How to draw BrF5 Lewis Structure? - Science Education and Tutorials

Q: What is the BrF5 Lewis structure?

BrF 5 Lewis structure is dot representation

Q: What is the formal charge on the BrF5 Lewis structure?

Zero charges on the BrF 5 molecular structure

Bromine pentafluoride chemical formula is BrF5. Drawing BrF5 Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct BrF5 Lewis Structure. The bromine and fluorine elements come as the member of the halogen family group from the periodic table. The valence electrons in bromine and fluorine are seven. The branch of halogen chemistry is used to make chemicals reagents for fluorination reactions.

Table of Contents

Key Points To Consider When Drawing The BrF5 Electron Dot Structure

A three-step approach for drawing the BrF5 Lewis structure can be used. The first step is to sketch the Lewis structure of the BrF5 molecule, to add valence electrons around the bromine atom; the second step is to add valence electrons to the five fluorine atoms, and the final step is to combine the step1 and step2 to get the BrF5 Lewis Structure.

The BrF5 Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the BrF5 molecule. The geometry of the BrF5 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the BrF5 geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the five Br-F bonds (dipole moment properties of the BrF5 molecule). The bromine-fluorine bonds in bromine pentafluoride(BrF5), for example, are polarised toward the more electronegative fluorine in BrF5molecule, and because both bonds have the same size and located around five fluorine terminals of the square pyramidal with one lone pair (in total two-electrons) on the bromine atom, their sum of dipole moment is nonzero due to the BrF5 molecule’s bond dipole moment and more electron polarity to the fluorine atoms. Because each five Br-F bonds polarity not canceled each other in the BrF5 molecule. The Penta fluoro bromine or bromine pentafluoride(BrF5) molecule is classified as a polar molecule.

The molecule of bromine pentafluoride (with square pyramidal molecular geometry) is tilted, the bond angles between bromine and fluorine are 90 degrees. It has a difference in electronegativity values between bromine and fluorine atoms, with central bromine’s pull being less than terminal fluorine’s in the BrF5 molecule. But they not canceled each other due to the asymmetrical molecular geometry of the BrF5 molecule.

As a result, it has the nonzero dipole moment. The BrF5 molecule has a nonzero dipole moment due to an unequal charge distribution of negative and positive charges. But both these atoms fall on the halogen family group. The fluorine atom is a more electronegative value than bromine in the BrF5 molecule. The BrF5 molecule has the net dipole moment.

Molecules can be classified as polar or nonpolar. The molecule polar behaves in a different manner as compared to nonpolar.

Overview: BrF5 Lewis Structure

The central atom is bromine, which is bordered on five terminals with fluorine atoms( in square pyramidal), and one lone pair on the central in the square pyramidal geometry. Bromine has seven outermost valence electrons, indicating that it possesses seven electrons in its outermost shell, whereas fluorine also has seven valence electrons in its outermost shell. To complete the octet of the bromine and fluorine atoms requires one valence electron on each of their outermost shell.

Five fluorine atoms establish covalent connections with the central bromine atom as a result, leaving the bromine atom with two lone pairs. There are two lone pairs of electrons on the bromine central atom that resists the bond pairs of the five Br-F. According to VSEPR theory, the Br-F bond pairs polarity lead the BrF5 molecule to take on the square pyramidal geometry structure.

The BrF5 molecule’s five Br-F bonds are arranged in unsymmetrical polarity order around the square pyramidal molecular geometry, giving rise to the BrF5 molecular shape. The BrF5 molecule has a square pyramidal molecular geometry because there is electrical repulsion between the lone pairs of electrons in fluorine and five bond pairs(Br-F) of the BrF5 molecule.

Lewis structure of BrF5 has dot electron representative structure. Valence electrons of atoms undergo orbitals mixing in the chemical reactions, gives new types of molecular species. The molecule is nothing but a bundle of valence electrons from the atoms. But it is converted to bond pairs and lone pairs in the molecular structure.

Electronegative value Difference Calculation of BrF5 Molecule:

Bromine and Fluorine Electronegative difference in BrF5:

The bromine atom has an electronegativity of 2.96, while fluorine has an electronegativity of 3.98 in the BrF5 molecule. The difference in electronegativity of bromine and fluorine can be estimated using the method below.

The electronegative value difference between bromine and fluorine in BrF5 molecule
Electronegativity value of bromine = 2.96
Electronegativity value of fluorine= 3.98
Difference of electronegativity value between bromine and fluorine in BrF5 molecule = 3.98 – 2.96=1.02
Electronegativity difference between Br-F bond calculation of BrF5 molecule

The electronegative difference between bromine and fluorine is greater than 0.5. This indicated the bond polarity moves near to polar nature. Br-F bond polarity in the BrF5 molecule is polar.

