Valence Bond Theory & Magnetism
Decode the geometry of complexes. Master the pairing rules of Strong Field Ligands, inner vs. outer orbital complexes, and the mathematics of the magnetic moment.
Module Focus: The Empty Orbital Model
Valence Bond Theory (VBT) assumes that the central metal ion makes available a number of empty orbitals equal to its Coordination Number. These empty orbitals undergo hybridization to yield a set of equivalent orbitals of definite geometry. The ligands then donate electron pairs into these hybrid orbitals. To predict the correct geometry, you must know whether the ligand is strong enough to force the metal's unpaired electrons to pair up!
1. The Role of Ligands (Strong vs. Weak)
In VBT, the nature of the ligand dictates what happens to the unpaired d-electrons of the central metal atom.
- Force Pairing: They have a strong tendency to force the unpaired electrons of the central metal to pair up against Hund's Rule, vacating inner d-orbitals.
- Examples: $CN^-, CO, en, NH_3$ (usually).
- Result: Often leads to Diamagnetic complexes.
- No Pairing: They are weak and cannot force the unpaired electrons to pair up. The electrons remain in their original states.
- Examples: Halides ($F^-, Cl^-, Br^-, I^-$), $H_2O, OH^-, SCN^-$.
- Result: Often leads to Paramagnetic complexes.
2. Coordination Number 4 (Tetrahedral vs Square Planar)
If C.N. = 4, the complex can be either Tetrahedral ($sp^3$) or Square Planar ($dsp^2$).
- Central Ion: $Ni^{2+} \ (3d^8)$
- Ligand: $Cl^-$ (Weak Field $\rightarrow$ NO PAIRING)
- Configuration: Two unpaired electrons remain in 3d.
- Orbitals Used: One 4s and three 4p.
- Hybridization: $sp^3$ (Tetrahedral)
- Paramagnetic ($n=2$)
- Central Ion: $Ni^{2+} \ (3d^8)$
- Ligand: $CN^-$ (Strong Field $\rightarrow$ PAIRING OCCURS)
- Configuration: Electrons pair up, vacating one 3d orbital.
- Orbitals Used: One 3d, one 4s, two 4p.
- Hybridization: $dsp^2$ (Square Planar)
- Diamagnetic ($n=0$)
Carbonyl ($CO$) is a neutral, strong field ligand. The oxidation state of Ni is 0 (configuration $3d^8 4s^2$).
Because $CO$ is a very strong ligand, it forces the two 4s electrons to jump into the 3d subshell, pairing up all electrons ($3d^{10}$). Now the 3d is totally full!
Hybridization = $sp^3$ (Tetrahedral)
Magnetism = Diamagnetic
3. Coordination Number 6 (Octahedral)
If C.N. = 6, the geometry is always Octahedral. The only question is which 'd' orbitals are used: the inner $(n-1)d$ or the outer $nd$.
| Property | Inner Orbital Complex | Outer Orbital Complex |
|---|---|---|
| Hybridization | $d^2sp^3$ | $sp^3d^2$ |
| Orbitals Used | Two $3d$, one $4s$, three $4p$ | One $4s$, three $4p$, two $4d$ |
| Ligand Type | Strong Field (forces pairing) | Weak Field (no pairing) |
| Spin Nature | Low Spin / Spin-Paired | High Spin / Spin-Free |
| Examples | $[Fe(CN)_6]^{4-}, [Co(NH_3)_6]^{3+}$ | $[Fe(H_2O)_6]^{2+}, [CoF_6]^{3-}$ |
4. Magnetic Properties
The magnetic moment provides experimental proof for the VBT configurations. It is calculated using the "spin-only" formula.
BM = Bohr Magneton (unit)
The integer part of the magnetic moment is ALWAYS equal to the number of unpaired electrons ($n$).
- $n=1 \rightarrow \mu \approx 1.73 \text{ BM}$
- $n=2 \rightarrow \mu \approx 2.83 \text{ BM}$
- $n=3 \rightarrow \mu \approx 3.87 \text{ BM}$
- $n=4 \rightarrow \mu \approx 4.90 \text{ BM}$
- $n=5 \rightarrow \mu \approx 5.92 \text{ BM}$
- It does not explain the color exhibited by coordination compounds.
- It does not quantitatively explain the magnetic data (temperature dependence).
- It gives no thermodynamic or kinetic stabilities.
- It does not distinguish explicitly between weak and strong field ligands (CFT does this).
NEET Grand Test: VBT & Magnetism
15 High-Yield Questions testing hybridization logic, spin-only calculations, and classic transition metal traps.
Join the Ultimate Chemistry Crash Course
Master Inorganic Chemistry, Coordination Compounds, and Block Elements. Get access to our full suite of Rapid Revision modules, formula sheets, and mock tests specifically designed for the NTA NEET pattern.
Explore All NEET Modules →
No comments:
Post a Comment