Question

The type of hybridization of P atom in ${\mathrm{PCl}}_5$, ${{\mathrm{PCl}}_4}_{}^{+}$, ${{\mathrm{PCl}}_6}_{}^{-}$ is

Answer 1

Hybridization

• The mixing of two atomic orbitals to form hybrid orbitals is called hybridization
• For eg:- s and p orbitals combine to form sp hybrid orbital having characteristics of both s and p orbital
• Atomic orbitals that are to be mixed to form hybrid orbitals should have comparable energies
• Steric number = No of sigma bonds formed by central atom + No of lone pairs around the central atom
• For steric number 1, the hybridization is sp and geometry is linear

In case of ${\mathrm{PCl}}_5$:-

• The number of sigma bonds formed by the phosphorus atom is five and no lone pair is attached to the central atom so the steric number of ${\mathrm{PCl}}_5$ is 5
• For steric number 5, the type of hybridization is ${\mathrm{sp}}^3\mathrm{d}$
• Therefore the type of hybridization of the P atom in ${\mathrm{PCl}}_5$ is ${\mathrm{sp}}^3\mathrm{d}$

In case of ${{\mathrm{PCl}}_4}_{}^{+}$:-

• The number of sigma bonds formed by the phosphorus atom is four and no lone pair is attached to the central atom so the steric number of ${{\mathrm{PCl}}_4}_{}^{+}$ is 4
• For steric number 4, the type of hybridization is ${\mathrm{sp}}^3$
• Therefore the type of hybridization of the P atom in ${{\mathrm{PCl}}_4}_{}^{+}$ is ${\mathrm{sp}}^3$.

In case of ${{\mathrm{PCl}}_6}_{}^{-}$:-

• The number of sigma bonds formed by the phosphorus atom is six and no lone pair is attached to the central atom so the steric number of ${{\mathrm{PCl}}_6}_{}^{-}$ is 6
• For steric number 6, the type of hybridization is ${\mathrm{sp}}^3{\mathrm{d}}^2$
• Therefore the type of hybridization of the P atom in ${{\mathrm{PCl}}_6}_{}^{-}$ is ${\mathrm{sp}}^3{\mathrm{d}}^2$

Therefore the type of hybridization of P atom in ${\mathrm{PCl}}_5$, ${{\mathrm{PCl}}_4}_{}^{+}$and ${{\mathrm{PCl}}_6}_{}^{-}$ is ${\mathrm{sp}}^3\mathrm{d}$, ${\mathrm{sp}}^3$ and ${\mathrm{sp}}^3{\mathrm{d}}^2$.