow\ \uparrow}_{3d^6}$ (4 unpaired)
- F⁻ is a weak field ligand → no pairing of 3d electrons
- Hybridisation uses outer orbitals: sp³d² (outer orbital)
- 6 F⁻ donate electron pairs into 6 sp³d² hybrid orbitals
- Geometry: Octahedral; Paramagnetic (4 unpaired electrons)
- $\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM

Crystal Field Splitting in Octahedral Field:

In a free metal ion, all five d orbitals are degenerate (equal energy).

When six ligands approach along the ±x, ±y, ±z axes (octahedral field), the d orbitals split into two sets:

Higher energy set — $e_g$ (2 orbitals):
- $d_{x^2-y^2}$ and $d_{z^2}$
- These orbitals point directly towards the ligands → greater repulsion → higher energy
- Energy raised by $+\frac{3}{5}\Delta_o$ (or $+0.6\Delta_o$) above the barycentre

Lower energy set — $t_{2g}$ (3 orbitals):
- $d_{xy}$, $d_{yz}$, $d_{xz}$
- These orbitals point between the ligands → less repulsion → lower energy
- Energy lowered by $-\frac{2}{5}\Delta_o$ (or $-0.4\Delta_o$) below the barycentre

Energy diagram:

$$\underbrace{\boxed{d_{x^2-y^2} \quad d_{z^2}}}_{e_g \text{ (+0.6}\Delta_o\text{)}} \quad \xrightarrow{\Delta_o} \quad \underbrace{\boxed{d_{xy} \quad d_{yz} \quad d_{xz}}}_{t_{2g} \text{ (−0.4}\Delta_o\text{)}}$$

The energy gap between $t_{2g}$ and $e_g$ is called the crystal field splitting energy $\Delta_o$ (or $10Dq$).

$$\Delta_o = E(e_g) - E(t_{2g}) = 10Dq$$

The barycentre (weighted average energy) remains unchanged:
$$3 \times (-0.4\Delta_o) + 2 \times (+0.6\Delta_o) = 0$$

Spectrochemical Series:

The spectrochemical series is an empirical series that arranges ligands in order of their increasing ability to cause crystal field splitting ($\Delta_o$) of the d orbitals of the central metal ion.

\text{I}^- &lt; \text{Br}^- &lt; \text{SCN}^- &lt; \text{Cl}^- &lt; \text{S}^{2-} &lt; \text{F}^- &lt; \text{OH}^- &lt; \text{C}_2\text{O}_4^{2-} &lt; \text{H}_2\text{O} &lt; \text{NCS}^- &lt; \text{edta}^{4-} &lt; \text{NH}_3 &lt; \text{en} &lt; \text{CN}^- &lt; \text{CO}

(Increasing $\Delta_o$ →)

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Weak Field Ligands:
- Ligands at the lower end of the spectrochemical series.
- Cause small crystal field splitting (\Delta_o &lt; P, where P = pairing energy).
- Electrons prefer to occupy all d orbitals singly before pairing → high spin complexes.
- Examples: I⁻, Br⁻, Cl⁻, F⁻, OH⁻, H₂O
- Result in outer orbital complexes (sp³d² hybridisation in VBT)

Strong Field Ligands:
- Ligands at the higher end of the spectrochemical series.
- Cause large crystal field splitting (\Delta_o &gt; P).
- Electrons pair up in lower energy $t_{2g}$ orbitals before occupying $e_g$ → low spin complexes.
- Examples: CN⁻, CO, en, NH₃
- Result in inner orbital complexes (d²sp³ hybridisation in VBT)

Crystal Field Splitting Energy ($\Delta_o$):

When a metal ion is placed in an octahedral crystal field (created by six ligands), the five degenerate d orbitals split into two sets:
- Lower energy $t_{2g}$ set (3 orbitals): stabilised by $0.4\Delta_o$ each
- Higher energy $e_g$ set (2 orbitals): destabilised by $0.6\Delta_o$ each

The energy difference between the $e_g$ and $t_{2g}$ sets is called the crystal field splitting energy, denoted $\Delta_o$ (or $10Dq$).

$$\Delta_o = E(e_g) - E(t_{2g})$$

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How $\Delta_o$ decides d orbital configuration:

