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On a test question, my organic chemistry professor gave us a really big molecule and told us to give list the formal charges on each atom. There was a $\ce{Br}$ with 3 lone pairs and a double bond to carbon (which then had a single bond to a $ \ce{H}$ and a $\ce{N}$). On the key, he labeled it with a + charge and I was wondering how is that possible (or if it is a mistake)? Shouldn't a $\ce{Br}$ with that many electrons be highly negative or does it go onto the next orbital allowing it to accommodate for this?

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  • $\begingroup$ Sounds like a typo to me. Probably should have been two lone pairs, double bond, and positive charge. $\endgroup$ – jerepierre Apr 28 '15 at 22:25
  • $\begingroup$ It is also highly unlikely, that Br can have 10 electrons. $\endgroup$ – Martin - マーチン May 13 '15 at 11:04
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I suspect that the issue here is with counting the number of lone pairs. I think the structure you are referring to looks something like the right hand one shown below.

As you can see, this structure can be described by the two resonance forms shown. The double bond arises because the bromine 'donates' one of its lone pairs to form another bond to the carbon, shifting the positive charge onto the more electronegative bromine and hence stabilising the ione. Since this is the 'more stable' resonance contributor it is usually shown as the structure of the ion in drawings but the actual structure is more accurately represented as a resonance hybrid of the two.

enter image description here

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Good question! As you state, normally bromine forms negative ions with a charge of -1, as all orbitals are complete except for the 4p5, able to accept one electron. However, there is a positive bromonium ion, first suggested by Irving Roberts and George Kimball in 1937. The generic term for a positive halogen ion is halonium ion.

The first reference has a nice electron models of this odd ion.

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