Question

The hybridization in $ P{F_3} $ is:
(A) $ s{p^3} $
(B) $ s{p^2} $
(C) $ ds{p^3} $
(D) $ {d^2}s{p^3} $

Answer 1

Hint: In Chemistry , hybridization or the orbital hybridization is defined as the concept of mixing or merging atomic orbitals to form a new hybrid orbital which is also known as the hybridized orbital. With the help of electronic configuration, valency and the presence of lone pairs on the central atom we can predict the hybridization of any molecule.

Complete step by step answer:
So now we are somewhat aware of the term hybridization. Now to predict or figure out the hybridization of $ P{F_3} $ we will study the chemistry of the molecule. The central atom in $ P{F_3} $ molecule is phosphorus. Now we know that the electronic configuration phosphorus is $ P \to [Ne]3{s^2}3{p^3} $ . The electronic configuration of the central atom helps us to find the number of valence electrons. So here according to the electronic configuration of phosphorus we can observe that there are five valence electrons. So five electrons will participate in hybridization of $ P{F_3} $ . Now we will study the chemical structure of $ P{F_3} $ .
Now from the structure we can observe that out of the five valence electrons of phosphorus. The three electrons are bonded with fluorine and the rest of the two electrons form a lone pair. Now for hybridization we count the sum of the total number of atoms bonded and the lone pair on the central atom. So, we have three bonded atoms and one lone pair so the total is four. Now if we check the option the sum of the orbital is four in $ s{p^3} $ which means $ 1s $ and $ 3p $ .

Therefore, the correct option is (A).

Note:
The hybridization helps in predicting the geometry or the shape of the molecule. If the hybridization of a molecule is $ s{p^3} $ then the shape of the molecule is tetrahedral. The shape of $ P{F_3} $ is tetrahedral.