Tetrafluoroborate (BF4-) ion Lewis structure, molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges, polar or nonpolar
BF4– represents the extremely versatile tetrafluoroborate ion. It is a highly stable anion, resistant to hydrolysis and thermal degradation. It is used in synthesizing metal complexes and as a catalyst in organic reactions.
The tetrafluoroborate ion is also making its place in the modern scientific world by being used as an electrolyte in lithium-ion batteries and for preparing novel extraction solvents called ionic liquids.
This article is very important for you if you want to learn how to draw the Lewis dot structure of BF4–. We will also discuss the molecular geometry or shape, electron geometry, bond angles, hybridization, formal charges and polarity of BF4–.
The Lewis dot structure of the tetrafluoroborate (BF4–) ion comprises a boron (B) atom at the center. It is surrounded by 4 fluorine (F) atoms at the sides. There is no lone pair of electrons on the central B-atom; however, each peripheral F-atom carries 3 lone pairs.
Drawing the Lewis structure of BF4– is quite easy if you follow the simple steps given below.
Steps for drawing the Lewis dot structure of BF4–
1. Count the total valence electrons present in BF4–
BF4– consists of two distinct elements, i.e., boron and fluorine.
Boron (B) is present in Group III A (or 13) of the Periodic Table of Elements. Thus, it has a total of 3 valence electrons in each atom.
In contrast, fluorine (F) is a halogen present in Group VII A (or 17), containing 7 valence electrons in each atom.
Total number of valence electrons in boron = 3
Total number of valence electrons in fluorine = 7
The BF4– ion comprises 1 B-atom and 4 F-atoms.
An important point to remember is that the BF4– ion carries a negative (-1) charge, which means 1 extra valence electron is added in this Lewis structure.
∴ Therefore, the total valence electrons available for drawing the Lewis dot structure of BF4– = 1(3) + 4(7) = 31 + 1 =32 valence electrons.
2. Find the least electronegative atom and place it at the center
By convention, the least electronegative atom out of all those available is chosen as the central atom while drawing the Lewis structure of a molecule or molecular ion.
The least electronegative atom can easily form covalent bonds with other atoms by sharing its electrons.
Among the two types of atoms present in BF4–, fluorine (E.N = 3.98) is the most electronegative element in the Periodic Table; therefore, it cannot be chosen as the central atom.
Boron (E.N = 2.04), in contrast, is certainly less electronegative than fluorine. Therefore, it is placed as the central atom in the BF4– Lewis structure while the 4 F-atoms occupy terminal positions, as shown below.
3. Connect the outer atoms with the central atom
In this step, all the outer F-atoms are joined to the central B-atom using single straight lines.
A straight line represents a single covalent bond, i.e., a bond pair containing 2 electrons.
In the above structure, there are 4 single bonds, i.e., 4(2) = 8 valence electrons are already consumed out of the 32 initially available.
Now let’s explore where we can place the remaining valence electrons.
4. Complete the octet of the outer atoms
An F-atom needs a total of 8 valence electrons to gain a full octet configuration.
A B-F bond represents 2 valence electrons surrounding each F-atom, which denotes a deficiency of 6 valence electrons.
Therefore, 3 lone pairs are placed around each F-atom in the BF4– Lewis structure.
5. Complete the octet of the central atom
Total valence electrons used till step 4 = 4 single bonds + 4 (electrons placed around each F-atom, shown as dots) = 4(2) + 4(6) = 32 valence electrons.
Total valence electrons – electrons used till step 4 = 32 – 32 = 0 valence electrons.
Hence there is no lone pair of electrons on the central B-atom in the BF4– Lewis dot structure.
As a final step, we just need to check the stability of the BF4– Lewis structure obtained above. Let’s do that by applying the formal charge concept.
6. Check the stability of Lewis’s structure using the formal charge concept
The less the formal charge on the atoms of a molecule or molecular ion, the better the stability of its Lewis structure.
The formal charges can be calculated using the formula given below.
Formal charge = [valence electrons- nonbonding electrons- ½ (bonding electrons)].
Now let us use this formula and the Lewis structure obtained in step 5 to determine the formal charges present on the BF4– bonded atoms.
As per the above calculation, zero or no formal charges are present on either of the F-atoms, while the central B-atom carries a -1 formal charge which is also the charge present on BF4– ion overall.
This proves we have drawn the correct Lewis structure for the tetrafluoroborate (BF4–) ion.
Now let’s move ahead and discuss its electron and molecular geometry or shape.
What are the electron and molecular geometry of BF4-?
The tetrafluoroborate (BF4–) ion possesses an identical electron and molecular geometry or shape, i.e., tetrahedral. There is no lone pair of electrons on the central B-atom; thus, no distortion is witnessed in its shape and/or geometry.
Molecular geometry of BF4–
The molecular geometry or shape of the tetrafluoroborate (BF4–) ion is tetrahedral.
There is no lone pair of electrons on the central B-atom in BF4– thus, no lone pair-lone pair or lone pair-bond pair electronic repulsions exist in the molecular ion.
The four F-atoms occupy the four vertices of a tetrahedron in a perfectly symmetrical manner. Hence forming a tetrahedral shape.
Electron geometry of BF4–
According to the valence shell electron pair repulsion (VSEPR) theory of chemical bonding, the ideal electron geometry of a molecule or molecular ion containing a total of 4 electron density regions around the central atom is tetrahedral.
In BF4–, the B-atom at the center is surrounded by 4 bond pairs, i.e., four B-F bonds, and it has no lone pair of electrons, making a total of 4 electron density regions only.
