Lewis Structure of SeI2 (With 6 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of SeI2?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of SeI2 (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 SeI2 contains two single bonds between the Selenium (Se) atom and each Iodine (I) atom. The Selenium atom (Se) is at the center and it is surrounded by 2 Iodine atoms (I). The Selenium atom has 2 lone pairs and 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 SeI2).

6 Steps to Draw the Lewis Structure of SeI2

Step #1: Calculate the total number of valence electrons

Here, the given molecule is SeI2. In order to draw the lewis structure of SeI2, first of all you have to find the total number of valence electrons present in the SeI2 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 SeI2

• For Selenium: 

Selenium is a group 16 element on the periodic table. ^[1]

Hence, the valence electrons present in selenium is 6 (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 SeI2 molecule, 

Valence electrons given by Selenium (Se) atom = 6
Valence electrons given by each Iodine (I) atom = 7
So, total number of Valence electrons in SeI2 molecule = 6 + 7(2) = 20

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). ^[1]

Here in the SeI2 molecule, if we compare the selenium atom (Se) and iodine atom (I), then the selenium is less electronegative than iodine.

So, selenium 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 SeI2 molecule, put the two electrons (i.e electron pair) between each selenium atom and iodine atom to represent a chemical bond between them.

These pairs of electrons present between the Selenium (Se) and Iodine (I) atoms form a chemical bond, which bonds the selenium and iodine atoms with each other in a SeI2 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 SeI2, the outer atoms are iodine atoms.

So now, you have to complete the octet on these iodine atoms (because iodine requires 8 electrons to have a complete outer shell).

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

Also, only 16 valence electrons of SeI2 molecule are used in the above structure.

But there are total 20 valence electrons in SeI2 molecule (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 20 – 16 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) on the central atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not. If it does not have an octet, then move the electron pair from the outer atom to form a double bond or triple bond

In this step, we have to check whether the central atom (i.e selenium) has an octet or not. 

In simple words, we have to check whether the central Selenium (Se) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e selenium), has 8 electrons. So it fulfills the octet rule and the selenium atom 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 selenium atom (Se) 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 Selenium:
  Valence electron = 6 (as it is in group 16)
  Nonbonding electrons = 4
  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
Se	=	6	–	4	–	4/2	=	0
I	=	7	–	6	–	2/2	=	0

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

Hence, there will not be any change in the above structure and the above lewis structure of SeI2 is the final stable structure only.

Each electron pair (:) in the lewis dot structure of SeI2 represents the single bond ( | ). So the above lewis dot structure of SeI2 can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of H2Te
Lewis Structure of TeCl2
Lewis Structure of CH2I2
Lewis Structure of GaI3
Lewis Structure of SeO4 2- 

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