Answer
The element here is fluorine ($F$). Its ground-state condensed electron configuration is $$F: [He]2s^22p^5$$
Work Step by Step
$1s^22s^22p^43s^1$
In the excited state, normally there is no loss or addition of electrons. Therefore, to find out which element it is, we still calculate the number of electrons and deduce the atomic number from there.
Here, there are $2+2+4+1=9$ electrons. So, the atomic number of the element is 9. This element is fluorine ($F$).
The nearest noble gas of the lower atomic number is $He$. For these 9 electrons:
- The first 2 will occupy subshell $1s$.
- The next 2 will occupy subshell $2s$.
- The last 5 will occupy subshell $2p$.
Therefore, the ground-state condensed electron configuration is $$F: [He]2s^22p^5$$
