The hybridisation and magnetic nature of [Ni(CO)4] and [Ni(CN)4]2- respectively are:

🧠 Same Two Complexes, Hybridisation Tag

Restating CORD-033's physics in VBT (valence-bond theory) language:

• $[\mathrm{Ni(CO)_4}]$: Ni⁰, $\mathrm{d^{10}}$ → $\mathrm{sp^3}$ tetrahedral, diamagnetic.
• $[\mathrm{Ni(CN)_4}]^{2-}$: Ni²⁺, $\mathrm{d^8}$ low-spin → $\mathrm{dsp^2}$ square planar, diamagnetic.

Both diamagnetic, but the hybridisations differ — that's option (3).

🗺️ VBT Walk-Through

$[\mathrm{Ni(CO)_4}]$. Ground state Ni: $3\mathrm{d^8}\,4\mathrm{s^2}$. CO drives all 10 electrons into 3d as $\mathrm{d^{10}}$, freeing 4s and 4p for sp³ hybridisation. Four CO lone pairs slot in. All paired → diamagnetic.

$[\mathrm{Ni(CN)_4}]^{2-}$. Ni²⁺ has $\mathrm{d^8}$. $\mathrm{CN^-}$ (strong field) pairs the unpaired 3d electron, leaving one inner 3d orbital empty. Hybridise: 3d (1) + 4s (1) + 4p (2) = dsp². Geometry: square planar. All electrons paired → diamagnetic.

⚡ The "Inner vs Outer" Tag

VBT classifies octahedral and four-coordinate complexes as inner-orbital (uses (n−1)d, e.g. $\mathrm{d^2sp^3}$ or $\mathrm{dsp^2}$) when ligand is strong-field, and outer-orbital (uses nd, e.g. $\mathrm{sp^3d^2}$) when ligand is weak-field. For Ni²⁺ + CN⁻: inner. For Ni⁰ + CO: 3d is full ($\mathrm{d^{10}}$), so no d-orbital is needed in the hybrid — sp³ alone.

⚠️ The "Always sp³d² for Octahedral / Tetrahedral" Reflex

There's no octahedral Ni complex here. Don't reach for sp³d² when the geometry is tetrahedral or square planar. Match the hybridisation to the coordination number and geometry first, not to a memorised metal-ligand combo.

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