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

I’m super excited to teach you the lewis structure of HI in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of HI molecule.

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

Lewis structure of HI (or Hydrogen iodide) contains one single bond between the Hydrogen (H) and Iodine (I) atom. The Iodine atom has 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 HI).

6 Steps to Draw the Lewis Structure of HI

Step #1: Calculate the total number of valence electrons

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

• For Hydrogen: 

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

Hence, the valence electron present in hydrogen is 1 (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 HI molecule, 

Valence electrons given by Iodine (I) atom = 7
Valence electron given by Hydrogen (H) atom = 1
So, total number of Valence electrons in HI molecule = 1 + 7 = 8

Step #2: Select the center atom (H is always outside)

While selecting the atom, you have to put the least electronegative atom at the center. 

But here in the HI molecule, there are only two atoms. So you can consider any of the atoms as a center atom.

So, let’s assume that the iodine atom is a central atom. (Because hydrogen is always placed outside in any lewis structure.)

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

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

These pair of electrons present between the Iodine (I) and Hydrogen (H) atoms form a chemical bond, which bonds the iodine and hydrogen atoms with each other in a HI 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 HI, the outer atom is hydrogen atom.

So now, you have to check whether this hydrogen atom is forming a duplet or not! (because hydrogen requires only 2 electrons to have a complete outer shell).

You can see in the above image that the hydrogen atom forms a duplet.

Also, only 2 valence electrons of HI molecule are used in the above structure.

But there are total 8 valence electrons in HI molecule (as calculated in step #1).

So the number of electrons left to be kept on the atom (i.e iodine atom) = 8 – 2 = 6.

So let’s keep these six electrons (i.e 3 electron pairs) on the iodine atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not

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

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

As you can see from the above image, the central atom (i.e iodine), is having 8 electrons. So it fulfills the octet rule and the iodine 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 iodine atom (I) as well as hydrogen atom (H).

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

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

• For Iodine:
  Valence electrons = 7 (as it is in group 17)
  Nonbonding electrons = 6
  Bonding electrons = 2
• For Hydrogen:
  Valence electron = 1 (as it is in group 1)
  Nonbonding electrons = 0
  Bonding electrons = 2

Formal charge	=	Valence electrons	–	Nonbonding electrons	–	(Bonding electrons)/2
I	=	7	–	6	–	2/2	=	0
H	=	1	–	0	–	2/2	=	0

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

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

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

Related lewis structures for your practice:
Lewis Structure of PO3-
Lewis Structure of BBr3
Lewis Structure of IF2-
Lewis Structure of BrF2-
Lewis Structure of P2 

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