How many unpaired electrons are present in the high spin form of [CoF₆]^3- complex and which metal orbitals are used in bonding?

4 unpaired electrons and 4s, 4p and 4d orbitals to give sp³d² hybridisation.

The correct answer is option 2. 4 unpaired electrons and 4s, 4p and 4d orbitals to give sp³d² hybridisation.

To determine the number of unpaired electrons in the high spin form of the \([CoF_6]^{3-}\) complex and the orbitals used in bonding, let's analyze the given options and the characteristics of the complex.

Complex: \([CoF_6]^{3-}\)

High Spin Configuration:  For a high spin configuration, the complex has a large crystal field splitting energy (Δ) such that the pairing energy is not sufficient to pair all the electrons in the lower energy orbitals before populating higher energy orbitals. This results in maximum number of unpaired electrons.

Cobalt (Co): Cobalt in the +3 oxidation state (\(Co^{3+}\)) has the electron configuration [Ar] 3d^6.

Orbitals Involved: In a high spin complex, the electrons fill the orbitals to maximize unpaired electrons. For cobalt in a high spin state, the electrons are distributed across the \(3d\), \(4s\), and \(4p\) orbitals primarily.

Let us match the options with the correct analysis:

Option 1: \(0\) unpaired electrons and \(4s\), \(4p\), and \(4d\) orbitals with \(sp^3d^2\) hybridization is incorrect because it suggests no unpaired electrons, which contradicts the high spin configuration expected for \([CoF_6]^{3-}\).

Option 2: \(4\) unpaired electrons and \(4s\), \(4p\), and \(4d\) orbitals with \(sp^3d^2\) hybridization is plausible because it aligns with the high spin configuration where cobalt would have \(4\) unpaired electrons due to the filling of \(4s\), \(4p\), and \(4d\) orbitals.

Option 3: \(0\) unpaired electrons and \(3d\), \(4s\), and \(4p\) orbitals with \(d^2sp^3\) hybridization is incorrect because it suggests \(0\) unpaired electrons, which again contradicts the high spin configuration.

Option 4: \(4\) unpaired electrons and \(3d\), \(4s\), and \(4p\) orbitals with \(d^2sp^3\) hybridization is incorrect because it states \(4\) unpaired electrons with \(d^2sp^3\) hybridization, which is not typical for high spin complexes.

The correct answer is: \(4 \text{ unpaired electrons and 4s, 4p and 4d orbitals to give sp}^3\text{d}^2 \text{ hybridisation.} \)

This option correctly identifies the high spin configuration of \([CoF_6]^{3-}\) with \(4\) unpaired electrons, utilizing the \(4s\), \(4p\), and \(4d\) orbitals for bonding, resulting in \(sp^3d^2\) hybridization.