Question

Account for the following
$C{{u}^{2+}}$ salts are colored, while $Z{{n}^{2+}}$ salts are white.

Answer 1

Hint: $Cu$ and $Zn$ are transition metals, in which $Cu$ has capability of showing various oxidation states.
$Cu$ and $Zn$ have a $3d$ orbital present in their electronic configuration.

Complete step by step answer:
So the question is asked to find the reason why the $C{{u}^{2+}}$ salts are colored, while $Z{{n}^{2+}}$ salts are white.
We know that, the colour is imparted in a complex or in a salt due to the electronic transitions happening in the compound in which it absorbs radiation of a certain wavelength and shows the complementary colour of the absorbed wavelength.
So elementary particles electrons are the reason for the different colours for different complexes.
And we know that $Cu$ and $Zn$ are elements which belong to the first series of the transition metals with atomic numbers 29 and 30 respectively.
We know that transition metals have a characteristic property of showing variable oxidation state, in the question we are only bothered about $Cu$ with oxidation state +2 and zinc is one of the exceptional cases and shows only +2 oxidation state.
Another information is that the transition metals consist of 3d orbitals and, here d-d transitions of electrons are possible which is the key reason why the transition metal salts are colored.
And to occur the d-d transition phenomenon, there should be unpaired electrons.
So let’s find whether unpaired electrons are present in $C{{u}^{2+}}$ and $Z{{n}^{2+}}$ or not.
The atomic number of $Cu$ is 29 with 29 electrons in them with an electronic configuration,
$Cu=\left[ Ar \right]3{{d}^{10}}4{{s}^{1}}$
For $C{{u}^{2+}}$, there is only 27 electrons and two electrons from the valence shell is removed and the electronic configuration is,
$C{{u}^{2+}}=\left[ Ar \right]3{{d}^{9}}$
In$C{{u}^{2+}}$ there is one unpaired electron, hence d-d transition is possible for $C{{u}^{2+}}$ system, which imparts the colour for its salts. As it absorbs the light from a particular wavelength and shows the colour complementary to the absorbed wavelength.
The electronic configuration for $Zn$, with atomic number 30 and 30 electrons are present in them.
$Zn=\left[ Ar \right]3{{d}^{10}}4{{s}^{2}}$
For$Z{{n}^{2+}}$, the electronic configuration is,
$Z{{n}^{2+}}=\left[ Ar \right]3{{d}^{10}}$
Here $Z{{n}^{2+}}$ is having a completely filled d-orbitals, so d-d transitions are not possible, it transmits all the light and will appear as white in colour.

Note: We must know the atomic number of each element to write its electronic configuration and should have an idea about the valence shell from where the electrons are removed, we may get confused for d-orbitals that have pseudo electronic configuration to obtain half-filled or completely filled d-orbital.
We should know the energy order of the orbitals.