Key idea: colour in coordination/transition-metal compounds

Compounds of transition metals are usually coloured when the metal ion has partially filled $d$-orbitals, so that $d\!\to d$ transitions are possible.

• $\mathrm{Cu^+}$ has electronic configuration: $\mathrm{Cu^+} : [Ar]3d^{10}$ Since $d$-orbitals are completely filled, such compounds are generally colourless.
• $\mathrm{Cu^{2+}}$ has electronic configuration: $\mathrm{Cu^{2+}} : [Ar]3d^9$ Since the $d$-subshell is incomplete, compounds are generally coloured.

Check each option

Option A: $\mathrm{CuCl}$

Here copper is in +1 oxidation state. $\mathrm{Cu^+} = 3d^{10}$ Hence this is expected to be colourless/white.

Option B: $\mathrm{K_3[Cu(CN)_4]}$

The complex ion is $\mathrm{[Cu(CN)_4]^{3-}}$.

Let oxidation state of Cu be $x$: $x + 4(-1) = -3$ $x = +1$

So copper is again $\mathrm{Cu^+}$: $\mathrm{Cu^+} = 3d^{10}$ Therefore this compound is expected to be colourless.

Option C: $\mathrm{CuF_2}$

Fluoride is $-1$, so copper oxidation state is: $x + 2(-1)=0 \Rightarrow x=+2$ Thus copper is $\mathrm{Cu^{2+}}$: $\mathrm{Cu^{2+}} = 3d^9$ This has an incompletely filled $d$-subshell, so the compound is coloured.

Option D: $\mathrm{[Cu(CH_3CH)_4]BF_4}$

The cation is $\mathrm{[Cu(CH_3CH)_4]^+}$ (since $\mathrm{BF_4^-}$ is the counter ion). The ligand shown is a neutral alkene-type ligand, so Cu is in +1 oxidation state.

Therefore: $\mathrm{Cu^+} = 3d^{10}$ Hence this should be colourless.

Conclusion

Only the compound containing $\mathrm{Cu^{2+}}$ is coloured.

Therefore, the coloured compound is: $\boxed{\mathrm{CuF_2}}$

Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They match.