Problem: Consider the following ionization energies for aluminum:Al(g) → Al+(g) + e-          I1 = 580 kJ/molAl+(g) → Al2+(g) + e-          I2 = 1815 kJ/molAl2+(g) → Al3+(g) + e-          I3 = 2740 kJ/molAl3+(g) → Al4+(g) + e-          I4 = 11,600 kJ/molAccount for the trend in the values of the ionization energies.

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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 Trends: Ionization Energy concept. You can view video lessons to learn Periodic Trends: Ionization Energy. Or if you need more Periodic Trends: Ionization Energy practice, you can also practice Periodic Trends: Ionization Energy 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 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.