An octahedral complex of $Co^{3+}$ is diamagnetic. The hybridisation involved in the formation of the complex is

  • A
    $sp^3d^2$
  • B
    $dsp^2$
  • C
    $d^2sp^3$
  • D
    $sp^3d$

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Given below are two statements :
Statement $I$: $[Ni(CN)_4]^{2-}$ is square planar and diamagnetic complex with $dsp^2$ hybridization for $Ni$ but $[Ni(CO)_4]$ is tetrahedral,diamagnetic and with $sp^3$ hybridization for $Ni$.
Statement $II$: $[NiCl_4]^{2-}$ and $[Ni(CO)_4]$ both have same $d$-electron configuration,have same geometry and are paramagnetic.
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