Question

A certain first-row transition metal ion forms many different coloured solutions. When four coordination compounds of this metal, each having the same coordination number, are dissolved in water, the colours of the solutions are red, yellow, green, and blue. Further experiments reveal that two of the complexions are paramagnetic with four unpaired electrons and the other two are diamagnetic. What can be deduced from this information about the four complexes.

Short answer

The metal ion must be a Co³⁺ ion, and the four metal complexes were ${\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}\mathbf{,}\mathbf{}{\mathbf{\left[}{\mathbf{CoF}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}\mathbf{,}\mathbf{}{\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{+}}\mathbf{,}\mathbf{}{\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{+}}$. Among these four, ${\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{+}}$are diamagnetic and they exhibit yellow and red colour respectively, while ${\mathbf{\left[}\mathbf{Co}{\mathbf{F}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\left[}\mathbf{Co}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{+}}$are paramagnetic with four unpaired electrons and they exhibit green and blue colour respectively.

Step-by-step solution

Why Co3+ ion?

Among the first-row transition metals, Sc, Ti, and V cannot attain four unpaired electrons evenin their lowest oxidation state. Meanwhile, Ni, Cu, and Zn cannot attain four unpaired electrons even in their highest oxidation state. Among Cr, Mn, Fe, and Co, the former three cannot simultaneously satisfy the conditions of no. of valance electrons, coordination number and colours as mentioned in the above question.

How should we proceed to find the complexes?

• As mentioned in the question two of the complexes are diamagnetic, i.e., they do not have any unpaired electron, so, the metal must have formed complexes with strong field ligands. CN and NH₃ being strong field ligands satisfy the condition of diamagnetism.
• · Similarly, F and H₂O being weak field ligands allow the electrons to jump to the higher t2g orbital before pairing, allowing them to satisfy the condition of paramagnetism.
• · Also, these complexes absorb light of specific wavelengths to meet the colour demand of the given question e.g.,${\left[{\mathrm{CoF}}_6\right]}^{\mathbf{3}\mathbf{-}}$ absorb light of wavelength 770nm which falls in the red region thus exhibiting green colour in aqueous solution.