Coordination Compounds
Coordination Compounds for JEE Main & Advanced
Nomenclature and Bonding
Ligands, Coordination Number, IUPAC NomenclatureTopic 1
Coordination Compound: Compound where a central metal atom/ion is bonded to surrounding species (ligands) via coordinate bonds.
Example: $K_4[Fe(CN)_6]$ (potassium ferrocyanide); $[Cu(NH_3)_4]SO_4$.
Terminology:
- Coordination entity: $[Fe(CN)_6]^{4-}$ — the part in square brackets
- Central metal atom/ion: Fe in $[Fe(CN)_6]^{4-}$
- Ligands: Molecules/ions bonded to metal (donating electron pairs)
- Coordination sphere: Inside square brackets
- Counter ion: Outside square brackets (K⁺ in $K_4[Fe(CN)_6]$)
- Coordination number (CN): Number of donor atoms attached to metal
Types of Ligands:
| Type | Description | Examples |
|---|---|---|
| Monodentate | One donor atom | $H_2O$ (aqua), $NH_3$ (ammine), $CN^-$ (cyano), $Cl^-$ (chloro), $CO$ (carbonyl) |
| Bidentate | Two donor atoms | Ethylenediamine (en, $NH_2CH_2CH_2NH_2$), oxalate ($C_2O_4^{2-}$) |
| Polydentate | Multiple donor atoms | EDTA (hexadentate), DMG (bidentate) |
| Ambidentate | Can bind through different atoms | $-NO_2$ (nitro, via N) vs $-ONO$ (nitrito, via O); $-SCN$ (thiocyanato-S) vs $-NCS$ (thiocyanato-N or isothiocyanato) |
| Chelating | Bidentate or polydentate forming ring | en, ox, EDTA |
Coordination Numbers: Common: 2, 4, 6. Less common: 3, 5, 7, 8.
| CN | Geometry |
|---|---|
| 2 | Linear |
| 4 | Tetrahedral OR Square planar |
| 6 | Octahedral |
IUPAC Nomenclature Rules:
- Cation first, then anion in name.
- Within complex ion: ligands alphabetically, then metal.
- Greek prefixes for multiple identical ligands: di, tri, tetra (for simple); bis, tris, tetrakis (for complex ones like en).
- Anionic ligands end in -o: chloride → chloro, hydroxide → hydroxo, sulfate → sulfato, cyanide → cyano.
- Neutral ligands keep their name; exceptions: H₂O = aqua, NH₃ = ammine, CO = carbonyl, NO = nitrosyl.
- Cationic ligands end in -ium: NO⁺ = nitrosylium.
- Oxidation state of metal in Roman numerals in parentheses, after metal name.
- If complex is anionic, metal name ends in -ate: Fe → Ferrate, Cu → Cuprate, Au → Aurate, Ag → Argentate, Pb → Plumbate.
Examples:
| Formula | IUPAC Name |
|---|---|
| $[Co(NH_3)_6]Cl_3$ | Hexaamminecobalt(III) chloride |
| $K_4[Fe(CN)_6]$ | Potassium hexacyanoferrate(II) |
| $K_3[Fe(CN)_6]$ | Potassium hexacyanoferrate(III) |
| $[Pt(NH_3)_2Cl_2]$ | Diamminedichloroplatinum(II) |
| $[Cu(NH_3)_4]SO_4$ | Tetraamminecopper(II) sulfate |
| $[Cr(H_2O)_6]Cl_3$ | Hexaaquachromium(III) chloride |
| $Na_3[Fe(C_2O_4)_3]$ | Sodium trioxalatoferrate(III) |
| $[Pt(en)_2Cl_2]$ | Dichlorobis(ethylenediamine)platinum(IV) (if +4) |
Oxidation State (OS) of Metal: Calculated from total charge of complex: $OS_M + \sum OS_{ligands} = $ overall charge of complex.
Name: $[Co(NH_3)_4Cl_2]^+$.
