Writing Ionic Formulas

Key Questions

How do you write an ionic formula given an anion and a cation?

Lets take the ionic formula for Calcium Chloride is $C a C l_{2}$ $CaCl_2$

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 $C a^{+ 2}$ $Ca^(+2)$ 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 $C l^{- 1}$ $Cl^(−1)$ anion.

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

This makes the formula for calcium chloride, $C a C l_{2}$ $CaCl_2$ .

For the example Aluminum Oxide $A l_{2} O_{3}$ $Al_2O_3$

Aluminum has an oxidation state of +3 or Al+3
Oxygen has an oxidation state of -2 or $O^{-} 2$ $O^−2$

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 $A l_{2} O_{3}$ $Al_2O_3$ .

I hope this is helpful.
SMARTERTEACHER?

BRIAN M. · 1 · Mar 6 2014
How do you write the formula of an ionic compound?

What you want to do is make the compound neutral.

Let's take the following example:

$N a^{+}$ $Na^(+)$ + $S O_{4}^{2 -}$ $SO_4^(2-)$

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 $N a$ $Na$ ion has a $+ 1$ $+1$ charge and the $S O_{4}$ $SO_4$ ion has a $- 2$ $-2$ charge. In order to give balance, we must have two Na ions to give an overall $+ 2$ $+2$ with regards to Na: this, thus, neutralises the compound. Therefore, the formula is:

$N a_{2} S O_{4}$ $Na_2SO_4$

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 $3 +$ $3+$ charge and the hydroxide ion (OH) has a $1 -$ $1-$ charge - as a result, the compounds in their individualised forms are:

$F e^{3 +}$ $Fe^(3+)$ and $O H^{-}$ $OH^(-)$

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

$F e {(O H)}_{3}$ $Fe(OH)_3$

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

Paul · 1 · Jun 6 2014

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Ionic Bonds

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Writing Ionic Formulas

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Ionic Bonds