JEE Main 2024 — Coordination Compounds Question with Solution

Among the given options, the compounds that show color due to d-d transitions are those that have partially filled d-orbitals when in the form of complex ions or compounds.

Let's examine each option:

Option A: ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

Chromium in ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ exists in the +6 oxidation state as the dichromate ion (${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$). In this oxidation state, chromium does not have any electrons in the d-orbitals (it has a $[\mathrm{Ar}]3{\mathrm{d}}^0$ configuration) so there is no possibility for d-d transitions. Instead, the color is due to charge transfer transitions.

Option B: ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$

Copper(II) sulfate pentahydrate, ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$, contains copper in a +2 oxidation state. In this state, copper has a $[\mathrm{Ar}]3{\mathrm{d}}^9$ electron configuration, which means there is one unpaired electron in the d-orbitals. Due to the presence of this unpaired electron, d-d transitions can occur, and this is why ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$ is blue in color.

Option C: ${\mathrm{KMnO}}_4$

In potassium permanganate (${\mathrm{KMnO}}_4$), manganese is in the +7 oxidation state, and it has a $[\mathrm{Ar}]3{\mathrm{d}}^0$ electron configuration. Like chromium in the +6 state, manganese in the +7 state does not have any d-electrons available for d-d transitions. The deep purple color of ${\mathrm{KMnO}}_4$ is also due to charge transfer transitions.

Option D: ${\mathrm{K}}_2{\mathrm{CrO}}_4$

Potassium chromate (${\mathrm{K}}_2{\mathrm{CrO}}_4$) contains chromium in the +6 oxidation state as the chromate ion (${\mathrm{CrO}}_4^{2-}$). Like dichromate, chromium does not have d-electrons and therefore cannot undergo d-d transitions in this compound. The color is due to charge transfer transitions.

So the only compound from the options listed that shows color due to d-d transitions is Option B: ${\mathrm{CuSO}}_4\cdot 5{\mathrm{H}}_2\mathrm{O}$.