# Why do BF3 and NH3 react when B has an expanded octet in the adduct?

I know that some elements like sulfur are able to have an expanded octet due to vacant $\mathrm{d}$ orbitals. However, boron and nitrogen do not have any such vacant place so how is it that $\ce{NH3}$ with its extra electron pair is able to share it with $\ce{BF3}$ and form a compound?

This is an example of a Lewis acid-base reaction that I read in my textbook.

• Draw the Lewis structures for $\ce{BF3}$, $\ce{NH3}$, and the adduct $\ce{F3B.NH3}$. Are you really sure boron has an expanded octet in the adduct? Are you sure that boron even has a full octet in $\ce{BF3}$? – orthocresol Nov 22 '15 at 14:50