Identify Coloured Compound Among Copper Complexes

To determine which compound is coloured, we need to analyze the oxidation state and the electronic configuration of the copper ($\text{Cu}$) ion in each compound:

1. In $\text{CuCl}$ (Copper(I) chloride):

   • The oxidation state of copper is $+1$ ($\text{Cu}^+$).
   • Electronic configuration of $\text{Cu}^+$: $[\text{Ar}]\, 3d^{10}$.
   • Since the $3d$ subshell is completely filled ($3d^{10}$), no $d\text{-}d$ transitions are possible.
   • Therefore, $\text{CuCl}$ is colourless.

2. In $\text{K}_3 [\text{Cu}(\text{CN})_4]$ (Potassium tetracyanocuprate(I)):

   • Let $x$ be the oxidation state of copper:
     $3(+1) + x + 4(-1) = 0 \implies x = +1$
   • Electronic configuration of $\text{Cu}^+$: $[\text{Ar}]\, 3d^{10}$.
   • With a completely filled $3d^{10}$ subshell, $d\text{-}d$ transitions cannot occur.
   • Therefore, $\text{K}_3 [\text{Cu}(\text{CN})_4]$ is colourless.

3. In $\text{CuF}_2$ (Copper(II) fluoride):

   • The oxidation state of copper is $+2$ ($\text{Cu}^{2+}$).
   • Electronic configuration of $\text{Cu}^{2+}$: $[\text{Ar}]\, 3d^9$.
   • Since it has an incompletely filled $3d$ subshell ($3d^9$) with one unpaired electron, $d\text{-}d$ transitions can take place upon absorbing visible light.
   • Therefore, $\text{CuF}_2$ is coloured (typically blue/blue-green).

4. In $[\text{Cu}(\text{CH}_3\text{CN})_4]\text{BF}_4$:

   • $\text{CH}_3\text{CN}$ is a neutral ligand and $\text{BF}_4^-$ is a univalent anion, so the oxidation state of copper is $+1$ ($\text{Cu}^+$).
   • Electronic configuration of $\text{Cu}^+$: $[\text{Ar}]\, 3d^{10}$.
   • Due to the filled $3d^{10}$ shell, $d\text{-}d$ transitions do not occur.
   • Therefore, $[\text{Cu}(\text{CH}_3\text{CN})_4]\text{BF}_4$ is colourless.

Thus, the coloured compound among the choices is $\text{CuF}_2$.

Correct Option: (C)