Ionization energy (I. E.) is the energy required to remove an electron from a gaseous atom or ion.
• First I.E. corresponds to removing the first valence electron from a neutral atom
• Second I.E. refers to removing a second electron
• Third I.E. refers to removing the third electron
Consider the following ionization energies for aluminum:
Al(g) → Al+(g) + e- I1 = 580 kJ/mol
Al+(g) → Al2+(g) + e- I2 = 1815 kJ/mol
Al2+(g) → Al3+(g) + e- I3 = 2740 kJ/mol
Al3+(g) → Al4+(g) + e- I4 = 11,600 kJ/mol
Account for the trend in the values of the ionization energies.
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Our tutors have indicated that to solve this problem you will need to apply the Periodic Trend: Successive Ionization Energies concept. You can view video lessons to learn Periodic Trend: Successive Ionization Energies. Or if you need more Periodic Trend: Successive Ionization Energies practice, you can also practice Periodic Trend: Successive Ionization Energies practice problems.
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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.