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
a. Account for the increasing trend in the values of the ionization energies.
b. Explain the large increase between I3 and I4.
c. Which one of the four ions has the greatest electron affinity? Explain.
d. List the four aluminum ions given in the preceding reactions in order of increasing size, and explain your ordering. (Hint: Remember that most of the size of an atom or ion is due to its electrons.)

  • CreatedMay 05, 2015
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