Question

Based on the VSEPR theory, what is the molecular geometry of a molecule of $ P{I_3} $ ?

Answer 1

Hint: Three iodine atoms combine with one atom of phosphorus resulting in the formation of phosphorus triiodide with the chemical formula of $ P{I_3} $ with three bonds and lone pair of electrons results in $ s{p^3} $ hybridization containing one lone pair of electrons.

Complete answer:
Phosphorus is an element with atomic number $ 15 $ and has five valence electrons. phosphorus belongs to the nitrogen family or pnictogens. Iodine is an element with atomic number $ 53 $ and has seven valence electrons and tries to gain one electron from other atoms to attain stability like inert or noble gases.
Out of the five valence electrons in a phosphorus atom, three electrons involved in the bond formation with three iodine atoms by sharing results in the formation of three phosphorus-iodine bonds. Whereas the remaining two electrons exist as lone pairs of electrons on a phosphorus atom.
The hybridization of a molecule can be written from the sum of the bonds formed and the lone pair of electrons. Thus, the hybridization of $ P{I_3} $ molecules is $ s{p^3} $ .
The molecule with $ s{p^3} $ hybridization has electron geometry of tetrahedral and the molecular geometry is trigonal planar.
Based on the VSEPR theory, the molecular geometry of a molecule of $ P{I_3} $ is trigonal planar.

Note:
VSEPR theory stands for valence shell electron pair repulsion. Based on the valence electrons the molecular geometry will be determined. For some compounds or molecules, the electron and geometry were the same. Molecular geometry will be written based on the lone pair of electrons.