Which of the following have tetrahedral structure?

In $[Ni(CN)_4]^{2-}$, the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration). Since $CN^-$ is a strong field ligand, it causes pairing of electrons in the $3d$ orbitals. This leaves one $3d$ orbital empty, resulting in $dsp^2$ hybridization and a square planar geometry.

In $[Ni(CO)_4]$, the oxidation state of $Ni$ is $0$ ($3d^8 4s^2$ configuration). $CO$ is a strong field ligand and forces the $4s$ electrons to pair up in the $3d$ orbitals, giving a $3d^{10}$ configuration. The empty $4s$ and $4p$ orbitals undergo $sp^3$ hybridization, resulting in a tetrahedral geometry.

In $[NiCl_4]^{2-}$, the oxidation state of $Ni$ is $+2$ ($3d^8$ configuration). Since $Cl^-$ is a weak field ligand, no pairing of electrons occurs. The complex undergoes $sp^3$ hybridization, resulting in a tetrahedral geometry.

In $CrO_4^{2-}$, the oxidation state of $Cr$ is $+6$ ($3d^0$ configuration). It forms four bonds with oxygen atoms, utilizing $d^3s$ hybridization, which gives it a tetrahedral geometry.

Thus, $[Ni(CO)_4]$, $[NiCl_4]^{2-}$, and $CrO_4^{2-}$ have tetrahedral structures.

Answer: $[Ni(CO)_4]$; $[NiCl_4]^{2-}$; $CrO_4^{2-}$