Question

Which of the following states of hybridization of the central metal atom are not the same as in the others?
(A) $B$ in $B{F_3}$
(B) $O$ in ${H_3}{O^ + }$
(C) $N$ in $N{H_3}$
(D) $P$ in $PC{l_3}$

Answer 1

The hybridization will be determined by the steric number. The steric number is the sum of the bonding pair of electrons and the lone pair of electrons. if the steric number is two, it is $sp$ hybridization. If the steric number is three, it is $s{p^2}$ hybridization. If the steric number is four, it will be $s{p^3}$ with tetrahedral or pyramidal geometry.

Complete answer:
Let us consider $B{F_3}$ , boron is the central metal atom with three valence electrons. These three valence electrons are involved in bonding with three fluorine atoms and form $s{p^2}$ hybridization with trigonal planar geometry.
Let us consider ${H_3}{O^ + }$ , oxygen is the central metal atom with six valence electrons. These six valence electrons as the positive charge are there, the valence electrons will be five. These five valence electrons are involved in bonding with three hydrogen atoms and the remaining two electrons exist as lone pairs forming $s{p^3}$ hybridization with pyramidal geometry.
Let us consider $N{H_3}$ , nitrogen has five valence electrons. Out of these five valence electrons, three are bonding and the remaining two are lone pairs of electrons. Thus, the hybridization is $s{p^3}$ with pyramidal geometry.
Let us consider $PC{l_3}$ , phosphorus has five valence electrons. Out of these five valence electrons, three are bonding and the remaining two are lone pairs of electrons. Thus, the hybridization is $s{p^3}$ with pyramidal geometry.
Thus, $B$ in $B{F_3}$ have a state of hybridization of the central metal atom, not the same as in the others.
Option A is the correct one.

Note:
When the electron geometry is tetrahedral with one lone pair of electrons, the molecular geometry will be pyramidal. When the electron geometry is tetrahedral with two lone pairs of electrons, the molecular geometry will be bent.