Bromine pentafluoride (BrF5) lewis dot structure, molecular geometry, polar or non-polar, hybridization
Bromine pentafluoride is a pale yellow color liquid with the chemical formula BrF5. It is an interhalogen compound uses as a fluorinating reagent. It is used in various rockets and chemicals.
In this article, we will study BrF5 lewis dot structure, molecular geometry, polar or non-polar, its hybridization, etc.
BrF5 is very toxic to inhale and severely corrosive to the skin. Long-term exposure to this can cause liver failure or kidney damage. It can explode upon contact with organic materials.
Properties of Bromine pentafluoride
It can react with water.
It is a powerful oxidizer that can cause severe hazards.
It has a molar mass of 174.894 g.mol−1.
It has a boiling point of 40.25 °C and a melting point of −61.30 °C.
BrF5 lewis structure comprises five fluorine (F) atoms and one bromine (Br) atom. The bromine (Br) atom is kept at the central position and the fluorine (F) atoms are in the surrounding position. The lewis dot structure of BrF5 has a total of five Br-F bonds.
Follow some steps for drawing the lewis dot structure of BrF5
1. Count total valence electron in BrF5
In the very first step, we need to determine how many valence electrons are available for BrF5. A valence electron is the outermost shell electron associated with an atom. It is represented as dots in the lewis diagram.
For finding the valence electron of BrF5, look out for the periodic group of each atom present in it. We know both bromine and fluorine belong to the halogen family which is the 17th group in the periodic table.
So, the valence electron for bromine is 7 and for fluorine, it is also 7 as both belong to the same group in the periodic table.
∴ Total valence electron available for BrF5 lewis structure = 7 + 7*5 = 42 electrons [∴BrF5 has 5 fluorine atom and 1 bromine ]
2. Find the least electronegative atom and placed it at center
In this step, we have to find the least electronegative atom in BrF5 for placing at its center in the lewis diagram. As electronegativity increase from left to right in the periodic table and decreases from top to bottom in periodic groups.
So, bromine is less electronegative than fluorine. Also, Fluorine is the most electronegative element in chemistry so it can never be a central atom in any lewis diagram.
Therefore, put bromine at the center in the lewis diagram and fluorine spaced evenly around it.
3. Connect outer atom to central atom with a single bond
In this step, connect all outer atoms (Fluorine) to the central atom(Bromine) with the help of a single bond.
Now count how many valence electrons we used in the above BrF5 structure till now. As we used 5 single bonds to connect each bromine to a fluorine atom and one bond contain 2 electrons.
So, 5 single bonds mean 10 electrons we used from the total of 42 valence electrons available for BrF5 lewis structure.
∴ (42 – 10) = 32 valence electron
We are still left with 32 valence electrons more.
4. Place remaining valence electrons starting from the outer atom first
In lewis’s diagram, we always start putting valence electrons from the outer atom first. So, put the remaining valence electron around each fluorine atom first.
Fluorine needs 8 electrons in its outermost shell to complete its octet. And each fluorine already sharing 2 electrons with the help of a single bond.
So, simply put 6 electrons around each fluorine atom.
As you see, we have placed 30 electrons represented as a dot around the fluorine atoms in the above structure. So, each fluorine comfortably completes its octet as all fluorine atoms have 8 electrons in their valence shell.
Now again count how many valence electrons are used till now with the help of the above structure.
30 electrons are represented as dots in the above structure and 5 single bonds that each contain 2 electrons. So, 30 + 10 = 40 valence electrons.
So, from the total 42 valence electron available for the BrF5 lewis structure, we used 40 electrons till now.
Still, we are left with 2 valence electrons more.
5. Complete central atom octet and use covalent bond if necessary
We know, that bromine is the central atom and it is connected with 5 single bonds to the fluorine atoms. It means it already shares 10 electrons with the help of 5 single bonds.
So, bromine is violating the rule of the octet as it has more than 8 electrons around it.
In chemistry, an atom that holds more than 8 electrons around them falls into the category of the expanded octet state.
Fluorine can never ever holds more than 8 electrons around it. So, place those 2 remaining valence electrons around the bromine central atom as it is playing the role of expanded octet here.
BrF5 lewis dot structure
So, after using all valence electrons and completing each atom octet, we got our stable Bromine pentafluoride lewis structure.
What are the electron and molecular geometry of BrF5
According to the VSEPR theory, the molecular geometry of BrF5 is square pyramidal and its electron geometry is octahedral because bromine being the central atom has five bonds connected with surrounding fluorine atoms. Each F-Br-F bond making an angle of 90º in the same plane. As four fluorine is in the same plane so they make square planar.
And one fluorine is present above the plane. After connecting upper fluorine to square planar, it holds the pyramidal-like geometry.
So, the overall molecular geometry of BrF5 is square pyramidal. And after considering the effect of lone pair also present on central atom then the geometry of BrF5 become octahedral which is also called electron geometry.
Image credit: Byjus.com
Let’s see how to find the molecular geometry of BrF5 step by step.
Follow three steps to find BrF5 molecular geometry
1. Find the Number of lone pairs present on the central atom of the BrF5 lewis dot structure
Here we need to determine how many lone pairs are on the central atom of the BrF5 lewis structure because the lone pair is mostly responsible for the distortion of shape in the molecule.
