Alfred Werner proposed that metals in coordination compounds show two types of valency: primary valency (the metal's ordinary oxidation state, satisfied by negative ions and ionizable) and secondary valency (the fixed number of ligands directly bonded to the metal, corresponding to its coordination number, non-ionizable). Secondary valency is always directed towards fixed positions in space, giving the complex a definite geometry.
For $\text{CoCl}_3 \cdot 6\text{NH}_3$, explain why all 3 chloride ions are precipitated immediately by AgNO₃, but for $\text{CoCl}_3 \cdot 5\text{NH}_3$, only 2 chlorides precipitate.
In $[\text{Co(NH}_3)_6]\text{Cl}_3$, cobalt's secondary valency (coordination number 6) is fully satisfied by 6 NH₃ ligands, leaving all 3 Cl⁻ as free, ionisable counter-ions — so all 3 precipitate immediately with AgNO₃.
In $[\text{Co(NH}_3)_5\text{Cl}]\text{Cl}_2$, one Cl⁻ has entered the coordination sphere itself (satisfying part of the secondary valency alongside 5 NH₃), leaving only 2 Cl⁻ as free ionisable counter-ions — so only 2 precipitate.
An ion or molecule bonded to the central metal atom/ion in a coordination entity, donating at least one lone pair of electrons to form a coordinate (dative) bond.
| Ligand Type | Description | Example |
|---|---|---|
| Unidentate | Donates through only 1 atom | NH₃, Cl⁻, H₂O |
| Didentate | Donates through 2 atoms simultaneously | Ethylenediamine (en) |
| Polydentate | Donates through several atoms at once | EDTA (hexadentate) |
| Ambidentate | Can attach through either of two different atoms | NO₂⁻ (via N or O), SCN⁻ (via S or N) |
A chelating ligand is a di/polydentate ligand that forms a ring structure by binding to the same metal atom at two or more points — chelate complexes are generally more stable than similar complexes with unidentate ligands (chelate effect). Coordination number is the total number of donor atoms bonded to the central metal atom.
Determine the coordination number of the central metal in $[\text{Co(en)}_3]^{3+}$.
Ethylenediamine (en) is didentate, and there are 3 en ligands: coordination number $= 3 \times 2 = 6$.
Write the IUPAC name of $[\text{Co(NH}_3)_4\text{Cl}_2]\text{Cl}$.
Ligands: 4 NH₃ (ammine) and 2 Cl⁻ (chlorido), in alphabetical order — chlorido before ammine.
Complex ion is a cation, so no special ending for cobalt. Cobalt's oxidation state: overall charge is +1 (one free Cl⁻ outside), and 2 Cl⁻ inside contribute −2, so Co must be +3.
Name: Tetraamminedichloridocobalt(III) chloride
| Type | Cause | Example |
|---|---|---|
| Ionisation isomerism | Different ions produced when dissolved, due to swapping of a ligand and counter-ion | [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br |
| Linkage isomerism | An ambidentate ligand attaches through a different donor atom | [Co(NH₃)₅(NO₂)]²⁺ vs [Co(NH₃)₅(ONO)]²⁺ |
| Coordination isomerism | Distribution of ligands differs between cationic and anionic complex parts of the same compound | [Co(NH₃)₆][Cr(CN)₆] vs [Cr(NH₃)₆][Co(CN)₆] |
| Solvate (hydrate) isomerism | Water molecules differ between being inside the coordination sphere or outside as free water of crystallisation | [Cr(H₂O)₆]Cl₃ vs [Cr(H₂O)₅Cl]Cl₂·H₂O |
Explain how [Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br are ionisation isomers, and how you could distinguish them experimentally.
In the first, Br⁻ is inside the coordination sphere (bonded to Co) and SO₄²⁻ is the free counter-ion; in the second, it's the reverse. Testing with BaCl₂ solution: the first compound would give no immediate precipitate with Ba²⁺ (since SO₄²⁻ is free and would form BaSO₄ precipitate) — actually the first would give a BaSO₄ precipitate readily (free SO₄²⁻), while the second would not react with Ba²⁺ (since SO₄²⁻ is locked inside the coordination sphere), instead reacting with AgNO₃ to precipitate AgBr (since Br⁻ is free in the second compound).
