Identify from the following species in which d²sp³ hybridization is shown by central atom:

🧠 $\mathrm{d^2sp^3}$ = Inner-Orbital Octahedral

The hybridisation $\mathrm{d^2sp^3}$ specifically means: two inner-d orbitals (e.g. 3d) + s + 3p → 6 hybrid orbitals → octahedral, low-spin (since the d-electrons must pair to vacate the two used d-orbitals).

So we need:

• An octahedral complex
• With a low-spin (strong-field) configuration that frees up two inner-d orbitals.

🗺️ Test Each Option

(1) $[\mathrm{Co(NH_3)_6}]^{3+}$: Co(III) $\mathrm{d^6}$ + strong-field NH₃ → low-spin. Six d-electrons pair into three $\mathrm{t_{2g}}$ orbitals; two inner d's are now empty. Hybridisation = $\mathrm{d^2sp^3}$. ✓

(2) $\mathrm{BrF_5}$: 5 bond pairs + 1 lone pair = 6 hybrids → square-pyramidal → $\mathrm{sp^3d^2}$ (outer-orbital, since Br has no inner d to use). ✗

(3) $[\mathrm{Pt(Cl)_4}]^{2-}$: Pt(II) $\mathrm{d^8}$ + Cl⁻ → square-planar $\mathrm{dsp^2}$ (4 bonds, not 6). ✗

(4) $\mathrm{SF_6}$: octahedral, 6 bonds → $\mathrm{sp^3d^2}$ (S is main-group, no inner d). ✗

⚡ The "Inner d²sp³ = Transition Metal + Strong Field" Anchor

$\mathrm{d^2sp^3}$ is exclusively for transition metals with low-spin configurations. Main-group elements (Br, S) use $\mathrm{sp^3d^2}$ (outer-orbital, with d's from the same shell).

⚠️ Same Geometry, Different Hybridisation

$\mathrm{[Co(NH_3)_6]^{3+}}$ and $\mathrm{SF_6}$ are both octahedral, but their hybridisations differ ($\mathrm{d^2sp^3}$ vs $\mathrm{sp^3d^2}$). Geometry alone doesn't decide hybridisation — the electronic source of the d orbitals does.

$\boxed{\text{Answer: (1) } [\mathrm{Co(NH_3)_6}]^{3+}}$