CBSE · Class 12 · Chemistry · Coordination CompoundsDiscuss Valence Bond Theory (VBT) of coordination compounds with its main postulates. Explain the formation, hybridization, geometry, and magnetic behavior of $Co(F6)^{3-}$ and $Co(NH3)6^{3-}$ complexes using VBT.

Introduction to Valence Bond Theory (VBT)\nValence Bond Theory was proposed by Linus Pauling to explain the bonding and magnetic properties of coordination compounds. It primarily focuses on coordinate covalent bond formation between the central metal ion and ligands.

Main Postulates

• Metal Ligand Interaction: The central metal atom/ion makes available a number of vacant orbitals equal to its coordination number for accommodating electron pairs donated by ligands.
• Hybridization: The vacant orbitals of the metal undergo hybridization to yield a set of equivalent orbitals of definite geometry.
• Overlap: The empty hybrid orbitals of the metal overlap with filled orbitals of ligands containing lone pairs of electrons to form coordinate covalent bonds.
• Inner/Outer Orbital Complexes: If inner $d$-orbitals $($(n-1)d$)$ are used for hybridization, it is called an inner orbital complex (low spin). If outer $d$-orbitals ($ns, np, nd$) are used, it is called an outer orbital complex (high spin).

Application to $[Co(F_6)]^{3-}$

• Oxidation state of Co is $+3$. Electronic configuration of Co ($Z = 27$) is $[Ar] 3d^7 4s^2$.
• $Co^{3+}$ configuration: $[Ar] 3d^6$.
• Fluoride ($F^-$) is a weak field ligand and cannot force electron pairing in the $3d$ orbitals.
• Therefore, $Co^{3+}$ utilizes one $4s$, three $4p$, and two $4d$ orbitals for hybridization.
• Hybridization: $sp^3d^2$ (Outer orbital complex).
• Geometry: Octahedral.
• Magnetic Behavior: It contains 4 unpaired electrons, hence it is paramagnetic in nature with a high spin configuration.

Application to $[Co(NH_3)_6]^{3-}$

• Oxidation state of Co is $+3$. $Co^{3+}$ configuration: $[Ar] 3d^6$.
• Ammonia ($NH_3$) is a strong field ligand and causes pairing of electrons in the $3d$ orbitals against Hund's rule.
• Two $3d$ orbitals become vacant, allowing $d^2sp^3$ hybridization.
• Hybridization: $d^2sp^3$ (Inner orbital complex).
• Geometry: Octahedral.
• Magnetic Behavior: It contains no unpaired electrons, hence it is diamagnetic in nature with a low spin configuration.