Question

Which of the following is not a coloured cation?
A) $F{e^{ + 2}}$
B) $C{a^{ + 2}}$
C) $F{e^{ + 3}}$
D) $N{i^{ + 2}}$

Answer 1

We know that transitions metals are placed in d-block. The presence of unpaired electrons in d-block shows d-d transition in d-orbital emit colour. Anyway when the metal begins holding with different ligands, this changes. Because of the d-orbital and inductive impacts of the ligands on the electrons. The d-orbitals split apart and became non-degenerate (have distinct energy levels).

Complete step by step answer:
Reason of colour of ions= Compound of the s and p-block elements are colourless as white if anion is not coloured while compounds of d-block elements are coloured.
It is due to the fact that under the influence of the magnetic field of ligands (like water, ammonia etc.) d subshell is split into two groups namely ${t_{2g}}$ and $eg$. This is called d-d splitting.

d-d transition= ${t_{2g}}$ and $eg$ orbitals have different energies and this energy difference lies in the visible region. Thus in a d-d transition (${t_{2g}} \rightleftarrows eg$), some wavelengths of visible region are absorbed by the ions and complementary colour of refractive wavelengths is seen.
S and p-orbital are symmetrical in geometry thus their splitting does not occur and their compounds are colourless.
The ions having ${d^0}$ or ${d^{10}}$ configuration do not show orbital splitting, hence, they are colourless. For the same reason $C{a^{ + 2}}$ ion are colourless.
Colour of metal ions are given below:
Hydrated $F{e^{ + 2}}$= green
Hydrated $F{e^{ + 3}}$= yellow
Hydrated $N{i^{ + 2}}$=green
$C{a^{ + 2}}$ Is colourless. Other cations are coloured.
Hence, option B is correct.

Additional information: This forms the basis crystal field theory (CFT). How these d-orbitals split depends on the geometry of the compound that is formed.

Note:
Transition metals are special in that the energy is different between the non-degenerate, d- orbitals corresponding to the energy of radiation of the visible light spectrum. Due to the presence of unpaired electrons in $F{e^{ + 2}}$, $F{e^{ + 3}}$ and $N{i^{ + 2}}$ ions it shows d-transition.