Problem: Under appropriate conditions, I2 can be oxidized to l2+, which is bright blue. The corresponding anion, I2-, is not known. Use the molecular orbital diagram below to explain why I2+ is more stable than I2-

FREE Expert Solution

Use bond orders to determine relative stability. 


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

Bond order =  (#e- in bond MO - #e - in antibond MO)2



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Under appropriate conditions, I2 can be oxidized to l2+, which is bright blue. The corresponding anion, I2-, is not known. Use the molecular orbital diagram below to explain why I2+ is more stable than I2-

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