The hybridisation of [Cr(NH3)6]3+ and [NiCl4]2- are respectively:

🧠 Two Different d-Counts, Two Different Geometries

The trick is that the two complexes have different d-electron counts and different coordination numbers, so they end up with two unrelated hybridisations:

• $[\mathrm{Cr(NH_3)_6}]^{3+}$: $\mathrm{Cr^{3+}}$ is $\mathrm{d^3}$ — only 3 d-electrons, one in each $\mathrm{t_{2g}}$ orbital. With 6-coordinate $\mathrm{NH_3}$ (moderate field), inner d-orbitals should be available — but the half-filled $\mathrm{t_{2g}^3}$ blocks two of them. So Cr(III) typically uses outer 4d → $\mathrm{sp^3d^2}$.

Wait — actually for $\mathrm{Cr^{3+}\,d^3}$, the $\mathrm{t_{2g}}$ orbitals each hold one electron leaving zero vacancies in the inner d set. NCERT/JEE convention here: hybridisation is $\mathrm{sp^3d^2}$ (outer-orbital) because no two inner 3d orbitals are simultaneously empty.

• $[\mathrm{NiCl_4}]^{2-}$: $\mathrm{Ni^{2+}}$ is $\mathrm{d^8}$. $\mathrm{Cl^-}$ is weak field and the coordination number is 4 → tetrahedral, $\mathrm{sp^3}$ hybridisation.

So the answer is $\mathrm{sp^3d^2}$ and $\mathrm{sp^3}$.

🗺️ Configuration Tags

$\mathrm{Cr^{3+}}\,(\mathrm{d^3})$ + 6 $\mathrm{NH_3}$. $\mathrm{t_{2g}^3}\,\mathrm{e_g^0}$ — three unpaired. With three of the inner d-orbitals each holding one electron, the standard NCERT convention says outer-orbital $\mathrm{sp^3d^2}$, paramagnetic.

$\mathrm{Ni^{2+}}\,(\mathrm{d^8})$ + 4 $\mathrm{Cl^-}$. Tetrahedral geometry → $\mathrm{sp^3}$ hybridisation. $\mathrm{Cl^-}$ is weak → no pairing in d → 2 unpaired electrons.

⚡ The "CN = 4 + Weak Ligand" Reflex

For 4-coordinate complexes:

• Strong field ($\mathrm{CN^-}$, $\mathrm{CO}$) + d⁸ → square planar, $\mathrm{dsp^2}$.
• Weak field ($\mathrm{Cl^-}$, $\mathrm{Br^-}$, $\mathrm{I^-}$) + d⁸ → tetrahedral, $\mathrm{sp^3}$.

This single rule decides $[\mathrm{NiCl_4}]^{2-}$ (sp³) vs $[\mathrm{Ni(CN)_4}]^{2-}$ (dsp²) instantly.

⚠️ Don't Default $\mathrm{Cr^{3+}}$ to Inner Orbital

The reflex "Cr(III) is always $\mathrm{d^2sp^3}$" is wrong. Because $\mathrm{d^3}$ leaves no two empty inner d-orbitals (each $\mathrm{t_{2g}}$ holds an electron), Cr(III) octahedral complexes often use outer-orbital $\mathrm{sp^3d^2}$ in JEE conventions. Watch the d-count first.

$\boxed{\text{Answer: (4) } \mathrm{sp^3d^2 \text{ and } sp^3}}$