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# Explain the hybridization, geometry and magnetic property of ${\left[ {Co{F_6}} \right]^{3 - }}$ on the basis of valence bond theory. [ $Z$ of $Co = 27$ ]

Last updated date: 22nd Jun 2024
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Hint:Based on the valence bond theory the hybridization of the ${\left[ {Co{F_6}} \right]^{3 - }}$ is determined by the electrons that are involved in the formation of the structure. The coordination number can also be determined from the given compound from which the chemical properties can be determined.
The atomic number of $Co$ is $27$ which is why the number of carbon residues which are part of $3d$ orbital. The ligand here which is associated with the central atom is $F$ as there are six $F$ residues which are associated with the central $Co$. Since the $F$ is a halogen which forms the bonded structure of the complex, the six pairs of electrons fill the next orbitals after $3d$. Six there are six pairs, according to the next orbitals the hybridization of ${\left[ {Co{F_6}} \right]^{3 - }}$ is $s{p^3}{d^2}$ which defines that there are six pairs of electrons in the given six orbitals. The hybridization of the ${\left[ {Co{F_6}} \right]^{3 - }}$ molecule is associated with geometry of the specific structure. The ${\left[ {Co{F_6}} \right]^{3 - }}$ has the octahedral shape which is based on the hybridisation of $s{p^3}{d^2}$. The magnetic property of the ${\left[ {Co{F_6}} \right]^{3 - }}$ is that the orbitals have some of the orbitals filled with only a single spin of electrons. This is why the property of such electronic configuration determines the paramagnetic property of the compound. The paramagnetic property of the compound is slightly attracted to the presence of a magnetic field because of this electronic configuration.