Hence, the ideal electron pair geometry of BF4– is tetrahedral.
An easy trick to finding the electron and molecular geometry of a molecular ion is using the AXN method.
AXN is a simple formula representing the number of bonded atoms and lone pairs on the central atom.
It is used to predict the shape and geometry of a molecular ion using the VSEPR concept.
AXN notation for BF4–
A in the AXN formula represents the central atom. In BF4–, a boron (B) atom is present at the center, so A = B.
X denotes the atoms bonded to the central atom. In BF4–, 4 F-atoms are directly bonded to the central B-atom. So, X = 4 for BF4–.
N stands for the lone pairs present on the central atom. As per the Lewis structure of BF4–, the central B-atom has no lone pairs of electrons. Thus, N = 0 for BF4–.
As a result, the AXN generic formula for BF4– is AX4N0 or simply AX4.
Now, you may have a look at the VSEPR chart below.
The VSEPR chart confirms that the electron and molecular geometry or shape of a molecule or molecular ion having an AX4 generic formula is tetrahedral, as we already noted down for BF4–.
Hybridization of BF4–
The central B-atom is sp3 hybridized in BF4–.
The electronic configuration of boron (B) is 1s2 2s2 2p1.
A -1 formal charge indicates that the electronic configuration of boron becomes 1s2 2s2 2p2 during the excited state.
During chemical bonding in BF4–, one of the two 2s electrons of boron shifts to its empty 2p atomic orbital. These one 2s and three 2p orbitals consequently hybridize to produce four sp3 hybrid orbitals.
Each sp3 hybrid orbital of boron overlaps with the half-filled p-orbitals of fluorine to form the B-F sigma (σ) bonds.
Refer to the figure drawn below.
A shortcut to finding the hybridization present in a molecule or molecular ion is using its steric number against the table shown below.
The steric number of the central B-atom in BF4– is 4, so it has sp3 hybridization.
Steric number
Hybridization
2
sp
3
sp2
4
sp3
5
sp3d
6
sp3d2
7
sp3d3
The bond angles of BF4–
As per the symmetrical tetrahedral shape and geometry of the tetrafluoroborate (BF4–) ion, each F-B-F bond angle is equal to 109.5°.
As per Pauling’s electronegativity scale, a polar covalent bond is formed between two dissimilar atoms with an electronegativity difference between 0.4 and 1.6 units.
In BF4–, each B-F bond is strongly polar, with a high electronegativity difference of 1.94 units present between a boron (E.N = 2.04) and a fluorine (E.N = 3.98) atom.
The extremely electronegative F-atoms attract the shared electron cloud away from the central B-atom in each B-F bond. The central B-atom thus gains a partial positive charge (δ+), while the terminal F-atoms obtain partial negative charges (δ–).
However, it is due to the symmetrical tetrahedral shape of BF4– that the net effect of three downwards-pointing B-F dipole moments gets canceled with an upwards-pointing B-F dipole moment.
The Lewis dot structure for the tetrafluoroborate (BF4–) ion displays a total of 32 valence electrons, i.e., 32/2 = 16 electron pairs.
Out of the 16 electron pairs, there are 4 bond pairs and 12 lone pairs of electrons.
A boron (B) atom is present at the center of the BF4– Lewis structure, while the F-atoms occupy terminal positions.
There is no lone pair of electrons on the central B-atom, while 3 lone pairs are present on each F-atom.
What is the molecular geometry or shape of BF4–?
The molecular geometry or shape of BF4– is tetrahedral. The central B-atom is bonded to four F-atoms like four corners of a tetrahedron, while there is no lone pair of electrons present on it.
Is the molecular shape of BF4– the same or different from its electronic geometry?
Same. BF4– possesses an identical electron and molecular geometry or shape, i.e., tetrahedral. The absence of any lone pairs of electrons on the central B-atom implies that there is no distortion seen in its molecular shape.
How is the shape of BF3 different from that of BF4–?
The molecular shape of BF3 is trigonal planar, while that of BF4– is tetrahedral.
In BF3, 3 F-atoms are bonded to the central B-atom like three corners of an equilateral triangle.
Contrarily, in BF4–, 4 F-atoms are covalently bonded to the central B-atom like four vertices of a tetrahedron.
How are the shapes of the following molecular ions the same or different?
Why is the shape of XeF4 different from that of BF4– although both possess 4 F-atoms?
In xenon tetrafluoride (XeF4), the central Xe-atom possesses 2 lone pairs of electrons.
Lone pair-lone pair and lone pair-bond pair repulsions make the molecule occupy a squareplanar shape in which the 4 F-atoms form a square base. The lone pairs lie above and below the central Xe-atom at equal distances.
In comparison, there is no lone pair of electrons on the central B-atom in BF4–. The central B-atom bonded to four F-atoms forms a tetrahedral shape and molecular geometry.
The total number of valence electrons available for drawing the tetrafluoroborate (BF4–) ion Lewis structure is 32.
BF4– possesses an identical electron and molecular geometry or shape, i.e., tetrahedral.
The central B-atom is sp3 hybridized in BF4–.
Each F-B-F bond angle is equal to 109.5° in the tetrafluoroborate ion.
BF4– is overall non-polar as the dipole moments of individually polar B-F bonds get canceled equally in opposite directions.
Zero or no formal charges are present on the F-atoms, while the central B-atom carries a -1 formal charge which is also the charge present on the tetrafluoroborate ion overall.
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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