what is the element symbol of this excited state configuration 1s22s22p43s1 ??
F*. It is an excited fluorine atom.
You know the atom is in an excited state, but it is not ionised. Therefore, simply count the number of electrons and see which element that corresponds with. 2+2+4+1 = 9, which is fluorine. And because the atom is in an excited state, you add a superscripted * to the symbol.