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To determine the Lewis structure for HOCO, I first counted the electrons to get 17. Then, I linked single bonds so the structure looked like H-O-C-O. I added a double bond between C and the second O and then filled in 2 lone pairs for each oxygen. At this point, 16 electrons have been filled in and I put the last one on C. This makes all the formal charges zero and also shows that HOCO is radical because of the unpaired electron on C.
Yes, I agree with the above reply. I also did the same thing. If you are wondering why the unpaired electron would go on the Carbon, there are two main reasons. One is that by giving Oxygen two bonded pairs and two unbonded pairs, both Oxygens get a FC of zero, which is ideal. Second, Carbon is less electronegative than Oxygen and therefore would be the atom most likely to have an unpaired electron.
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