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How to draw BBr3 Lewis Structure? - Science Education and Tutorials

How to draw BBr3 Lewis Structure?

The boron tribromide chemical formula is BBr3. Drawing BBr3 Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct BBr3 Lewis Structure. The boron and bromine elements come as the member of the boron and halogen family groups from the periodic table respectively. The valence electrons in boron and bromine are three and seven respectively.

Key Points To Consider When Drawing The BBr3 Structure

A three-step approach for drawing the BBr3 Lewis structure can be used. The first step is to sketch the Lewis structure of the BBr3 molecule, to add valence electrons around the boron atom; the second step is to add valence electrons to the three bromine atoms, and the final step is to combine the step1 and step2 to get the BBr3 Lewis Structure.

The BBr3 Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the BBr3 molecule. The geometry of the BBr3 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the BBr3 geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the B-Br bond (dipole moment properties of the BBr3 molecule). The boron-bromine bonds in boron tribromide(BBr3), for example, are polarised toward the more electronegative bromine in BBr3 molecule, and because both bonds have the same size and located around three bromine terminals of the trigonal planar with no lone pairs of electrons just out of the plan, their sum is zero due to the BBr3 molecule’s bond dipole moment and no lone pairs of electrons on the boron atom. Because each B-Br bond polarity canceled each other in the BBr3 molecule. The boron tribromide(BBr3) molecule is classified as a nonpolar molecule.

How to draw BBr3 Lewis Structure?

The molecule of boron tribromide(with trigonal planar molecular geometry) is tilted, the bond angles between boron and bromine are 120 degrees. It has a difference in electronegativity values between boron and bromine atoms, with boron’s pull being less than bromine’s terminal in the BBr3 molecule. But they canceled each other due to the symmetrical molecular geometry of BBr3. As a result, it has the zero dipole moment. The BBr3 molecule has a zero dipole moment due to an equal charge distribution of negative and positive charges. The net dipole moment of the BBr3 molecule is 0 D.

Overview: BBr3 Lewis Structure

The central atom is boron, which is bordered on three terminals with three bromine atoms( in trigonal planar geometry), and no lone pairs of electrons on the boron in the trigonal planar geometry. Boron has three outermost valence electrons, indicating that it possesses three electrons in its outermost shell, whereas bromine only has seven valence electrons in its outermost shell. To complete the octet of the bromine atom, a bromine terminal atom requires one valence electron. If you’re interested in learning more about the bromine octet rule, please see in our previous post.

Three bromine atoms establish covalent connections with the central boron atom as a result, leaving the boron atom with no lone pairs. There are no lone pair of electrons on the boron central atom that resist the bond pairs of the three B-Br. According to VSEPR theory, the bond pairs lead the BBr3 molecule to take on a trigonal planar molecular geometry shape.

The BBr3 molecule’s B-Br bonds are arranged in symmetrical order around the trigonal planar molecular geometry, giving rise to the BBr3 molecular shape. The BBr3 molecule has a trigonal planar molecular geometry because there is no electrical repulsion between lone pairs and three bond pairs(B-Br) of the BBr3 molecule.

Electronegative Difference Calculation of BBr3 Molecule:

The boron atom has an electronegativity of 2.04, while bromine has an electronegativity of 2.96 in the BBr3 molecule. The difference in electronegativity can be estimated using the method below.

The electronegative value difference between boron and bromine

Electronegativity value of boron = 2.04

Electronegativity value of bromine= 2.96

Difference of electronegativity value between boron and bromine= 2.96 – 2.04=0.92

Electronegativity difference between B-Br bond calculation of BBr3 molecule

Due to the difference in electronegativity value of greater than 0.5, the B-Br bond of the BBr3 molecule becomes polar. Because of this difference in electronegativity, the BBr3 molecule’s B-Br bond becomes polar. The total net dipole moment of the BBr3 molecule is zero due to the cancellation of the bond dipole moment in the trigonal planar geometry. The electronegativity of an atom is the strength with which it may attract bound electron pairs to its side. The polarity of BF3 is discussed in our previous post.

As a result, the B-Br bond’s dipole moment is high due to the polarization of the bonds and no lone pairs of electrons, and all B-Br bonds’ dipoles are arranged in the symmetrical BBr3 molecular geometry. The BBr3 molecule’s total dipole moment is predicted to be 0 D. It has a partial negative charge for the terminal bromine atoms and a partial positive charge for the central boron atom.

