Valence Bond Theory & Magnetic Properties
Module 2 | CBSE Class 12 Chemistry | Coordination Compounds
1. Valence Bond Theory (VBT) Postulates
While Werner's theory explained the structure of coordination compounds, it could not explain why only certain elements form them, or why they exhibit characteristic magnetic and optical properties. To answer these, Linus Pauling developed the Valence Bond Theory.
- The central metal ion makes available a number of empty orbitals equal to its coordination number to accept electron pairs from ligands.
- These empty atomic orbitals (s, p, and d) hybridize to form a new set of equivalent orbitals of definite geometry (e.g., octahedral, tetrahedral, square planar).
- The ligands have at least one orbital (of donor atom) containing a lone pair of electrons.
- A coordinate covalent bond is formed by the overlap of the empty hybridized orbital of the metal ion with the filled orbital of the ligand.
| Coordination Number | Type of Hybridization | Geometry of Complex |
|---|---|---|
| 4 | sp3 | Tetrahedral |
| 4 | dsp2 | Square planar |
| 5 | sp3d | Trigonal bipyramidal |
| 6 | sp3d2 | Octahedral (Outer orbital) |
| 6 | d2sp3 | Octahedral (Inner orbital) |
2. Application of VBT to Coordination Number 6 (Octahedral)
Complexes with a coordination number of 6 are invariably octahedral. However, they can be formed using either the inner (n-1)d orbitals or the outer nd orbitals for hybridization.
2.1 Inner Orbital Complexes (Low Spin Complexes)
When the complex is formed by utilizing inner d orbitals for hybridization (d2sp3), it is called an inner orbital complex. These are generally formed under the influence of strong field ligands (like CN-, CO, NH3).
NCERT Example: [Co(NH3)6]3+ (Hexaamminecobalt(III) ion)
1. Cobalt (Z=27) ground state: [Ar] 3d7 4s2.
2. Co3+ ion state: [Ar] 3d6.
3. NH3 is a strong field ligand. It forces the six 3d electrons to pair up into three orbitals.
4. Two 3d orbitals become empty. They hybridize with one 4s and three 4p orbitals to form six d2sp3 hybrid orbitals.
5. Since all electrons are paired, the complex is diamagnetic.
2.2 Outer Orbital Complexes (High Spin Complexes)
When the complex is formed by utilizing outer d orbitals for hybridization (sp3d2), it is called an outer orbital complex. These are formed under the influence of weak field ligands (like F-, Cl-, H2O).
NCERT Example: [CoF6]3- (Hexafluoridocobaltate(III) ion)
1. Co3+ ion state: [Ar] 3d6.
2. F- is a weak field ligand. No pairing of 3d electrons occurs.
3. To accommodate six F- ions, it uses one 4s, three 4p, and two 4d orbitals, undergoing sp3d2 hybridization.
4. Since there are 4 unpaired electrons in the 3d subshell, the complex is highly paramagnetic.
3. Application of VBT to Coordination Number 4
Complexes with a coordination number of 4 can be either tetrahedral or square planar, depending on the hybridization.
3.1 Tetrahedral Complexes (sp3 Hybridization)
These involve the hybridization of one s and three p orbitals. They are usually formed by weak field ligands.
NCERT Example: [NiCl4]2- (Tetrachloridonickelate(II) ion)
1. Ni (Z=28) ground state: [Ar] 3d8 4s2.
2. Ni2+ ion state: [Ar] 3d8. (It has 2 unpaired electrons).
3. Cl- is a weak ligand. No pairing occurs.
4. The empty 4s and three 4p orbitals hybridize (sp3).
5. The geometry is tetrahedral, and due to 2 unpaired electrons, it is paramagnetic.
3.2 Square Planar Complexes (dsp2 Hybridization)
These involve the hybridization of one inner d, one s, and two p orbitals. They are generally formed under the influence of strong field ligands.
