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

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


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