JEE Main & Advanced

Coordination Compounds

Coordination Compounds for JEE Main & Advanced

1
Module 1

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:

TypeDescriptionExamples
MonodentateOne donor atom$H_2O$ (aqua), $NH_3$ (ammine), $CN^-$ (cyano), $Cl^-$ (chloro), $CO$ (carbonyl)
BidentateTwo donor atomsEthylenediamine (en, $NH_2CH_2CH_2NH_2$), oxalate ($C_2O_4^{2-}$)
PolydentateMultiple donor atomsEDTA (hexadentate), DMG (bidentate)
AmbidentateCan bind through different atoms$-NO_2$ (nitro, via N) vs $-ONO$ (nitrito, via O); $-SCN$ (thiocyanato-S) vs $-NCS$ (thiocyanato-N or isothiocyanato)
ChelatingBidentate or polydentate forming ringen, ox, EDTA

Coordination Numbers: Common: 2, 4, 6. Less common: 3, 5, 7, 8.

CNGeometry
2Linear
4Tetrahedral OR Square planar
6Octahedral

IUPAC Nomenclature Rules:

  1. Cation first, then anion in name.
  2. Within complex ion: ligands alphabetically, then metal.
  3. Greek prefixes for multiple identical ligands: di, tri, tetra (for simple); bis, tris, tetrakis (for complex ones like en).
  4. Anionic ligands end in -o: chloride → chloro, hydroxide → hydroxo, sulfate → sulfato, cyanide → cyano.
  5. Neutral ligands keep their name; exceptions: H₂O = aqua, NH₃ = ammine, CO = carbonyl, NO = nitrosyl.
  6. Cationic ligands end in -ium: NO⁺ = nitrosylium.
  7. Oxidation state of metal in Roman numerals in parentheses, after metal name.
  8. If complex is anionic, metal name ends in -ate: Fe → Ferrate, Cu → Cuprate, Au → Aurate, Ag → Argentate, Pb → Plumbate.

Examples:

FormulaIUPAC 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.

Worked Examples
1

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.

2

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: $Pt^{+4}$.

✎ Self-Check — 5 questions0 / 5
Q1.

Ligand always:

Q2.

EDTA is:

Q3.

Coordination number of Co in $[Co(NH_3)_6]^{3+}$:

Q4.

Name of $K_3[Fe(CN)_6]$:

Q5.

Ambidentate ligand:

Werner's Theory, VBT, HybridizationTopic 2

Werner's Theory (1893): First explanation of coordination compounds.

Postulates:

  1. 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
  1. Primary valences are non-directional; secondary are directional → geometry
  2. 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:

  1. Determine OS of metal and electronic configuration of ion
  2. Identify hybridization of metal orbitals to accept ligand electron pairs
  3. 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:

CNHybridizationGeometryExample
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.
Worked Examples
1

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.

2

$[NiCl_4]^{2-}$: 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.

✎ Self-Check — 5 questions0 / 5
Q1.

Werner's primary valence corresponds to:

Q2.

$[Ni(CN)_4]^{2-}$ is:

Q3.

Strong field ligand:

Q4.

Hybridization of $[Fe(CN)_6]^{4-}$:

Q5.

CFT explains:

2
Module 2

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:

GeometryCis-trans?
Linear (CN 2)No
TetrahedralNo (all positions equivalent)
Square planarYes
OctahedralYes

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.

Worked Examples
1

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).

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

cisplatin is:

Q2.

Linkage isomerism is observed with:

Q3.

$[Cr(H_2O)_6]Cl_3$ and $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ are:

Q4.

Optical isomers are observed in:

Q5.

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:

ApplicationCompound
Cisplatin (anti-cancer)cis-$[Pt(NH_3)_2Cl_2]$
Carrier in biologyHemoglobin ($Fe$ porphyrin), chlorophyll ($Mg$ porphyrin), vitamin B12 ($Co$ corrin)
Photography$Na_3[Ag(S_2O_3)_2]$ in fixing
Water hardness removalEDTA complexes Ca²⁺, Mg²⁺
Electroplating$K[Au(CN)_2]$ gold plating; $K[Ag(CN)_2]$ silver
CatalysisWilkinson'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 NiDMG + Ni²⁺ → rose-red complex
Worked Examples
1

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).

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

In octahedral field, d orbitals split into:

Q2.

Strong field ligand causes:

Q3.

$\Delta_t$ vs $\Delta_o$:

Q4.

Color of $[Cu(H_2O)_4]^{2+}$:

Q5.

Hemoglobin contains:

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