Question

Question #de62a

Answer 1

The obvious answer is `"Na"^(+)`, since losing an electron would complete sodium's octet.

On the other hand, you simply cannot have an `"Na"^(-7)` anion, since that would imply that sodium takes on 7 extra electrons in order to complete its octet.

Sodium is a group 1 metal with a very low electronegativity value - 0.93 on the Pauling scale - which means that it tends to lose an electron in order to complete its octet- hence the `"Na"^(+)` cation.

There is nothing that can react with sodium which will allow sodium to take on 7 more electrons because it would be much easier to lose 1 than to gain 7.

Another reason that stops sodium from ever adding 7 electrons in order to complete its octet its the fact that it wouldn't have anywhere to put those electrons. Sodium's electron configuration is

`"Na": 1s^(1)2s^(2) 2p^(6)3s^(1)`

The maxium number of electrons it could theoretically add to its configuration is 1, but that's not going to happen because of the very low electronegativity value sodium has.