The electron dot structure of the BBr3 molecule is also known as the BBr3 Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the BBr3 molecule’s bond formation. The outermost valence electrons of the BBr3 molecule must be understood while considering the Lewis structure of the molecule.

The boron atom is the middle element in BBr3 molecular geometry, with three electrons in its outermost valence electron shell, whereas the bromine atom has seven electrons in its outermost valence electron shell.

The BBr3 molecule has a total of 24 valence electrons as a result of the foregoing above said reasoning. With the core central boron atom, the three terminal with three bromine atoms forms covalent bonds, leaving the boron atom with no lone pairs of electrons on the top and bottom of trigonal planar geometry.

Because no lone pairs of electrons on the central boron atom create interaction with B-Br bond pairs. The bond angle of the Br-B-Br bond in the trigonal planar molecular geometry is approximately 120 degrees. This angle is greater than the CH4 molecule bond angle due to the no lone pairs of electrons on the BBr3 molecule.

There are two types of bonds in the BBr3 molecule. Due to the resonance structure of the BBr3 molecule creates a double bond and a single bond. The double bond of B-Br is shorter as compared to the single bond of B-Br. The B-Br bond length of the single bond of BBr3 is longer than the B-Cl bond length of BCl3.

To sketch the BBr3 Lewis structure by following these instructions:

Step-1: BBr3 Lewis dot Structure by counting valence electrons on the boron atom

To calculate the valence electron of each atom in BBr3, look for its periodic group from the periodic table. The halogen and boron families, which are the 17th and 13th groups in the periodic table, are both made up of bromine and boron atoms. In their outermost shells, bromine and boron have seven and three valence electrons, respectively.

Because bromine and boron are members of the periodic table’s halogen and boron family groups, their valence electrons are seven and three, respectively.

Calculate the total number of valence electrons in the BBr3 molecule’s outermost valence shell. The first step is to determine how many electrons are in the BBr3 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 BBr3 Lewis diagram. The BBr3 molecule’s core boron atom can be represented as follows:

Total outermost valence shell electron of boron atom in BBr3= 3

Total outermost valence shell electron of the bromine atom in BBr3= 7

The BC3 molecule has one central boron and three bromine atoms. Then the total outermost valence shell electrons can be calculated as follows

∴ Total outermost valence shell electrons available for BBr3 Lewis structure( dot structure) = 3+3*7= 24 valence electrons  in BBr3.  

calculation of total valence electron of BBr3 molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of BBr3. We’ll choose the least electronegative value atom in the BBr3 molecule to place in the center of the BBr3 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.

How to draw BBr3 Lewis Structure?

Step-2: Lewis Structure of BBr3 for counting valence electrons around the terminal bromine atom

As a result, boron is the first atom in the periodic table’s boron family group. Boron is the first member of the boron family. The electronegative value of the boron atom is lower than that of the bromine atom in the BBr3 molecule. Furthermore, boron has a three electrons limit since bromine is the most electronegative element in the BBr3 molecule.

In the BBr3 Lewis structure diagram, the boron atom can be the center atom of the molecule. As a result, central boron in the BBr3 Lewis structure, with all three bromine arranged in the trigonal planar geometry.

How to draw BBr3 Lewis Structure?

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

Step-3: Lewis dot Structure for BBr3 generated from step-1 and step-2

Connect the exterior and core central atom of the BBr3 molecule with three single bonds (B-Br). In this stage, use three bromine atoms on the outside of the BBr3 molecule to the central boron atom in the middle.

Count how many electrons from the outermost valence shell have been used in the BBr3 structure so far. Each B-Br bond carries two electrons because each boron atom is connected to three bromine atoms by three B-Br bonds. Bond pairings of B-Br are what they’re called.

So, out of the total of 24 valence electrons available for the BBr3 Lewis structure, we used 6 for the BBr3 molecule’s three B-Br bonds. The BBr3 molecule has no lone pairs of electrons in the central boron atom. We need to put extra electrons in the molecular geometry of BBr3. Where to place the extra electron in the BBr3 molecule?

Place the valence electrons in the B-Br bond pairs starting with the core boron, three bromine, and no lone pairs of electrons in the BBr3 molecule. In the BBr3 Lewis structure diagram, we always begin by introducing valence electrons from the central boron atom(in step1). As a result, wrap around the central boron atom’s bond pair valence electrons first (see figure for step1).

