Why is vicinal diiodopropane unstable, leading to deiodination?
That 1,2-diodopropane is unstable does not indicate whether it is as the neat halide or in solution. Breaking (BDE) two $\ce{C-I}$ bonds costs $\pu{+114 kcal/mol}$ while forming the $\pi$-bond of propene is worth approximately $\pu{-62 kcal/mol}$ and the formation of iodine, $\pu{-36 kcal/mol}$. The net reaction is endothermic by $\pu{+16 kcal/mol}$. The reaction is entropically favored and the products are volatile. Therefore, the unfavorable equilibrium can be shifted to the right.