How to find out oxidation state when compound has multiple electronegativities

Question

How do I solve for oxidation state in something like CH3-O-F? The fluorine is more electronegative than the oxygen which is more electronegative than the carbon which is more electronegative than the hydrogen. In this case would everything except the fluorine be oxidized?

Answer 1

You can assign oxidation states with a method similar to assigning formal charges.

1. Draw the Lewis structure.

2. Assign each atom all of its lone pair electrons. For bonding electrons assign all of the electrons to the more electronegative element.

3. For each atom perform the following calculation: number of electrons in the atom in its elemental state - assigned electrons = oxidation state.

In your example, the oxidation number of each $\ce{H}$ is +1 and of $\ce{F}$ is -1. The oxidation number of $\ce{C}$ is -2 and of $\ce{O}$ is 0.

Answer 2

To successfully assign oxidation state you should define what oxidation is. Oxidation is defined as the loss of electrons.

This, along with the fact that oxidation state is represented as a charge - i.e. something may have a "+2" oxidation state or a "-8" oxidation state, it follows that oxidation state is the charge that an atom possesses if all bonding were ionic. Ionic interactions involve the taking of electrons; covalent bonds involve the sharing of electrons.

Now, what "takes" the electrons? Clearly, the element which can best stabilize the electrons. This would be the most electronegative element.

So I would start by drawing out the Lewis structure of $\ce{CH3OF}$. Assign electrons to element based on relative electronegativities, and then think about the charges. I.e. if an element normally has 7 electrons in its valence (such as fluorine) - but is assigned 8 electrons, what do you think the charge on the fluorine will be? It has surfeit of electrons. This should suggest to you that the fluorine has a negative 1 oxidation state.