What is the #"oxidation state"# of an element in a compound?

1 Answer
Nov 3, 2015

Answer:

The oxidation state of an element in a compound is the charge left on a central atom when the bonding pairs are broken, with the charge going to the most electronegative atom. Element-element bonds are conceived to share the charge.

Explanation:

Of course, oxidation state is a formalism. Consider the water molecules, #OH_2#. We break an #O-H# bond (i.e. 1 bond, so 2 electrons), the charge goes to the most electronegative atom; we get #O^(2-)# #+# #2 xx H^+#. So the oxidation state of #H# in water is #+I#, and the oxidation state of #O# is #-II# (because we have an #O^(2-)# species). Now break the #O-O# bond in hydrogen peroxide, #H-O-O-H#; the 2 electrons are shared between each oxygen atom atom to give 2 radical species, i.e. #2 xx H-O*#. This species is neutral because of course oxygen atoms have equal electronegativity, and when we break the remaining #H-O# we get #H^+#, and #O^-#; remember this is a formalism, but this means that oxygen atom has a formal oxidation state of #-I#. This is the same when we assign oxidation states to #C# in a chain, #H_3C-CH_3#, the #C# atoms have a #-III# oxidation state. Are we clear?

Now tell me the oxidation state of oxygen in #OF_2#? This is a real molecule. Which is the most electronegative atom that will get the charge from the #O-F# bonds? What about the oxidation state of oxygen in #O=O(g)#?