lewis acid

[mundane]

So, i just learned that because of backbonding in BF3, BF3 is a weak acid. This is due to the fact that B orbital if filled internally with F electrons. Now, if we bring  NH3 to make a coordinate bond with BF3, it does. Why? Wasn't the vacant orbital of B already filled during backbonding? I am confused. 

[exchemist]

8 minutes ago, mundane said:

So, i just learned that because of backbonding in BF3, BF3 is a weak acid. This is due to the fact that B orbital if filled internally with F electrons. Now, if we bring  NH3 to make a coordinate bond with BF3, it does. Why? Wasn't the vacant orbital of B already filled during backbonding? I am confused. 

As so often in chemistry, it's not a black and white situation. You don't have either "a bond" or "no bond", but bonds, sometimes "partial" in character,  that can vary in strength and completeness. BF3 is indeed the weakest Lewis acid of the various trihalides, which is attributed to F providing more effective  π overlap than its congeners. This would be because it would involve 2p subshells on both atoms, which are of similar size: overlap between orbitals of different shells is commonly not so effective.

But there is still a gain in stability in going from BF3 with 3 partial  π-bonds to 4 full  σ-bonds in an sp3 hybridised adduct with NH3. In part this will be because the back-donation from F requires a degree of polarisation against the electronegativity of the atoms, i.e. with a δ- on B and a   δ+ on F. So the extra stability from the extra bonding won't be that great.  

The above is all a bit handwavy, I know. There is a more advanced discussion of this issue using Molecular Orbital theory here: https://chemistry.stackexchange.com/questions/80247/molecular-orbital-diagram-for-bf3. which is really a more proper way to analyse it. (I'm afraid I've forgotten my Group Theory, so I can't guide you all the way through this.)  But you will see there is a low-lying antibonding LUMO (Lowest Unoccupied Molecular Orbital) into which a pair of electrons can be accepted, at the expense only of reversing out the π-bonding contribution. 

You will see from that that it is a quite a complicated story, so your question is by no means a trivial one.