# When a sodium atom reacts with a chlorine atom to form a compound, the electron configuration the ions forming the compound are the same as those in which noble gases?

Jan 1, 2017

Metals are electron rich..............

#### Explanation:

The valence electrons of metals tend to be lost, and for this reason metals TEND to be reducing. And non-metals are electron poor. And as a result they tend to be oxidizing. (This tendency is another manifestation of electronic structure.)

Sodium metal and dichlorine thus tend to be involved in redox phenomena:

$N a \left(s\right) + \frac{1}{2} C {l}_{2} \rightarrow N {a}^{+} C {l}^{-}$

Given the electronic configuration of the reactants, this redox transfer means that each species shares the electronic configuration of it nearest inert gas neighbourL i.e. $N {a}^{+} \equiv \left[N e\right]$; $C {l}^{-} \equiv \left[A r\right]$. Each species is still a charged ion, clearly capable of electrostatic interaction, yets its valence electronic shell has been filled and stabilized.

And please look at the Periodic Table to confirm these configurations.