The actual filling of electrons depends on the competition between $\Delta_o$ and the pairing energy (P):

Case 1: Weak field ligands (\Delta_o &lt; P)
- It costs less energy to place an electron in the higher $e_g$ orbital than to pair it in $t_{2g}$.
- Electrons spread over all five d orbitals (Hund's rule applies).
- Result: High spin configuration (maximum unpaired electrons)
- Example: $[\text{Fe}(\text{H}_2\text{O})_6]^{3+}$: $t_{2g}^3\ e_g^2$ (5 unpaired)

Case 2: Strong field ligands (\Delta_o &gt; P)
- It costs less energy to pair electrons in $t_{2g}$ than to promote them to $e_g$.
- Electrons fill $t_{2g}$ completely before entering $e_g$.
- Result: Low spin configuration (minimum unpaired electrons)
- Example: $[\text{Fe}(\text{CN})_6]^{3-}$: $t_{2g}^5\ e_g^0$ (1 unpaired)

Summary:
\Delta_o &lt; P \Rightarrow \text{High spin (weak field)}
\Delta_o &gt; P \Rightarrow \text{Low spin (strong field)}

[Cr(NH₃)₆]³⁺ — Paramagnetic:

- Cr³⁺: [Ar] 3d³ (3 electrons in d orbitals)
- NH₃ is a strong field ligand, but with only 3 d electrons, they occupy the three $t_{2g}$ orbitals singly (no pairing needed regardless of field strength):
$$t_{2g}^3\ e_g^0: \underbrace{\uparrow\ \uparrow\ \uparrow}_{t_{2g}} \underbrace{\phantom{x}\ \phantom{x}}_{e_g}$$
- 3 unpaired electrons → Paramagnetic
- $\mu = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ BM
- Hybridisation: d²sp³ (inner orbital, octahedral)

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[Ni(CN)₄]²⁻ — Diamagnetic:

- Ni²⁺: [Ar] 3d⁸ (8 electrons in d orbitals)
- CN⁻ is a strong field ligand → forces pairing of 3d electrons
- One 3d orbital is emptied by pairing:
$$\underbrace{\uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \uparrow\downarrow\ \boxed{}}_{3d^8 \rightarrow \text{one empty}}$$
- Hybridisation: dsp² (square planar)
- 0 unpaired electrons → Diamagnetic

Conclusion: Cr³⁺ (3d³) has 3 unpaired electrons even with strong-field NH₃ (no pairing needed for 3 electrons in 3 orbitals), making [Cr(NH₃)₆]³⁺ paramagnetic. Ni²⁺ (3d⁸) with strong-field CN⁻ undergoes forced pairing, making [Ni(CN)₄]²⁻ diamagnetic.

Concept: Colour in coordination compounds arises from d-d transitions — absorption of visible light causes electrons to jump from lower energy d orbitals to higher energy d orbitals. The complementary colour is observed.

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[Ni(H₂O)₆]²⁺ — Green:

- Ni²⁺: 3d⁸ configuration
- H₂O is a weak field ligand → small $\Delta_o$
- The energy gap $\Delta_o$ corresponds to a photon in the red/orange region of visible light
- Red light is absorbed → complementary colour green is transmitted/observed
- d-d transition is possible because the $t_{2g}$ and $e_g$ levels are partially filled
- Solution appears green ✓

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[Ni(CN)₄]²⁻ — Colourless:

- Ni²⁺: 3d⁸; CN⁻ is a strong field ligand → large $\Delta_o$
- In the square planar [Ni(CN)₄]²⁻, the dsp² hybridisation results in all electrons being paired (no unpaired electrons)
- The energy gap between filled and empty d orbitals is very large (in the UV region)
- No d-d transition occurs in the visible region
- No visible light is absorbed → solution appears colourless ✓

Conclusion: [Ni(H₂O)₆]²⁺ absorbs visible light (d-d transition possible with small Δ), while [Ni(CN)₄]²⁻ has all electrons paired and the energy gap falls outside the visible range, so no colour is observed.

Both complexes contain Fe²⁺ (3d⁶), but they have different colours due to different crystal field splitting energies ($\Delta_o$).