The boron atom in the molecule gets only six electrons around its molecular structure. This central boron atom is octet deficient. BBr3 molecule goes on resonance. A bromine atom has three pairs of lone pairs of electrons. Due to the back bonding mechanism of B-Br of the BBr3 molecule, bromine gives one lone pair of electrons to the bond pairs of B-Br bond. This makes a single bond into a double bond.

This makes boron negative in charge and bromine makes positive in charge. The double bond resonated around the BBr3 molecule. The double bond of B-Br is shorter as compared to the single B-Br bond. In this way, Boron gets its octet stability.

Boron requires 6 electrons in its outermost valence shell to complete the molecular stability, 6 electrons bond pairs in B-Br bonds. Then no electrons as a lone pair of electrons on the boron atom of the BBr3 molecule are placed in a trigonal planar geometry. Boron already shares 6 electrons to the three B-Br bonds. Then place the valence electron in the bromine atom, it placed around seven electrons(step-2). Totally, 21 valence electrons placed on the three bromine atoms of the BBr3 molecule.

We’ve positioned 6 electrons around the central bromine atom(step-3), which is represented by a dot, in the BBr3 molecular structure above. The boron atom completes its molecular stability in the BBr3 molecule because it possesses 6 electrons in its (B-Br bond pairs with three bromine in the outermost valence shell.

How to draw BBr3 Lewis Structure?

Count how many outermost valence shell electrons have been used so far using the BBr3 Lewis structure. Three electron bond pairs are shown as dots in the BBr3 chemical structure, whereas three single bonds each contain two electrons. The outermost valence shell electrons of the BBr3 molecule(bond pairs) are six as a result of the calculation.

So far, we’ve used six of the BBr3 Lewis structure’s total 6 outermost valence shell electrons. No lone pairs of electrons on the boron atom in the trigonal planar of the BBr3 molecule.

Complete the middle boron atom stability and, if necessary, apply a covalent bond. The central boron atom undergoes octet stability(due to resonance structure). Because it has a total of 8 electrons in the outermost valence shell. Eight electrons come from one double bond pair of B-Br and two single bond pairs of B-Br on the boron central atom of BBr3.

The core atom in the BBr3 Lewis structure is boron, which is bonded to the three bromine atoms by single bonds (B-Br). With the help of three single bonds, it already shares six electrons. As a result, the boron follows the octet rule and has 8 electrons (due to resonance structure) surrounding it on the three terminals of the BBr3 molecule’s trigonal planar geometry.

How to calculate the formal charge on a boron atom in BBr3 Lewis Structure?

The formal charge on the BBr3 molecule’s boron central atom often corresponds to the actual charge on that boron central atom. In the following computation, the formal charge will be calculated on the central boron atom of the BBr3 Lewis dot structure.

To calculate the formal charge on the central boron atom of the BBr3 molecule by using the following formula:

The formal charge on the boron atom of BBr3 molecule= (V. E(B)– L.E(B) – 1/2(B.E))

V.E (B) = Valence electron in a boron atom of BBr3 molecule

L.E(B) = Lone pairs of an electron in the boron atom of the BBr3molecule.

B.E = Bond pair electron in B atom of BBr3 molecule

calculation of formal charge on boron atom in BBr3 molecule

The boron core atom (three single bonds connected to three bromine atoms ) of the BBr3 molecule has 3 valence electrons, no lone pairs of electrons(zero electrons), and six bonding pairing valence electrons. Put these values for the boron atom in the formula above.

Formal charge on boron atom of BBr3 molecule = (3- 0-(6/2)) =0

In the Lewis structure of BBr3, the formal charge on the central boron atom is zero.

Summary:

In this post, we discussed the method to construct the BBr3 Lewis structure. First, the valence electrons are placed around the boron atom. Second, place the valence electron on the bromine atom. Finally, when we combined the first and second steps. It gives BBr3 Lewis structure. Need to remember that, if you follow above said method, you can construct molecular dot structure very easily.

What is the BBr3 Lewis structure?

BBr3 Lewis structure is dot representation

What is the formal charge on the BBr3 Lewis structure?

Zero charge on the BBr3 molecular structure

The polarity of the molecules

The polarity of the molecules are listed as follows

Lewis Structure and Molecular Geometry

Lewis structure and molecular geometry of molecules are listed below

External References

Information on boron tribromide(BBr3)

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