The elements present in the 3rd period comprise d-orbitals along with the s and p orbitals. The energy of the 3d orbitals is more close to the energy of 3s and 3p orbitals as well. Also, the energy of 3d orbitals is equivalent to the 4s and 4p orbitals as well. Resultantly, the hybridization including either 3d or 3d; 3s, 3p, 4s, and the 4p is feasible. Because of the difference in the energies of 4s and 3p orbitals, no hybridization including 3d, 3p, and 4s orbitals is possible.
Let us discuss the essential hybridizations, including the s, p, d orbitals below.
The excited state and ground state of outer electronic configurations for Z=15 (phosphorus) are represented below.
The sp3d hybridization can be represented as follows.
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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). The below diagram represents the same.
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It is prominent that the entire bond angles present in the trigonal bipyramidal geometry are unidentical. 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.
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.
In solid-phase, the PCl5 molecule remains as 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-, 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.
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 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 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.
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 enough polar 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.
1. 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.
2. 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.
3. 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.
4. 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.