Show solution
Ligands: 4 NH₃ (tetraammine), 2 Cl (dichloro). Alphabetical: ammine before chloro. Total charge: $+1$. Charge on Co: $+1 = x + 4(0) + 2(-1) \implies x = +3$.
Name: Tetraamminedichlorocobalt(III) ion.
Final Answer: Tetraamminedichlorocobalt(III) ion.
Determine OS of Pt in $K_2[PtCl_6]$.
Show solution
$K^+ \times 2 = +2$. So $[PtCl_6]$ has charge $-2$. $x + 6(-1) = -2 \implies x = +4$. Pt is in $+4$ oxidation state.
Final Answer: $Pt^{+4}$.
Ligand always:
EDTA is:
Coordination number of Co in $[Co(NH_3)_6]^{3+}$:
Name of $K_3[Fe(CN)_6]$:
Ambidentate ligand:
Werner's Theory, VBT, HybridizationTopic 2
Werner's Theory (1893): First explanation of coordination compounds.
Postulates:
- Metals have two types of valences:
- Primary valence (ionizable): Equivalent to oxidation state; satisfied by anions
- Secondary valence (non-ionizable): Equivalent to coordination number; satisfied by ligands; fixed for each metal in a given complex
- Primary valences are non-directional; secondary are directional → geometry
- Ligands occupy specific positions around metal at fixed geometry
For $CoCl_3 \cdot 6NH_3$:
- Primary valence: $3$ (from $Co^{3+}$)
- Secondary valence: $6$ (from 6 NH₃)
- Structure: $[Co(NH_3)_6]^{3+}\,3Cl^-$
- All 3 Cl⁻ are ionizable (give AgCl with AgNO₃)
For $CoCl_3 \cdot 5NH_3$:
- Structure: $[Co(NH_3)_5Cl]^{2+}\,2Cl^-$
- Only 2 Cl⁻ ionize (other is in coord. sphere)
For $CoCl_3 \cdot 4NH_3$:
- Structure: $[Co(NH_3)_4Cl_2]^+\,Cl^-$
- Only 1 Cl⁻ ionizes
Valence Bond Theory (VBT): Explains shape and magnetic properties.
Steps:
- Determine OS of metal and electronic configuration of ion
- Identify hybridization of metal orbitals to accept ligand electron pairs
- Predict geometry from hybridization
Strong vs Weak Field Ligands (for VBT):
Strong field ligands (cause pairing of d electrons): $CN^-, CO, en, NH_3, NO_2^-$ Weak field ligands (don't cause pairing): $F^-, Cl^-, Br^-, I^-, OH^-, H_2O$ (intermediate sometimes)
Hybridization-Geometry:
| CN | Hybridization | Geometry | Example |
|---|---|---|---|
| 2 | $sp$ | Linear | $[Ag(NH_3)_2]^+$, $[Cu(NH_3)_2]^+$ |
| 4 | $sp^3$ | Tetrahedral | $[NiCl_4]^{2-}, [Zn(NH_3)_4]^{2+}$ |
| 4 | $dsp^2$ | Square planar | $[Ni(CN)_4]^{2-}$, $[Pt(NH_3)_2Cl_2]$ |
| 5 | $sp^3d$ | TBP | $[Fe(CO)_5]$ |
| 6 | $sp^3d^2$ (outer; weak field) | Octahedral | $[FeF_6]^{3-}, [Fe(H_2O)_6]^{2+}$ |
| 6 | $d^2sp^3$ (inner; strong field) | Octahedral | $[Fe(CN)_6]^{4-}, [Co(NH_3)_6]^{3+}$ |
Inner vs Outer Orbital Complex:
- Inner orbital (low spin): Uses inner $d$ orbitals ($(n-1)d$); requires strong field; pairing of e⁻ occurs.
- Outer orbital (high spin): Uses outer $d$ orbitals ($nd$); weak field; no pairing.
Examples:
- $[Fe(CN)_6]^{4-}$: Fe²⁺ ($d^6$). $CN^-$ strong field → pair up → low spin → $d^2sp^3$ → octahedral diamagnetic.
