The hybridization and magnetic behaviour of cobalt ion in [Co(NH3)6]3+ complex, respectively is:

🧠 Co(III) + Strong N-Donor = Low-Spin Inner-Orbital

$[\mathrm{Co(NH_3)_6}]^{3+}$: Co(III) is $\mathrm{d^6}$. NH₃ is strong enough on Co(III) to force pairing → low-spin.

Configuration: $\mathrm{t_{2g}^6 e_g^0}$. All 6 electrons paired in three lower orbitals.

Hybridisation: low-spin uses inner d-orbitals (3d) → $\mathrm{d^2sp^3}$ (octahedral, inner-orbital).

Magnetic: 0 unpaired → diamagnetic.

Match: $\mathrm{d^2sp^3}$ + diamagnetic → option (3).

🗺️ Inner-Orbital Recipe

For VBT octahedral hybridisation:

• Inner-orbital ($\mathrm{d^2sp^3}$): two 3d orbitals + 4s + three 4p. Used when d-electrons can be paired (low-spin). Diamagnetic if all paired.
• Outer-orbital ($\mathrm{sp^3d^2}$): 4s + three 4p + two 4d. Used when d-electrons stay unpaired (high-spin). Always paramagnetic.

⚡ The Co(III)-NH₃ Constant

$[\mathrm{Co(NH_3)_6}]^{3+}, [\mathrm{Co(en)_3}]^{3+}, [\mathrm{Co(CN)_6}]^{3-}$ — all $\mathrm{d^2sp^3}$, all 0 unpaired, all diamagnetic. Memorise.

⚠️ Don't Pair Diamagnetic With Outer-Orbital

$\mathrm{sp^3d^2}$ uses 4d outer orbitals — d-electrons stay in the inner 3d, unpaired → always paramagnetic. Diamagnetic + outer-orbital is a logical contradiction.

$\boxed{\text{Answer: (3) } d^2sp^3 \text{ and diamagnetic}}$