Lewis Structure of BeI2 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of BeI2?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of BeI2 (along with images).

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of BeI2 contains two single bonds between the Beryllium (Be) atom and each Iodine (I) atom. The Beryllium atom (Be) is at the center and it is surrounded by 2 Iodine atoms (I). The Beryllium does not have lone pairs while both the Iodine atoms have 3 lone pairs.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of BeI2).

6 Steps to Draw the Lewis Structure of BeI2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is BeI2. In order to draw the lewis structure of BeI2, first of all you have to find the total number of valence electrons present in the BeI2 molecule.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in BeI2

• For Beryllium: 

Beryllium is a group 2 element on the periodic table. ^[1]

Hence, the valence electron present in beryllium is 2 (see below image).

• For Iodine:

Iodine is a group 17 element on the periodic table. ^[2]

Hence, the valence electrons present in iodine is 7 (see below image).

Hence in a BeI2 molecule, 

Valence electrons given by Beryllium (Be) atom = 2
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in BeI2 molecule = 2 + 7(2) = 16

Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center. 

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). ^[3]

Here in the BeI2 molecule, if we compare the beryllium atom (Be) and iodine atom (I), then the beryllium is less electronegative than iodine.

So, beryllium should be placed in the center and the remaining 2 iodine atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of BeI2 molecule, put the two electrons (i.e electron pair) between beryllium atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Beryllium (Be) and Iodine (I) atoms form a chemical bond, which bonds the beryllium and iodine atoms with each other in a BeI2 molecule.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of BeI2, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms.

Now, you can see in the above image that all the iodine atoms form an octet.

Also, all the 16 valence electrons of BeI2 molecule (as calculated in step #1) are used in the above structure. So there are no remaining electron pairs.

Hence there is no change in the above sketch of BeI2.

Let’s move to the next step.

Step #5: Check whether the central atom is stable or not

In this step, we have to check whether the central atom (i.e beryllium) is stable or not. 

The beryllium atom does not require 8 electrons in its outer orbit to become stable.

As you can see from the above image, the central atom (i.e beryllium), has 4 electrons. This indicates that the s-orbitals of beryllium are completely filled. So it is stable.

Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on beryllium atom (Be) as well as each iodine atom (I).

For that, you need to remember the formula of formal charge;

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

• For Beryllium:
  Valence electron = 2 (as it is in group 2)
  Nonbonding electrons = 0
  Bonding electrons = 4
• For Iodine:
  Valence electron = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2

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

So you can see above that the formal charges on beryllium as well as iodine are “zero”.

Hence, there will not be any change in the above structure and the above lewis structure of BeI2 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of BeI2 represents the single bond ( | ). So the above lewis dot structure of BeI2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of CHBr3
Lewis Structure of SiCl2Br2
Lewis Structure of SbF5
Lewis Structure of ClBr3
Lewis Structure of GeH4 

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