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The color of ${\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$is violet.

Every color has its complementary color: Red - Green Yellow - Violet Orange Blue When $(∆)$there is a same metal ion in different compounds (different ligands), the color of the compound depends on the ligands. When a strong-field ligand is replaced with a weak-field ligand, the color changes to its complementary color. Spectrochemical series (from weak field to strong field ligands):

${\mathbf{I}}^{\mathbf{-}}\mathbf{<}{\mathbf{Cl}}^{\mathbf{-}}\mathbf{<}{\mathbf{F}}^{\mathbf{-}}\mathbf{<}{\mathbf{OH}}^{\mathbf{-}}\mathbf{<}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{<}{\mathbf{SCN}}^{\mathbf{-}}\mathbf{<}{\mathbf{NH}}_{\mathbf{3}}\mathbf{<}\mathbf{en}\mathbf{<}{\mathbf{NO}}_{\mathbf{2}}^{\mathbf{-}}\mathbf{<}{\mathbf{CN}}^{\mathbf{-}}\mathbf{<}\mathbf{CO}$

The weaker the field is, the lower the splitting energy is $(∆)$and the longer the absorbed wavelengths are.

The stronger the field is, the higher the splitting energy is $(∆)$and the shorter the absorbed wavelengths are.

${\left[\mathrm{Cr}{\left({\mathrm{H}}_2\mathrm{O}\right)}_6\right]}^{2+}$is violet, which means the absorbed light is yellow. ${\mathrm{H}}_2\mathrm{O}$is a weak-field ligand, which means the$∆$is lower and the range of the absorbed wavelengths is longer.

${\mathrm{CrL}}_6$is green, which means the absorbed light is red. When the red light is absorbed, the range of the absorbed wavelengths is long, which means the ligand must be a weak field.

In spectrochemical series ${\mathrm{CN}}^{-}$is a strong-field ligand and ${\mathrm{Cl}}^{-}$is a weak-field ligand, so the answer would be${\mathrm{Cl}}^{-}$.