To determine which of the given complexes have tetrahedral structures, we need to consider the hybridization state of the central metal atom in each complex. The geometry of these complexes is largely determined by the hybridization and coordination number of the central atom. Let’s analyze each option:

1. $[Ni(CN)_4]^{2-}$: Nickel in this complex has a coordination number of 4 and forms a square planar structure. This is because $CN^-$ is a strong field ligand, causing the $3d$ electrons to pair up, leading to $dsp^2$ hybridization which results in a square planar geometry.
2. $[Ni(CO)_4]$: In this complex, Nickel is in the zero oxidation state and carbon monoxide is a neutral, strong field ligand. The $3d$ electrons pair up, utilizing $sp^3$ hybridization, which results in a tetrahedral structure.
3. $[NiCl_4]^{2-}$: Chloride is a weak field ligand, and Ni in this complex uses $sp^3$ hybridization due to its electronic configuration, leading to a tetrahedral structure.
4. $CrO_4^{2-}$: The chromate ion features a $d^0$ electron configuration in $Cr^{6+}$. The $Cr–O$ bonds are highly covalent and chromate ion inherently adopts a tetrahedral structure due to the $sp^3$ hybridization from the overlap with $p$ orbitals of oxygen.

From this analysis, the complexes that have tetrahedral structures are:

• \([Ni(CO)_4]\)
• \([NiCl_4]^{2-}\)
• \(CrO_4^{2-}\)

Therefore, the correct answer is $[Ni(CO)_4]$, $[NiCl_4]^{2-}$, $CrO_4^{2-}$.