Introduction to PCl₅ Hybridization
Hybridization is the mixing of atomic orbitals to form a new orbital. The main three types of hybridization possible involving only s and p orbitals are sp3, sp2, sp. Sp3 results in single bonds, sp2 for double bonds, and sp for triple bonds. The new hybridized orbital has properties and energy which is almost an average of the unhybridized orbitals which took part in hybridization. Hybridization of elements of the third period includes the involvement of d orbitals which otherwise remain vacant in Phosphorus. Phosphorus is an element of the third period and is a p-block element. Phosphorus has five valence electrons which can take part in hybridization with one electron taken by each Chlorine atom and the resultant hybridization for PCl5 is thus sp3d. Chlorine is also a p-block element with vacant d orbitals and has seven valence electrons with a tendency to take one electron to fill its valence shell. The sp3d hybridization results in trigonal bipyramidal shape following the VSEPR theory with 5 chlorine atoms occupying the five corners of the trigonal bipyramid. PCl5 is a useful compound with chlorinating properties.
Let us discuss the essential hybridizations, including the s, p, d orbitals below.
Important Hybridizations
Formation of PCl5
The excited state and ground state of outer electron configurations for Z=15 (phosphorus) are represented below.
The sp3d hybridization can be represented as follows.
The five orbitals, viz 1s, 3p, and 1d orbitals, are free for the hybridization process. Thus, it can obtain a 5sp3d hybrid orbital set, which is directed to the 5 corners of a trigonal bipyramidal (according to the VSEPR theory).
It is prominent that the entire bond angles present in the trigonal bipyramidal geometry are identical. The 5sp3d orbitals present in the PCl5 of phosphorus overlap with the p chlorine atom’s orbitals, where the p orbitals are singly occupied. They form 5 P–Cl sigma bonds together.
Types of Bonds Formed During PCl5 Hybridization
Axial Bonds: 2 P–Cl bonds at which one lies above the equatorial plane and the other bond below the plane to form an angle with the plane. The angle made with the plane is given as 90°.
Equatorial Bonds: 3 P–Cl bond at which lies in one plane to form an angle with each other. The angle made between them is given as 120°.
Because the axial bond pairs agonize massive repulsive interaction from the equatorial bond pairs, the axial bonds tend to be a bit longer. Thus, it makes it a bit weaker than the equatorial bonds, resulting in a more reactive PCl5 molecule.
Hybridization of PCl5 Central Atom
In solid-phase, the PCl5 molecule remains as an ion pair of PCl4+ and PCl6-. Whereas, in PCl4+, the central atom P contains 4 bonding electron pairs and zero lone pair electrons. Thus, it can be given as sp3 hybridized, and PCl4+ is represented in tetrahedral in shape. Moreover, in PCl6-, the P atom contains 6 bonding electron pairs and zero lone pair electrons, and PCl6- becomes octahedral in shape.
In the gaseous phase, this compound remains as PCl5. Here, the central atom P is bonded to 5 Cl atoms via 5 sigma bonds, and there is zero lone electron pair. Thus, in the gaseous phase, P can be given as PCl5 is sp3d hybridized, and the shape of the molecule is trigonal bipyramidal.
Calculation of Hybridization
One of the best and easiest methods to calculate the hybridization is to count the surrounding atom leaving the primary atom. For example, in the NH4+ compound, N is the primary atom, and there are 4 surrounding atoms. So, according to the formula,
H = SA+½ (G-V+E for negative charge and -E for a positive charge )
Where H is the hybridization,
SA is the surrounding atom,
G is the valence electron for the primary atom,
V is the valency of all surrounding atoms,
E is the number of charges.
In NH4+ compound, H is 4+ ½ (5-4-1) = 4+0.
Where 4 is the bond pair, 0 is the lone pair = 4.