- $[Fe(H_2O)_6]^{2+}$: Fe²⁺ ($d^6$). $H_2O$ weak (for Fe²⁺) → no pairing → high spin → $sp^3d^2$ → octahedral paramagnetic (4 unpaired).
- $[Ni(CN)_4]^{2-}$: Ni²⁺ ($d^8$). $CN^-$ strong → pair to free up $d$ → $dsp^2$ → square planar diamagnetic.
- $[NiCl_4]^{2-}$: Ni²⁺ ($d^8$). $Cl^-$ weak → no pairing → $sp^3$ → tetrahedral paramagnetic.
Determine hybridization, geometry, magnetic nature of $[Co(NH_3)_6]^{3+}$.
Show solution
Co³⁺ has $d^6$ config: $[Ar]\,3d^6$. $NH_3$ is strong field → pairing → 0 unpaired e⁻. Inner orbital: 2 inner d + s + 3p = $d^2sp^3$ hybridization. Geometry: octahedral. Diamagnetic (all paired).
Final Answer: $d^2sp^3$; octahedral; diamagnetic.
$[NiCl_4]^{2-}$: predict everything.
Show solution
Ni²⁺: $d^8$. $Cl^-$ weak field → no pairing → 2 unpaired e⁻. Hybridization: $sp^3$ (using outer 4s, 4p). Geometry: tetrahedral. Paramagnetic (2 unpaired).
Final Answer: $sp^3$; tetrahedral; paramagnetic.
Werner's primary valence corresponds to:
$[Ni(CN)_4]^{2-}$ is:
Strong field ligand:
Hybridization of $[Fe(CN)_6]^{4-}$:
CFT explains:
Isomerism, CFT and Applications
Isomerism in Coordination CompoundsTopic 1
Isomerism: Compounds with same molecular formula but different structure/properties.
Structural Isomerism:
1. Ionization Isomerism: Same formula, different ions outside coordination sphere.
- $[Co(NH_3)_5SO_4]Br$ vs $[Co(NH_3)_5Br]SO_4$
- First gives Br⁻ (test with AgNO₃ → AgBr), second gives SO₄²⁻ (test with BaCl₂ → BaSO₄).
2. Hydrate (Solvate) Isomerism: Water inside vs outside coordination sphere.
- $[Cr(H_2O)_6]Cl_3$ (violet) vs $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ (blue-green) vs $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$ (dark green)
3. Linkage Isomerism: Ambidentate ligands attached through different atoms.
- $[Co(NH_3)_5(NO_2)]^{2+}$ (nitro, via N) vs $[Co(NH_3)_5(ONO)]^{2+}$ (nitrito, via O)
- $[Co(NH_3)_5(SCN)]^{2+}$ (S-attached) vs $[Co(NH_3)_5(NCS)]^{2+}$ (N-attached)
4. Coordination Isomerism: Different metals or distributions in cation and anion.
- $[Co(NH_3)_6][Cr(CN)_6]$ vs $[Cr(NH_3)_6][Co(CN)_6]$
- Requires complex with both cationic and anionic parts.
Stereoisomerism:
1. Geometrical (cis-trans) Isomerism:
| Geometry | Cis-trans? |
|---|---|
| Linear (CN 2) | No |
| Tetrahedral | No (all positions equivalent) |
| Square planar | Yes |
| Octahedral | Yes |
Square planar $[Ma_2b_2]$:
- cis: same groups adjacent (90° apart)
- trans: same groups opposite (180° apart)
- Example: $[Pt(NH_3)_2Cl_2]$ — both cis and trans forms exist; cis is cisplatin (anti-cancer drug)
Octahedral:
- $[Ma_4b_2]$: cis (b's adjacent) or trans (b's opposite)
- $[Ma_3b_3]$: facial (fac, three a's on triangular face) or meridional (mer, three a's in plane)
- $[M(en)_2Cl_2]$: cis and trans
2. Optical Isomerism:
A complex shows optical isomerism if it's chiral (non-superimposable on its mirror image).
