# Boron triiodide (BI3) lewis dot structure, molecular geometry, polar or nonpolar, hybridization

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Boron triiodide appears as a crystalline solid composed of boron and iodine with the chemical formula BI3. It has a hexagonal crystal structure.

In this tutorial, we will discuss BI3 lewis structure, molecular geometry, polar or nonpolar, etc.

Bromide triiodide can react with water to form hydroiodic acid and boric acid.

 Name of Molecule Boron triiodide Chemical formula BI3 Molecular geometry of BI3 Trigonal planar Electron geometry of BI3 Trigonal planar Hybridization Sp2 Nature Nonpolar Total Valence electron for BI3 24
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## How to draw lewis structure of BI3?

BI3 lewis structure contains three B-I bonds with boron in a central position and all three iodine atoms in an outside position. The lewis structure of BI3 contains a total of 3 bond pairs and 9 lone pairs.

The drawing of the BI3 lewis’s structure is very easy and simple. Let’s see how to do it.

## Follow some steps for drawing the Lewis dot structure for BI3

1. Count total valence electron in BI3

First of all, determine the valence electron that is available for drawing the lewis structure of BI3 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 BI3 molecule is, just to look at the periodic group of boron and iodine atoms.

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

⇒ Total number of the valence electrons in boron = 3

⇒ Total number of the valence electrons in iodine = 7

∴ Total number of valence electrons available for the BI3 Lewis structure = 3 + 7×3 = 24 valence electrons         [∴BI3 molecule has one boron and three iodine 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.

In the case of the BI3 molecule, the boron atom is less electronegative than the iodine atom.

Hence, put the boron atom at the central position of the lewis diagram and all three iodine atoms outside to 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, BI3 molecule, iodine 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 BI3 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 BI3 molecule, iodine 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 iodine atoms as dots till they complete their octet.

So, all iodine 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 BI3.

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.

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.

Therefore, the boron central atom in the BI3 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 iodine atom

⇒ Valence electrons of iodine = 7

⇒ Nonbonding electrons on iodine = 6

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

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

### BI3 lewis structure

Hence, in the above structure, all atoms get a formal charge equal to zero. Therefore, the above lewis structure of BI3 is most stable and appropriate.

Also check –

## What is the molecular geometry of BI3?

The molecular geometry of BI3 is trigonal planar because the central atom boron is bonded with three iodine atoms and it contains no lone pair that means, it is an AX3 type molecule.

A represent central atom

X represent the number of bonded atom to central atom

According to VSEPR theory or chart, the AX3 type molecule forms trigonal planar molecular geometry.

## Hybridization of BI3

The hybridization of BI3 is Sp2 because the steric number of the boron central atoms is three.

The calculation of steric number is done by adding a number of bonded atoms attached to the central atom and lone pair on the central atom.

In the case of the BI3 molecule, boron is the central atom that is attached to the three bonded atoms(iodine) and it has no lone pairs.

Hence, (3 + 0) = 3 is the steric number of central atom boron in the BI3 molecule that gives Sp2 hybridization.

## Is BI3 polar or nonpolar?

So, Is BI3 polar or nonpolar? Well, it is obvious that BI3 is a nonpolar molecule because each B-I 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 BI3 molecule, hence, it is nonpolar in nature.

Also check-

## FAQ

### The total number of lone pairs present in the lewis structure of BI3?

The lone pair are represented as dots in the lewis diagram. According to the BI3 lewis structure, there is a total of 18 dots present(6 dots on each iodine atom).

Hence, the number of lone pairs in the BI3 lewis structure is 9.

### The total number of bond pairs present in the lewis structure of BI3?

The bond pairs simply are the bonding electrons that are found between the atoms, generally in the covalent bond.

Two bonding electron between the atoms forms a single covalent bond.

Now, as per BI3 lewis’s 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.

Properties of Boron triiodide

• It has a boiling point of 210 °C  and a melting point of 49.9 °C.
• It is soluble in CCl4, CS2, benzene, chloroform.
• Its dielectric constant is 5.38.
• Its dipole moment is zero.

## Summary

• The total valence electron is available for drawing the BI3 Lewis structure is 24.
• The molecular geometry or shape of BI3 is trigonal planar.
• BI3 is nonpolar and has Sp2 hybridization.
• In the BI3 Lewis structure, a total of 18 nonbonding electrons and 6 bonded electrons are present.
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