Question

Which of the following molecules is formed by \[\;s{p^2}\] hybridized orbital?
A. \[\;C{H_4}\]
B. \[B{F^3}\]
C. \[\;\;{O_2}\]
D. \[\;Be{F_2}\]

Answer 1

Hint: in the given question \[\;s{p^2}\] hybridized orbital of a molecule is discussed. \[\;s{p^2}\] hybrid molecule consists of two \[p\]orbital and one\[\;s\] orbital. Before going for the answer we should know the exact hybridization states of all the compounds in the given options.

Complete Step by step answer: The \[\;s{p^2}\] hybridization is the combination of one s and two p atomic orbital which involves the promotion of one electron in the s orbital to one of the \[2p\] atomic orbital. The combination of these two atomic orbitals creates three new hybrid orbitals which are equal in energy-level.
The electronic configuration of boron is \[1{s^2}2{s^2} and {\text{ }}2{p^1}\] in the excited state it has three electrons and becomes unpaired and undergoes a \[\;s{p^2}\] hybridization with fluorine.
One out of the three boron electrons are unpaired into the ground state with the configuration 1s2 2s2 and 2p1. In the excited state, one electron from the \[2s\] orbital hop to the \[\;2p\]orbital, and then the three \[\;2p\] orbitals and one\[2s\] orbital combine and give rise to four\[\;s{p^2}\] hybridized orbitals, one of which is unused in \[B{F^3}\].
In the ground state of boron has two electrons in \[s\] orbital and one electron in \[\;p\] orbital, in excited state thus one electron jumps from \[s\]to \[p\] orbital and thus we get three atomic orbitals for bonding which have unpaired electrons.
\[{1^s}\]and\[\;{2^p}\] Thus \[B{F^3}\] forms three \[\;s{p^2}\] hybridized orbitals.

Hence the correct option is B.

Note: The \[\;s{p^2}\] hybridized orbitals are of the same in size, same energy shape but are in the different spatial orientation and this unique orientation is very important and that is what can characterize a \[\;s{p^2}\] hybridized orbital from other hybridized orbitals.