Question

Which of the following compounds is not colored yellow?

• A. ${\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$
• B. ${\mathrm{BaCrO}}_4$
• C. ${\mathrm{K}}_3\left[\mathrm{Co}{\left({\mathrm{NO}}_2\right)}_6\right]$
• D. ${\left({\mathrm{NH}}_4\right)}_3\left[\mathrm{As}{\left({\mathrm{Mo}}_3{\mathrm{O}}_{10}\right)}_4\right]$

Answer 1

The correct option is A

${\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$

The explanation for the correct option

(A) ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{\left[}\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}$

• Oxidation number of $\mathrm{Fe}$ is $+2$ here so the outer electronic configuration of $\mathrm{Fe}$ will be $3{\mathrm{d}}^6$
• $\mathrm{CN}$ is a strong ligand. Thus there will be a paring up of electrons. Therefore there are no unpaired electrons to show $\mathrm{d}-\mathrm{d}$ transition.
• The compound to show color there should be either $\mathrm{d}-\mathrm{d}$ the transition of charge transfer.
• Thus here ${\mathrm{Zn}}_2\left[\mathrm{Fe}{\left(\mathrm{CN}\right)}_6\right]$ will not show a strong $\mathrm{d}-\mathrm{d}$ due to the absence of unpaired electrons. And therefore the compound is white-bluish in color.

The explanation for the incorrect options

(B) ${\mathbf{BaCrO}}_{\mathbf{4}}$

• This compound is not a coordination compound.
• $\mathrm{d}-\mathrm{d}$ transition is absent here.
• But this compound shows yellow color due to charge transfer

(C) ${\mathbf{K}}_{\mathbf{3}}\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}{\mathbf{NO}}_{\mathbf{2}}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}\mathbf{}$

• ${\mathrm{NO}}_2$ is not a very strong ligand.
• So there will not be any pairing of electrons. This unpaired electron will cause $\mathrm{d}-\mathrm{d}$ transition. And thereby show yellow color.
• Here since the ligand is not that much strong, the change in energy will be less.
• The energy is inversely proportional to wavelength. Thus this compound is showing a color of more wavelength which belongs to yellow.

(D) ${\mathbf{\left(}{\mathbf{NH}}_{\mathbf{4}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\left[}\mathbf{As}{\mathbf{\left(}{\mathbf{Mo}}_{\mathbf{3}}{\mathbf{O}}_{\mathbf{10}}\mathbf{\right)}}_{\mathbf{4}}\mathbf{\right]}$

• Here also the ligand is not that strong to cause the pairing up of electrons.
• The existence of unpaired electrons will show $\mathrm{d}-\mathrm{d}$ transition. Thus the compound will show color.
• Since the ligand is not strong the energy in transition will be less. So the color of the compound will be in the higher range of wavelength that is yellow.

Therefore, the correct answer is option (A) ${\mathbf{Zn}}_{\mathbf{2}}\mathbf{\left[}\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}$