Based on electron affinity, predict whether $B{F_3}$ is ionic or covalent.

Hint: We know that ionic compounds are generally formed when a metal reacts with a non-metal. In these types of bonds the metal donates an electron and the non-metal accepts the donated electrons. In covalent compounds, covalent bonds are present which are formed when the electrons are shared between two atoms which have a negligible electronegativity difference. To answer this question, we will see the nature of the atoms present in the molecule and also the nature of the bond between them.

Complete answer:
In $ B{F_3}$ , we know that the atomic number of Boron is $5$ and three electrons are present in its outermost shell. Fluorine (F) is halogen and its atomic number is $9$ and seven electrons are present in its outermost shell. The formation of boron trifluoride is because of the presence of nine electrons in fluorine.
The electronic configuration of Boron is:
$1{s^2}2{s^2}2{p^1}$
This is because the boron is an electron deficient atom and it will easily attack the fluorine to complete its octet and become stable (the total electrons present in the outermost shell is eight).
So, the three electrons in the outermost shell of boron form three bonds with the fluorine atoms and share the three valence electrons equally among themselves to complete the octet. So now, a total of six electrons are present in the valence shell orbital of boron.
Therefore, we can see that the electrons are being shared between the boron and fluorine atoms. So, $ B{F_3}$ is a covalent compound.

Note:
In Boron trifluoride, boron has six electrons in its valence shell and hence, we can say that it is an electron deficient compound. It is a Lewis acid because it can accept the incoming electrons from a nucleophile to complete its octet. The geometry of $ B{F_3}$ is trigonal planar. It has a pungent smell and it is a toxic colourless gas.