Colour of coordination compounds

A complex absorbs visible light (and so appears coloured) when it has unpaired d electrons that can undergo d–d transitions of suitable energy.
$[Ni(CN)_4]^{2-}$: $Ni^{2+}$ is $3d^8$; the strong field ligand $CN^-$ pairs all the electrons ($dsp^2$, square planar), leaving no unpaired electron. With the large splitting it does not absorb in the visible region.
$[Ni(H_2O)_6]^{2+}$: $Ni^{2+}$ ($3d^8$) with a weak field ligand has 2 unpaired electrons; it is green.
$[Cr(NH_3)_6]^{3+}$: $Cr^{3+}$ ($3d^3$) has 3 unpaired electrons; it is coloured.
$[Fe(H_2O)_6]^{2+}$: $Fe^{2+}$ ($3d^6$, high spin) has 4 unpaired electrons; it is pale green.
Hence $[Ni(CN)_4]^{2-}$ is not expected to absorb visible light.

A complex absorbs visible light when the metal ion has a partly filled d subshell, so that d–d transitions are possible.
$Zn^{2+}$: $3d^{10}$, completely filled; no d–d transition, colourless.
$Sc^{3+}$: $3d^0$; no d electrons, colourless.
$Ti^{4+}$: $3d^0$; no d electrons, colourless.
$Cr^{3+}$: $3d^3$, with three unpaired electrons in $t_{2g}$; $t_{2g} \rightarrow e_g$ transitions absorb visible light.
Hence $[Cr(NH_3)_6]^{3+}$ is expected to absorb visible light.