Question

According to VSEPR theory, what is the shape of the \[P{H_3}\] molecule best described as?

Answer 1

Hint: The phosphine molecule with the formula \[P{H_3}\] is very similar to the molecule of ammonia \[N{H_3}\] . Both these molecules have a central atom that belongs to the fifteenth group and the same electronic configuration but the size difference between phosphorus and nitrogen is an important factor.

Complete answer:
The valence shell electron pair repulsion (VSEPR) theory helps in predicting the shapes and geometries of different covalently bonded molecules based on the interaction of valence electrons of the central atom with those of the bonded atoms.
A covalent molecule is formed as a result of sharing of electrons between two atoms. A molecule consisting of more than two atoms can have different angular arrangements of atoms with respect to each other which is known as their geometry. The central atom is decided to be used as reference and the positions of all other atoms around it is determined.
In \[P{H_3}\] molecule, the phosphorus atom is the central atom and has a total of five valence electrons. Three hydrogen atoms bond with phosphorus atoms by sharing one electron each. Thus phosphorus has six shared pairs and one lone pair of electrons in the \[P{H_3}\] molecule.
The molecule should have been tetrahedral due to the four pairs of electrons around the phosphorus atom but one of the electron pairs is a lone pair that strongly repels the three equivalent bond pairs resulting in a trigonal pyramidal arrangement.

Thus, the \[P{H_3}\] molecule has a trigonal pyramidal shape due to lone pair bond pair repulsion.

Note:
Hybridization can also be used to determine the shape of molecules. But the phosphorus atoms remain unhybridized in the \[P{H_3}\] molecule as the hydrogen atoms are not electronegative enough to contract the large phosphorus atom. The pure p-orbitals of phosphorus are nearly placed at right angles and can easily accommodate the bonded pairs of electrons.