The hybridisation and magnetic behaviour of [Fe(CN)6]4- are:

🧠 The Strong-Field Counterpart

$[\mathrm{Fe(CN)_6}]^{4-}$: $\mathrm{Fe^{2+}}$ is $\mathrm{d^6}$, and $\mathrm{CN^-}$ is one of the strongest field ligands. Strong field forces pairing — all six d-electrons collapse into the lower three orbitals. Hybridisation uses the inner 3d set: $\mathrm{d^2sp^3}$, diamagnetic.

🗺️ Walk Through the VBT Picture

$\mathrm{Fe^{2+}}$: $[\mathrm{Ar}]\,3\mathrm{d^6}$.

$\mathrm{CN^-}$ pairs all six d-electrons into three orbitals → $\mathrm{t_{2g}^6}$, $\mathrm{e_g^0}$. Two of the inner 3d orbitals are now vacant.

Hybridisation: 2 × 3d (vacant) + 1 × 4s + 3 × 4p = $\mathrm{d^2sp^3}$ (octahedral, inner-orbital).

All electrons paired → diamagnetic.

⚡ The "$\mathrm{Fe^{2+}}$ + $\mathrm{CN^-}$" Anchor

$[\mathrm{Fe(CN)_6}]^{4-}$ (ferrocyanide) — d²sp³, diamagnetic — and its $\mathrm{Fe^{3+}}$ cousin $[\mathrm{Fe(CN)_6}]^{3-}$ (ferricyanide) — d²sp³, one unpaired electron — are the two most-tested $\mathrm{Fe}$/$\mathrm{CN}$ pairs in JEE coordination chemistry. Memorise both.

⚠️ Don't Default to sp³d²

Students who saw $[\mathrm{CoF_6}]^{3-}$ answered with sp³d² and might re-apply the same reasoning here. Stop: $\mathrm{CN^-}$ is a strong-field ligand, F⁻ is weak. Strong → inner ($\mathrm{d^2sp^3}$). Weak → outer ($\mathrm{sp^3d^2}$). The ligand identity is what flips the hybridisation, not the geometry.

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