Arises in complexes where two identical ligands can be positioned either adjacent (cis, at 90°) or opposite (trans, at 180°) to each other. Common in square planar $[MA_2B_2]$ complexes and octahedral $[MA_4B_2]$ complexes.
Arises when a complex and its mirror image are non-superimposable on each other (chiral) — the two forms are called enantiomers, and rotate plane-polarised light in opposite directions. Common in octahedral complexes with 3 didentate ligands (e.g., $[\text{Co(en)}_3]^{3+}$).
Does the square planar complex $[\text{Pt(NH}_3)_2\text{Cl}_2]$ show geometrical isomerism? Describe both forms.
Yes. In the cis form, the two NH₃ groups are adjacent (90° apart), as are the two Cl⁻ ligands. In the trans form, the two NH₃ groups are opposite each other (180° apart), as are the two Cl⁻ ligands. These are the two distinct geometrical isomers of this complex.
Valence Bond Theory (VBT) explains bonding in complexes through hybridisation of the metal's vacant orbitals, which then accept lone pairs donated by ligands, forming coordinate bonds. The type of hybridisation determines the complex's geometry.
| Coordination Number | Hybridisation | Geometry |
|---|---|---|
| 4 | sp³ | Tetrahedral |
| 4 | dsp² | Square planar |
| 6 | sp³d² | Octahedral (outer orbital) |
| 6 | d²sp³ | Octahedral (inner orbital) |
An inner orbital complex uses the metal's inner $(n-1)d$ orbitals for hybridisation (d²sp³), typically formed with strong-field ligands that force electron pairing, giving a low-spin complex. An outer orbital complex uses the outer $nd$ orbitals (sp³d²), typically formed with weak-field ligands that don't force pairing, giving a high-spin complex.
Predict the hybridisation and magnetic nature of $[\text{Ni(CN)}_4]^{2-}$ (Ni²⁺ is $3d^8$).
CN⁻ is a strong-field ligand that pairs up electrons within the d-orbitals, freeing one d-orbital for hybridisation — Ni²⁺'s 8 d-electrons pair up completely, leaving all electrons paired. Hybridisation: dsp² (square planar). Since all electrons are paired, the complex is diamagnetic.
Crystal Field Theory (CFT) treats the bonding in a complex as purely electrostatic — ligands are treated as point negative charges (or dipoles) that repel the metal's d-orbital electrons unequally, depending on orbital orientation, causing the originally degenerate (equal-energy) d-orbitals to split into sets of different energies.
In an octahedral field, the d-orbitals split into two sets: the higher-energy $e_g$ set (orbitals pointing directly at the ligands, experiencing more repulsion) and the lower-energy $t_{2g}$ set (orbitals pointing between the ligands, experiencing less repulsion). The energy gap between them is called crystal field splitting energy, $\Delta_o$.
If $\Delta_o$ is small (weak-field ligand), electrons prefer to occupy all 5 orbitals singly first (following Hund's rule) before pairing — giving a high-spin complex, with more unpaired electrons. If $\Delta_o$ is large (strong-field ligand), electrons preferentially pair up in the lower $t_{2g}$ set before occupying the higher $e_g$ set — giving a low-spin complex, with fewer unpaired electrons.
Ligands towards the right cause large splitting (strong field, favour low spin); ligands towards the left cause small splitting (weak field, favour high spin).
CFSE is the net energy stabilisation gained by the electron configuration in a split field compared to if the d-orbitals had remained unsplit — it explains extra stability trends and colour intensity in transition metal complexes.
Predict whether $[\text{Co(NH}_3)_6]^{3+}$ (Co³⁺, $3d^6$) is high-spin or low-spin, given NH₃ is a moderately strong field ligand and Co³⁺ complexes with NH₃ are known to be diamagnetic.
Since the complex is diamagnetic, all 6 d-electrons must be paired. This is only possible in the low-spin arrangement: $t_{2g}^6 e_g^0$ — all 6 electrons paired in the lower $t_{2g}$ set, none in the higher $e_g$ set. This confirms $[\text{Co(NH}_3)_6]^{3+}$ is a low-spin complex.