To find the lone pair on the central atom of any molecule use the formula given below.
∴ L.P = (V.E. – N.A.)/2
⇒ L.P. = Lone pair on the central atom
⇒ V.E. = valence electron of that central atom
⇒ N.A. = Number of atoms attached to that central atom
∴ So, in the case of BrF5, bromine which is the central atom has 7 valence electrons and 5 atoms attached to it.
Put these values in the above formula-
So, L.P. = (7 – 5)/2
= 1 is the lone pair present on the central atom of the BrF5 lewis dot structure.
2. Find hybridization number of BrF5
Now we have to determine what is the hybridization number of BrF5.
Here’s the formula for this-
∴ H = N.A. + L.P.
where H = hybridization number
N.A. = Number of atoms attached to the central atom
L.P. = lone pairs on that central atom
So, bromine is a central atom that has 5 fluorine atoms attached to it and it contains one lone pair.
∴ H = 5 + 1
= 6 is the hybridization number of BrF5 that means it has Sp³d² hybridization.
Hybridization number also called steric number.
Steric number
Hybridization
1
S
2
Sp
3
Sp²
4
Sp³
5
Sp³d
6
Sp³d²
3. Use AXN method to determine BrF5 molecular shape
Generally, the AXN method is used when VSEPR theory is applied to determine the shape of any molecule.
AXN notation follow as-
A represents the central atom.
X represents the bonded pairs of electrons to the central atom.
N represents the lone pairs of electrons on the central atom
With the help of the BrF5 lewis structure, we know bromine is the central atom that has 5 bonded pairs of electrons and one lone pair.
∴ The generic formula for BrF5 is AX5N1.
So, according to the VSEPR chart, if the molecule has an AX5N1 generic formula then the molecular geometry for that molecule will be square pyramidal, and electron geometry is octahedral.
Molecular geometry of BrF5
Bonded atoms
Lone pair
Generic formula
Hybridization
Molecular geometry
Electron geometry
1
0
AX
S
Linear
Linear
2
0
AX2
Sp
Linear
Linear
1
1
AXN
Sp
Linear
Linear
3
0
AX3
Sp²
Trigonal planar
Trigonal planar
2
1
AX2N
Sp²
Bent
Trigonal planar
1
2
AXN2
Sp²
Linear
Trigonal planar
4
0
AX4
Sp³
Tetrahedral
Tetrahedral
3
1
AX3N1
Sp³
Trigonal pyramid
Tetrahedral
2
2
AX2N2
Sp³
Bent
Tetrahedral
1
3
AXN3
Sp³
Linear
Tetrahedral
3
2
AX3N2
Sp³d
T-shaped
Trigonal bipyramidal
5
1
AX5N
Sp³d²
Square pyramidal
Octahedral
Bromine pentafluoride polarity: is BrF5 polar or nonpolar
BrF5 is a polar molecule because of its asymmetric shape as it causes unequal charge distribution around atoms that make it difficult to cancel out the dipole along with them which ultimately gives some net dipole moment for it, therefore it makes the overall structure of BrF5 is polar.
Any molecule that has some net dipole moment falls into the category of the polar molecule and if zero dipole moment present in the molecule then it lies in the nonpolar molecule categories.
The presence of lone pair on the central atom in the BrF5 molecule cause distortion in its geometry as lone cause repulsion between lone pair and bond pair. And for this reason, a negative charge is not distributed equally around the molecule.
Also, the electronegativity difference between the atoms plays a huge role to predict whether a molecule is polar or non-polar. Higher the difference of electronegativity between the two different atoms more is the strength of polarity between them.
The electronegativity of bromine is 2.96 and for fluorine, it is 3.98. The difference of electronegativity between these is greater than 0.5, that show they make polar covalent bond.
As per the Pauling scale, if the difference of electronegativity between the atoms is greater than 0.5 then they behave as polar.
What is the formal charge in BrF5 lewis structure and how to calculate it?
The formal charge generally represents the actual charge on an individual atom of any molecule. We will find the formal charge on the central atom of the BrF5 which is bromine.
To calculate the formal charge. Use this equation:
So, for bromine, we have 7 valence electrons, 2 lone pair electrons, and 10 bonding electrons(5 single bonds attached to fluorine)
Now put these values in the above formula
∴ (7 – 2 – 10/2)
= 0 is the formal charge on bromine in the BrF5 molecule.
FAQ
How many sharing electrons are present in the BrF5 lewis dot structure?
Sharing electrons is electrons that are available for making the chemical bond. As in the BrF5 lewis structure, bromine is sharing 5 bonds that contain 10 electrons with fluorine atoms. Also, one lone pair is present on bromine but they are not-shareable electrons. Hence we will not count it.
So, the total number of sharing electrons present in the BrF5 dot structure is 10.
Why does BrF5 lewis’s structure violate the octet rule?
Generally, the octet rule said an atom is most stable when it has 8 electrons around it, more than this an atom is called a member of the hypervalent compound or known as expanded octet.
So, in the case of Bromine pentafluoride lewis’s structure, bromine holds more than 8 electrons around it for making the bond with fluorine atoms present surrounding it.
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
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