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This is the Lewis structure for IBr2-. Iodine, 7 valence electrons; Bromine has 7, but we
have two Bromines; and then we have this negative up here, that means we have an extra valence
electron; for a total of 22 valence electrons for IBr2-. Iodine's the least electronegative,
we'll put that right here in the center. And then the Bromines will go on the outside like
that. We'll put two valence electrons between atoms to form the chemical bonds. And then
around the outside, so we have 2, 4, 6, 16, and then we have six left over. So we're going
to put three more pairs on the central Iodine. We'll put a pair right there, a pair right
there, and then down here. So we've used all 22 valence electrons. Both of the Bromine
atoms have octets, they have 8 valence electrons. And the Iodine, it has 10, but Iodine is in
period 5 on the periodic table. That means it can hold more than eight valence electrons.
So that's it: that is the IBr2- Lewis structure. One last thing we need to do: we do need to
put brackets around to show that it is an ion. And then we'll put the negative charge
on the outside like we have here. And now we're done with the Lewis structure for IBr2-.
This is Dr. B., and thanks for watching.