The electronic structure of molecule \[{\rm{O}}{{\rm{F}}_{\rm{2}}}\] is a hybrid of
A) \[sp\]
B) \[s{p^2}\]
C) \[s{p^3}\]
D) \[s{d^3}\]

Hint: The electronic structure of a molecule can be found out if the central atom of the molecule's hybridization is known. Here, we will find Oxygen atom's hybridization.

Formula used:
\[H = \dfrac{{V + X - C + A}}{2}\]; Here, H is for hybridization, V is count of valence electrons, X is the count of monovalent groups bonded to the central atom, C stands for cationic charge and A stands for charge of anion.

Complete Step by Step Answer:
In the \[{\rm{O}}{{\rm{F}}_{\rm{2}}}\] molecule, oxygen is the central atom. The valence electrons of O atom=6, count of monovalent atoms of the molecule =2
So, the hybridization of O is,
\[H = \dfrac{{6 + 2}}{2} = 4\]

As the value of hybridization is 4, the central atom, that is, oxygen is \[s{p^3}\] hybridised.
Hence, option C is right.

Additional Information: The hybridization value of 3 indicates the central atom nature of \[s{p^2}\]hybridization and the molecular shape is trigonal planar. And the hybridization value of 2 indicates the sp hybridised central atom and of the molecular shape linear.

Note: The short trick to identify hybridization of a molecule is to count the surrounding groups of the central atom. The surrounding groups include the atoms hat form bonds with the central atom and the lone pairs. If we follow this rule, in the molecule of \[{\rm{O}}{{\rm{F}}_{\rm{2}}}\], two fluorine atoms form bonds with the O atom and there are two lone pairs in the oxygen atom. So, there are four groups surrounding the central atom. So, the hybridization of \[{\rm{O}}{{\rm{F}}_{\rm{2}}}\]molecule is \[s{p^3}\].