BeI2 Lewis structure

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BeI₂ (beryllium iodide) has one beryllium atom and two iodine atoms.

In the BeI₂ Lewis structure, there are two single bonds around the beryllium atom, with two iodine atoms attached to it, and on each iodine atom, there are three lone pairs.

Steps

Use these steps to correctly draw the BeI₂ Lewis structure:

#1 First draw a rough sketch
#2 Mark lone pairs on the atoms
#3 Calculate and mark formal charges on the atoms, if required

Let’s discuss each step in more detail.

#1 First draw a rough sketch

• First, determine the total number of valence electrons

In the periodic table, beryllium lies in group 2, and iodine lies in group 17.

Hence, beryllium has two valence electrons and iodine has seven valence electrons.

Since BeI₂ has one beryllium atom and two iodine atoms, so…

Valence electrons of one beryllium atom = 2 × 1 = 2
Valence electrons of two iodine atoms = 7 × 2 = 14

And the total valence electrons = 2 + 14 = 16

Learn how to find: Beryllium valence electrons

• Second, find the total electron pairs

We have a total of 16 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 = 16 ÷ 2 = 8

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

#2 Mark lone pairs on the atoms

Here, we have a total of 8 electron pairs. And two Be — I bonds are already marked. So we have to only mark the remaining six electron pairs as lone pairs on the sketch.

Also remember that beryllium 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 beryllium, there is zero lone pair because all six electron pairs are over.

Mark the lone pairs on the sketch as follows:

#3 Calculate and mark formal charges on the atoms, if required

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

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

For beryllium atom, formal charge = 2 – 0 – ½ (4) = 0

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

Here, both beryllium 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 (beryllium) doesn’t form an octet. But, beryllium 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 BeI₂.

Next: AsCl₃ Lewis structure

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External links

• https://techiescientist.com/bei2-lewis-structure/
• https://lambdageeks.com/bei2-lewis-structure/
• https://sciedutut.com/bei2-lewis-structure/
• https://oneclass.com/homework-help/chemistry/7044984-bei2-lewis-structure.en.html