What is the electron configuration of #"Na"^(+)#? How many paired electrons does it have?

Question

Owen Ambrose · Accepted Answer

Well, the typical ion that sodium (Na) forms, as an alkali metal (first column in the periodic table) is a #+1# cation.

As a neutral atom, its electron configuration was:

#[Ne] 3s^1#

As a #+1# ion, its electron configuration involves one less valence electron, giving it a noble gas core electronic structure:

#[Ne]#

Or in longer-hand notation:

#color(blue)ul(1s^2 2s^2 2p^6)#

All "noble gas cores" have all electrons paired. We say that #"Na"^(+)# is isoelectronic with (having the same electronic structure as) #"Ne"#.

Therefore, all the electrons in #bb("Na"^(+))# are paired.

(I'll leave it as an exercise for you to count how many electrons are paired from the electron configuration given by #1s^2 2s^2 2p^6#. Hint: It's no more than #11#.)

I hope that helps!

Ellie Andrews · Answer

undefined A neutral sodium (Na) atom has the electron configuration 1s² 2s² 2p⁶ 3s¹. When sodium loses an electron to form Na⁺, it loses the one electron in the 3s orbital, resulting in the electron configuration of 1s² 2s² 2p⁶, which is the same as that of the noble gas neon (Ne). This electron configuration means that all of the electrons in Na⁺ are paired; therefore, Na has ten electrons in total.