Borane (BH3) lewis dot structure, molecular geometry, polar or nonpolar, hybridization, Bond angle
Borane is also known as boron trihydride is an unstable and colorless gas with the chemical formula BH3. It is a highly reactive molecule.
In this tutorial, we will discuss Borane (BH3) lewis structure, molecular geometry, Bond angle, hybridization, polar or nonpolar, etc.
Borane is a strong lewis acid and its conjugate acid is Boronium ion.
|Name of Molecule||Borane (BH3)|
|Molecular geometry of BH3||Trigonal planar|
|Electron geometry OF BH3||Trigonal planar|
|Bond angle (H-B-H)||120º|
|Total Valence electrons of BH3||24|
|Overall formal charge in BH3||Zero|
How to draw lewis structure of BH3 (Borane)?
Borane (BH3) lewis structure is made up of three B-H bonds, with boron (B) in a central position and all three hydrogens (H) as outer atoms in the lewis diagram. The lewis structure of BH3 contains a total of 3 bond pairs and 0 lone pairs.
The drawing of the BH3 lewis’s structure is very easy and simple. Let’s see how to do it.
Steps for drawing the Lewis dot structure for BH3
1. Count total valence electron in BH3
First of all, determine the valence electron that is available for drawing the lewis structure of BH3 because the lewis diagram is all about the representation of valence electrons around atoms.
So, an easy way to find the valence electron of atoms in the BH3 molecule is, just to look at the periodic group of boron and hydrogen atoms.
As the boron atom belongs to the 13th group in the periodic table and hydrogen is situated in the 1st group, hence, the valence electron for the boron is 3, and for the hydrogen atom, it is only 1.
⇒ Total number of the valence electrons in boron = 3
⇒ Total number of the valence electrons in hydrogen = 1
∴ Total number of valence electrons available for the BH3 Lewis structure = 3 + 1×3 = 6 valence electrons [∴ BH3 molecule has one boron and three hydrogen atoms]
2. Find the least electronegative atom and place it at center
An atom with a less electronegative value is preferable for the central position in the lewis diagram because they are more prone to share the electrons with surrounding atoms.
It should be noted that “Hydrogen always go outside in lewis diagram” Because, Hydrogen atom can form only one single bond.
Hence, put the boron atom at the central position of the lewis diagram and all three hydrogen atoms outside to it.
3. Connect outer atoms to central atom with a single bond
In this step, join all outer atoms to the central atom with the help of a single bond.
In, the BH3 molecule, Hydrogen is the outer atom, and boron is the central atom. Hence, joined them as shown in the figure given below.
Count the number of valence electrons used in the above structure. There are 3 single bonds used in the above structure, and one single bond means 2 electrons.
Hence, in the above structure, (3 × 2) = 6 valence electrons are used.
And we have a total of 6 valence electrons available for drawing the BH3 Lewis structure.
∴ (6 – 6) = 0 valence electrons
So, we are left with 0 valence electrons.
4. Check the octet of all atoms
In the 3rd step structure, the hydrogen atoms completed their octet since they have 2 electrons(one single bond means 2 electrons) in their outer shell.
Hydrogen only need 2 electrons to fulfill the outer shell.
We don’t have any extra valence electrons left and the central atom boron has only 6 electrons(3 single bonds) in its valence shell.
It should be noted that Boron is exceptional to the octet rule as it can have 8 electrons or less than 8 electrons in the outermost shell to attain stability. Boron is an exception just like aluminum where it can be octet deficient.
Octet deficient molecules are the molecules that can attains the stability by having less than 8 electrons around the atoms. Some examples – Boron, beryllium, aluminium, hydrogen, lithium, helium
But boron and aluminium is two most common element that can fail to complete the octet as they attains stability having only 6 valence electrons.
Let’s check the formal charge for the above structure to verify it’s stable or not.
5. Check the stability with the help of a formal charge concept
The lesser the formal charge on atoms, the better is the stability of the lewis diagram.
