# Writing Ionic Formulas

## Key Questions

• 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 has 2 valence electrons it readily gives away in order to seek the stability of the octet. This makes calcium a $C {a}^{+ 2}$ cation.

Chlorine is a Halogen in the 17th column or p5 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 has an oxidation state of +3 or Al+3
Oxygen has 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 $A {l}_{2} {O}_{3}$.

SMARTERTEACHER?

• What you want to do is make the compound neutral.

Let's take the following example:

$N {a}^{+}$ + $S {O}_{4}^{2 -}$

We need to balance the charges, the easiest way to balance this charge is by looking at the overall charge of the ions involved. The $N a$ ion has a $+ 1$ charge and the $S {O}_{4}$ ion has a $- 2$ charge. In order to give balance, we must have two Na ions to give an overall $+ 2$ with regards to Na: this, thus, neutralises the compound. Therefore, the formula is:

$N {a}_{2} S {O}_{4}$

If you're asked to balance an ionic compound such as Iron(III) Hydroxide, write down the formula. We know that Fe (Iron) has a $3 +$ charge and the hydroxide ion (OH) has a $1 -$ charge - as a result, the compounds in their individualised forms are:

$F {e}^{3 +}$ and $O {H}^{-}$

In order to balance this, we need to add brackets around the hydroxide ion to give:

$F e {\left(O H\right)}_{3}$

This balances the charges and, thus, an accurate ionic formula has been given.