Conditions:
- No improper rotation axis (e.g., plane of symmetry, center of symmetry)
- Common in octahedral with bidentate ligands: $[M(en)_3]^{n+}$ has $\Delta$ (right-handed) and $\Lambda$ (left-handed) forms.
Examples:
- $[Co(en)_3]^{3+}$: only optical isomers (no cis-trans)
- cis-$[Co(en)_2Cl_2]^+$: optical isomers exist (chiral)
- trans-$[Co(en)_2Cl_2]^+$: not chiral (has plane of symmetry)
Tetrahedral with 4 different ligands can be chiral, but few examples.
How many geometric isomers does $[Pt(NH_3)_2Cl_2]$ have?
Show solution
Square planar (Pt²⁺ has $d^8$, typically square planar with strong field). $Ma_2b_2$ square planar: 2 isomers (cis with $NH_3$'s adjacent, and trans with them opposite).
- cis-$[Pt(NH_3)_2Cl_2]$: cisplatin (anti-cancer drug, FDA approved)
- trans-$[Pt(NH_3)_2Cl_2]$: inactive
Final Answer: 2 isomers (cis and trans).
Distinguish ionization isomers $[Co(NH_3)_5SO_4]Br$ vs $[Co(NH_3)_5Br]SO_4$.
Show solution
Test for free Br⁻: Add AgNO₃ solution.
- $[Co(NH_3)_5SO_4]Br$: Br⁻ outside sphere → reacts → white-yellow AgBr ppt.
- $[Co(NH_3)_5Br]SO_4$: Br⁻ inside sphere → no AgBr.
Test for free SO₄²⁻: Add BaCl₂ solution.
- First: no ppt.
- Second: white BaSO₄ ppt.
Final Answer: Ionic Br⁻ vs ionic SO₄²⁻; distinguish with AgNO₃ and BaCl₂ tests.
cisplatin is:
Linkage isomerism is observed with:
$[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ are:
Optical isomers are observed in:
fac and mer isomers occur in:
Crystal Field Theory, Color, Magnetism, ApplicationsTopic 2
Crystal Field Theory (CFT): Treats ligands as point charges (negative) producing electrostatic field around metal. The field splits $d$-orbitals.
Octahedral Field:
- 5 $d$-orbitals split into:
- $e_g$ set: $d_{x^2-y^2}, d_{z^2}$ (higher energy; lobes point at ligands)
- $t_{2g}$ set: $d_{xy}, d_{yz}, d_{xz}$ (lower energy; lobes between ligands)
Crystal Field Splitting Energy (CFSE): $\Delta_o$ = energy gap between $e_g$ and $t_{2g}$.
For octahedral:
- $e_g$ at $+0.6\Delta_o$ above mean
- $t_{2g}$ at $-0.4\Delta_o$ below mean
Total CFSE = (electrons in $t_{2g}$)($-0.4\Delta_o$) + (electrons in $e_g$)($+0.6\Delta_o$) + (number of pairs caused by strong field)(P), where P = pairing energy.
Tetrahedral Field:
- 5 $d$-orbitals split into:
- $e$ set ($d_{x^2-y^2}, d_{z^2}$): lower energy
- $t_2$ set ($d_{xy}, d_{yz}, d_{xz}$): higher energy
- $\Delta_t \approx (4/9)\Delta_o$ (smaller splitting)
- No low-spin tetrahedral complexes (splitting too small)
Square Planar: Similar to octahedral but $z$-axis ligands removed.