NCERT Example: [Ni(CN)4]2- (Tetracyanidonickelate(II) ion)
1. Ni2+ ion state: [Ar] 3d8.
2. CN- is a strong ligand. It forces the two unpaired 3d electrons to pair up.
3. One 3d orbital is now empty. It hybridizes with one 4s and two 4p orbitals to give dsp2 hybridization.
4. The geometry is square planar. Since all electrons are now paired, it is diamagnetic.
Nickel is in the 0 oxidation state: [Ar] 3d8 4s2. CO is a very strong ligand. It forces the two 4s electrons to fall back into the 3d subshell, pairing up the electrons to make 3d10. Now the 4s is empty. It uses 4s and 4p for sp3 hybridization. It is tetrahedral and diamagnetic.
4. Magnetic Properties of Coordination Compounds
The magnetic properties of coordination compounds depend on the presence of unpaired electrons in the d-orbitals of the central metal ion.
- Diamagnetic: All electrons are paired. Repelled by a magnetic field.
- Paramagnetic: Has one or more unpaired electrons. Attracted by a magnetic field.
μ = √[n(n+2)] B.M.
Where:
n = Number of unpaired electrons.
B.M. = Bohr Magneton (unit of magnetic moment).
Predicting VBT from Magnetic Data:
If experimental data shows a complex [MnBr4]2- has a magnetic moment of 5.9 B.M., this indicates n=5 (five unpaired electrons). Since Mn2+ is 3d5, it means no pairing occurred. Thus, Br- is a weak ligand, and the hybridization must be sp3 (Tetrahedral).
5. Limitations of Valence Bond Theory
While VBT successfully explains geometry and magnetic behavior, it suffers from severe drawbacks which necessitated the development of Crystal Field Theory (CFT).
- It involves a number of assumptions.
- It does not give quantitative interpretation of magnetic data.
- It does not explain the colour exhibited by coordination compounds (cannot explain d-d transitions).
- It does not give a quantitative interpretation of the thermodynamic or kinetic stabilities of coordination compounds.
- It cannot systematically distinguish between weak and strong ligands.
6. Previous Year Questions (PYQs) & Exhaustive Question Bank
Part A: Conceptual (1-2 Marks)
Q1. [NiCl4]2- is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
Q2. Predict the geometry and magnetic behavior of [Fe(CN)6]3- on the basis of VBT.
1. Oxidation state of Fe: x - 6 = -3 &implies; Fe is +3.
2. Electronic configuration of Fe3+: [Ar] 3d5.
3. CN- is a strong field ligand. It forces the 5 electrons to pair up in the inner 3d orbitals (two pairs and one unpaired).
4. This leaves two 3d orbitals empty. Hybridization is d2sp3.
5. Geometry is Octahedral. Since it has 1 unpaired electron, it is paramagnetic (with a low spin magnetic moment of √3 ≈ 1.73 B.M.).
Part B: Application Based Problems (3 Marks)
Q3. Compare the following complexes with respect to their shape, magnetic behavior and the hybrid orbitals involved:
(i) [CoF6]3-
(ii) [Co(NH3)6]3+
In both complexes, Cobalt is in the +3 oxidation state, meaning the configuration is [Ar] 3d6.
(i) [CoF6]3-:
F- is a weak field ligand. It does not force electron pairing. The 3d electrons remain with 4 unpaired electrons. The complex uses outer 4d orbitals for hybridization (sp3d2).
Shape: Octahedral.
Magnetic Behavior: Strongly Paramagnetic (High Spin).
(ii) [Co(NH3)6]3+:
NH3 is a strong field ligand. It forces the six 3d electrons to pair up, vacating two inner 3d orbitals. The hybridization is d2sp3.
Shape: Octahedral.
Magnetic Behavior: Diamagnetic (Low Spin, zero unpaired electrons).
Q4. The magnetic moment of [MnBr4]2- is 5.9 B.M. Predict the geometry of the complex ion.
Oxidation state of Mn is +2. Electronic configuration of Mn2+ is 3d5.
We are given μ = 5.9 B.M.
Using the formula μ = √[n(n+2)], a value of 5.9 B.M. corresponds to n = 5 (five unpaired electrons).
Since all 5 electrons in the 3d subshell remain unpaired, it proves that no pairing occurred. Therefore, the inner 3d orbitals are occupied. To accommodate four Br- ligands, the metal ion must use one 4s and three 4p orbitals, resulting in sp3 hybridization.
Hence, the geometry of the complex is Tetrahedral.
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