An octahedral complex of Co$$^{3+}$$ is diamagnetic. The hybridisation involved in the formation of the complex is:

An octahedral complex of Co³⁺ is diamagnetic, meaning it has no unpaired electrons. To determine the hybridization, we start by examining the electron configuration of cobalt.

Cobalt has an atomic number of 27. The electron configuration of neutral cobalt (Co) is: $$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^7$$. For Co³⁺, three electrons are removed. Electrons are removed first from the 4s orbital and then from the 3d orbital. Removing two electrons from 4s² gives Co²⁺: $$[Ar] 3d^7$$. Removing one more electron from 3d⁷ gives Co³⁺: $$[Ar] 3d^6$$. So, the configuration of Co³⁺ is $$1s^2 2s^2 2p^6 3s^2 3p^6 3d^6$$.

In an octahedral complex, the d-orbitals split into a lower-energy t₂g set (dxy, dyz, dzx) and a higher-energy eg set (dx²-y², dz²). The magnetic properties depend on whether the complex is high-spin or low-spin.

A diamagnetic complex has all electrons paired. For Co³⁺ (d⁶), we consider two cases:

1. High-spin complex: The pairing energy is greater than the crystal field splitting energy (Δₒ). Electrons occupy orbitals singly before pairing. For d⁶, the configuration is t₂g⁴ eg². This has four unpaired electrons (two in t₂g and two in eg), making it paramagnetic.

2. Low-spin complex: The crystal field splitting energy (Δₒ) is greater than the pairing energy. Electrons pair in the t₂g orbitals before occupying eg. For d⁶, all six electrons occupy the t₂g orbitals: t₂g⁶ eg⁰. Since t₂g has three orbitals, each holds two electrons: ↑↓, ↑↓, ↑↓. This results in zero unpaired electrons, making it diamagnetic.

Given that the complex is diamagnetic, it must be low-spin. Therefore, the eg orbitals (dx²-y² and dz²) are empty.

In octahedral complexes, hybridization involves two d orbitals, one s orbital, and three p orbitals. The type of hybridization depends on whether the inner (n-1)d or outer nd orbitals are used:

- Inner orbital hybridization (d²sp³) uses two inner d orbitals from the (n-1)d set.
- Outer orbital hybridization (sp³d²) uses two outer d orbitals from the nd set.

For Co³⁺ (in the fourth period), n=4. The inner d orbitals are 3d. In the low-spin complex, the eg orbitals (dx²-y² and dz²) are empty and available for hybridization. Thus, the hybridization uses these two 3d orbitals (from the (n-1)d set), the 4s orbital, and the three 4p orbitals, forming d²sp³ hybridization.

Now, evaluating the options:

A. dsp²: This is for square planar geometry, not octahedral.
B. d²sp³: This matches inner orbital hybridization for octahedral complexes.
C. dsp³d: This notation is non-standard and typically implies a mix of inner and outer d orbitals, which is not applicable here.
D. sp³d²: This is outer orbital hybridization, used in high-spin octahedral complexes where inner d orbitals are occupied.

Since the complex is diamagnetic (low-spin), inner orbital hybridization (d²sp³) is involved.

Hence, the correct answer is Option B.