Drawing and predicting the BF3 molecular geometry is very easy by following the given method. Here in this post, we described step by step to construct BF3 molecular geometry. Boron comes from the 13th family group in the periodic table. Boron has three valence electrons. It forms different types of polymeric inorganic compounds.
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Key Points To Consider When drawing The BF3 Molecular Geometry
A three-step approach for drawing the BF3 molecular can be used. The first step is to sketch the molecular geometry of the BF3 molecule, to calculate the lone pairs of the electron in the central boron atom; the second step is to calculate the BF3 hybridization, and the third step is to give perfect notation for the BF3 molecular geometry.
The BF3 molecular geometry is a diagram that illustrates the number of valence electrons and bond electron pairs in the BF3 molecule in a specific geometric manner. The geometry of the BF3 molecule can then be predicted using the Valence Shell Electron Pair Repulsion Theory (VSEPR Theory) and molecular hybridization theory, which states that molecules will choose the BF3 geometrical shape in which the electrons have from one another in the specific molecular structure.
Finally, you must add their bond polarities characteristics to compute the strength of the B-F bond (dipole moment properties of the BF3 molecular geometry). The Boron-fluorine bonds in the boron trifluoride(BF3), for example, are polarised toward the more electronegative value fluorine atom, and because all (B-F) bonds have the same size and polarity, their sum is zero due to the BF3 molecule’s bond dipole moment due to it oppose to each other in the trigonal planar geometry, and the BF3 molecule is classified as a polar molecule.
The molecule of boron trifluoride (with trigonal planar shape BF3 molecular geometry) is tilted at 120 degrees bond angle of F-B-F. It has a difference in electronegativity values between boron and fluorine atoms, with fluorine’s pull the electron cloud being greater than boron’s. But bond polarity of B-F canceled to each other in the trigonal planar geometry. As a result, it has no permanent dipole moment in its molecular structure. The BF3 molecule has no dipole moment due to an equal charge distribution of negative and positive charges.
Overview: BF3 electron and molecular geometry
According to the VSEPR theory, BF3 possesses trigonal planar molecular geometry. Because the center atom, boron, has three B-F bonds with the fluorine atoms surrounding it. The F-B-F bond angle is 120 degrees in the trigonal planar molecular geometry. The BF3 molecule has a trigonal planar geometry shape because it contains three fluorine atoms.
There are three B-F bonds at the BF3 molecular geometry. After linking the three fluorine atoms and no lone pairs of electrons in the trigonal planar form, it maintains the planar-T-like structure. In the BF3 molecular geometry, the B-F bonds have stayed in the three terminals and no lone pairs of electrons in the top and bottom of the trigonal planar molecule.
The center boron atom of BF3 has no lone pairs of electrons, resulting in trigonal planar electron geometry. However, the molecular geometry of BF3 looks like a trigonal planar and no lone pairs on the top and bottom of the BF3 geometry. It’s the BF3 molecule’s symmetrical geometry. As a result, the BF3 molecule is nonpolar.
How to find BF3 hybridization and molecular geometry
Calculating lone pairs of electrons on boron in the BF3 geometry:
1.Determine the number of lone pairs on the core boron atom of the BF3 Lewis structure. Because the lone pairs on boron are mostly responsible for the BF3 molecule geometry distortion, we need to calculate out how many there are on the central boron atom of the Lewis structure.
Use the formula below to find the lone pair on the boron atom of the BF3 molecule.
L.P(B) = V.E(B) – N.A(B-F)/2
Lone pair on the central boron atom = L.P(B)
The core central boron atom’s valence electron = V.E(B)
Number of B-F bonds = N.A (B-F)calculation for boron atom lone pair in BF3 molecule
For instance of BF3, the central atom, boron, has three electrons in its outermost valence shell, three B-F bond connections.
As a result of this,L.P(B) = (3 –3)/2=0
In the BF3 electron geometry structure, the lone pair on the central boron atom is zero. It means there are no lone pairs of electrons in the core boron atom. This makes BF3 molecule electron deficient. It makes coordination with ammonia. No lone pairs of electrons on the central boron atom are responsible for the planar nature of BF3 molecular geometry.
If you imagine, there is no lone pair on the boron atom of the BF3 molecule. Then, electronic repulsion of B-F bonds pair and zero lone pair of electrons in the BF3. That gives stable trigonal planar geometry. No lone pairs of electrons are located on the top and bottom of the geometry. It makes a stable trigonal planar structure.
But in reality, the BF3 molecule undergoes distortion in its geometry due to the polarity of the B-F bond and no lone pairs of electrons in the trigonal planar geometry. This leads to a trigonal planar for the BF3 molecule.
Calculate the number of molecular hybridizations of the BF3 molecule
What is BF3 hybridization? This is a very fundamental question in the field of molecular chemistry. All the molecules made by atoms. In chemistry, atoms are the fundamental particles. There are four different types of orbitals in chemistry. They are named as s, p, d, and f orbitals.
The entire periodic table arrangement is based on these orbital theories. Atoms in the periodic table are classified as follows:
s- block elements
p- block elements
f-block elementsAtoms are classified in the periodic table
BF3 molecule is made of one boron and three fluorine atoms. The boron atom has s and p orbitals. Fluorine comes as the first element from the halogen family in the periodic table. The fluorine atom has s and p orbitals.
When these atoms combine to form the BF3 molecule, its atomic orbitals mixed and form unique molecular orbitals due to hybridization.
How do you find the BF3 molecule’s hybridization? We must now determine the molecular hybridization number of BF3.
The formula of BF3 molecular hybridization is as follows:
No. Hyb of BF3= N.A(B-F bonds) + L.P(B)
No. Hyof BF3= the number of hybridizations of BF3
Number of B-F bonds = N.A (B-F bonds)
Lone pair on the central boron atom = L.P(B)Calculation for hybridization number for BF3 molecule
In the BF3 molecule, the boron is a core central atom with three fluorine atoms connected to it and no lone pairs of electrons. The number of BF3 hybridizations (No. Hyb of BF3) can then be estimated using the formula below.
See more: How Do You Draw A Parallelogram With One Ruler, How To Draw A Parallelogram (With Pictures)
No. Hyb of BF3= 3+0 =3
The BF3 molecule hybridization is three. The boron atom has s and p orbitals. The fluorine atom has s and p orbital. The sp2 hybridization of the BF3 molecule is formed when one S orbital and two p orbitals join together to form a molecular orbital.