Lewis structure of PBr3 - How to draw Lewis structure of PBr3, Polarity, Molecular Geometry and Hybridization along with FAQs (2024)

In the trigonal pyramidal molecular geometry, the central atom is phosphorus, which is bordered on three terminals by bromine atoms (in trigonal pyramidal geometry), and one lone pair on the central phosphorus atom. Phosphorus has five outermost valence electrons, indicating that it has five electrons in its outermost shell, whereas bromine has seven outermost valence electrons. Three and one valence electrons are required on each of the phosphorus and bromine atoms outermost shells, respectively, to complete the octet.

Table of Contents

• How to draw the Lewis Structure of PBr3?
• Molecular Geometry of PBr3
• Hybridization of PBr3
• Polarity of PBr3
• Frequently Asked Questions – FAQs

How to Draw the Lewis Structure of PBr₃

Lewis dot structures, also known as Lewis structures, are diagrams that enable us to understand the bonding of atoms as well as the lone pairs present in a molecule. The valence electrons of atoms form bonds, which are represented by straight lines. Each bond absorbs two valence electrons, and the excess electrons are represented as lone pairs.

Bonding pairs of electrons participate in bond formation, whereas lone pairs or nonbonding pairs of electrons do not participate in bond formation.

Steps to draw Lewis structure ofPBr_3,

1. Determine the total number of electrons in the valence (outer shell).

In the periodic table, phosphorus is in the 15th group, and bromine is in the 17th.

As a result, the valence electron of-

Phosphorus = 5

Bromine = 7

∴ Total available valence electrons = 5 + 3(7) = 26.

2. Determine the total number of electron pairs that exist as lone pairs and bonds.

Total electron pairs are calculated by dividing the total valence electron count by two. In the valence shells of the PBr₃ molecule, there are 13 pairs of electrons.

3. Make a skeleton structure for the molecule by arranging the atoms around a central atom.

The central atom will be phosphorus as it is the least electronegative among phosphorus and bromine. A single bond will connect each atom to the central atom (one electron pair).

4. Mark atom with lone pairs.

A total of 13 electron pairs should be present as lone pairs and bonds in valence shells.

• There are already three bonds. As a result, there are ten more lone pairs to mark on Phosphorous and Bromine atoms.
• First, mark any remaining lone pairs on the outside atoms. As a result, the nine remaining electron pairs are marked.
• The remaining one electron pair will be marked on the central atom.

5. If there are charges on atoms, mark them.

There are no charges present on any of the atoms.

6. To obtain the best Lewis structure, minimise charges on atoms by converting lonepairs to bonds.

Since there are no charges on atoms, there is no need to reduce charges as part of the process of drawing the best Lewis structure. We already have the best Lewis structure for PBr₃.

7. Check the stability of the structure

It can be checked by using the formula-

Formal charge = Valence Electrons – Unbonded Electrons – ½ Bonded Electrons

Since the overall formal charge is zero, the above Lewis structure of PBr₃ is most appropriate, reliable, and stable in nature.

Molecular Geometry of PBr₃

There are three bonding pairs of electrons and one lone pair of electrons in PBr₃. The molecule will form a geometry in such a way that the repulsive forces are minimised.

Phosphorus is the central atom in PBr₃, and three bromine atoms form bonds with it. There is one lone pair of electrons. As a result, it has a tetrahedral geometry. Hence, the required shape is trigonal bipyramidal.

Bond angles of 109.5° are common in molecules with tetrahedral electron geometry and one central atom forming bonds with three atoms. However, the atom arrangement is asymmetrical here, and there is a single pair of electrons on the central atom; the bond angle will not be exactly the same, but it will be around 109.5°.

Hybridization of PBr₃

Phosphorus Tribromide contains three bonding pairs and one nonbonding or lone pair of electrons.

To accommodate all four pairs of electrons, the central atom forms four hybridised orbitals. Since four hybrid orbitals are formed, its hybridization is sp³.

Polarity of PBr₃

Phosphorus has an electronegativity of 2.19, while Bromine has an electronegativity of 2.96. The disparity in electronegativities is quite large. As a result, there will be a dipole moment between the atoms of phosphorus and bromine. Since the bromine atom is more electronegative, the direction of this dipole moment will be towards it.

Since there are three Bromine atoms in this molecule, dipole moments do not cancel each other out, and the molecule has a net dipole moment. As a result, it is a polar molecule with a partially positively charged region surrounding the Phosphorus atom and a partially negatively charged region surrounding the Bromine atoms.