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Question: The hybridization state of the central atom in \({\rm{PC}}{{\rm{l}}_{\rm{5}}}\) is: A. \(s{p^3}d\...

The hybridization state of the central atom in PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} is:
A. sp3ds{p^3}d
B. sp3d2s{p^3}{d^2}
C. sp3s{p^3}
D. d2sp3{d^2}s{p^3}

Explanation

Solution

Phosphorus pentachloride is the commonly abbreviated as PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}}. PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} has a trigonal bipyramidal structure.

Complete step by step answer:
We know that, in phosphorus pentachloride, the phosphorus is the central atom. The electronic configurations of phosphorus and chloride atoms is as follows:
Phosphorous- (1s22s22p63s23p3)\left( {{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{6}}}{\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^{\rm{3}}}} \right) and chloride- (1s22s22p63s23p5)\left( {{\rm{1}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{s}}^{\rm{2}}}{\rm{2}}{{\rm{p}}^{\rm{6}}}{\rm{3}}{{\rm{s}}^{\rm{2}}}{\rm{3}}{{\rm{p}}^5}} \right)
So, one of the “s”, three of the “p” and of the “d” orbitals participate to give a 3pz{\rm{3}}{{\rm{p}}_{\rm{z}}} hybridisation. The five orbitals of the sp3d{\rm{s}}{{\rm{p}}^{\rm{3}}}{\rm{d}} hybrid is occupies singly, so it overlaps with the 3pz{\rm{3}}{{\rm{p}}_{\rm{z}}} orbital of the chlorine atom forming five sigma P-Cl bonds. This gives rise to the trigonal bipyramidal shape of the PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} molecule having the bond angles of 90{90^\circ } and 120{120^\circ }.
We can draw the structure of PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} as follows:

                                        ![](https://www.vedantu.com/question-sets/978a3f50-856d-48b6-83a7-f43d9ce8ea603792009202129999888.png)  

The above PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} structure that we have drawn is a trigonal bipyramidal structure. Due to the trigonal bipyramidal structure, the two Cl{\rm{C}}{{\rm{l}}^ - } atoms of PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}} molecule are stretched and lie along the axis and it becomes little longer and so these bonds called as axial bonds, while the other three Cl{\rm{C}}{{\rm{l}}^ - } atoms lie on equator and these bonds are called as equatorial bonds.

So, out of the given four options, A is the correct option.

Note:
Students may get confused in interpreting the structure of the PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}}. It should be noted that in PCl5{\rm{PC}}{{\rm{l}}_{\rm{5}}}, two PCl{\rm{P}} - {\rm{Cl}} bonds remains as axial bonds along the axis and the other three PCl{\rm{P}} - {\rm{Cl}} bonds remains as equatorial bonds.