Use bond orders to determine relative stability.

**For bond orders: ↑ BO, ↑ stability ↑ bond strength**

$\overline{){\mathbf{Bond}}{\mathbf{}}{\mathbf{order}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{}}\frac{\mathbf{(}\mathbf{\#}\mathbf{e}\mathbf{-}\mathbf{}\mathbf{in}\mathbf{}\mathbf{bond}\mathbf{}\mathbf{MO}\mathbf{}\mathbf{-}\mathbf{}\mathbf{\#}\mathbf{e}\mathbf{}\mathbf{-}\mathbf{}\mathbf{in}\mathbf{}\mathbf{antibond}\mathbf{}\mathbf{MO}\mathbf{)}}{\mathbf{2}}}$

Under appropriate conditions, I_{2} can be oxidized to l_{2}^{+}, which is bright blue. The corresponding anion, I_{2}^{-}, is not known. Use the molecular orbital diagram below to explain why I_{2}^{+} is more stable than I_{2}^{-}

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