Because of this difference in electronegativity of bromine and fluorine atoms, the BrF5 molecule’s Br-F bond becomes polar. The total net dipole moment of the BrF5 molecule is nonzero due to the no cancellation of the bond dipole moment in the square pyramidal geometry. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side. The polarity of BrF5 is discussed in our previous post.

As a result, the Br-F bond’s dipole moment is high due to the polarization of the bonds and one lone pair of electrons on bromine, and all Br-F bonds’ dipoles are arranged in the asymmetrical BrF5 molecular geometry. The BrF5 molecule has a total dipole moment.

The electron dot structure of the BrF5 molecule is also known as the BrF5 Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the BrF5 molecule’s bond formation. The outermost valence electrons of the BrF5 molecule must be understood while considering the Lewis structure of the molecule.

The bromine atom is the middle element in BrF5 molecular geometry, with seven electrons in its outermost valence electron shell, whereas the fluorine atom has seven electrons in its outermost valence electron shell. The fluorine atom has seven valence electrons.

The BrF5 has a total of 42 valence electrons as a result of the foregoing above-said reasoning. With the core central bromine atom, the five terminals with five fluorine atoms form covalent bonds, leaving the bromine atom with one lone pair in the middle of square pyramidal geometry.

Because lone pair on the terminal fluorine atom creates interaction with Br-F bond pairs(but it is negligible). The bond angle of the F-Br-F bond in the square pyramidal molecular geometry is approximately 90 degrees. This angle is greater than the BrF3 molecule bond angle. The Br-F bond length is 176 pm(picometer).

To sketch the BrF5 Lewis structure by following these instructions:

Step-1: BrF5 Lewis dot Structure by counting valence electrons on the bromine atom

To calculate the valence electron of each atom in BrF5, look for its periodic group from the periodic table. The halogen group families, which are the 17th group in the periodic table, are both made up of bromine and fluorine atoms. In their outermost shells, fluorine and bromine have seven valence electrons.

Calculate the total number of valence electrons in the BrF5 molecule’s outermost valence shell. The first step is to determine how many electrons are in the BrF5 Lewis structure’s outermost valence shell. An electron in an atom’s outermost shell is known as a valence electron. It is represented by dots in the BrF5 Lewis diagram. The BrF5 molecule’s core bromine atom can be represented as follows:

Total outermost valence shell electron of bromine atom in BrF5= 7
Total outermost valence shell electron of fluorine atom in BrF5= 7
The BrF5 molecule has one central bromine and five fluorine atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for BrF5 Lewis structure( dot structure) = 7+5*7=42 valence electrons in BrF5.
calculation of total valence electron of BrF5 molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of BrF5. We’ll choose the least electronegative value atom in the BrF5 molecule to place in the center of the BrF5 Lewis structure diagram in this phase. The electronegativity value in periodic groups grows from left to right in the periodic table and drops from top to bottom.

The first step is to put seven valence electrons around the bromine atom as given in the figure.

Step-2: Lewis Structure of BrF5 for counting valence electrons around the terminal fluorine atoms

As a result, bromine is the first atom in the periodic table’s halogen family group. Fluorine is the first member of the halogen family. The electronegative value of the fluorine atom is higher than that of the bromine atom in the BrF5 molecule. Furthermore, fluorine has a seven electrons limit since bromine is the less electronegative element in the BrF5 molecule.

In the BrF5 Lewis structure diagram, the bromine atom can be the center atom of the molecule. As a result, central bromine in the BrF5 Lewis structure, with all five fluorine atoms arranged in square pyramidal geometry.

Add valence electrons around the fluorine atom, as given in the figure.

Step-3: Lewis dot Structure for BrF5 generated from step-1 and step-2

Connect the exterior and core central atom of the BrF5 molecule with five single Br-F bonds. In this stage, use five fluorine atoms on the outside of the BrF5 molecule to the central bromine atom in the middle.

Count how many electrons from the outermost valence shell have been used in the BrF5 structure so far. Each Br-F bond carries two electrons because each bromine atom is connected to five fluorine atoms by five Br-F bonds. Bond pairings of Br-F are what they’re called.

So, out of the total of 42 valence electrons available for the BrF5 Lewis structure, we used 10 electrons for the BrF5 molecule’s five Br-F bonds. The BrF5 molecule has one lone pair of electrons in the central bromine atom.

Place the valence electrons in the Br-F bond pairs starting with the core bromine, five fluorine atoms in the BrF5 molecule. In the BrF5 Lewis structure diagram, we always begin by introducing valence electrons from the central bromine atom(in step1). As a result, wrap around the central bromine atom’s bond pair valence electrons first (see figure for step1).

