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Question: Which of the following compounds are coloured due to charge transfer spectra? (A) \({{K}_{2}}C{{r...

Which of the following compounds are coloured due to charge transfer spectra?
(A) K2Cr2O7{{K}_{2}}C{{r}_{2}}{{O}_{7}}
(B) KMnO4KMn{{O}_{4}}
(C) CuSO4.5H2OCuS{{O}_{4}}.5{{H}_{2}}O
(D) Both A and B

Explanation

Solution

Several metal complexes are coloured due to the d-d electronic transitions and the Charge-transfer complexes do not experience d-d transitions. Hence the reason for color in charge transfer complexes will be charge transfer spectra.

Complete step by step answer:
- As we know, an electron-donor-acceptor complex or charge-transfer complex (CT complex) is an association of two or more molecules, or of different parts of one large molecule, in which a fraction of electric charge is transferred between the molecular entities and this charge transfer will give rise to colour of certain compounds.
(i) K2Cr2O7{{K}_{2}}C{{r}_{2}}{{O}_{7}}
- In potassium dichromate or K2Cr2O7{{K}_{2}}C{{r}_{2}}{{O}_{7}} the central chromium atom has an oxidation state of +6. As we know, chromium has the atomic number of 24 and since the charge is +6, the number of total electrons will be 18. The electronic configuration can be written as 1s22s22p63s23p61{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}} or as[Ar]3d04s0\left[ Ar \right]3{{d}^{0}}4{{s}^{0}}.
- Thus the number of unpaired electrons and d electrons is zero .As a result d-d transition can’t occur. The oxygen will donate electrons and the vacant orbitals in Chromium will accept the electrons and from this charge transfer spectra the colour of K2Cr2O7{{K}_{2}}C{{r}_{2}}{{O}_{7}} arises.
(ii) KMnO4KMn{{O}_{4}}
- In KMnO4KMn{{O}_{4}} or potassium permanganate, the central Manganese atom has a +7 oxidation state. As we know, Manganese has the atomic number of 25 and since the charge is +7, the number of total electrons will be 18. The electronic configuration can be written as 1s22s22p63s23p61{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}} or as [Ar]3d04s0\left[ Ar \right]3{{d}^{0}}4{{s}^{0}}. Here also there are no unpaired electrons and hence d-d transition can’t occur.
- The oxygen in KMnO4KMn{{O}_{4}} will donate electrons to the vacant orbitals of manganese and as a result a charge transfer spectra occurs and therefore we can say that the colour of KMnO4KMn{{O}_{4}} also arises from charge transfer spectra.
(iii) CuSO4.5H2OCuS{{O}_{4}}.5{{H}_{2}}O
In hydrated copper sulphate o rCuSO4.5H2OCuS{{O}_{4}}.5{{H}_{2}}O, the water molecules acts as ligands and will cause the splitting of d orbitals. This will facilitate the d-d transition and this causes the colour of the complex CuSO4.5H2OCuS{{O}_{4}}.5{{H}_{2}}O, not charge transfer spectra.

Therefore the compounds are coloured due to charge transfer spectra are K2Cr2O7{{K}_{2}}C{{r}_{2}}{{O}_{7}} and KMnO4KMn{{O}_{4}}. Thus the answers are options (A) and (B).

Note: The colours which are formed by d-d transition are usually quite faint as a result of the selection rules such as spin rule and Laporte’s rule. Since the charge-transfer complexes do not undergo d-d transitions and thus, these selection rules do not apply and, in common, these absorptions are very intense.