how would i write the formula for the following compound?
a) iodine heptafluoride
i'm having trouble with this one because both iodine and fluoride have 7 valence electrons and i dont know what to do.
can someone try to at least explain, not just give me the answer? much thanks!
- William QLv 51 decade agoFavorite Answer
This compound is covalent. Iodine does not gives up any electrons to fluorine, rather it shares them with the seven fluorine atoms. The bonds are polarized, but IF7 is not ionic.
Its structure is a pentagonal bipyramid. You can see it in this link: http://en.wikipedia.org/wiki/Iodine_heptafluoride.
What happens is that Iodine can expand its octet by using d orbitals to accomodate the extra electrons coming from the fluorine. Since the electron configuration of Iodine is [Kr]4d^105s^25p^5, close to the 7 valence electrons in the 5s and 5p orbitals you have 5d orbitals, combining them with the 5s and 5p orbitals, you get 7 orbitals (hybrid), which are d^3sp^3, which adopt a pentagonal bipyramidal arrangement.Source(s): Me (University Chem Professor)
- Daniel CLv 41 decade ago
Hepta means 7 so the answer is IF7. It would be an ionic compound where the iodine loses all 7 outer electrons and each fluorine gains one electron to make a complete shell.
- 1 decade ago
Sometimes an non-metallic element can act as though it were a metal when reacting with more reactive element. Both of these elements are halides but fluorine is far more reactive so iodine assumes the role of the metal in the compound and "gives" up it's 7 shell electrons. Note that fluorine is one of the most reactive elements and will form similar compounds with the other halides.