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[Fe(H₂O)₆]²⁺:
- H₂O is a weak field ligand → small $\Delta_o$
- The energy gap corresponds to absorption in the red region of visible spectrum
- Complementary colour (green-blue) is transmitted
- Appears pale green in solution
- High spin: $t_{2g}^4\ e_g^2$

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[Fe(CN)₆]⁴⁻:
- CN⁻ is a strong field ligand → large $\Delta_o$
- The energy gap corresponds to absorption in a different region of visible spectrum (higher energy, towards blue/violet)
- A different complementary colour is transmitted
- Appears yellow in solution
- Low spin: $t_{2g}^6\ e_g^0$

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Conclusion: The two complexes absorb different wavelengths of visible light because $\Delta_o$ is different (H₂O gives small $\Delta_o$, CN⁻ gives large $\Delta_o$). Since different wavelengths are absorbed, different complementary colours are observed, resulting in different colours for the two complexes.

Bonding in Metal Carbonyls:

Metal carbonyls are compounds in which CO (carbonyl) ligands are bonded to transition metals. The bonding involves a unique synergic (synergistic) mechanism with two components:

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1. σ-Bond (Ligand → Metal donation):
- The carbon atom of CO has a lone pair in a σ-type orbital.
- This lone pair is donated to an empty d orbital (or hybrid orbital) of the metal.
- This forms a σ-bond: CO → Metal (ligand to metal donation)
- This increases electron density on the metal.

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2. π-Bond (Metal → Ligand back-donation):
- The metal has filled d orbitals with electron density.
- These electrons are donated back into the **empty π* (antibonding) orbital of CO.
- This forms a π-bond: Metal → CO (metal to ligand back-donation)
- This is called back-bonding or π-back donation.
- It reduces the electron density on the metal (relieving the excess from σ-donation).

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Synergic Effect:
The σ-donation and π-back donation reinforce each other:
- σ-donation increases electron density on metal → metal donates back via π
- π-back donation strengthens the M–C bond and weakens the C≡O bond
- This mutual reinforcement is called the synergic effect

Consequence:
- The M–C bond is strengthened (has both σ and π character)
- The C–O bond is weakened compared to free CO (C≡O bond order decreases)
- This is evidenced by the lower C–O stretching frequency in metal carbonyls compared to free CO

Examples:**
- $[\text{Ni}(\text{CO})_4]$: tetrahedral, Ni(0), $d^{10}$
- $[\text{Fe}(\text{CO})_5]$: trigonal bipyramidal, Fe(0), $d^8$
- $[\text{Cr}(\text{CO})_6]$: octahedral, Cr(0), $d^6$

(i) K₃[Co(C₂O₄)₃]

- 3 K⁺ outside → complex ion has charge −3
- C₂O₄²⁻: −2 each; 3 oxalates = −6
- Let oxidation state of Co = x: $x + (-6) = -3 \Rightarrow x = +3$
- Oxidation state: +3
- Co³⁺: [Ar] 3d⁶ → d orbital occupation: d⁶
- 3 bidentate oxalate ligands → 6 donor atoms → Coordination number: 6

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(ii) cis-[CrCl₂(en)₂]Cl

- Cl⁻ outside → complex ion has charge +1
- 2 Cl⁻ inside (−2) + 2 en (0)
- Let oxidation state of Cr = x: $x + (-2) + 0 = +1 \Rightarrow x = +3$
- Oxidation state: +3
- Cr³⁺: [Ar] 3d³ → d orbital occupation: d³
- 2 Cl⁻ (2 donor atoms) + 2 en (4 donor atoms) = 6 donor atoms → Coordination number: 6

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(iii) (NH₄)₂[CoF₄]

- 2 NH₄⁺ outside → complex ion has charge −2
- 4 F⁻ inside (−4)
- Let oxidation state of Co = x: $x + (-4) = -2 \Rightarrow x = +2$
- Oxidation state: +2
- Co²⁺: [Ar] 3d⁷ → d orbital occupation: d⁷
- 4 F⁻ ligands → Coordination number: 4

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(iv) [Mn(H₂O)₆]SO₄

- SO₄²⁻ outside → complex ion has charge +2
- 6 H₂O (neutral)
- Let oxidation state of Mn = x: $x + 0 = +2 \Rightarrow x = +2$
- Oxidation state: +2
- Mn²⁺: [Ar] 3d⁵ → d orbital occupation: d⁵
- 6 H₂O ligands → Coordination number: 6

