Question

Which of the following compounds show colour due to d-d transition?

• A. $\text{CuSO}_4.5\text{H}_2\text{O}$
• B. $\text{K}_2\text{Cr}_2\text{O}_7$
• C. $\text{K}_2\text{CrO}_4$
• D. $\text{KMnO}_4$

Answer 1

Answer: (A)

$\text{CuSO}_4.5\text{H}_2\text{O}$

Answer 2

Explanation: CuSO$_4$ $\cdot$ 5H$_2$O (copper(II) sulfate pentahydrate) contains the Cu$^{2+}$ ion, which has a partially filled $d$-orbital. The electronic configuration of Cu$^{2+}$ is [Ar] $3d^9$. In an aqueous environment, the Cu$^{2+}$ ion forms a complex with water molecules, creating a distorted octahedral geometry. The $d$-orbitals of the Cu$^{2+}$ ion split into two energy levels due to the ligand field created by the surrounding water molecules. The absorption of visible light causes electrons to transition between these $d$-orbital energy levels ($d$-$d$ transitions), resulting in the characteristic blue colour of the compound.
Other Options: K$_2$Cr$_2$O$_7$ (potassium dichromate), K$_2$CrO$_4$ (potassium chromate), and KMnO$_4$ (potassium permanganate) show colour due to charge transfer transitions, not $d$-$d$ transitions. In these compounds, the chromate and permanganate ions are highly coloured due to the transfer of electrons between the metal and oxygen atoms.
Conclusion: The compound CuSO$_4$ $\cdot$ 5H$_2$O shows colour due to $d$-$d$ transitions.