Molecular Geometry, Hybridization and Polarity of BrF5 Lewis Structure

An interhalogen chemical, the fluoride of bromine, BrF5 is built up entirely of halogen atoms. Colorless, seething, and pungently scented, it is in the liquid stage. Humans should avoid inhaling bromine pentafluoride at all costs because it is damaging to the eyes and internal membranes.

Combustible materials used in rockets can be burned more quickly if this compound is employed in the production of rocket chemicals. When bromine pentafluoride combines with water present in the wet air, it produces hydrofluoric acid smoke.

Bromine pentafluoride is kept in only in case it bursts noisily and ruptures the vessel if left open. For these reasons, the aeronautical industry has examined this chemical extensively. The fluorinating agent bromine pentafluoride is well-known for its usage in the production of fluorocarbons.

Bromine pentafluoride’s atomic structure is seen here (BrF5)

The Lewis diagrams of the atoms involved in bromine pentafluoride must first be studied in order to begin researching the Lewis structure of this compound. The electrical configuration of bromine is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5, which is the atomic number 35.

The electrons from the shells with the highest energy levels are involved in the bond creation here. The total number of valence electrons in bromine is 7 because of the 4s2 and 4p5 shells.

Along with its atomic number 9, Fluorine possesses a chemical structure shown by the symbols 1S2 2S2 2P5. Electrons in the sublevels of these primary high energy levels can be added to determine their total amount of valence electrons, which can be found at the highest principle energy level.

Because of this, we’ll need to put in two of each of the two valence electrons from the 2s and 5 from the 2p.

Let’s take a closer look at how to create a Lewis structure for BrF5.

In order to begin, we must first determine how many valence electrons a molecule of BrF5 contains: The fluorine and bromine atoms contribute 7 atoms each, for a total of 42 atoms.

Step 2: Determine how many additional valence electrons are needed for one molecule of BrF5: Each participating atom needs a valence electron, hence the number is 6.

Step 3: Identify the BrF5 molecule’s core atom. The core atom will be bromine because it is the only chemical element existing as a single entity.

To find out what kind of link is being formed, proceed to step 4: The fluorine and bromine atoms are creating single covalent bonds in this reaction.

Assemble the above-mentioned points to draw the structure. The framework will be appropriately drawn if you understand what we’re talking about now.

Why does BrF5 have a Lewis dot structure that’s out of whack?

A bromine nucleus is surrounded by five fluorine atoms as soon as you begin drawing the structure. Each Br-F interaction draws a single bond, exhausting 10 of the 42 available valence electrons in the process.

Following the completion of each fluorine atom’s octet, the total number of valence electrons spent is now 40. There are still two valence electrons left, and these will be attached to the bromine atom as an isolated pair.

Nevertheless, according to the octet rule, the outermost shell can only have eight valence electrons, whereas the Lewis structure for BrF5 shows that bromine has 12 valence electrons.

Sulfur and bromine can both expand their octet in order to accommodate more valence electrons, however this is a rare occurrence in chemistry.

Period four of the periodic table contains bromine, an element that, when excited to a higher energy level, can easily expand its octet.

Molecular Structure of BrF5

When the structure of BrF5 is square, the bond angle between each atom is 90 degrees. The overall shape of a molecule is determined by the total amount of bonded and non-bonding electrons as well as their orientation around the central atom, according to the Valence Shell Electron Pair Repulsion (VSEPR) theory.

There is a minimum amount of repulsion between these atoms due to their arrangement. In order to understand the electrons around the centre atom, it is necessary that you know that they tend to stay far apart. Due to the presence of a single pair of electrons, the octahedron-like BrF5 structure is shifted into a square pyramidal shape.

Because the lone pair of electrons are closer to the centre atom, they exert the largest repelling force according to the VSEPR hypothesis.

This is the reason that their repulsive impact is far stronger than the pair of electrons bonding.

Hybridization of BrF5

There are four different types of orbitals involved in the mixing and overlapping that creates new hybrid orbitals in the BrF5 hybridization process: one 4s, three 4p, and two 4d. Some of the valence electrons in bromine must shift to the 4d orbitals in order for the atom to attain pentavalent status.

An unpaired orbital in bromine causes it to enter an excited state, making it more reactive. This is the point at which the organisms begin to interbreed. For each of the newly produced hybrid orbitals, there are two pairs of electrons left over from the previously formed hybrid orbitals; these electrons can be found in one newly formed hybrid orbital.

Understanding that the atomic orbitals of similar energy mix and overlap in a manner such as, s orbital with p orbital or s orbital with a d orbital, is essential to the process of hybridization

New hybrid orbitals with the same energy as the original orbitals are formed when the energy of individual atoms is redistributed among their orbitals of similar energy.

Inverse Polarity of BrF5

There is polarity in the separation of electric charge, which results in one negatively charged end and one positively charged one. Because of the difference in electronegativity values of the individual atoms, the entire molecule experiences a net dipole moment.

The strength with which a molecule attracts a shared pair of electrons towards itself depends on the value of this net dipole moment, which can be positive or negative.

Due to a lone pair of electrons present in bromine pentafluoride’s molecular structure, the centre atom will have an asymmetric charge distribution. Because of this, the BrF5 molecule has a polar structure.

The electron geometry of BrF5 is octahedral, as can be seen from the Lewis dot structure, with bromine and fluorine having electronegativity values of 2.96 and 3.98, respectively. The BrF5 molecule is polar because the difference between the two values is bigger than 0.4.

The article on BrF5 polarity is also an excellent resource for more in-depth information.

An increase in polarity can be predicted by increasing electronegativity, which is defined as an increase in the tendency of an atom or molecule to attract a shared pair of electrons toward itself.

In the BrF5 molecule, there is a lone pair of electrons that are negatively charged and not associated with the positively charged nucleus, which causes a net dipole moment that makes the molecular polar in nature.

Conclusion

Figured out what’s going on with bromine pentafluoride by looking at the Lewis structure. The abnormality is that the octet has been expanded to allow 12 valence electrons instead of just eight. It was because of this that only one covalent link could be formed, leaving behind a pair of electrons for the bromine atom to take on the 5 fluoride atoms.

To modify the chemical structure, this lone pair was responsible for making the bromine pentafluoride octahedral. Remember that the electron geometry is an octahedron, and do not confuse it with other octahedra.

Furthermore, the bromine hybridization in BrF5 is sp3d2, as one 4s, three 4p, and two 4d orbitals participate in the mixing and overlapping that leads to new hybrid orbitals being formed. As a last point, the bromine pentafluoride molecule has a net dipole moment due to the presence of a pair of electrons on the bromine atom, which makes it polar.

Read more: The Lewis structure of H2O2, Molecular Geometry, Hybridization, and Polarity of H2O2 are discussed.

Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.

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