(i) K[Cr(H₂O)₂(C₂O₄)₂]·3H₂O

- IUPAC Name: Potassium diaquadioxalatochromate(III) trihydrate
- Oxidation state: K⁺ outside; complex charge = −1; 2 H₂O (0) + 2 C₂O₄²⁻ (−4); x − 4 = −1 → Cr = +3
- Electronic configuration: Cr³⁺: [Ar] 3d³
- Coordination number: 2 (H₂O) + 4 (from 2 oxalates) = 6
- Stereochemistry: Octahedral; cis and trans geometric isomers possible; cis form shows optical isomers
- Magnetic moment: 3d³, 3 unpaired electrons; $\mu = \sqrt{3(5)} = \sqrt{15} \approx 3.87$ BM

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(ii) [Co(NH₃)₅Cl]Cl₂

- IUPAC Name: Pentaamminechloridocobalt(III) chloride
- Oxidation state: 2 Cl⁻ outside; complex charge = +2; 5 NH₃ (0) + Cl⁻ (−1); x − 1 = +2 → Co = +3
- Electronic configuration: Co³⁺: [Ar] 3d⁶; with NH₃ (strong field) → low spin: $t_{2g}^6\ e_g^0$
- Coordination number: 5 + 1 = 6
- Stereochemistry: Octahedral; no geometric isomerism (MA₅B type)
- Magnetic moment: 0 unpaired electrons; $\mu = 0$ BM → Diamagnetic

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(iii) [CrCl₃(py)₃]

- IUPAC Name: Trichloridotripyridinechromium(III)
- Oxidation state: Neutral complex; 3 Cl⁻ (−3) + 3 py (0); x − 3 = 0 → Cr = +3
- Electronic configuration: Cr³⁺: [Ar] 3d³
- Coordination number: 3 + 3 = 6
- Stereochemistry: Octahedral; two geometric isomers: fac and mer
- Magnetic moment: 3 unpaired electrons; $\mu = \sqrt{3(5)} \approx 3.87$ BM → Paramagnetic

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(iv) Cs[FeCl₄]

- IUPAC Name: Caesium tetrachloridoferrate(III)
- Oxidation state: Cs⁺ outside; complex charge = −1; 4 Cl⁻ (−4); x − 4 = −1 → Fe = +3
- Electronic configuration: Fe³⁺: [Ar] 3d⁵; Cl⁻ weak field → high spin: all 5 d electrons unpaired
- Coordination number: 4
- Stereochemistry: Tetrahedral
- Magnetic moment: 5 unpaired electrons; $\mu = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM → Strongly paramagnetic

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(v) K₄[Mn(CN)₆]

- IUPAC Name: Potassium hexacyanidomanganate(II)
- Oxidation state: 4 K⁺ outside; complex charge = −4; 6 CN⁻ (−6); x − 6 = −4 → Mn = +2
- Electronic configuration: Mn²⁺: [Ar] 3d⁵; CN⁻ strong field → low spin: $t_{2g}^5\ e_g^0$ → 1 unpaired electron
- Coordination number: 6
- Stereochemistry: Octahedral
- Magnetic moment: 1 unpaired electron; $\mu = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM → Weakly paramagnetic

Given: [Ti(H₂O)₆]³⁺ appears violet in solution.

Step 1: Electronic configuration of Ti³⁺
Ti: [Ar] 3d² 4s²; Ti³⁺: [Ar] 3d¹ (one electron in d orbitals)

Step 2: Crystal field splitting
In the octahedral field of 6 H₂O ligands, the d orbitals split:
- Lower energy: $t_{2g}$ (3 orbitals)
- Higher energy: $e_g$ (2 orbitals)
- Energy gap: $\Delta_o$

The single d electron occupies the lower $t_{2g}$ level in the ground state:
$$\text{Ground state: } t_{2g}^1\ e_g^0$$

Step 3: d-d Transition
When visible light of appropriate energy falls on the complex, the electron absorbs a photon and is excited from $t_{2g}$ to $e_g$:
$$t_{2g}^1\ e_g^0 \xrightarrow{h\nu = \Delta_o} t_{2g}^0\ e_g^1$$

The energy absorbed corresponds to:
$$\Delta_o = h\nu = \frac{hc}{\lambda}$$

Step 4: Colour observed
For [Ti(H₂O)₆]³⁺, the absorbed wavelength corresponds to the yellow-green region (~500 nm) of visible light.

