Question

In\[O{F_2}\], oxygen has hybridisation of:
A. $sp$
B. $s{p^2}$
C. $s{p^3}$
D. $s{p^3}d$

Answer 1

Hint: To find out the hybridisation of any molecules, first we need to find out the central atom and find out the number of valence shell electrons of the central atom. In the next step, find out the steric number and there is specific hybridisation for each steric number.

Complete step by step answer:
The central atom given in this molecule of \[O{F_2}\] is oxygen and oxygen has eight valence shell electrons. The two substituent atoms attached to the oxygen central atoms are fluorine. The fluorine is a monovalent atom and it shares one electron with the oxygen to complete its octet. So, two fluorine atoms will take two electrons from the oxygen from its eight electrons. The remaining four electrons will exist as two lone pairs. Therefore, two bond pairs and two lone pairs gives the steric number four. If the steric number is four, the hybridisation is $s{p^3}$.

So, the correct answer is Option C .

Additional Information:
The hybridisation for the steric numbers 2, 3, 4, 5, 6 are $sp$,$s{p^2}$,$s{p^3}$, $s{p^3}d$ and $s{p^3}{d^2}$ respectively and the shapes corresponding to them are linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral.

Note:
There is another method to find out the steric number using formula. The formula for hybridisation is $\dfrac{1}{2}\left[ {V + M - C + A} \right]$ where V is the number of valence electrons of central atom, M is number of monovalent atom, C is the total cation charge and A is the total anion charge.