Spectrochemical Series: Ligands arranged by increasing $\Delta$: $$I^- < Br^- < S^{2-} < SCN^- < Cl^- < NO_3^- < F^- < OH^- < H_2O < NCS^- < NH_3 < en < NO_2^- < CN^- < CO$$
(weak field) ←————————————————→ (strong field)
Spin State for $d^4$ to $d^7$ Octahedral:
| $d^n$ | Weak field (high spin) | Strong field (low spin) |
|---|---|---|
| $d^4$ | $t_{2g}^3\,e_g^1$ (4 unpaired) | $t_{2g}^4$ (2 unpaired) |
| $d^5$ | $t_{2g}^3\,e_g^2$ (5 unpaired) | $t_{2g}^5$ (1 unpaired) |
| $d^6$ | $t_{2g}^4\,e_g^2$ (4 unpaired) | $t_{2g}^6$ (0 unpaired, diamagnetic) |
| $d^7$ | $t_{2g}^5\,e_g^2$ (3 unpaired) | $t_{2g}^6\,e_g^1$ (1 unpaired) |
Color of Coordination Compounds:
- Complementary to absorbed light (d-d transition)
- $\Delta_o$ depends on metal, ligand, geometry
- Larger $\Delta_o$ → higher energy light absorbed → bluer/redder shift
Examples:
- $[Ti(H_2O)_6]^{3+}$ ($d^1$): absorbs in yellow region → violet color
- $[Cu(H_2O)_4]^{2+}$ ($d^9$): blue
- $[Cu(NH_3)_4]^{2+}$ ($d^9$): deep blue (greater $\Delta_o$, deeper blue)
Crystal Field Stabilization Energy (CFSE) Calculations:
Example: $[Fe(CN)_6]^{4-}$, Fe²⁺ $d^6$ low spin: $t_{2g}^6 e_g^0$. CFSE = $6 \times (-0.4) \Delta_o + 0 \times 0.6 \Delta_o = -2.4 \Delta_o$ (extra stability).
Limitations of CFT:
- Ignores covalent character (ligands treated as pure point charges)
- Cannot explain $\pi$-bonding (e.g., why CO is a strong field ligand)
- More accurate: Ligand Field Theory (LFT) or MOT
Applications of Coordination Compounds:
| Application | Compound |
|---|---|
| Cisplatin (anti-cancer) | cis-$[Pt(NH_3)_2Cl_2]$ |
| Carrier in biology | Hemoglobin ($Fe$ porphyrin), chlorophyll ($Mg$ porphyrin), vitamin B12 ($Co$ corrin) |
| Photography | $Na_3[Ag(S_2O_3)_2]$ in fixing |
| Water hardness removal | EDTA complexes Ca²⁺, Mg²⁺ |
| Electroplating | $K[Au(CN)_2]$ gold plating; $K[Ag(CN)_2]$ silver |
| Catalysis | Wilkinson's catalyst $[RhCl(PPh_3)_3]$ in hydrogenation |
| Qualitative analysis | $[Fe(SCN)]^{2+}$ deep red (Fe³⁺ test); $[Cu(NH_3)_4]^{2+}$ deep blue (Cu²⁺) |
| Estimation of Ni | DMG + Ni²⁺ → rose-red complex |
Calculate CFSE for $[Fe(CN)_6]^{4-}$.
Show solution
Fe²⁺: $d^6$. $CN^-$ is strong field → low spin. Config: $t_{2g}^6\,e_g^0$. CFSE = $6 \times (-0.4\Delta_o) + 0 \times (+0.6\Delta_o) = -2.4\Delta_o$. Plus pairing energy: 2 extra pairs formed compared to free ion (which had 4 unpaired in $d^6$).
Final Answer: CFSE = $-2.4\Delta_o + 2P$ (where P = pairing energy).
Why is $[Co(NH_3)_6]^{3+}$ diamagnetic but $[CoF_6]^{3-}$ paramagnetic?
Show solution
Both have Co³⁺ ($d^6$). $NH_3$ is strong field → low spin → $t_{2g}^6\,e_g^0$ → all paired → diamagnetic. $F^-$ is weak field → high spin → $t_{2g}^4\,e_g^2$ → 4 unpaired → paramagnetic.
Final Answer: Different ligand field strength causes different electronic configurations.
In octahedral field, d orbitals split into:
Strong field ligand causes:
$\Delta_t$ vs $\Delta_o$:
Color of $[Cu(H_2O)_4]^{2+}$:
Hemoglobin contains:
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