Question

write the electron configuration of sodium using noble - gas notation. 250 words remaining

Explanation:

Sodium has an atomic number of 11. The noble - gas before sodium is neon (Ne) which has 10 electrons. So, we use [Ne] to represent the first 10 electrons. The remaining electron (11 - 10 = 1) goes into the 3s orbital.

Answer:

$$ [\text{Ne}]3s^{1} $$