Periodic Trend: Successive Ionization Energies Video Lessons

Concept: Periodic Trend: Successive Ionization Energies

# 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.