Question #03ed1

Jul 11, 2016

Atomic number of Sodium, $\text{Na}$ is $11$. Orbital filling of its eleven electrons is shown below

$1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$

In other words its electronic arrangement is $2 , 8 , 1$. Meaning it has single electron in the valence shell.

The completely filled shells corresponding to principal quantum number $n = 1 \mathmr{and} 2$ are very stable.

We have observed that elements, while forming chemical bonds, follow the octet thumb rule. Atoms of main-group tend to combine in such a way that each atom has eight electrons in its valence shell, the same electronic configuration of a noble gas.

Sodium atom can either gain $7$ electrons or lose$1$ to attain this stable configuration. It is easier for a sodium atom to lose its extra electron and become a sodium cation having charge $+ 1$ in the process
$\text{ } N a \to N {a}^{+} + {e}^{-}$