BeI2 Lewis Structure in 6 Steps (With Images)

So you have seen the above image by now, right?

Let me explain the above image in short.

BeI2 lewis structure has a Beryllium atom (Be) at the center which is surrounded by two Iodine atoms (I). There are 2 single bonds between the Beryllium atom (Be) and each Iodine atom (I). There are 3 lone pairs on both the Iodine atoms (I).

If you haven’t understood anything from the above image of BeI2 lewis structure, then just stick with me and you will get the detailed step by step explanation on drawing a lewis structure of BeI2.

So let’s move to the steps of drawing the lewis structure of BeI2.

Steps of drawing BeI2 lewis structure

Step 1: Find the total valence electrons in BeI2 molecule

In order to find the total valence electrons in a BeI2 molecule, first of all you should know the valence electrons present in beryllium atom as well as iodine atom.
(Valence electrons are the electrons that are present in the outermost orbit of any atom.)

Here, I’ll tell you how you can easily find the valence electrons of beryllium as well as iodine using a periodic table.

Total valence electrons in BeI2 molecule

→ Valence electrons given by beryllium atom:

Beryllium is a group 2 element on the periodic table. ^[1] Hence the valence electrons present in beryllium is 2.

You can see the 2 valence electrons present in the beryllium atom as shown in the above image.

→ Valence electrons given by iodine atom:

Iodine is a group 17 element on the periodic table. ^[2] Hence the valence electrons present in iodine is 7.

You can see the 7 valence electrons present in the iodine atom as shown in the above image.

Hence, 

Total valence electrons in BeI2 molecule = valence electrons given by 1 beryllium atom + valence electrons given by 2 iodine atoms = 2 + 7(2) = 16.

Step 2: Select the central atom

For selecting the center atom, you have to remember that the atom which is less electronegative remains at the center.

Now here the given molecule is BeI2 and it contains beryllium atom (Be) and iodine atoms (I).

You can see the electronegativity values of beryllium atom (Be) and iodine atom (I) in the above periodic table.

If we compare the electronegativity values of beryllium (Be) and iodine (I) then the beryllium atom is less electronegative.

So here the beryllium atom (Be) is the center atom and the iodine atoms (I) are the outside atoms.

Step 3: Connect each atoms by putting an electron pair between them

Now in the BeI2 molecule, you have to put the electron pairs between the beryllium atom (Be) and iodine atoms (I).

This indicates that the beryllium (Be) and iodine (I) are chemically bonded with each other in a BeI2 molecule.

Step 4: Make the outer atoms stable

Now in this step, you have to check the stability of the outer atoms.

Here in the sketch of BeI2 molecule, you can see that the outer atoms are iodine atoms.

These outer iodine atoms are forming an octet and hence they are stable.

Also, in step 1 we have calculated the total number of valence electrons present in the BeI2 molecule.

The BeI2 molecule has a total 16 valence electrons and all these valence electrons are used in the above sketch of BeI2. 

Hence there are no remaining electron pairs to be kept on the central atom. 

So now let’s proceed to the next step.

Step 5: Check the stability of the central atom

In this step, you have to check whether the central beryllium atom (Be) is stable or not.

Now beryllium requires only 4 electrons to become stable. The s-orbitals of beryllium get completely filled by these 4 electrons.

You can see from the above picture that the beryllium atom has 4 electrons and hence it is stable.

Now let’s proceed to the final step to check whether the lewis structure of BeI2 is stable or not.

Step 6: Check the stability of lewis structure

Now you have come to the final step in which you have to check the stability of lewis structure of BeI2.

The stability of lewis structure can be checked by using a concept of formal charge.

In short, now you have to find the formal charge on beryllium (Be) atom as well as iodine (I) atoms present in the BeI2 molecule.

For calculating the formal charge, you have to use the following formula;

Formal charge = Valence electrons – (Bonding electrons)/2 – Nonbonding electrons

You can see the number of bonding electrons and nonbonding electrons for each atom of BeI2 molecule in the image given below.

For Beryllium (Be) atom:
Valence electrons = 2 (because beryllium is in group 2)
Bonding electrons = 4
Nonbonding electrons = 0

For Iodine (I) atom:
Valence electron = 7 (because iodine is in group 17)
Bonding electrons = 2
Nonbonding electrons = 6

Formal charge	=	Valence electrons	–	(Bonding electrons)/2	–	Nonbonding electrons
Be	=	2	–	4/2	–	0	=	0
I	=	7	–	2/2	–	6	=	0

From the above calculations of formal charge, you can see that the beryllium (Be) atom as well as iodine (I) atom has a “zero” formal charge.

This indicates that the above lewis structure of BeI2 is stable and there is no further change in the above structure of BeI2.

In the above lewis dot structure of BeI2, you can also represent each bonding electron pair (:) as a single bond (|). By doing so, you will get the following lewis structure of BeI2.

I hope you have completely understood all the above steps.

For more practice and better understanding, you can try other lewis structures listed below.

Try (or at least See) these lewis structures for better understanding:

SiCl2Br2 Lewis Structure	SbF5 Lewis Structure
ClBr3 Lewis Structure	GeH4 Lewis Structure
FCN Lewis Structure	HClO2 Lewis Structure