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Identify the incorrect order of ionization energy among the following:

1. Pb (IE) > Sn (IE)
2. Na+ (IE) > Mg+ (IE)
3. Li+ (IE) < O+ (IE)
4. Be+ (IE) < C+ (IE)

Correct answer: Na+ (IE) > Mg+ (IE)

Solution

Li+ has only 2 electrons in the 1s orbital with Z=3. Its ionization energy (removing the 3rd electron of Li, 2nd electron of Li+) is approximately 7298 kJ/mol (75.6 eV). O+ has 7 electrons (configuration 1s2 2s2 2p3) with Z=8. Its next IE (removing from 2p) is approximately 3388 kJ/mol (35.1 eV). Therefore IE(Li+) > IE(O+), making the statement 'Li+(IE) < O+(IE)' in option C INCORRECT.

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