Textbook Question 2.61

[Adam_Ventura_1H]

After calculating that the compound must have 17e-, I expected the compound to be a radical, but my question is why would the 1 lone e- go to the C even though there is no central atom, why would it not g to one of the O. I know it is usually the carbon that does not fill up its octet in this situations, but why?

Question: Methane, the most abundant hydrocarbon in the atmosphere and a potent greenhouse gas, is slowly oxidized in the air to carbon dioxide. An intermediate in the oxidation of methane to carbon dioxide is HOCO. (a) Draw the Lewis structure for this compound. (b) Decide whether the compound is a radical.

[Isabel_Eslabon_2G]

I am not sure if this is correct, but I tried looking at alternate structures where the lone electron is not on the C. The structure where the lone electron is on C is the favored structure because that structure has all atoms with a formal charge of 0. If the lone electron is on an oxygen, the formal charges are no longer 0 (formal charges are C=-1 & O=+1).

[Michael Iter 2F]

I also think electronegativity plays a role here. Since oxygen is more electronegative than carbon it is more likely to fill its octet

[emilyyribarren1k]

I'm not sure if this is true all the time, but when I was deciding where to put the unpaired e- I looked for which would have the lowest formal charge. When you put it on the C, each of the atoms has a FC of 0. If you put it on one of the O and gave C the complete lone pair, the C would have an FC of -1 and the O would have an FC +1. Therefore, it is more favorable for the unpaired electron to be placed on the C atom in this case.