The complementary colour of yellow-green is violet/purple.

Therefore, the complex appears violet in solution.

Conclusion: The violet colour of [Ti(H₂O)₆]³⁺ is due to the absorption of yellow-green light during the d-d transition of the single 3d electron from the $t_{2g}$ to the $e_g$ level in the octahedral crystal field.

Chelate Effect:

The chelate effect refers to the enhanced stability of a complex containing chelate rings (formed by polydentate ligands) compared to a similar complex containing unidentate ligands.

When a polydentate (chelating) ligand replaces several unidentate ligands, the resulting chelate complex is thermodynamically more stable (higher stability constant) than the corresponding complex with unidentate ligands.

Reason:
1. Entropy factor: When a chelating ligand replaces unidentate ligands, the number of free particles in solution increases (positive $\Delta S$), making the process thermodynamically favourable.
2. Enthalpy factor: The ring formation also contributes to stability.

Example:

Compare the stability of:
$$[\text{Ni}(\text{NH}_3)_6]^{2+} \quad \text{vs} \quad [\text{Ni}(\text{en})_3]^{2+}$$

The reaction:
$$[\text{Ni}(\text{NH}_3)_6]^{2+} + 3\text{en} \rightarrow [\text{Ni}(\text{en})_3]^{2+} + 6\text{NH}_3$$

This reaction proceeds spontaneously to the right because:
- Reactants: 1 complex + 3 en = 4 species
- Products: 1 complex + 6 NH₃ = 7 species
- Increase in number of molecules → increase in entropy (\Delta S &gt; 0)
- \Delta G = \Delta H - T\Delta S &lt; 0 → reaction is spontaneous

$[\text{Ni}(\text{en})_3]^{2+}$ (log K ≈ 18.3) is far more stable than $[\text{Ni}(\text{NH}_3)_6]^{2+}$ (log K ≈ 8.6).

(i) Biological Systems:

Coordination compounds play vital roles in biological processes:

1. Haemoglobin: A coordination compound of iron (Fe²⁺) with a porphyrin ring (haem) and a protein (globin). It acts as an oxygen carrier in blood. The Fe²⁺ centre coordinates with O₂ reversibly.

2. Chlorophyll: A coordination compound of magnesium (Mg²⁺) with a porphyrin ring. It is the green pigment responsible for photosynthesis in plants.

3. Vitamin B₁₂ (Cyanocobalamin): A coordination compound of cobalt — used in treatment of pernicious anaemia.

4. Carbonic anhydrase: An enzyme containing zinc as a coordination centre — catalyses CO₂ hydration in biological systems.

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(ii) Medicinal Chemistry:

1. Cisplatin [$\text{cis}$-[PtCl₂(NH₃)₂]]: Used as an anticancer drug — inhibits the growth of tumours by binding to DNA.

2. EDTA complexes: EDTA (ethylenediaminetetraacetic acid) forms stable complexes with Pb²⁺ and is used in the treatment of lead poisoning (chelation therapy).

3. D-penicillamine and desferrioxime B: Chelating agents used to remove excess copper and iron from the body in cases of metal toxicity.

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(iii) Analytical Chemistry:

1. Detection of metal ions: Many metal ions are detected using specific ligands that form coloured complexes. For example, $\text{Fe}^{3+}$ is detected using SCN⁻ (forms blood-red $[\text{Fe}(\text{SCN})]^{2+}$).

2. EDTA titrations: EDTA forms stable 1:1 complexes with most metal ions and is used in complexometric titrations to determine the hardness of water (Ca²⁺, Mg²⁺ estimation).

3. Gravimetric analysis: Ni²⁺ is precipitated as $[\text{Ni}(\text{dmg})_2]$ (nickel dimethylglyoximate) for gravimetric determination.

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(iv) Extraction/Metallurgy of Metals:

1. Extraction of silver and gold (hydrometallurgy): Ag and Au are extracted using NaCN solution, forming soluble complex ions:
$$4\text{Au} + 8\text{NaCN} + 2\text{H}_2\text{O} + \text{O}_2 \rightarrow 4\text{Na}[\text{Au}(\text{CN})_2] + 4\text{NaOH}$$
Gold is then recovered by displacement with zinc.

