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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?

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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.

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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.

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  • $\begingroup$ okay so if I start with methanol(CH3OH) and end with CH3OF(Methoxy Flouride I think is what it is) it occurs in 2 steps. Carbon starts off in a -2 oxidation state and O in a -2 oxidation state. The O gets deprotonated by a hydroxide anion. The oxidation state does not change at either the O or the C. Fluorination of the alkoxide oxidizes the O to a -1 oxidation state. The C does not change. It doesn't get reduced or oxidized. $\endgroup$ – Caters Aug 16 '14 at 15:07
  • $\begingroup$ of course the deprotonation could happen at the C and give you a carbanion which then gets oxidized by the flourine and the oxygen does not change in either the bonding or the oxidation state in this case. $\endgroup$ – Caters Aug 16 '14 at 15:12
  • $\begingroup$ and the difference between fluorination and fluoridation is that fluoridation means you use the fluoride anion whereas fluorination means you use some source of fluorine which could be the fluoride anion or a compound with fluorine in it. $\endgroup$ – Caters Aug 16 '14 at 15:17

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