Question

Why is \( \mathrm{[Sc(H_2O)_6]^{3+}} \) colourless ?

Hint:

Sc3+ is 3d0, so it has no d electron to undergo a d-d transition. It does not absorb visible light.

Answer 1

Step 1: Know the origin of colour in complexes:
Colour in a transition metal complex comes from d-d transitions.
An electron in a lower d level absorbs visible light and jumps to a higher d level.
The light that is left over is the colour we see.

Step 2: Find the d configuration of scandium(III):
Scandium has atomic number 21 and the configuration \( \mathrm{[Ar]\,3d^1 4s^2} \).
It loses three electrons to form \( \mathrm{Sc^{3+}} \).
So \( \mathrm{Sc^{3+}} \) is \( \mathrm{[Ar]\,3d^0} \) with no electron in the d orbitals.

Step 3: Apply the idea to the complex:
Water splits the d orbitals into two sets, but with no d electron there is nothing to excite.
No d-d transition is possible, so no visible light is absorbed.
The complex transmits all visible light and looks colourless.

Final Answer:
\( \mathrm{Sc^{3+}} \) has a \( 3d^0 \) configuration, so no d-d transition occurs and the complex is colourless.