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. Explain the hybridized structure of PCl5 in detail.
PCl5 molecule has P as its central atom. The hybridization between P and Cl in PCl5 is sp3d. This is because P in its ground state has the valence electrons occupancy as two in s orbital, three in p orbitals since it has five valence electrons. But in an excited state one electron from the s orbital jumps to the d orbital which is a shell of a bit higher energy. So the hybridization involves sp3d orbitals with each of the sp3d orbital being occupied singly. The five chlorine atoms occupy these half-filled five orbitals to give the molecule an sp3d hybridized molecule with 5 P-Cl single bonds. Each occupies the five corners of the trigonal bipyramid. The shape is a trigonal bipyramid according to VSEPR theory.
2. How many and what type of bonds are formed in the hybridized PCl5 molecule?
Five chlorine atoms form five covalent bonds with the central P atom. However, the 5 P-Cl bonds are not all identical. Three of them are identical which lie on one plane and are referred to as the equatorial bonds making an angle of 120 degrees with each other. The other two P-Cl bonds are the axial bonds that remain out of the plane, one above the plane and the other below the plane making an angle of 90 degrees with the plane. The axial bonds are longer and weaker than the equatorial bonds because of the repulsion it faces from the equatorial bonds and this makes the molecule PCl5 highly reactive.
3. Comment on the dipole moment of the PCl5 molecule?
Dipole moments arise in a polar molecule due to the difference in electronegativity of the involved atoms of the molecule. In PCl5 the structure shows there are three identical P-Cl equatorial bonds each of which cancels the dipole moment of the other resulting in zero dipole moment. The other two P-Cl bonds which are axial are exactly opposite to each other, one being above the plane and the other below the plane thus completely nullifying the dipole effect. This makes the molecule PCl5 a nonpolar molecule with zero dipole moment. Apart from the geometry of the molecule, the similarity in electronegativity of P and Cl, absence of lone pairs and the lone pair on the central atom is situated exactly opposite to each other the molecule has a zero dipole moment.
4. Out of PCl3 and PCl5 which is more reactive and why?
PCl3 and PCl5 are important compounds. The geometrical structure of the two molecules makes it clear that PCl5 is more reactive than PCl3. PCl3 is pyramidal structured and hence a more stable compound while PCl5 being a trigonal pyramid one the two axial P-Cl bonds of the molecule are quite weak and long due to the constant repulsion from the equatorial bonds making the compound more susceptible to react and hence it is a more reactive molecule. PCl5 is more covalent than PCl3 and when subjected to heat it easily converts to the more stable form of PCl3.
5. What is the purpose of studying the hybridization of molecules in chemistry?
Hybridization is a very important concept in chemistry. All three major branches of chemistry depend on this concept for their advanced concepts. It is only because of hybridization we can get an idea about the energy and shape of the new molecule formed. Hybridization gives the reasoning and calculation behind the formation of the new molecule along with its geometry and properties. It is a mathematical concept in the field of chemistry which explains that all the molecules hybridize only to reach a more stable form.
6. Explain the hybridization of PCl5?
The P molecule hybridization in the PCl5 molecule is given as sp3d. Here, the P atom requires 5 orbitals to produce 5 P-Cl bonds. It contains 3s and 3 3p orbitals, so it must use one of its 3d orbitals to make the 5th bond. These orbitals are hybridized to make 5 sp3d orbitals and hence forming a trigonal bipyramid.
7. Explain the dipole moment of PCl5?
Since there is zero lone pair of electrons on the central atom, which means, P. It holds a trigonal bipyramidal structure with 3 P-Cl bonds in a place having an angle of 120° from each, which cancels each other dipole moment. Furthermore, 2 P-Cl bonds perpendicular to the plane are the opposite, thereby nullifying the effect.
8. Explain the permanent dipole moment?
This occurs when 2 atoms present in a molecule hold substantially different electronegativity, where one atom attracts electrons more than the other, becoming more negative. At the same time, the other atom becomes more positive. Hence, a molecule with a permanent dipole moment is referred to as a polar molecule.
9. Explain the zero dipole moment?
Zero dipole moment is that the particle does not contain any dipole interactions. Moreover, when placed between 2 charged plates, it does not increase the charge on any plate. A molecule holds zero dipole moment when the individual dipole moment’s vector sum adds up to zero. Symmetric molecules contain a zero dipole moment.