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(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In the complex $[CoF_6]^{3-}$,the oxidation state of $

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$4$,Octahedral

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(A) The atomic number of $Co$ is $27$. The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$.In the complex $[CoF_6]^{3-}$,the oxidation state of $Co$ is $x + 6(-1) = -3$,so $x = +3$.The electronic configuration of $Co^{3+}$ is $[Ar] 3d^6$.$F^-$ is a weak field ligand,so it does not cause pairing of electrons in the $3d$ orbitals.The $3d$ orbitals have $4$ unpaired electrons.The complex undergoes $sp^3d^2$ hybridization,which corresponds to an octahedral geometry.

Identify the number of unpaired electrons present and the geometry,respectively,of the complex $[CoF_6]^{3-}$ (given atomic number of $Co = 27$).

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