To calculate the formal charge on an atom. Use the formula given below-
⇒ Formal charge = (valence electrons – nonbonding electrons – 1/2 bonding electrons)
Let’s count the formal charge for the 3th step structure.
For hydrogen atom
⇒ Valence electrons of hydrogen = 1
⇒ Nonbonding electrons on hydrogen = 0
⇒ Bonding electrons around hydrogen (1 single bond) = 2
∴ (1 – 0 – 2/2) = 0 formal charge on the hydrogen atoms.
For boron atom
⇒ Valence electrons of boron = 3
⇒ Nonbonding electrons on boron = 0
⇒ Bonding electrons around boron (3 single bonds) = 6
∴ (3 – 0 – 6/2) = 0 formal charge on the boron central atom.
Borane (BH3) Lewis structure
Hence, in the above BH3 lewis structure, all atoms get a formal charge equal to zero. Even the boron central atom has only 6 electrons instead of 8 in the valence shell, it also gets a formal charge equal to zero.
We have no required to form multiple bonds and provide boron atom to 8 electrons in its valence shell.
The boron can achieve stability by just having 6 electrons in the valence shell.
Therefore, the above lewis structure of BH3 (Borane) is most stable and appropriate in nature.
What are the electron and molecular geometry of BH3?
In BH3, there are 3 regions of electron density around the central atom, all are the bonding regions. Therefore, its electron geometry is Trigonal planar, and its molecular geometry is Trigonal planar, too.
According to the VSEPR theory, the central atom with three regions of electron density adopts a trigonal planar geometry. Because repulsion is minimum in electron pairs at this position.
“A region of electron density means the group of bonding or nonbonding electrons that present around the atom. The single bond, double bond, or even triple bond around the atom will be counted as one region.”
The electron pair around the Boron central atom will repel each other and tried to go far from each other, they will take the position where repulsion becomes minimum between them.
According to the VSEPR theory, “the maximum distance three regions of electron density can get away from affords a geometry called Trigonal planar.”
Therefore, the molecular geometry or shape of BH3 is Trigonal planar.
We can also find the electron and molecular geometry of BH3 using the AXN method and VSEPR chart.
AXN is a simple formula that represents the number of the bonded atom and lone pair on the central atom to predict the shape of the molecule using the VSEPR chart.
AXN notation for BH3 molecule:
- A denotes the central atom, so, Boron is the central atom in BH3 molecule A = Boron
- X denotes the bonded atoms to the central atom, Boron is bonded with three hydrogens atoms. Therefore, X = 3
- N represents the lone pair on the central atom, as per BH3 Lewis structure, the Boron central atom has zero lone pair. Hence, N = 0
So, the AXN generic formula for the BH3 molecule becomes AX3N0 or AX3.
As per the VSEPR chart, if a molecule gets AX3 generic formula then its molecular geometry will be trigonal planar and electron geometry will also be trigonal planar.
Therefore, the molecular geometry for BH3 is trigonal planar and its electron geometry is also trigonal planar.
Hybridization of BH3
The hybridization of BH3 is Sp2 because the steric number of the boron central atom is three.
The formula for calculating the steric number is-
Steric number = (Number of bonded atoms attached to central atom + Lone pair on central atom)
In the case of the BH3 molecule, boron is the central atom that is attached to the three bonded atoms(hydrogen) and it has no lone pairs.
Hence, (3 + 0) = 3 is the steric number of central atom boron in the BH3 molecule that gives Sp2 hybridization.
The bond angle of BH3
Since, we know, the molecular geometry of BH3 is Trigonal planar, which means, all the atoms lie in the same plane. The three B-H bonds are arranged in the same plane with a 120º bond angle to each other.
∴ The H-B-H bond angle in BH3 is 120º.
Is BH3 polar or nonpolar?
