Q) Iodine dissolves in a variety of solvents, forming solutions that are purple, brown or any intermediate shade. This property is due to:

  1. Hydrolysis of iodine
  2. Formation of crystalline adducts
  3. Charge transfer between $\ce{I-I-}$ and the solvent
  4. Charge transfer between $\ce{I-I+}$ and the solvent

I am completely aware of which concept does option 3 and 4 allude to. I know about the formation of $\ce{I3-}$ when iodine is dissolved in solution of iodides. This fact, according to me, is what option 2 refers to.

But the answer given is 3

Can someone provide an explanation as to what concept is involved here and how can we predict the answer? I read about charge transfer complexes, but the Wikipedia page was too complicated for me to understand, and also the application of that to the question under consideration was not given.


1 Answer 1


Iodine in an inert solvent is purple. Electron donation alters that. Linear starch (amylose)-iodine is dark brown. Branched starch (amylopectin)-iodine is red. The Wikipedia page shows iodine in chlorinated solvent (very weak electron donation) then addition of strong electron donor triphenylphosphine.

If an electron donor is reactive, then the interacting iodine molecule looks like iodonium (elecron accepter). Initial interation is (4), post-mixing is (3).

Put a tiny iodine crystal into a melting point capillary, evacuate under vacuum, flame closed. You will see a very pale purple vapor at equilibrium. Crack the seal and the vapor is suddenly deeper purple. Why? Take a sealed cap and an evacuated sealed cap, and insert the vapor volumes into the beam of a green laser pointer. interesting. Take care that the optical geometry does not reflect back into your eyes.

  • $\begingroup$ Can you please expand on why this Charge transfer takes place? And why does it affect the perceived colours? $\endgroup$
    – user4059
    Feb 25, 2014 at 14:15

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