# Question #096b2

Dec 26, 2014

Actually, in an ionic bond, the metal is said to lose an electron, not gain one (this is the case for alkali metals) ; the nonmetal that bonds with such a metal can be said to gain an electron.

Let's take an easy example, sodium chloride - $N a C l$.

SInce sodium is an alkali metal - it's in group I of the periodic table, it has 1 electron in its outermost shell. SInce the goal of (almost) every atom is to reach a complete outer shell - that is ,to have 8 electrons in its outermost shell- it's easier for $N a$ to lose an electron than, to say, gain 7.

Since chlorine is a halogen (a group 17 nonmetal), it has 7 electrons in its outermost shell, so it would only need one electron to do that.

So, when an ionic bond is formed between $N a$ and $C l$, sodium practically gives its electron to chlorine; this makes sodium reach the stable configuration of neon ($N e$). Since chlorine now has 8 electrons, it reached the stable configuration of argon ($A r$).

This is true for alkali earth metals as well; metals found in group II of the periodic table will donate 2 electrons when forming an ionic bond with a nonmetal.