Question

What is the hybridization of the central atom in the following? \[P{{H}_{3}},PH_{4}^{+}\]

Answer 1

Hint: In this question, we will use the concept of hybridization. Hybridization is a method of combining two or more atomic orbitals to form an entirely new orbital different from its components. The bond angle decreases due to the presence of lone pairs of electrons and increases along with electronegativity.

Complete step by step answer:
There are certain molecules in which there is no hybridization occurring. This is known as Drago's rule. According to Drago’s rule, for hybridization the molecule must follow these conditions, the central atom should have at least one lone pair. The electronegativity of a terminal atom should be less than carbon. The central atom should belong to the third or higher period. Instead of hybridization, these atoms involve pure p orbitals in bond formation.
In, $P{{H}_{3}}$ nitrogen has a lone pair and it forms three bonds with a hydrogen atom. Hence the hybridization of ammonium molecules is $s{{p}^{3}}$. Similarly, phosphorus has one lone pair and forms three bonds with hydrogen atoms in $P{{H}_{3}}$ yet it does not show hybridization as it obeys all the conditions of Drago’s rule. And hence the bond angle of phosphine is not the same as that of ammonia. The bond angle of phosphine is ${{93.5}^{o}}$
Hybridizations is the concept of mixing atomic orbitals into new hybrid orbitals (with different energies, shapes, etc.) suitable for the pairing of electrons to form chemical bonds in valence bond theory. Lattice energy is usually the largest factor in determining the stability of an ionic solid. The extra energy gained by the lattice energy more than compensates for the energy needed to transfer a chloride ion from one. Here the cation is \[PH_{4}^{+}\]
So, here ‘P’ has five valence electrons. In the cation part it has $+1$ charge on ‘P’, so it has only four electrons in its valence shell; it has four electrons and these four electrons get paired with four chlorine atoms. And it has zero lone pair. So, the hybridization of \[PH_{4}^{+}\] is $s{{p}^{3}}$

Note: Phosphine is a colourless, flammable and toxic gas having a rotten fish-like smell. Ammonia is a colourless gas with a pungent smell. The prominent use of ammonia is in the preparation of fertilizers. Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space.