Question

Amongst $\text{TiF}_{\text{6}}^{\text{2-}}$ ,$\text{CoF}_{\text{6}}^{\text{2-}}$,$\text{C}{{\text{u}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ and $\text{NiCl}_{\text{4}}^{\text{2-}}$ the colour less species are-
(At. No. of $\text{Ti = 22,}\,\,\text{Co = 27,}\,\,\text{Cu = 29,}\,\,\text{Ni =}\,\text{28}$)
(A) $\text{CoF}_{\text{6}}^{\text{2-}}$ and $\text{NiCl}_{\text{4}}^{\text{2-}}$
(B) $\text{TiF}_{\text{6}}^{\text{2-}}$ and $\text{CoF}_{\text{6}}^{\text{2-}}$
(C) $\text{C}{{\text{u}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ and $\text{NiCl}_{\text{4}}^{\text{2-}}$
(D) $\text{TiF}_{\text{6}}^{\text{2-}}$ and $\text{C}{{\text{u}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$

Answer 1

Hint: Colour of transition metal ion salt is due to d-d transition of unpaired electron of d-orbital. The unpaired electron present in partially filled d-orbitals is excited into higher energy d-orbital by absorbing energy from visible light and thus exhibits the complementary colour.
- In the presence of strong field ligand (such as $\text{CN,}\,\text{N}{{\text{H}}_{\text{3}}}$ ) back pairing of unpaired electrons takes place, while in the presence of weak field ligand (such as $\text{C}{{\text{l}}^{\text{-}}}\text{,}\,\text{F}$ ) unpaired electrons are not paired.

Complete Solution :
For a colourless complex, there should not be any unpaired electron in the valence shell of the central metal ion.
- In $\text{TiF}_{\text{6}}^{\text{2-}}$, $\text{T}{{\text{i}}^{\text{+4}}}$ has ${{\text{d}}^{\text{0}}}$-electronic configuration and there is no unpaired electron is present in the valence shell of central metal, hence it is a colourless species.
-In $\text{CoF}_{\text{6}}^{\text{2-}}$, $\text{C}{{\text{o}}^{\text{+4}}}$ has ${{\text{d}}^{\text{5}}}$-electronic configuration and have five unpaired electron in the valence shell of the central metal ion. Since fluorine is a weak field ligand and thus unpaired electron of valence shell will not pair up, hence it is a coloured species.
-In $\text{C}{{\text{u}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$, $\text{C}{{\text{u}}^{+1}}$ has ${{\text{d}}^{\text{10}}}$-electronic configuration and all the electrons in the valence shell of the central metal ion are paired up so there is no unpaired electron. Hence it will be a colourless species.
-In $\text{NiCl}_{\text{4}}^{\text{2-}}$, $\text{N}{{\text{i}}^{\text{+2}}}$ has ${{\text{d}}^{\text{8}}}$-electronic configuration and there are two unpaired electrons in the valence shell of the central metal ion, hence it is a coloured species.
So, the correct answer is “Option D”.

Note: -All the diamagnetic species are colourless species while all the paramagnetic species are colourless species.
-In case of transition metal ions, the electron can be easily promoted from one energy level to another in the same d-subshell.