What type of hybridization is present in $[Co(NH_3)_6]^{3+}$ complex?

Match the following hybridization in Column $I$ with the corresponding coordination complexes in Column $II$.

$A. sp^3$	$(i). [Co(NH_3)_6]^{3+}$
$B. dsp^2$	$(ii). [Ni(CO)_4]$
$C. sp^3d^2$	$(iii). [Pt(NH_3)_2Cl_2]$
$D. d^2sp^3$	$(iv). [CoF_6]^{3-}$
	$(v). [Fe(CO)_5]$

Considering $H_2O$ as a weak field ligand,the number of unpaired electrons in $[Mn(H_2O)_6]^{2+}$ will be (Atomic number of $Mn = 25$).

Nickel $(Z = 28)$ combines with a uninegative monodentate ligand to form a diamagnetic complex $[NiL_4]^{2-}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively

The geometry and the number of unpaired electron$(s)$ of $[MnBr_4]^{2-}$,respectively,are

Assertion $:$ Complex ion $[Co(NH_3)_6]^{2+}$ is readily oxidized to $[Co(NH_3)_6]^{3+}$.
Reason $:$ Unpaired electron in complex ion $[Co(NH_3)_6]^{2+}$ is present in $4p$ orbital.