Question

In ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{3}}}$, hybridisation of P atom is:
A.${\rm{sp}}$
B.${\rm{s}}{{\rm{p}}^2}$
C.${\rm{s}}{{\rm{p}}^3}$
D.${\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}}$

Answer 1

Hint: : We know that hybridisation can be calculated on the basis of the number of bonds which are pi or sigma bonds. By finding out the number of bond pairs and lone pairs we can easily calculate hybridization of P atoms.

Step by step answer: The ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{3}}}$ is known as orthophosphorus acid. Orthophosphorus acid is also known as phosphorus acid, phosphorane, etc. It is a white or yellow crystalline solid. Actually, phosphoric acid is a conjugate of dihydrogen phosphite.
The ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{3}}}$ constitutes of a polyatomic ion named phosphite ion in which the central atom is the phosphorus atom. This central phosphorus atom is $s{p^3}$ hybridised. So it has a tetrahedral geometry.

Hybridisation is explained as the intermixing of two atomic orbitals bearing the same energy level to degenerate a new type of orbital. The concept of hybridisation is based on quantum mechanics.
The ${\rm{s}}{{\rm{p}}^3}$ hybridisation takes place when one s and three p orbitals of the same shell intermix to generate four new equivalent orbitals. These equivalent orbitals formed are known as ${\rm{s}}{{\rm{p}}^3}$ hybrid orbitals.

There is $25\% $ s character and $75\% $ p character in each ${\rm{s}}{{\rm{p}}^3}$ hybrid orbital. The geometry of ${\rm{s}}{{\rm{p}}^3}$ hybridised orbital is tetrahedral as the orbitals are directed towards the four corners of a regular tetrahedron at an angle of \[{109.28^o}\].

The ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{3}}}$ is an oxyacid of phosphorus and all the oxyacid of the phosphorus contain the ${\rm{s}}{{\rm{p}}^3}$ hybridised phosphorus atom.
As a result there is a tetrahedral geometry around the phosphorus atom.

Hence, the correct option is C that is ${\rm{s}}{{\rm{p}}^3}$.

Note: Thus, ${{\rm{H}}_{\rm{3}}}{\rm{P}}{{\rm{O}}_{\rm{3}}}$ is a diprotic acid as the hydrogen atom is not ionisable since it is directly attached to the central phosphorus atom.