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.

FREE Expert Solution

Ionization energy (I. E.) is the energy required to remove an electron from a gaseous atom or ion.

First I.E. corresponds to removing thfirst valence electron from a neutral atom

 Second I.E. refers to removing a second electron

 Third I.E. refers to removing the third electron

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Problem Details

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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What scientific concept do you need to know in order to solve this problem?

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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Based on our data, we think this problem is relevant for Professor Albright's class at UMICH.

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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.