Building the Periodic Table — Question 2

Question 2

A main-group configuration determines electron count and valence occupancy

A neutral atom has configuration $[\mathrm{Ne}]3s^23p^4$, where the neon core contains ten electrons. For the p subshell use three spatial orbitals and Hund filling: occupy each singly with parallel spins before pairing.

1. Find atomic number, principal valence shell, and valence-electron count.

2. Draw the p occupancy and find the number of unpaired electrons and maximal spin S for this configuration.

Question 2 - Solution

A main-group configuration determines electron count and valence occupancy

(a) Method and calculation

$$Z=N_e=10+2+4=\boxed{16}.$$$$\boxed{n_{\mathrm{val}}=3,\quad N_{\mathrm{val}}=2+4=6}.$$

(b) Interpretation and further calculation

$$3p^4:\quad(\uparrow\downarrow)(\uparrow)(\uparrow)\Rightarrow\boxed{N_{\mathrm{unpaired}}=2}.$$$$S_{\max}=\tfrac12+\tfrac12=\boxed{1}.$$

Physical check

The paired electrons contribute zero net spin. The answer assumes Hund ordering in a free-atom-like degenerate p subshell; a strong crystal field would require a separate model.