Question

Which of the following compounds is not coloured yellow?
(A) $Z{n_2}\left[ {Fe{{(CN)}_6}} \right]$
(B) ${K_3}\left[ {Co{{(N{O_2})}_6}} \right]$
(C) ${(N{H_4})_3}\left[ {As{{(M{o_3}{O_{10}})}_4}} \right]$
(D) $BaCr{O_4}$

Answer 1

Hint: Any compound or complex will show colour if there is presence of unpaired electrons and these electrons undergo d-d transition or charge transfer phenomenon. If the ligands attached to the metal ion are strong field ligands then they cause pairing of electrons and they cannot undergo d-d transition.

Complete step by step answer:
-First of all let us see what gives compounds a colour.
Most usually compounds show colour due to presence of unpaired electrons and their excitation or transition from a lower energy orbitals to higher energy orbital when visible lights falls on the compound or ion. Some part of this light gets absorbed while the rest gets transmitted and the transmitted light shows colour complementary to the absorbed colour.
-Now we will check each option to see which one of them has the ability to show yellow colour.

For (A) $Z{n_2}\left[ {Fe{{(CN)}_6}} \right]$: The oxidation state of Fe in this complex is (+2) and the electronic configuration of $F{e^{ + 2}}$ is $\left[ {Ar} \right]3{d^6}$. We know that the ligand $C{N^ - }$is a strong field ligand and so causes pairing of the 3d electrons. After thus there is no d-d transition possible and hence this complex will show white bluish colour.

For (B) ${K_3}\left[ {Co{{(N{O_2})}_6}} \right]$: The oxidation state of Co in this complex is (+3) and the electronic configuration of $C{o^{ + 3}}$ will be $\left[ {Ar} \right]3{d^6}$. Since the ligand $NO_2^ - $ is a medium field ligand there will be no pairing. So this will undergo d-d transition and hence the complex will show yellow colour.

For (C) ${(N{H_4})_3}\left[ {As{{(M{o_3}{O_{10}})}_4}} \right]$: The ligand involved here is also a weak field ligand and so there will be no pairing. Hence this complex will also undergo d-d transition and so it will show yellow colour.

For (D) $BaCr{O_4}$: This is not a complex but a compound. It shows charge transfer phenomenon and thus it will be yellow coloured.

-So, from the above discussion we can say that $Z{n_2}\left[ {Fe{{(CN)}_6}} \right]$ shows white colour while ${K_3}\left[ {Co{{(N{O_2})}_6}} \right]$, ${(N{H_4})_3}\left[ {As{{(M{o_3}{O_{10}})}_4}} \right]$ and $BaCr{O_4}$ show yellow colour.
So, the correct answer is “Option A”.

Note: The complex having a strong field ligand like $C{N^ - }$ shows white colour because the complex possessing strong field ligands have higher energy and hence will show a colour of lower wavelength. The weak field ligands make the complex show colour of higher wavelength. The strong and weak field ligands are shown by the spectrochemical series:
${I^ - } < B{r^ - } < {S^{2 - }} < SC{N^ - } < C{l^ - } < NO_3^ - < N_3^ - < {F^ - } < O{H^ - } < {C_2}{O_4}^{2 - } = {H_2}O < NC{S^ - } < C{H_3}CN < py < N{H_3} < en < bipy < phen < NO_2^ - < pp{h_3} < C{N^ - } = CO$