Calculate the total number of electron for $Fe$ having $n+l+m=2$.

Solution:

$$\begin{gathered} 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} \\ n\to 1 2 2 3 3 4 3 \\ l\to 0 0 1 0 1 0 2 \\ m\to 0 0 -1 0 -1 0 -2 \\ 0 0 -1 \\ +1 +1 1 \\ 2 \end{gathered}$$