Is #"BI"_3# polar?

1 Answer
Truong-Son N.
Jan 9, 2018

Well, boron starts out with three valence electrons:

#cdot dot"B" cdot#

and each iodine atom starts out with seven:

#: ddot"I"cdot#
#""^(color(white)("...")"..")#

So hypothetically, #B# would interact with #3 xx I#... and one #B# could contribute #3# valence electrons to the structure...

But electrons are NOT transferred from boron to iodine. They are shared, but partially skewed towards iodine, in a slightly polar covalent bond within this nonpolar molecule:

https://upload.wikimedia.org/

The Pauling electronegativity of #"I"# is #2.5#, and that for #"B"# is #2.0#, so the difference is only #0.5#, meaning it is considered slightly polar, i.e. the valence electrons are almost perfectly evenly shared, but are pulled on more by iodine atom.

(Due to the symmetry of the molecule, it turns out that this molecule is still nonpolar.)

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