How do you calculate the oxidation number of an ion?

1 Answer
anor277
Aug 19, 2016

For simple ions, i.e. #Na^+#, #Cl^-#, #Fe^(2+)#, #Fe^(3+)#, the oxidation number of the element is simply the charge on the ion.

Explanation:

For more complex ions, the SUM of the oxidation numbers equals the charge on the ion. The sum of the oxidation numbers also equals the charge on the molecule in a NEUTRAL molecule.

Let's consider #Cl^-#, #ClO_2#, #ClO_4^-#. Typically the oxidation number of oxygen in its oxides is #-II#, and it is here in these examples.

Given what I have said the oxidation number of #Cl# in chloride ion, #Cl^-#, is #-I#. The oxidation number of #Cl# in #ClO_2# is #+IV#, and the oxidation number of #Cl# in perchlorate, #ClO_4^-#, is #+VII#.

I don't what level you are at. If you are at A level, you simply have to know what I wrote in the opening statement. If you are an undergrad you do have to know how to assign oxidation states in more complicated molecules and ions. Good luck.

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