Therefore, here, the hybridization is sp3, and the shape is tetrahedral. We have to count the full primary atom, simple valency, and atom valence electron in V for all surrounding atoms in this hybridization. For example, in the iodine compound, G is 7, whereas V is 1. But, in the NO2+ compound, H = 2+ ½ (5-4-1) = 2. It is also sp hybridized and linear. In the same way, we can try for more hybridization.
Polarity of PCl5
PCl5 is polar. The structure is not covalent, but it is ionic. It means PCl4+/Cl-. Phosphorus trichloride (PCl3) is a significantly less polar liquid, with a boiling point of 73°. If we are likely to form an acid chloride (for suppose, p-chlorobenzene acid with a melting point of 242°), we had better not try heating either with thionyl chloride (SOCl2) or PCl3. It is because they are not polar enough to dissolve the high-melting crystals even at their respective boiling points. To make that acid chloride, we need to melt the PCl5 compound and the chlorobenzene acid together ~170°, and the conversion will become smooth on continued heating.
FAQs on PCl₅ Hybridization
1. What is the hybridization of the central Phosphorus atom in PCl₅?
The central Phosphorus (P) atom in Phosphorus pentachloride (PCl₅) has sp³d hybridization. This means one 's' orbital, three 'p' orbitals, and one 'd' orbital from the phosphorus atom mix together to form five new, identical hybrid orbitals that are used for bonding.
2. What is the molecular geometry and shape of PCl₅ as a result of its hybridization?
Due to its sp³d hybridization, PCl₅ adopts a trigonal bipyramidal geometry. This shape consists of three chlorine atoms forming a flat triangle around the central phosphorus atom (equatorial positions) and two other chlorine atoms positioned directly above and below this triangle (axial positions).
3. What are the different bond angles in a PCl₅ molecule?
The PCl₅ molecule has two distinct bond angles because of its shape:
- The angle between the three equatorial P-Cl bonds is 120°.
- The angle between an axial P-Cl bond and any of the equatorial P-Cl bonds is 90°.
4. How does the orbital promotion in Phosphorus lead to sp³d hybridization?
In its ground state, Phosphorus has three unpaired electrons. To form five bonds in PCl₅, an electron from its filled 3s orbital is promoted to an empty 3d orbital. This creates five unpaired electrons. These five orbitals (one 3s, three 3p, one 3d) then mix to form five stable sp³d hybrid orbitals, ready for bonding.
5. Why are the axial bonds in PCl₅ longer and weaker than the equatorial bonds?
The axial bonds experience greater electron-pair repulsion compared to the equatorial bonds. Each axial bond is repelled by three equatorial bonds at a 90° angle. In contrast, equatorial bonds face less repulsion from other bonds. This stronger repulsion pushes the axial bonds further away from the center, making them longer and consequently weaker.
6. If the P-Cl bonds are polar, why does the PCl₅ molecule have an overall zero dipole moment?
Although each P-Cl bond is polar, the molecule's perfectly symmetrical trigonal bipyramidal shape causes the individual bond dipoles to cancel each other out. The three equatorial dipoles cancel out in their plane, and the two axial dipoles point in opposite directions and also cancel. This results in a net dipole moment of zero, making PCl₅ a nonpolar molecule.
7. Why is PCl₅ generally more reactive than PCl₃?
PCl₅ is more reactive because its two axial bonds are relatively unstable and easy to break. Due to this inherent instability, PCl₅ readily decomposes upon heating to form the more stable PCl₃ and chlorine gas (Cl₂). This tendency to lose its axial chlorines makes it a highly effective chlorinating agent and thus more reactive.
8. Why can Phosphorus form PCl₅ but Nitrogen, which is in the same group, cannot form NCl₅?
The key difference is the availability of d-orbitals. Phosphorus is in the third period and has empty d-orbitals. It can expand its valence shell by promoting an electron to a d-orbital to form five bonds. Nitrogen, being in the second period, lacks d-orbitals and cannot expand its octet beyond eight electrons, so it can only form a maximum of three bonds, as in NCl₃.