The bromine atom in the molecule gets only ten electrons around its molecular structure. This central bromine atom is extra octet stable. But it has one lone pair. Bromine(Br2) is a brownish liquid in nature. when bromine acted as brominating agent, it is used as a chemical reagent. But BrF5 is used as a fluorinating agent in organic chemistry.

Bromine requires 12 electrons in its outermost valence shell to complete the molecular extra octet stability, 10 electrons bond pairs in five Br-F bonds. Then lone pairs of electrons on the fluorine atoms of the BrF5 molecule are placed in a square pyramidal geometry. Bromine already shares ten electrons to the five Br-F bonds. Then place the valence electron in the fluorine atoms, it placed around seven electrons on each atom(step-2). 30 valence electrons placed around fluorine atoms as lone pairs of electrons.

We’ve positioned 30 electrons around the terminal fluorine atoms(step-3), which is represented by a dot, in the BrF5 molecular structure above. The bromine atom completes its molecular extra octet stability in the BrF5 molecule because it possesses ten electrons in its (five Br-F) bond pairs with five fluorine in the outermost valence shell.

Count how many outermost valence shell electrons have been used so far using the BrF5 Lewis structure. Five electron bond pairs are shown as dots in the BrF5 chemical structure, whereas five single bonds each contain two electrons. The outermost valence shell electrons of the BrF5 molecule(bond pairs) are ten as a result of the calculation.

So far, we’ve used 42 of the BrF5 Lewis structure’s total 42 outermost valence shell electrons. One lone pair of electrons on the bromine atom in the square pyramidal of the BrF5 molecule.

Complete the middle bromine atom stability and, if necessary, apply a covalent bond. The central bromine atom undergoes extra octet stability(due to one lone pair of electrons).

The core atom in the BrF5 Lewis structure is bromine, which is bonded to the five fluorine atoms by single bonds (five Br-F). With the help of five single bonds, it already shares ten electrons. As a result, the bromine follows the extra octet rule and has ten electrons surrounding it on the five terminals of the BrF5 molecule’s square pyramidal geometry.

How to calculate the formal charge on bromine and fluorine atoms in BrF5 Lewis Structure?

Calculating formal charge on the bromine of BrF5 molecule:

The formal charge on the BrF5 molecule’s bromine central atom often corresponds to the actual charge on that bromine central atom. In the following computation, the formal charge will be calculated on the central bromine atom of the BrF5 Lewis dot structure.

To calculate the formal charge on the central bromine atom of the BrF5 molecule by using the following formula:
The formal charge on the bromine atom of BrF5 molecule= (V. E(Br)– L.E(Br) – 1/2(B.E))
V.E (Br) = Valence electron in a bromine atom of BrF5 molecule
L.E(Br) = Lone pairs of an electron in the bromine atom of the BrF5 molecule.
B.E = Bond pair electron in Br atom of BrF5 molecule
calculation of formal charge on bromine atom in BrF5 molecule

The bromine core atom (five single bonds connected to five fluorine atoms ) of the BrF5 molecule has seven valence electrons, one lone pair of electrons(two electrons), and ten bonding pairing valence electrons. Put these values for the bromine atom in the formula above.

Formal charge on bromine atom of BrF5 molecule = (7- 2-(10/2)) =0

In the Lewis structure of BrF5, the formal charge on the central bromine atom is zero.

Calculating formal charge on the fluorine of BrF5 molecule:

The formal charge on the BrF5 molecule’s fluorine terminal atom often corresponds to the actual charge on that fluorine terminal atom. In the following computation, the formal charge will be calculated on the terminal fluorine atom of the BrF5 Lewis dot structure.

To calculate the formal charge on the terminal fluorine atom of the BrF5 molecule by using the following formula:
The formal charge on the fluorine atom of BrF5 molecule= (V. E(F)– L.E(F) – 1/2(B.E))
V.E (F) = Valence electron in a fluorine atom of BrF5 molecule
L.E(F) = Lone pairs of an electron in the fluorine atom of the BrF5 molecule.
B.E = Bond pair electron in F atom of BrF5 molecule
calculation of formal charge on fluorine atom in BrF5 molecule

The fluorine terminal atom of the BrF5 molecule has seven valence electrons, three lone pairs of electrons(six electrons), and two bonding pairing valence electrons. Put these values for the fluorine atom in the formula above.

Formal charge on fluorine atom of BrF5 molecule = (7- 6-(2/2)) =0

In the Lewis structure of BrF5, the formal charge on the terminal fluorine atom is zero.

Summary:

In this post, we discussed the method to construct the BrF5 Lewis structure. First, the valence electrons are placed around the bromine atom. Second, place the valence electron on the fluorine atoms. Finally, when we combined the first and second steps. It gives BrF5 Lewis structure. Need to remember that, if you follow the above-said method, you can construct molecular dot structure very easily.