Electron orbitals are filled in the order $n = 0\ l = 0$ , $n = 1\ l = 0$ , $n = 1\ l = 1$ , $n = 2\ l = 0$ , $n = 2\ l = 1$ , $n = 2\ l = 2$ , etc until the fourth row of the periodic table (starting with potassium (Z = 19)). Sodium comes before this, so its orbitals are filled by first filling $n = 0\ l = 0$ (2 electrons), then filling $n = 1\ l = 0$ (2 more electrons), then filling $n = 1\ l = 1$ (6 more electrons), then filling the first electron in the $n = 2\ l = 0$ subshell. This electron configuration can be written as $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}$ . Therefore, answer (C) is correct.

Solution to 1992 Problem 58