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(B) The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{2+}$,the configuration is $3d^7$.In a tetrahedral field,the $d-$ orbitals split

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$Co^{2+}$ (tetrahedral field)

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(B) The electronic configuration of $Co$ is $[Ar] 3d^7 4s^2$. For $Co^{2+}$,the configuration is $3d^7$.In a tetrahedral field,the $d-$ orbitals split into $e_g$ (lower energy,doubly degenerate) and $t_{2g}$ (higher energy,triply degenerate) sets.The $7$ electrons fill as follows: $e_g^4 t_{2g}^3$.The $t_{2g}$ orbitals are triply degenerate and contain $3$ electrons,meaning they are exactly half-filled ($3/6$ capacity).Thus,$Co^{2+}$ in a tetrahedral field satisfies the condition.

Triple degenerate $d-$ orbitals are exactly half-filled in

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