Question
$[CoCl_4]^{2⊕}$ is a tetrahedral complex. Draw its box orbital diagram. State which orbitals participate in hybridization.

Answer

$_{27}Co [Ar] 3d^74s^2$
Oxidation state of $Co = +2 Co^{2+} [Ar] 3d^7 4s^\circ$
Since $CI^–$​​​​​​​​​​​​​​ is a weak ligand, there is no pairing of electrons. Since C.N. is $4$, there is $sp^3​​​​​​​$​​​​​​​ hybridisation.
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