# How do you determine the electron configuration of Na?

It has $11$ electrons. How did I know this? How will you know this if it is asked in your exam? And we write the electronic configuration in the usual $\text{aufbau}$ way...
$N a : 1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$
The $3 s$ electron is READILY LOST, and this sodium metal is a very capable REDUCTANT....