Re: Why is the compound BF3 nonpolar instead of ionic?

It’s important to distinguish between a polar BOND and a polar MOLECULE.  
Since boron and fluorine have different electronegativities,  a BOND 
between B and F will be polar.

Whether or not a MOLECULE is polar depends on at least two factors:  1) 
whether there are polar bonds in the molecule; 2) the geometry of the 
molecule.  In order for a molecule to be polar, the first factor is a 
necessary, but not sufficient, condition.  That is, there must be polar 
bonds present for a molecule to be polar, but their presence doesn’t 
guarantee a polar molecule.

So, boron trifluoride meets the first condition--it might be polar, because 
it contains polar bonds.  Now we need to look at the second condition, 
molecular geometry.  Predictions of geometry are often made on the basis of 
a theory called Valence-Shell Electron-Pair Repulsion Theory (VSEPR), which 
basically states that pairs of electrons, whether present in bonds or as 
unshared electron pairs, will choose a geometry around a central atom that 
will minimize the repulsion they feel for each other because of their 
negative charges.  That is, they will get as far away from each other as 
possible.

Since the neutral boron atom has only three valence electrons, when it is 
bonded to three fluorine atoms it shares one of these electrons with each 
of the fluorines (and each fluorine contributes its odd seventh valence 
electron to the bond).  This uses up all of boron’s valence electrons, so 
there are no unshared pairs of electrons on boron in boron trifluoride.  
Therefore, there are three pairs of electrons around the boron (in the 
three bonds), and in order for them to be as far away from each other as 
possible, they will form a trigonal planar geometry:  all four atoms are in 
a plane, and the angle between pairs of adjacent bonds is 120°.

Hence, boron is surrounded by three equally spaced fluorine atoms, all with 
a higher electronegativity than boron.  Since the electrons are pulled away 
from boron equally in all directions by these fluorines, the net result is 
that they are pulled nowhere--that is, there is no net separation of charge 
in the molecule, and it is not polar.  Another way of looking at this is to 
say that there is no negative or positive “end” to the molecule. 

There are many sites on the Internet with information about VSEPR and 
molecular geometries.  One you might like to look at is

http://www.shef.ac.uk/~chem/vsepr/index-chime-false.html

It even includes specific information about boron trifluoride.