Coordination Compounds — NCERT Solutions

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.

(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).

(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).

(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)

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

[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 ✓

[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)

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

• H₂O is a weak field ligand → small crystal field splitting energy $\Delta_o$
• Since $\Delta_o < P$ (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

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

• CN⁻ is a strong field ligand → large crystal field splitting energy $\Delta_o$
• Since $\Delta_o > P$, 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

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

(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

(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

(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

(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

(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

(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

(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

(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

(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)

(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

(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°)

(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.

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)

(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 $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$

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

(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)

(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.

(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)

(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)

(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)