HI 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.

HI (hydrogen iodide) lewis structure has one Hydrogen atom (H) and one Iodine atom (I) which contain a single bond between them. There are 3 lone pairs on the Iodine atom (I).

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

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

Steps of drawing HI lewis structure

Step 1: Find the total valence electrons in HI molecule

In order to find the total valence electrons in HI (hydrogen iodide) molecule, first of all you should know the valence electrons present in a single hydrogen 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 hydrogen as well as iodine using a periodic table.

Total valence electrons in HI molecule

→ Valence electrons given by hydrogen atom:

Hydrogen is group 1 element on the periodic table. ^[1] Hence the valence electron present in hydrogen is 1.

You can see that only 1 valence electron is present in the hydrogen 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 HI molecule = valence electrons given by 1 hydrogen atom + valence electrons given by 1 iodine atom = 1 + 7 = 8.

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 the given molecule is HI (hydrogen iodide). It has only two atoms, so you can select any of the atoms as a center atom.

Let’s assume the iodine atom as a central atom (Because we have to keep hydrogen outside in any lewis structure).

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

Now in the HI molecule, you have to put the electron pairs between the hydrogen atom (H) and iodine atom (I).

This indicates that the hydrogen (H) atom and iodine (I) atom are chemically bonded with each other in a HI molecule.

Step 4: Make the outer atoms stable. Place the remaining valence electrons pair on the central atom.

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

Here in the sketch of HI molecule, we have assumed the iodine atom as a center atom. So the hydrogen is the outer atom.

Hence you have to make the hydrogen atom stable.

You can see in the below image that the hydrogen atom is forming a duplet and hence it is stable.

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

The HI molecule has a total 8 valence electrons and out of these, only 2 valence electrons are used in the above sketch.

So the number of electrons which are left = 8 – 2 = 6.

You have to put these 6 electrons on the iodine atom in the above sketch of HI molecule.

Now let’s proceed to the next step.

Step 5: Check the octet on the central atom

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

In order to check the stability of the central iodine (I) atom, we have to check whether it is forming an octet or not.

You can see from the above picture that the iodine atom is forming an octet. That means it has 8 electrons.

And hence the central iodine atom is stable.

Now let’s proceed to the final step to check whether the lewis structure of HI 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 HI.

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 hydrogen (H) atom as well as iodine (I) atoms present in the HI 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 HI molecule in the image given below.

For Hydrogen (H) atom:
Valence electron = 1 (because hydrogen is in group 1)
Bonding electrons = 2
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
H	=	1	–	2/2	–	0	=	0
I	=	7	–	2/2	–	6	=	0

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

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

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

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:

PO3- Lewis Structure	BBr3 Lewis Structure
IF2- Lewis Structure	BrF2- Lewis Structure
P2 Lewis Structure	IBr2- Lewis Structure