Chapter 12, Problem 112E

Consider the following ionization energies for aluminum.

$\begin{array}{l}\text{Al(}g\text{)}\to {\text{Al}}^{\text{+}}{\text{(g)+e}}^{\text{–}}{I}_{\text{1}}\text{=580 kJ/mol}\\ {\text{Al}}^{\text{+}}\text{(}g\text{)}\to {\text{Al}}^{\text{2+}}\text{(}g{\text{)+e}}^{\text{–}}{I}_{\text{2}}\text{=1815 kJ/mol}\\ {\text{Al}}^{\text{2+}}\text{(}g\text{)}\to {\text{Al}}^{\text{3+}}\text{(}g{\text{)+e}}^{\text{–}}{I}_{\text{3}}\text{=2740 kJ/mol}\\ {\text{Al}}^{\text{3+}}\text{(}g\text{)}\to {\text{Al}}^{\text{4+}}\text{(}g{\text{)+e}}^{\text{–}}{I}_{\text{4}}\text{=11,600 kJ/mol}\end{array}$
a. Account for the increasing trend in the values of theionization energies.
b. Explain the large increase between $I_3$ and $I_4$ .
c. Which one of the four ions has the greatest electronaffinity? Explain.
d. List the four aluminum ions given in the precedingreactions in order of increasing size, and explainyour ordering. (Hint: Remember that most of thesize of an atom or ion is due to its electrons.)