This electronic configuration shows element of: \[\underset{1S}{\mathop{}}\,\,\,\,\,\,\,\,\,\,\,\underset{2S}{\mathop{}}\,\,\,\,\,\,\,\,\,\,\,\,\,\underset{2P}{\mathop{}}\,\]
A) oxygen
B) neon
C) fluorine
D) nitrogen
Correct Answer:
C
Solution :
This electronic configuration shows atomic number 9. So that element is fluorine.