Transition Elements: Question 5

Worked solution

Step 1: Determine the coordination number and shape of $[\text{CuCl}_4]^{2-}$

$\text{Cl}^-$ ions are considerably larger than $\text{H}_2\text{O}$ molecules, so fewer of them can pack around the central $\text{Cu}^{2+}$ ion. Instead of the six water ligands in $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (coordination number 6, octahedral), only four chloride ions can coordinate: $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CuCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)$

So $[\text{CuCl}_4]^{2-}$ has coordination number 4 and is tetrahedral.

Step 2: Explain the colour change in terms of d orbitals

$\text{Cu}^{2+}$ has the electron configuration $[\text{Ar}]3d^9$, a partly-filled d subshell, in both complexes, so d–d electron transitions remain possible throughout.

In a free $\text{Cu}^{2+}$ ion, the five 3d orbitals are degenerate (equal energy). When ligands surround the ion, their electric field splits the d orbitals into two sets of different energy, and the size of this splitting, $\Delta E$, depends on both the identity of the ligand and the geometry of the complex (octahedral and tetrahedral fields split the d orbitals by different amounts, and even in opposite senses). An electron can absorb a photon of visible light with energy exactly equal to $\Delta E$ and be promoted from the lower to the higher set of orbitals (a d–d transition). The light that is not absorbed is transmitted/reflected, and its complementary colour is what we observe.

Because changing from six $\text{H}_2\text{O}$ ligands (octahedral) to four $\text{Cl}^-$ ligands (tetrahedral) changes both the ligand and the geometry, it changes $\Delta E$, so a different frequency of visible light is absorbed, and a different complementary colour (yellow-green instead of blue) is seen.

Step 3: Evaluate each option

• A: correctly identifies the tetrahedral geometry (coordination number 4, due to the larger size of $\text{Cl}^-$) and correctly explains the colour change via the change in $\Delta E$ between the two sets of d orbitals. This matches the reasoning above.
• B: incorrect on both facts. $\text{Cl}^-$ is larger, not smaller, than $\text{H}_2\text{O}$ (so coordination number decreases, it does not stay at 6), and colour in these complexes arises from electron transitions on the metal ion, not absorption by the ligand itself.
• C: incorrect. The geometry does change (octahedral to tetrahedral), and it is the change in $\Delta E$ from both the different ligand and the different geometry that shifts the absorbed frequency, not simply a difference in ligand charge.
• D: incorrect. $\text{Cu}^{2+}$ is $3d^9$, not $3d^{10}$ (that configuration belongs to $\text{Cu}^+$), so d–d transitions are indeed possible; also, copper remains in the $+2$ oxidation state throughout this ligand-exchange reaction.

Final answer

• $[\text{CuCl}_4]^{2-}$ is tetrahedral (coordination number 4); the colour change from blue to yellow-green occurs because the change in ligand and geometry alters the d-orbital splitting, $\Delta E$, changing the frequency of light absorbed, option A.