What is the electron configuration (arrangement) by spdf notation of a sodium ion?

2 Answers
Feb 24, 2018

Answer:

#1s^2 2s^2 2p^6#

Explanation:

Sodium loses 1 electron from its outermost shell to get
the configuration of neon, or form the sodium ion.

  • The first shell has #1s^2#
  • The second shell has #2s^2, 2p^6#
Feb 24, 2018

Answer:

#1s^2\2s^2\2p^6# or just #[Ne]#

Explanation:

A neutral sodium atom has an electron configuration of #[Ne]3s^1#.

Since it has #11# electrons, it would want to lose one electron to remove its third shell, while keeping the other #2# electron shells full. So, in a sodium ion, it becomes #Na^+# and has only #10# electrons.

So, it will remove #1# electron, and lose its #3s# orbital. The electron configuration therefore just becomes #[Ne]# or if you want the full configuration, it is #1s^2\2s^2\2p^6#.