Question

Which among the following qualifies as a Lewis acid?
A. $NaF$
B. $NaCl$
C. $B{F_3}$
D. $MgC{l_2}$

Answer 1

For a compound to behave as a Lewis acid, it must satisfy two basic conditions i.e., either the compound exists in the form of cations or an atom, ion or molecule is present with an incomplete octet of electrons. The cations or molecules with incomplete octets have a tendency to accept electrons and thus act as a Lewis acid.

Complete Step By Step Answer:
Acids and bases are an important part of chemistry and the Lewis acid-base theory is one of the most applicable theories which explains the concept of an acid and a base as follows:
Lewis Acid: A species which accepts an electron pair, i.e., it acts as an electrophile. It may either have a positive charge or vacant orbitals.
Lewis Base: A species which donates an electron pair, i.e., it acts as a nucleophile. It may either have a negative charge or lone pair of electrons.
The reaction of a Lewis acid and a Lewis base will produce a salt which is bonded via coordinate covalent bond. The resulting product is known as an addition compound or more commonly known as an adduct.
Now, among given options let us check the property of each compound as follows:
Compound-A: $NaF$
The formation of $NaF$ takes place when sodium ion (Lewis acid) reacts with fluoride ion (Lewis base). The reaction is shown as follows:
$N{a^ + } + {F^ - } \to NaF$
Thus, it is an adduct.
Compound-B: $NaCl$
The formation of $NaCl$ takes place when sodium ion (Lewis acid) reacts with chloride ion (Lewis base). The reaction is shown as follows:
$N{a^ + } + C{l^ - } \to NaCl$
Thus, it is an adduct.
Compound-C: $B{F_3}$
The electronic configuration of boron is $1{s^2}2{s^2}2{p^1}$ and after forming bond with three fluorine atoms, it only has six electrons in the outermost shell and thus its octet is incomplete. Due to the presence of vacant p-orbitals, it has a tendency to accept electron pairs and thus, $B{F_3}$ molecule acts as a Lewis acid.
Compound-D: $MgC{l_2}$
The formation of $MgC{l_2}$ takes place when magnesium ion (Lewis acid) reacts with chloride ion (Lewis base). The reaction is shown as follows:
$M{g^{2 + }} + 2C{l^ - } \to MgC{l_2}$
Thus, it is an adduct.
Hence, among given options, the compound which qualifies as a Lewis acid is $B{F_3}$ .
So, option (C) is the correct answer.

Note:
It is important to note that the molecules in which the central atom have more than eight valence electrons or the molecules which have multiple bonds between two atoms of different electronegativities are also classified as the Lewis acids. For example: $SiB{r_4}$ , $C{O_2}$ , etc.