The color in MnO 4- arises from an electronic transition, but it is actually not a d-d transition, since the Mn in this compound has no d electrons. It arises from a charge transfer reaction within the molecule, in which photons promote an electron from the highest energy molecular orbital in one of the Mn-O bonds to an empty d orbital on the ...

• Tell the students that the pink MnO 4-ion will be reduced to colorless Mn2+ as it oxidizes NO 2-to NO 3-. • Pour the pink solution into the large flask, expecting it to become colorless – it will stay pink. • In an effort to discover the problem, go back and balance the equation with the students. You may want to balance the half-

Limiting reagent can be computed for a balanced equation by entering the number of moles or weight for all reagents. The limiting reagent row will be highlighted in pink. A chemical equation represents a chemical reaction. It shows the reactants (substances that start a reaction) and products (substances formed by the reaction).

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Explain why Mn (H 2 O) 6 2 + is pale pink. MnO 2 is black and MnO 4 - is intensely coloured purple. When ammonium vanadate is heated with an oxalic acid solution, a compound (Z) is formed. A sample of (Z) was titrated with KMnO 4 solution in hot acidic solution.

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The correct answer is Since charge transfer occurs from ligand to metal, i.e. from oxygen to Mn, reducing it to +7 to +6, resulting in the intense pink color.Hence option C is correct.

MnO 4– is of intense pink colour, though Mn is in +VII oxidation state. This is due : Know your College Admission Chances Based on your Rank/Percentile, Category and Home State. Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA. Please subscribe our Youtube channel to unlock this solution. Was this answer helpful?

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MnO-4 is of intense pink colour, though Mn is in (+7) oxidation state. It is due to (A) oxygen gives colour to it (B) None of the above is correct (C) Tardigrade