How to draw BF3 Lewis Structure? - Science Education and Tutorials

Drawing BF3 Lewis Structure is very easy to by using the following method. Here in this post, we described step by step method to construct BF3 Lewis Structure. The boron element comes the first member of the boron family from the periodic table. The valence electrons boron is 3.

Table of Contents

Key Points To Consider When Drawing The BF3 Structure

A three-step approach for drawing the BF3 Lewis structure can be used. The first step is to sketch the Lewis structure of the BF3 molecule, to add valence electrons around the boron atom; the second step is to add valence electrons to the three fluorine atoms, and the final step is to combine the step1 and step2 to get the BF3 Lewis Structure.

The BF3 Lewis structure is a diagram that illustrates the number of valence electrons and bond electron pairs in the BF3 molecule. The geometry of the BF3 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory), which states that molecules will choose the BF3 geometrical shape in which the electrons have from one another.

Finally, you must add their bond polarities to compute the strength of the B-F bond (dipole moment properties of the BF3 molecule). The boron-fluorine bonds in boron trifluoride(BF3), for example, are polarised toward the more electronegative fluorine in BF3 molecule, and because both bonds have the same size and located around three fluorine terminals of the trigonal planar with no lone pairs of electrons just out of the plan, their sum is zero due to the BF3 molecule’s bond dipole moment and no lone pairs of electrons on the boron atom. Because each B-F bond polarity canceled each other in the BF3 molecule. The boron trifluoride(BF3) molecule is classified as a nonpolar molecule.

The molecule of boron trifluoride (with trigonal planar molecular geometry) is tilted, the bond angles between boron and fluorine are 120 degrees. It has a difference in electronegativity values between boron and fluorine atoms, with boron’s pull being less than fluorine’s terminal in the BF3 molecule. But they canceled each other due to the symmetrical molecular geometry of BF3. As a result, it has the zero dipole moment. The BF3 molecule has a zero dipole moment due to an equal charge distribution of negative and positive charges. The net dipole moment of the BF3 molecule is 0 D.

Overview: BF3 Lewis Structure

The central atom is boron, which is bordered on three terminals with three fluorine 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 fluorine only has seven valence electrons in its outermost shell. To complete the octet of the fluorine atom, a fluorine terminal atom requires one valence electron. If you’re interested in learning more about the fluorine octet rule, please see in our previous post.

Three fluorine 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-F. According to VSEPR theory, the bond pairs lead the BF3 molecule to take on a trigonal planar molecular geometry shape.

The BF3 molecule’s B-F bonds are arranged in symmetrical order around the trigonal planar molecular geometry, giving rise to the BF3 molecular shape. The BF3 molecule has a trigonal planar molecular geometry because there is no electrical repulsion between lone pairs and three bond pairs(B-F) of the BF3 molecule.

Electronegative Difference Calculation of BF3 Molecule:

The boron atom has an electronegativity of 2.04, while fluorine has an electronegativity of 3.98 in the BF3 molecule. The difference in electronegativity can be estimated using the method below.

The electronegative value difference between boron and fluorine
Electronegativity value of boron = 2.04
Electronegativity value of fluorine= 3.98
Difference of electronegativity value between boron and fluorine= 3.98 – 2.04=1.94
Electronegativity difference between B-F bond calculation of BF3 molecule

Due to the difference in electronegativity value of greater than 0.5, the B-F bond of the BF3 molecule becomes polar. Because of this difference in electronegativity, the BF3 molecule’s B-F bond becomes polar. The total net dipole moment of the BF3 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-F bond’s dipole moment is high due to the polarization of the bonds and no lone pairs of electrons, and all B-F bonds’ dipoles are arranged in the symmetrical BF3 molecular geometry. The BF3 molecule’s total dipole moment is predicted to be 0 D. It has a partial negative charge for the terminal fluorine atoms and a partial positive charge for the central boron atom.

The electron dot structure of the BF3 molecule is also known as the BF3 Lewis structure. It determines the number of outermost valence electrons as well as the electrons engaged in the BF3 molecule’s bond formation. The outermost valence electrons of the BF3 molecule must be understood while considering the Lewis structure of the molecule.

The boron atom is the middle element in BF3 molecular geometry, with three electrons in its outermost valence electron shell, whereas the fluorine atom has seven electrons in its outermost valence electron shell.

The BF3 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 fluorine 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-F bond pairs. The bond angle of the F-B-F 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 BF3 molecule.

There are two types of bonds in the BF3 molecule. Due to the resonance structure of the BF3 molecule creates a double bond and a single bond. The double bond of B-F is shorter as compared to the single bond of B-F. The B-F bond length of the double bond is 130 pm(picometer). The B-F bond length of the single bond is 143 pm(picometer).

To sketch the BF3 Lewis structure by following these instructions:

Step-1: BF3 Lewis dot Structure by counting valence electrons on the boron atom

To calculate the valence electron of each atom in BF3, 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 fluorine and boron atoms. In their outermost shells, fluorine and boron have seven and three valence electrons, respectively.