So, Is BH3 polar or nonpolar? Well, it is obvious that BH3 is a nonpolar molecule, because each B-H bond is directed at the angle of 120° to each other in a plane, hence, canceling of dipole moment generated along these bonds is very easy.
Therefore, no dipole moment is generated in the BH3 molecule, hence, it is nonpolar in nature.
Also, the molecular geometry of BH3 is very symmetrical since no lone pair is present on the central atom that can cause distortion in a molecule, so, the charges are distributed uniformly all over the atoms.
Hence, the dipole generated in the BH3 molecule will easily cancel out each other leaving this molecule nonpolar in nature.
How many lone pairs are present in the lewis structure of BH3?
Lone pairs are those represented as dots in the lewis diagram that do not take part in the formation of bonds and are also called nonbonding electrons.
By looking at the BH3 Lewis structure, we see, that there are no dot electrons present, only single bonds are present.
So, in the BH3 Lewis structure, there is no lone pair, not on outer atoms, and not on the central atoms as well.
The total number of bond pairs present in the lewis structure of BH3?
Bonding pairs are the pair of electrons that are in a bond. A single bond has one bond pair means 2 bonding electrons.
Two bonding electron between the atoms forms a single covalent bond.
Now, as per the BH3 Lewis structure, the central atom boron is attached with three single covalent bonds, and one single covalent bond means 2 bonding electrons.
Hence, the total bonding electrons is (3 × 2) = 6 bonding electrons that make 3 bond pairs.
∴ In the BH3 Lewis structure, a total of 3 bond pairs are present.
Why electron and molecular geometry of BH3 are same?
Two types of geometry can be predicted with the help of VSEPR theory- (a). Electron geometry (b). Molecular geometry
Electron geometry considers all electrons(Bonding and Lone pair electrons) whereas molecular geometry considers only Bonding pairs to determine the geometry of any molecule.
As we know, the molecular geometry of BH3 is trigonal planar and electron geometry is also trigonal planar.
Hence, the molecular geometry and electron geometry of BH3 is the same.
Why Boron central atom of the BH3 lewis structure complete the octet in just 6 electrons?
In the case of the BH3 Lewis structure, the Boron central atom gets a formal charge equal to zero when it has 6 electrons around. But when a Boron central atom is distributed with 8 electrons it gets an uneven formal charge.
Hence, we have to choose the lewis diagram that has the least formal charge on each atom, Therefore, the Boron central atom is provided with only 6 electrons instead of 8 for completing the octet shell.
Why the molecular geometry of BH3 is Trigonal planar?
The molecular geometry of BH3 is Trigonal planar, because, the central atom Boron is attached to three bonded regions, they will repel each other as much as possible.
According to VSEPR theory, the three bonded regions can be at a maximum distance from each other when they afford a geometry called Trigonal planar.
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Properties of Borane
- It is strong Lewis acid.
- It is a highly unstable and colorless gas.
- Its B–H bond length is 119 pm.
- It is highly reactive in nature.
- It is stable at very low temperatures.
- The aqueous solution of BH3 is extremely unstable.
Reactions of Borane
Borane can react with itself to form diborane because of its high instability.
⇒ 2BH3 → B2H6
In reaction with water molecules, it forms boric acid.
⇒ BH3 + 3H2O → B(OH)3 + 3H2
- The total valence electron available for drawing the Borane (BH3) Lewis structure is 6.
- The molecular geometry or shape of BH3 is Trigonal planar.
- The electron geometry of BH3 is Trigonal planar as its central atom has 3 regions of electron density.
- Lewis dot structure of BH3 contains 1 lone pair on the central atom(boron) and 0 lone pairs on outer atoms(hydrogens).
- The hybridization of boron in BH3 is sp2. Since its steric number is 3.
- The bond angle in BH3 is 120º. According to VSEPR, the BH3 has a Trigonal planar geometry with each H-B-H bond angle equal to 120º.
- The net dipole moment of BH3 is zero, hence, it is nonpolar in nature.
- The overall formal charge in BH3 is zero.