What do valence electrons have to do with bonding?

May 19, 2014

The valence electrons are the electrons that determine the most typical bonding patterns for an element.

These electrons are found in the s and p orbitals of the highest energy level for the element.

Sodium $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{1}$
Sodium has 1 valence electron from the 3s orbital

Phosphorus $1 {s}^{2} 2 {s}^{2} 2 {p}^{6} 3 {s}^{2} 3 {p}^{3}$
Phosphorus has 5 valence electrons 2 from the 3s and 3 from the 3p

Lets take the ionic formula for Calcium Chloride is $C a C {l}_{2}$

Calcium is an Alkaline Earth Metal in the second column of the periodic table. This means that calcium ${s}^{2}$ has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a Ca+2 cation.

Chlorine is a Halogen in the 17th column or ${s}^{2} {p}^{5}$ group.
Chlorine has 7 valence electrons. It needs one electron to make it stable at 8 electrons in its valence shells. This makes chlorine a Cl^(−1) anion.

Ionic bonds form when the charges between the metal cation and non-metal anion are equal and opposite. This means that two Cl^(−1) anions will balance with one $C {a}^{+ 2}$ cation.

This makes the formula for calcium chloride, $C a C {l}_{2}$.

For the example Aluminum Oxide $A {l}_{2} {O}_{3}$

Aluminum ${s}^{2} {p}^{1}$ has 3 valence electrons and an oxidation state of +3 or $A {l}^{+ 3}$
Oxygen ${s}^{2} {p}^{4}$ has 6 valence electrons and an oxidation state of -2 or O^(−2)

The common multiple of 2 and 3 is 6.
We will need 2 aluminum atoms to get a +6 charge and 3 oxygen atoms to get a -6 charge. When the charges are equal and opposite the atoms will bond as $A {l}_{2} {O}_{3}$.

In molecular (covalent) compounds these same valence electrons are shared by atoms in order to satisfy the rule of octet.