Question #1a22d

You can think of atom stability from two perspectives: nucleus stability and electron configuration stability.

Nucleus stability is determined by the strong force, the force that overwhelms the powerful electrostatic repulsion that exists between protons (protons being positively charged particles) and keeps the nucleus together.

I assume you're more interested in electron configuration stability. As you know, all atoms strive to achieve a full octet on their outermost shell. This is done either by sharing electrons, as is the case with covalent bonding, or by gaining or losing electrons.

In your case, the neutral sodium atom has 11 protons and 11 electrons. Its electron configuration is

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

This electron configuration is very unstable, since the last electron is in the third energy level, and is what makes sodium highly reactive.

By losing its 3s electron, sodium now has all its 10 remaining electrons in lower energy levels, which makes the ion very stable; moreover, its newly obtained octet configuration will increase its stability and make it less reactive, since its electron configuration will match that of #"Ne"#, a noble gas.

As a conclusion, ions that have full octets will be more stable than atoms that don't have full octets from the perspective of their electron configurations.