Lewis structure of BeI2

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Lewis structure of BeI2
Lewis structure of BeI2 | Image: Root Memory

The Lewis structure of BeI2 contains two single bonds, with beryllium in the center, and two iodines on either side. There are three lone pairs on each iodine atom, and the beryllium atom does not have any lone pair.

Steps

By using the following steps, you can easily draw the Lewis structure of BeI2.

#1 Draw skeleton
#2 Show chemical bond
#3 Mark lone pairs
#4 Calculate formal charge and check stability (if octet is already completed on central atom)

Let’s one by one discuss each step in detail.

#1 Draw skeleton

In this step, first calculate the total number of valence electrons. And then, decide the central atom.

  • Let’s calculate the total number of valence electrons

We know that… beryllium is a group 2 element and iodine is a group 17 element. Hence, beryllium has two valence electrons and iodine has seven valence electrons.

Now BeI2 has one beryllium atom and two iodine atoms.

So the total number of valence electrons = valence electrons of beryllium atom + (valence electrons of iodine atom × 2)

Therefore, the total number of valence electrons = 2 + 14 = 16

  • Now decide the central atom

The atom with the least electronegative value is placed at the center. By looking at the periodic table, we get the electronegativity values for beryllium and iodine as follows:

Electronegativity value of beryllium = 1.57
Electronegativity value of iodine = 2.66

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

So now, put beryllium in the center and iodines on either side. And draw the rough skeleton structure for the Lewis structure of BeI2 something like this:

Skeleton structure for Lewis structure of BeI2 | Image: Root Memory

#2 Show chemical bond

Place two electrons between the atoms to show a chemical bond. Since beryllium is surrounded by two iodines, use four electrons to show two chemical bonds as follows:

Two chemical bonds shown between atoms | Image: Root Memory

#3 Mark lone pairs

As calculated earlier, we have a total of 16 valence electrons. And in the above structure, we have already used four valence electrons. Hence, twelve valence electrons are remaining.

Two valence electrons represent one lone pair. So twelve valence electrons = six lone pairs.

Note 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.

Also, make sure that you start marking these lone pairs on outside atoms first. And then, on the central atom.

The outside atoms are iodines, so each iodine will get three lone pairs. And the central atom (beryllium) will not get any lone pair, because all six lone pairs are used.

So the Lewis structure of BeI2 looks something like this:

Lone pairs marked on Lewis structure of BeI2 | Image: Root Memory

In the above structure, you can see that the octet is completed on outside atoms. But, the central atom (beryllium) doesn’t form an octet.

Now beryllium has an exception in BeI2, that it does not require eight electrons to complete its octet.

So one last thing we need to do is, calculate the formal charge and check the stability of the above structure.

#4 Calculate formal charge and check stability

The following formula is used to calculate the formal charges on atoms:

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

Collect the data from the above structure and then, write it down below as follows:

  • For beryllium atom

Valence electrons = 2
Nonbonding electrons = 0
Bonding electrons = 4

Formal charge = 2 – 0 – ½ (4) = 0

  • For each iodine atom

Valence electrons = 7
Nonbonding electrons = 6
Bonding electrons = 2

Formal charge = 7 – 6 – ½ (2) = 0

Mention the formal charges of atoms on the structure. So the Lewis structure of BeI2 looks something like this:

Formal charges are calculated, and got the stable Lewis structure of BeI2 | Image: Root Memory

In the above structure, you can see that the formal charges of both (beryllium and iodine) are zero. Therefore, this is the stable Lewis structure of BeI2.

And each horizontal line drawn in the above structure represents a pair of bonding valence electrons.

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Deep

Rootmemory.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.

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