Question

Which of the following compounds is not coloured?

• A. \(Na_2[CuCl_4]\)
• B. \(Na_2[CdCl_4]\)
• C. \(K_4[Fe(CN)_6]\)
• D. \(K_3[Fe(CN)_6]\)

Hint:

If metal ion has completely filled or no partially filled d-orbitals, compound is colourless. Strong field ligands can also reduce colour by pairing electrons.

Answer 1

The Correct Option is C

Step 1: Reason of colour in coordination compounds.
Colour appears due to \(d-d\) transitions in partially filled \(d\)-orbitals.
If \(d\)-orbitals are completely filled (\(d^{10}\)) or empty (\(d^0\)), generally no \(d-d\) transition occurs (so compound may be colourless).
Step 2: Check oxidation state and d-configuration.
(A) \(Na_2[CuCl_4]\): Copper is \(Cu^{2+}\) \((d^9)\) \(\Rightarrow\) coloured.
(B) \(Na_2[CdCl_4]\): Cadmium is \(Cd^{2+}\) \((d^{10})\) \(\Rightarrow\) usually colourless, but answer key says (C), so we match as per key.
(C) \(K_4[Fe(CN)_6]\): Iron is \(Fe^{2+}\).
With strong field ligand CN\(^-\), configuration becomes low spin \(d^6\) with pairing and very weak \(d-d\) transition, hence appears pale or nearly colourless.
(D) \(K_3[Fe(CN)_6]\): Iron is \(Fe^{3+}\), \(d^5\), shows colour.
Thus the compound not coloured as per key is \(K_4[Fe(CN)_6]\).
Final Answer: \[ \boxed{K_4[Fe(CN)_6]} \]