BI3 Lewis structure

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BI₃ (boron triiodide) has one boron atom and three iodine atoms.

In BI₃ Lewis structure, there are three single bonds around the boron atom, with three iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Steps

To properly draw the BI₃ Lewis structure, follow these steps:

#1 Draw a rough sketch of the structure
#2 Next, indicate lone pairs on the atoms
#3 Indicate formal charges on the atoms, if necessary

Let’s break down each step in more detail.

#1 Draw a rough sketch of the structure

• First, determine the total number of valence electrons

In the periodic table, boron lies in group 13, and iodine lies in group 17.

Hence, boron has three valence electrons and iodine has seven valence electrons.

Since BI₃ has one boron atom and three iodine atoms, so…

Valence electrons of one boron atom = 3 × 1 = 3
Valence electrons of three iodine atoms = 7 × 3 = 21

And the total valence electrons = 3 + 21 = 24

Learn how to find: Boron valence electrons

• Second, find the total electron pairs

We have a total of 24 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 24 ÷ 2 = 12

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since boron is less electronegative than iodine, assume that the central atom is boron.

Therefore, place boron in the center and iodines on either side.

• And finally, draw the rough sketch

#2 Next, indicate lone pairs on the atoms

Here, we have a total of 12 electron pairs. And three B — I bonds are already marked. So we have to only mark the remaining nine electron pairs as lone pairs on the sketch.

Also remember that boron is a period 2 element, so it can not keep more than 8 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are iodines.

So for each iodine, there are three lone pairs, and for boron, there is zero lone pair because all nine electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Indicate formal charges on the atoms, if necessary

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For boron atom, formal charge = 3 – 0 – ½ (6) = 0

For each iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both boron and iodine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (boron) doesn’t form an octet. But, boron has an exception that it does not require eight electrons to form an octet. So no need to worry about the octet rule here.

Therefore, this structure is the stable Lewis structure of BI₃.

FAQs

Next: H₃BO₃ Lewis structure

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External links

• https://topblogtenz.com/bi3-lewis-structure-molecular-geometry-polar-or-nonpolar-hybridization/
• https://lambdageeks.com/bi3-lewis-structure/
• https://homework.study.com/explanation/what-is-the-lewis-structure-for-bi3.html
• https://www.numerade.com/questions/draw-the-lewis-structure-of-a-mathrmni_3-and-b-mathrmbi_3-name-the-molecular-shape-and-indicate-whet/