2. Purification of nickel (Mond process): Impure nickel reacts with CO to form volatile $[\text{Ni}(\text{CO})_4]$ (tetracarbonylnickel), which is then decomposed at higher temperature to give pure nickel:
$$\text{Ni} + 4\text{CO} \xrightarrow{330-350 K} [\text{Ni}(\text{CO})_4] \xrightarrow{450-470 K} \text{Ni} + 4\text{CO}$$

Correct option: (iii) 3

Justification:

The complex is $[\text{Co}(\text{NH}_3)_6]\text{Cl}_2$.

In solution, it ionises as:
$$[\text{Co}(\text{NH}_3)_6]\text{Cl}_2 \rightarrow [\text{Co}(\text{NH}_3)_6]^{2+} + 2\text{Cl}^-$$

This produces 3 ions: one $[\text{Co}(\text{NH}_3)_6]^{2+}$ cation and two Cl⁻ anions.

The NH₃ ligands are inside the coordination sphere and do not dissociate.

Correct option: (ii) [Fe(H₂O)₆]²⁺

Justification:

Magnetic moment $\mu = \sqrt{n(n+2)}$ BM, where n = number of unpaired electrons.

- [Cr(H₂O)₆]³⁺: Cr³⁺ = 3d³ → 3 unpaired electrons; $\mu = \sqrt{3 \times 5} = \sqrt{15} \approx 3.87$ BM

- [Fe(H₂O)₆]²⁺: Fe²⁺ = 3d⁶; H₂O is weak field → high spin: $t_{2g}^4\ e_g^2$ → 4 unpaired electrons; $\mu = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90$ BM

- [Zn(H₂O)₆]²⁺: Zn²⁺ = 3d¹⁰ → 0 unpaired electrons; $\mu = 0$ BM

Highest magnetic moment = [Fe(H₂O)₆]²⁺ ≈ 4.90 BM

Correct option: (iii) [Fe(C₂O₄)₃]³⁻

Justification:

$[\text{Fe}(\text{C}_2\text{O}_4)_3]^{3-}$ contains oxalate ($\text{C}_2\text{O}_4^{2-}$), which is a bidentate chelating ligand.

Chelate complexes are significantly more stable than complexes with unidentate ligands due to the chelate effect:
- Formation of 5-membered chelate rings with oxalate
- Large positive entropy change upon chelation
- High stability constant (log K is very large)

The other complexes contain unidentate ligands (H₂O, NH₃, Cl⁻) which do not benefit from the chelate effect and are therefore less stable.

[Fe(C₂O₄)₃]³⁻ is the most stable due to the chelate effect of three bidentate oxalate ligands.

Concept: The wavelength of absorbed light is inversely related to the crystal field splitting energy $\Delta_o$:
$$\Delta_o = \frac{hc}{\lambda} \Rightarrow \lambda \propto \frac{1}{\Delta_o}$$

Stronger field ligands → larger $\Delta_o$ → shorter wavelength absorbed.

Spectrochemical series order of ligands:
\text{H}_2\text{O} &lt; \text{NH}_3 &lt; \text{NO}_2^-

Wait — from the spectrochemical series:
\text{H}_2\text{O} &lt; \text{NH}_3 &lt; \text{NO}_2^-

So: \Delta_o(\text{NO}_2^-) &gt; \Delta_o(\text{NH}_3) &gt; \Delta_o(\text{H}_2\text{O})

Therefore:
\lambda(\text{NO}_2^-) &lt; \lambda(\text{NH}_3) &lt; \lambda(\text{H}_2\text{O})

Correct order of wavelengths of absorption:
\boxed{[\text{Ni}(\text{NO}_2)_6]^{4-} &lt; [\text{Ni}(\text{NH}_3)_6]^{2+} &lt; [\text{Ni}(\text{H}_2\text{O})_6]^{2+}}

That is, $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ absorbs at the longest wavelength (lowest energy), and $[\text{Ni}(\text{NO}_2)_6]^{4-}$ absorbs at the shortest wavelength (highest energy) in the visible region.