Building the Periodic Table — Question 7

Question 7

A large ionization jump identifies the number of loosely bound electrons

A hypothetical main-group atom has successive ionization energies $6$, $15$, $80$, and $115\,\mathrm{eV}$. Assume the sharp jump indicates removal of a compact closed-shell core electron after all outer valence electrons have been removed.

1. Infer the valence-electron count and favored closed-core positive charge.

2. Find the work to make the favored ion and the fraction of the three-electron removal work spent removing the first core electron.

Question 7 - Solution

A large ionization jump identifies the number of loosely bound electrons

(a) Method and calculation

$$I_2/I_1=15/6=2.5,\quad I_3/I_2=80/15=5.33\Rightarrow\text{jump after two removals}.$$$$\boxed{N_{\mathrm{val}}=2,\quad q_{\mathrm{closed\ core}}=+2e}.$$

(b) Interpretation and further calculation

$$\boxed{W_{2+}=6+15=21\,\mathrm{eV}}.$$$$W_{3+}=101\,\mathrm{eV}\Rightarrow\boxed{I_3/W_{3+}=80/101=.7921}.$$

Physical check

The shell inference uses the stated sharp-jump model. The cumulative work is the sum of successive energies, and the favored charge statement concerns shell structure rather than a complete chemical thermodynamic calculation.