# Does the diiodide(1−) anion exist?

With respect to chemical bonding does the $\ce{I2-}$ ion exist?

• Probably not, is assume that the equilibrium is entirely on the right side: $\ce{2I2- <=> I- + I3-}$ Dec 28 '17 at 11:32
• It seems like it does (+refs 1-7 therein). Dec 28 '17 at 11:33

I found a paper$$\ce{^{[1]}}$$ regarding the formation of diiodide anion species($$\ce{I2^-}$$). It is assumed to be an unstable species which forms during the formation of triiodide anion (excess of iodide in solution which makes it brown). This is a part of the paper which describes the formation of the species. For more information, read the full paper. Also, you can check the abstract which @andselisk linked.

The $$\ce{I2^-}$$ species is formed by the 248-nm laser photolysis of iodide through the following reactions:

$$\ce{I- + hν -> I + e_s-}$$

$$\ce{I + I- ->[k2]I2^-}$$

$$\ce{I2- + I2- ->[k3]I3- + I-}$$

As discussed later, we found that photodetachment of electrons from $$\ce{I-}$$ ions produces solvated electrons ($$\ce{e_s-}$$) in ionic liquids as well as in molecular solvents. After photodetachment, iodine atoms react with $$\ce{I-}$$ to form diiodide anion radicals. In aqueous solution, the transient absorption maxima of $$\ce{I2-}$$ are located around 400 and 720 nm. The extinction coefficients of $$\ce{I2-}$$ in aqueous solution at 385 and 725 nm are 10000 and 2560 M-1cm-1, respectively.[...]

$$\ce{^{[1]}}$$: J. Phys. Chem. B, 2007, 111 (18), pp 4807–4811