1. Find the oxidation state and electronic configuration of cobalt

In $\mathrm{[Co(NH_3)_6]^{3+}}$, ammonia $\mathrm{NH_3}$ is a neutral ligand. So, oxidation state of Co is:

$x + 6(0) = +3 \Rightarrow x = +3$

Cobalt has atomic number $27$.

Ground-state electronic configuration of Co:

$\mathrm{Co}: [Ar] \, 3d^7 4s^2$

For $\mathrm{Co^{3+}}$, remove two $4s$ electrons and one $3d$ electron:

$\mathrm{Co^{3+}}: [Ar] \, 3d^6$

2. Determine the nature of ligand and spin state

$\mathrm{NH_3}$ is a strong-field ligand (for $\mathrm{Co^{3+}}$ it produces large splitting). Hence, the octahedral complex is low spin.

For an octahedral $d^6$ low-spin configuration:

$t_{2g}^6 e_g^0$

All six electrons are paired.

3. Magnetic behaviour

Since all electrons are paired, the complex is:

$\text{Diamagnetic}$

4. Hybridization

In a low-spin octahedral complex, pairing occurs in the $3d$ orbitals, leaving two vacant $3d$ orbitals available for bonding. Thus, the hybridization is inner-orbital:

$d^2sp^3$

5. Check options

• A: $\mathrm{sp^3d^2}$ and paramagnetic $\rightarrow$ incorrect
• B: $\mathrm{sp^3d^2}$ and diamagnetic $\rightarrow$ incorrect
• C: $\mathrm{d^2sp^3}$ and diamagnetic $\rightarrow$ correct
• D: $\mathrm{d^2sp^3}$ and paramagnetic $\rightarrow$ incorrect

Therefore, the correct answer is C.