is bbr3 polar or nonpolar?

Answer and Explanation: Boron tribromide (BBr ) is nonpolar. Given its symmetrical trigonal planar geometry, there is no dipole moment, making it nonpolar.

BBr3 (Boron Tri- bromide) is a incensed liquid with no colour or amber in colour. It is quite toxic to inhale. It has a sharp (irritating / pungent) smell. Boron tri- bromide has molecular weight 250.53. The IUPAC name of Boron tri- bromide is Tri- bromo borone. In this editorial we are learning about the BBr3 lewis structure and its various facts.

While drawing any lewis structure main points to be noted are valence electrons of molecule, bonding with central atom, octet rule follow, formal charge count. In BBr3 there is one boron and three bromine atoms are present.

In BBr3 lewis structure, the boron atom is situated centrally surrounded by three bromine atom. The atom should be at central position which has low electronegativity. Boron has electronegativity 2.04 and bromine has electronegativity 2.96. That’s why boron atom with lowest electronegativity should be at central position.

The BBr3 lewis structure has three B-Br bonds, thus it has three bond electron pairs and nine lone electron pairs. As we selected the boron as central atom, the three bromine atoms get linked to the boron atom.

To calculate the valence electrons in BBr3 lewis structure, first we have to check the positions of boron and bromine atom in the periodic table. As the boron atom belongs to 13th group of periodic table, the B atom contains three valence electrons in its outer orbital. Similarly Bromine atom belongs to the 17th group of periodic table, so it contains seven valence electrons in its outer orbital.

Total valence electrons in B atom = 3

Total valence electrons in Br atom = 7

Total valence electrons in BBr3 lewis structure = 3 (B) + 7 x 3 (Br) = 24

Thus, the BBr3 lewis structure has 24 total valence electrons.

If we are using six electrons in bonding between boron and bromine atom, so we are left with total eighteen valence electrons for distribution on three bromine atoms. Therefore, three bonds containing two electrons each ( 3 x 2 = 6 ), so we have 24 – 6 = 18 electrons for sharing.

The octet rule says there should be eight electrons present to complete the octet of any element or atom. Now, we have remaining eighteen valence electrons for sharing in BBr3 lewis structure. So, put the remaining 18 electrons first on outer three bromine atoms to complete its octet.

As we have put all the remaining 18 valence electrons on three bromine atoms, so the single bromine atom now contains 8 electrons i.e. two bond pair electron in each single B-Br bonds and six valence electrons on each single bromine atom. So, the outer three bromine atoms of BBr3 lewis structure has complete octet.

Now we have used all the eighteen electrons by sharing on three bromine atoms. So, we don’t have more valence electrons remain for sharing. Thus, the central boron atom has only six electrons i.e. only three bond pair electrons containing two valence electrons each. So, the Boron atom has incomplete octet. Hence, in BBr3 lewis structure, B atom has incomplete octet and three Br atoms has complete octet.

There is a formula to count formal charge on any lewis structure as follows:

Formal charge = (valence electrons – non-bonding electrons – ½ bonding electrons)

The calculation for formal charge on BBr3 molecule as follows:

Boron atom: Valence electrons on boron = 03

Lone pair electrons on boron = 00

Bonding electrons with boron = 06 (three single bond)

Formal charge on boron = (3 – 0 – 6/2) = 0

So, the boron atom has zero formal charge.

Bromine atom: Bromine atom have Valence electrons = 07

Bromine atom have Lone pair electrons = 06

Bromine atom have Bonding electrons = 2 (one single bond)

Formal charge on iodine = (7 – 6 – 2/2) = 0

So, all the three bromine atoms in BBr3 molecule have zero formal charges.

The BBr3 lewis structure contains total twenty four valence electrons, out of which six valence electrons are being bond pairs as they involved in bonding between three bromine atoms with the central boron atom. Thus, we are remained with eighteen valence electrons for further sharing on outer bromine atoms.

As we have put all the 18 electrons on three Br atoms, so each Br atom has complete octet with 8 electrons. Hence, each Br atom has one bond pair electron and three lone electron pairs. Therefore, in BBr3 lewis structure, B atom has no lone electron but Br atom has 9 lone electron pairs.

As per the VSEPR theory, the generic formula the molecular geometry of BBr3 molecule is AX3. As the central boron atom is linked with three bromine atoms which has more electron density on it, so the BBr3 lewis structure has trigonal planar shape or geometry.

Hybridization of any molecule or lewis structure is determined by its steric number. To calculate the steric number of any molecule there is a formula:

Steric number = sum of number of bonded atoms joined to central atom and presence of lone electron pair on central atom

Steric number for BBr3 = 3 + 0 = 3

As the BBr3 lewis structure has 3 steric numbers, it is sp2 hybridized. So, the BBr3 lewis structure has sp2 hybridization.

The BBr3 lewis structure has trigonal planar geometry and also it is sp2 hybridized. As the central boron atom is joined with three bromine atom surrounded to it thus having three B-Br bonds. So each bromine boron bromine bond ( Br-B-Br ) has 120 degree bond angle within it. Hence the BBr3 lewis structure has 120 degree bond angle in its structure.

