Question

Which of the following compounds is colourless?
A) \[C{{u}_{2}}{{(C{{H}_{3}}COO)}_{4}}.2{{H}_{2}}O\]
B) \[C{{u}_{2}}C{{l}_{2}}\]
C) \[CuS{{O}_{4}}.5{{H}_{2}}O\]
D) \[[Cu{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}.4{{H}_{2}}O\]

Answer 1

Hint: The transition metals show colours when the light is absorbed or energy is released in the visible region during d-d transitions. The electrons from the lower d orbital get excited and jump to the higher d orbital known as the d-d transition. This transition is only shown by the compounds having paramagnetic nature i.e., they have unpaired electrons and a tendency to get excited to move to the higher levels.

Complete answer:The d orbitals in the transition metal complexes do not consist of the same energy level, and the d orbital degenerates into further levels. The electrons filled in the d orbital jump from one d orbital energy level to the other. Due to these transitions, the energy absorbed or released lies in the visible band and the complex ions tend to show highly intense colours.

However, these transitions are only possible if the compound is paramagnetic in nature i.e., it must consist of unpaired electrons so that it can show transition. If the compound is diamagnetic in nature i.e., all the electrons are paired. In that case, no transition is observed as the paired electrons do not undergo transitions. Hence, transition metal ions which are diamagnetic in nature do not show any colors.

Now, considering each of the given compounds:

Compound A- \[C{{u}_{2}}{{(C{{H}_{3}}COO)}_{4}}.2{{H}_{2}}O\]
It is the hydrated form of copper (II) acetate in which the copper atom exists in its $+2$ oxidation state and its electronic configuration is $\left[ Ar \right]3{{d}^{9}}4{{s}^{0}}$i.e., it consist of 1 unpaired electron in its valence d orbital and thus, it is a coloured compound. It is usually found as a dark green coloured crystalline solid.

Compound B- \[C{{u}_{2}}C{{l}_{2}}\]
The name of the compound is cuprous chloride in which the copper atom exists in its $+1$ oxidation state and its electronic configuration is $\left[ Ar \right]3{{d}^{10}}4{{s}^{0}}$ i.e., there is no unpaired electron in the valence d orbital and thus, the compound is colourless.

Compound C- \[CuS{{O}_{4}}.5{{H}_{2}}O\]
It is the hydrated form of copper (II) sulphate in which the copper atom exists in its $+2$ oxidation state and its electronic configuration is $\left[ Ar \right]3{{d}^{9}}4{{s}^{0}}$i.e., it consist of 1 unpaired electron in its valence d orbital and thus, it is a coloured compound. It is usually found as a bright blue coloured crystalline solid.

Compound D- \[[Cu{{(N{{H}_{3}})}_{4}}]S{{O}_{4}}.4{{H}_{2}}O\]
In the given coordination complex, the copper atom exists in its $+2$ oxidation state, and its electronic configuration is $\left[ Ar \right]3{{d}^{9}}4{{s}^{0}}$i.e., it consist of 1 unpaired electron in its valence d orbital and thus, it is a coloured compound. It is a dark blue coordination complex.
Hence, among the given options, \[C{{u}_{2}}C{{l}_{2}}\] is colourless due to its diamagnetic nature.

option (B) is the correct answer.

Note: It is important to note that the d-d transition was originated from the crystal field splitting of the d orbitals and it operates on the principle of the quantization of energy which means the energy needed by the electrons to jump from one energy level to other is discrete and fixed.