To display d-d transition and paramagnetism, the transition metal ion must possess at least one unpaired electron in its d-orbital. Let's calculate the oxidation state and d-electron count for each:

• In $\text{CrO}_4^{2-}$ (chromate ion), Cr is in $+6$ oxidation state. Its electronic configuration is $3d^0$, so it has zero unpaired electrons and is diamagnetic. Its colour is due to ligand-to-metal charge transfer.
• In $\text{Cr}_2\text{O}_7^{2-}$ (dichromate ion), Cr is also in $+6$ oxidation state ($3d^0$) and is diamagnetic.
• In $\text{MnO}_4^-$ (permanganate ion), Mn is in $+7$ oxidation state. Its electronic configuration is $3d^0$, meaning it is diamagnetic and its colour is due to charge transfer.
• In $\text{MnO}_4^{2-}$ (manganate ion), Mn is in $+6$ oxidation state. Its electronic configuration is $3d^1$. Because it has one unpaired d-electron, it exhibits paramagnetism and displays colour due to d-d transition.