Determination of hybridization according to VBT:

• $\text{Co}^{3+}$ has ground-state configuration $3d^6 4s^0 4p^0 4d^0$.
• Under the strong ligand field of ammine ligands, the six $3d$ electrons pair up into three $3d$ orbitals ($t_{2g}^6$).
• This leaves two inner $(n-1)d$ orbitals vacant, along with one $4s$ and three $4p$ orbitals.
• These six orbitals undergo $d^2sp^3$ hybridization to accept six lone pairs of electrons from the five $\text{NH}_3$ ligands and one $\text{Cl}^-$ ligand.
• Therefore, the complex is an inner orbital octahedral complex with $d^2sp^3$ hybridization.

The coordination number of a central metal atom/ion in a complex is defined as the total number of ligand donor atoms directly bonded to the metal atom by coordinate bonds:

• There are five ammine ($\text{NH}_3$) ligands, each donating 1 electron pair (monodentate) $\rightarrow 5$ bonds.
• There is one chlorido ($\text{Cl}^-$) ligand donating 1 electron pair (monodentate) $\rightarrow 1$ bond.
• Total number of coordinate bonds formed with cobalt $= 5 + 1 = 6$.

According to Werner's coordination theory:

• Primary valences are ionizable and satisfied by negative counter-ions outside the coordination sphere.
• In $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$, when dissolved in water, the complex dissociates to yield two ionizable chloride ions: $[\text{Co}(\text{NH}_3)_5\text{Cl}]^{2+} + 2\text{Cl}^-$.
• The NTA official answer key accepted 2 as the primary valence corresponding to the charge on the coordination entity satisfied by the two ionizable chloride counter-ions outside the coordination sphere. (Note: In standard textbook definitions, total primary valence equals the overall oxidation state $+3$, of which 2 are outside the sphere and 1 is inside the sphere).