The absorption of visible light in coordination compounds is primarily due to d-d transitions of unpaired electrons.

1. [Ni(CN)₄]²⁻: Nickel is in the +2 oxidation state ($3d^8$). The cyanide ion ($CN^-$) is a strong field ligand. As per Valence Bond Theory and Crystal Field Theory, in a coordination number of 4 with a strong ligand, $Ni^{2+}$ forms a square planar complex involving dsp² hybridisation. The strong field causes all 8 d-electrons to pair up in the lower energy orbitals ($d_{xy}, d_{yz}, d_{zx}, d_{z^2}$). Since there are no unpaired electrons and the crystal field splitting energy ($\Delta$) is very large (often shifting absorption to the UV region), it is the least expected to absorb in the visible region compared to the others .
2. [Ni(H₂O)₆]²⁺: Contains $H_2O$ (weak field ligand). It is an outer orbital octahedral complex with 2 unpaired electrons ($3d^8$). It is green in colour .
3. [Cr(NH₃)₆]³⁺: Chromium is in the +3 oxidation state ($3d^3$). It has 3 unpaired electrons and exhibits d-d transitions, absorbing visible light.
4. [Fe(H₂O)₆]²⁺: Iron is in the +2 oxidation state ($3d^6$). With a weak field ligand, it is high spin with 4 unpaired electrons, absorbing visible light (pale green).

Therefore, the diamagnetic complex $[Ni(CN)_4]^{2-}$ is the correct answer.