Because fluorine 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 BF3 molecule’s outermost valence shell. The first step is to determine how many electrons are in the BF3 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 BF3 Lewis diagram. The BF3 molecule’s core boron atom can be represented as follows:

Total outermost valence shell electron of boron atom in BF3= 3
Total outermost valence shell electron of the fluorine atom in BF3= 7
The BF3 molecule has one central boron and three fluorine atoms. Then the total outermost valence shell electrons can be calculated as follows
∴ Total outermost valence shell electrons available for BF3 Lewis structure( dot structure) = 3+3*7= 24 valence electrons in BF3.
calculation of total valence electron of BF3 molecule

Choose the atom with the least electronegative value atom and insert it in the center of the molecular geometry of BF3. We’ll choose the least electronegative value atom in the BF3 molecule to place in the center of the BF3 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.

Step-2: Lewis Structure of BF3 for counting valence electrons around the terminal fluorine 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 fluorine atom in the BF3 molecule. Furthermore, boron has a three electrons limit since fluorine is the most electronegative element in the BF3 molecule.

In the BF3 Lewis structure diagram, the boron atom can be the center atom of the molecule. As a result, central boron in the BF3 Lewis structure, with all three fluorine arranged in the trigonal planar geometry.

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

Step-3: Lewis dot Structure for BF3 generated from step-1 and step-2

Connect the exterior and core central atom of the BF3 molecule with three single bonds (B-F). In this stage, use three fluorine atoms on the outside of the BF3 molecule to the central boron atom in the middle.

Count how many electrons from the outermost valence shell have been used in the BF3 structure so far. Each B-F bond carries two electrons because each boron atom is connected to three fluorine atoms by three B-F bonds. Bond pairings of B-F are what they’re called.

So, out of the total of 24 valence electrons available for the BF3 Lewis structure, we used 6 for the BF3 molecule’s three B-F bonds. The BF3 molecule has no lone pairs of electrons in the central boron atom. We need to put extra electrons in the molecular geometry of BF3. Where to place the extra electron in the BF3 molecule?

Place the valence electrons in the B-F bond pairs starting with the core boron, three fluorine, and no lone pairs of electrons in the BF3 molecule. In the BF3 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. BF3 molecule goes on resonance. A fluorine atom has three pairs of lone pairs of electrons. Due to the back bonding mechanism of B-F of BF3 molecule, fluorine gives one lone pair of electrons to the bond pairs of B-F bond. This makes a single bond into a double bond.

This makes boron negative in charge and fluorine makes positive in charge. The double bond resonated around the BF3 molecule. The double bond of B-F is shorter as compared to the single B-F 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-F bonds. Then no electrons as a lone pair of electrons on the boron atom of the BF3 molecule are placed in a trigonal planar geometry. Boron already shares 6 electrons to the three B-F bonds. Then place the valence electron in the fluorine atom, it placed around seven electrons(step-2). Totally, 21 valence electrons placed on the three fluorine atoms of the BF3 molecule.

We’ve positioned 6 electrons around the central fluorine atom(step-3), which is represented by a dot, in the BF3 molecular structure above. The boron atom completes its molecular stability in the BF3 molecule because it possesses 6 electrons in its (B-F) bond pairs with three fluorine in the outermost valence shell.

Count how many outermost valence shell electrons have been used so far using the BF3 Lewis structure. Three electron bond pairs are shown as dots in the BF3 chemical structure, whereas three single bonds each contain two electrons. The outermost valence shell electrons of the BF3 molecule(bond pairs) are six as a result of the calculation.

So far, we’ve used six of the BF3 Lewis structure’s total 6 outermost valence shell electrons. No lone pairs of electrons on the boron atom in the trigonal planar of the BF3 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-F and two single bond pairs of B-F on the boron central atom of BF3.

The core atom in the BF3 Lewis structure is boron, which is bonded to the three fluorine atoms by single bonds (B-F). 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 BF3 molecule’s trigonal planar geometry.

Watch video of BF3 Lewis Structure

How to calculate the formal charge on a boron atom in BF3 Lewis Structure?

The formal charge on the BF3 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 BF3 Lewis dot structure.

To calculate the formal charge on the central boron atom of the BF3 molecule by using the following formula:
The formal charge on the boron atom of BF3 molecule= (V. E(B)– L.E(B) – 1/2(B.E))
V.E (B) = Valence electron in a boron atom of BF3 molecule
L.E(B) = Lone pairs of an electron in the boron atom of the BF3molecule.
B.E = Bond pair electron in B atom of BF3 molecule
calculation of formal charge on boron atom in BF3 molecule

The boron core atom (three single bonds connected to three fluorine atoms ) of the BF3 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 BF3 molecule = (3- 0-(6/2)) =0

In the Lewis structure of BF3, the formal charge on the central boron atom is zero.

Summary:

In this post, we discussed the method to construct the BF3 Lewis structure. First, the valence electrons are placed around the boron atom. Second, place the valence electron on the fluorine atom. Finally, when we combined the first and second steps. It gives BF3 Lewis structure. Need to remember that, if you follow above said method, you can construct molecular dot structure very easily.