Writing Ionic Formulas
Key Questions
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Lets take the ionic formula for Calcium Chloride is
#CaCl_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
#Ca^(+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#Ca^(+2)# cation.This makes the formula for calcium chloride,
#CaCl_2# .For the example Aluminum Oxide
#Al_2O_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
#Al_2O_3# .I hope this is helpful.
SMARTERTEACHER? -
What you want to do is make the compound neutral.
Let's take the following example:
#Na^(+)# +#SO_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
#Na# ion has a#+1# charge and the#SO_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:#Na_2SO_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:#Fe^(3+)# and#OH^(-)# In order to balance this, we need to add brackets around the hydroxide ion to give:
#Fe(OH)_3# This balances the charges and, thus, an accurate ionic formula has been given.