With respect to chemical bonding does the $\ce{I2-}$ ion exist?
$\begingroup$
$\endgroup$
2
-
1$\begingroup$ Probably not, is assume that the equilibrium is entirely on the right side: $\ce{2I2- <=> I- + I3-}$ $\endgroup$– Martin - マーチン ♦Commented Dec 28, 2017 at 11:32
-
1$\begingroup$ It seems like it does (+refs 1-7 therein). $\endgroup$– andselisk ♦Commented Dec 28, 2017 at 11:33
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
0
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
answered Dec 28, 2017 at 13:32