Predict whether [CoF6]3- is diamagnetic or paramagnetic and why?…

Predict whether [CoF6]3- is diamagnetic or paramagnetic and why?[Atomic number: Co = 27]

Ans.  According to the valence bond theory, the nature of bonding and the magnetic properties of coordination compounds can be explained by considering the hybridization of atomic orbitals on the central metal ion.

In the case of [CoF6]^3-, cobalt (Co) has an atomic number of 27, and its electronic configuration is [Ar] 3d^7 4s^2.

For the [CoF6]^3- complex, cobalt loses 3 electrons to become Co^3+, resulting in the electronic configuration of [Ar] 3d^6. This implies that there are 6 electrons in the 3d orbitals of cobalt.

Now, to determine whether the complex is diamagnetic or paramagnetic, we need to consider the distribution of these 6 electrons in the 3d orbitals.

According to Hund’s rule, electrons will occupy individual orbitals with parallel spins before pairing up. In a high spin complex, electrons will fill up the orbitals individually before pairing up, leading to unpaired electrons. In a low spin complex, electrons will pair up in the same orbital before moving to the next one.

Since there are 6 electrons available and 6 ligands in an octahedral coordination geometry, the electronic configuration of [CoF6]^3- can either be high spin (outer orbital complex) or low spin (inner orbital complex).

In the case of [CoF6]^3-, if it is a high spin complex, all 6 electrons will occupy the 3d orbitals individually before pairing up. This would result in 3 unpaired electrons, making the complex paramagnetic.

If it is a low spin complex, the 6 electrons would pair up before occupying other orbitals, resulting in no unpaired electrons and making the complex diamagnetic.

Since fluoride ions are weak ligands, they would typically lead to high spin complexes.

Therefore, [CoF6]^3- is likely to be paramagnetic due to the presence of unpaired electrons in its high spin configuration.