(N/A) $1$. The central metal ion $Co^{3+}$ has an electronic configuration of $[Ar]3d^6$.
$2$. In the presence of the strong field ligand $NH_3$,the electrons in the $3d$ orbitals pair up,leaving two $3d$ orbitals vacant.
$3$. These two $3d$ orbitals,along with one $4s$ and three $4p$ orbitals,undergo $d^2sp^3$ hybridization to form six equivalent hybrid orbitals.
$4$. Six pairs of electrons from six $NH_3$ ligands occupy these hybrid orbitals,resulting in an octahedral geometry.
$5$. Since there are no unpaired electrons,the complex is diamagnetic.
$6$. Because the inner $3d$ orbitals are involved in hybridization,it is called an inner orbital complex or a low-spin complex.