Any molecule can show the resonance structure only if in the molecule there are multiple (double / triple) bonds present and also it has some formal ( positive or negative ) charge with the presence of lone electron pairs on atoms of molecule.

In BBr3 lewis structure, there is no multiple bonds are present. All the three bromine atoms get attached with the central boron atom with single covalent bonds i.e. three covalent (B-Br) bonds in BBr3 lewis structure. Also the formal charge on B atom and Br atom is zero. So, the resonance structure of BBr3 lewis structure is not possible.

BBr3 (boron tri- bromide) is soluble in:

No, BBr3 molecule is not ionic compound. BBr3 molecule consists of two elements i.e. boron and three bromine atoms with are connected to each other with the covalent bonds. Also there is no negative or positive formal charge is present on B and Br atoms. Even the B and Br atoms are not showing the characteristics of being cation or anion.

Both B and Br atoms contains zero formal charge with covalent bonds, which makes BBr3 molecule a covalent compound. So, the BBr3 molecule is not ionic but it is a covalent inorganic compound. Hence, BBr3 is not ionic but covalent compound.

The one B and all three Br atoms are attached with each other with single B-Br covalent bonds which is a strong bond. So, there is no formation of ions with positive or negative charge. So, it is not ionic but covalent in nature.

BBr3 molecule is non- polar in nature, because the BBr3 molecule has a symmetrical arrangement of atoms in its structure. So the dipole which creates on B-Br molecule gets cancel each other makes it a non- polar molecule.

The B atom has 3 valence electrons and Br atom has 7 valence electrons, So Br needs only 1 electron for octet completion. As they forming three bonds with each other B atom shares its three valence electrons to three Br atoms and form covalent bonds. So, BBr3 molecule has symmetrical structure as each bromine atom making 120 degree bond angle with other Br atoms. Hence, all three Br atoms lie in a similar plane forming trigonal planar geometry.

BBr3 is non- polar as each B- Br bonds having 120 degree bond angle within the molecule in same plane, so they are cancelling the dipole moment produced within the bonds. Hence, there is zero dipole moment creates in BBr3 molecule making it non- polar in nature. As the dipole get cancel in BBr3 molecule it is non- polar in nature.

BBr3 molecule shows the characteristic of lewis acid. So, it is acidic in nature and not basic in nature.

The acceptor of electron pair is known to be an acid compound. In boron halides like BBr3, they are good acceptor of electron cloud to form BBr3 molecule. BBr3 is a strong lewis acid as the bromine atom has much electron cloud to donate to other atoms, hence it is a strong lewis acid in nature.

The drawing of the BBr3 lewis’s structure is very easy and simple. Let’s see how to do it.

1. Count total valence electron in BBr3

First of all, determine the valence electron that is available for drawing the lewis structure of BBr3 because the lewis diagram is all about the representation of valence electrons on atoms.

So, an easy way to find the valence electron of atoms in the BBr3 molecule is, just to look at the periodic group of boron and bromine atoms.

As the boron atom belongs to the 13th group in the periodic table and bromine is situated in the 17th group, hence, the valence electron for the boron is 3, and for the bromine atom, it is 7.

⇒ Total number of the valence electrons in boron = 3

⇒ Total number of the valence electrons in bromine = 7

∴ Total number of valence electron available for the BBr3 Lewis structure = 3 + 7×3 = 24 valence electrons         

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.

In the case of the BBr3 molecule, the boron atom is less electronegative than the bromine atom.

Hence, put the boron atom at the central position of the lewis diagram and all three bromine atoms outside it.

3. Connect outer atoms to the 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 BBr3 molecule, bromine is the outer atom, and boron is the central atom. Hence, joined them.

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 from a total of 24 valence electrons available for drawing the BBr3 Lewis structure.

∴ (24 – 6) = 18 valence electrons

So, we are left with 18 valence electrons more.

4. Place remaining electrons on the outer atom first and complete their octet

Let’s start putting the remaining valence electrons on outer atoms first. In the case of the BBr3 molecule, bromine is the outer atom and each of them needs 8 electrons in its valence shell to complete the octet.

Start putting the remaining electrons on bromine atoms as dots till they complete their octet.

So, all bromine atoms in the above structure completed their octet, because all of them have 8 electrons(electrons represented as dots + 2 electrons in every single bond) in their outermost shell.

Now again count the valence electron in the above structure.

In the above structure, there is 18 electrons are represented as dots + three single bonds that contain 6 electrons means a total of 24 valence electrons is used in the above structure.

So, we have used all the valence electrons available for drawing the lewis structure of BBr3.

5. Complete the octet of the central atom

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.

Therefore, the boron central atom in the BBr3 lewis structure attains stability by just having 6 valence electrons around it.

Let’s check the formal charge for the 4th step structure to verify it’s stable or not.

6. 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 4th step structure.

For bromine atom

⇒ Valence electrons of bromine = 7

⇒ Nonbonding electrons on bromine = 6

⇒ Bonding electrons around bromine (1 single bond) = 2

∴ (7 – 6 – 2/2) = 0 formal charge on the bromine atom.

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.

Hence, in the above BBr3 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.