The molecule PCl5 has a trigonal bipyramidal structure; what is the hybridisation state of the central phosphorus atom in this molecule?

Introduction / Context:
This question deals with valence bond theory and hybridisation, concepts used to explain the shapes of molecules. Phosphorus pentachloride, PCl5, is a classic example of a molecule with trigonal bipyramidal geometry. The question asks for the hybridisation of the central phosphorus atom that leads to this specific shape. Understanding how different hybrid orbitals correspond to geometries such as tetrahedral, trigonal bipyramidal and octahedral is a fundamental skill in chemical bonding and molecular structure topics.

Given Data / Assumptions:
PCl5 has five P–Cl sigma bonds and no lone pairs on the central phosphorus atom in the valence shell model.
The molecular geometry of PCl5 is trigonal bipyramidal, which means three chlorine atoms occupy equatorial positions and two occupy axial positions.
Hybridisation schemes such as sp2, sp3, dsp2 and dsp3 are common ways of describing the mixing of atomic orbitals to form equivalent hybrid orbitals.
We assume standard valence shell electron pair repulsion and valence bond descriptions without invoking advanced molecular orbital theory.

Concept / Approach:
In valence bond theory, the hybridisation state is chosen to match the number of electron pairs and the observed geometry around the central atom. A trigonal bipyramidal geometry requires five electron domains arranged to minimise repulsions, corresponding to five hybrid orbitals. The notation traditionally used for this is dsp3, indicating that one d orbital, one s orbital and three p orbitals combine to form five equivalent hybrid orbitals. These hybrids are directed toward the five corners of a trigonal bipyramid. Other hybridisation schemes such as sp3 (four hybrids) or sp2 (three hybrids) cannot accommodate five equivalent bonding directions.

Step-by-Step Solution:
Step 1: Count the sigma bonds and lone pairs on phosphorus in PCl5. Phosphorus forms five sigma bonds with five chlorine atoms and has no lone pairs in the idealised structure. Step 2: This means phosphorus needs five equivalent hybrid orbitals to form these five sigma bonds. Step 3: Consider possible hybridisations. sp2 hybridisation gives three hybrid orbitals, suitable for trigonal planar geometry. sp3 hybridisation gives four hybrid orbitals, suitable for tetrahedral geometry. Step 4: To obtain five hybrid orbitals, one must combine one s orbital, three p orbitals and one d orbital, producing a dsp3 or sp3d set, which corresponds to trigonal bipyramidal geometry. Step 5: Therefore the correct hybridisation description for the central phosphorus atom in PCl5 within this valence bond framework is dsp3.

Verification / Alternative check:
An alternative way to check this is to match known hybridisation geometries: sp corresponds to linear (2 directions), sp2 to trigonal planar (3 directions), sp3 to tetrahedral (4 directions), dsp3 to trigonal bipyramidal (5 directions) and d2sp3 to octahedral (6 directions). Since PCl5 is experimentally known to have a trigonal bipyramidal shape and five bonded chlorine atoms, its central atom must use a hybridisation that generates five bonding directions, which is dsp3 in the traditional scheme. This pattern appears consistently in textbooks, confirming that dsp3 is the standard answer for PCl5.

Why Other Options Are Wrong:
sp2 is incorrect because it provides only three hybrid orbitals and produces a trigonal planar geometry, which cannot account for five P–Cl bonds in PCl5.
sp3 is wrong because it yields four hybrid orbitals and gives a tetrahedral geometry, again insufficient for bonding to five chlorine atoms.
dsp2 is associated with square planar geometry and four coordination sites, so it also cannot describe the trigonal bipyramidal structure of PCl5, making option C incorrect.

Common Pitfalls:
Students sometimes confuse dsp3 with d2sp3 or assume that any involvement of d orbitals automatically leads to octahedral geometry. Others incorrectly try to fit PCl5 into an sp3 framework because they remember that four electron pairs correspond to sp3 hybridisation but forget that five pairs require a different arrangement. Another common slip is ignoring the actual shape of the molecule and focusing only on the total number of bonds. Remembering the mapping between electron pair number, hybridisation and geometry is essential to avoid these mistakes.

Final Answer:
For PCl5 with trigonal bipyramidal geometry, the central phosphorus atom is described as having dsp3 hybridisation.