# HI Lewis structure

HI (hydrogen iodide) has one hydrogen atom and one iodine atom.

In HI Lewis structure, there is a single bond between the hydrogen and iodine atom, and on the iodine atom, there are three lone pairs.

Contents

## Steps

Here’s how you can easily draw the HI Lewis structure step by step:

#1 Draw a rough skeleton structure
#2 Mention lone pairs on the atoms
#3 If needed, mention formal charges on the atoms

Now, let’s take a closer look at each step mentioned above.

### #1 Draw a rough skeleton structure

• First, determine the total number of valence electrons

In the periodic table, hydrogen lies in group 1, and iodine lies in group 17.

Hence, hydrogen has one valence electron and iodine has seven valence electrons.

Since HI has one hydrogen atom and one iodine atom, so…

Valence electrons of one hydrogen atom = 1 × 1 = 1
Valence electrons of one iodine atom = 7 × 1 = 7

And the total valence electrons = 1 + 7 = 8

• Second, find the total electron pairs

We have a total of 8 valence electrons. And when we divide this value by two, we get the value of total electron pairs.

Total electron pairs = total valence electrons ÷ 2

So the total electron pairs = 8 ÷ 2 = 4

• Third, determine the central atom

Here hydrogen can not be the central atom. Because the central atom is bonded with at least two other atoms, and hydrogen has only one electron in its last shell, so it can not make more than one bond.

Hence, here we have to assume that the central atom is iodine.

• And finally, draw the rough sketch

### #2 Mention lone pairs on the atoms

Here, we have a total of 4 electron pairs. And one H — I bond is already marked. So we have to only mark the remaining three electron pairs as lone pairs on the sketch.

Also remember that hydrogen is a period 1 element, so it can not keep more than 2 electrons in its last shell. And iodine is a period 5 element, so it can keep more than 8 electrons in its last shell.

Always start to mark the lone pairs from outside atoms. Here, the outside atom is iodine.

So for iodine, there are three lone pairs, and for hydrogen, there is zero lone pair because all three electron pairs are over.

Mark the lone pairs on the sketch as follows:

### #3 If needed, mention formal charges on the atoms

Use the following formula to calculate the formal charges on atoms:

Formal charge = valence electrons – nonbonding electrons – ½ bonding electrons

For hydrogen atom, formal charge = 1 – 0 – ½ (2) = 0

For iodine atom, formal charge = 7 – 6 – ½ (2) = 0

Here, both hydrogen and iodine atoms do not have charges, so no need to mark the charges.

In the above structure, you can see that the central atom (iodine) forms an octet. And the outside atom (hydrogen) also forms a duet. Hence, the octet rule and duet rule are satisfied.

Therefore, this structure is the stable Lewis structure of HI.

## FAQs

### How many valence electrons are in the HI molecule?

To determine the number of valence electrons in HI, you need to add up the valence electrons of each atom in the molecule.

Hydrogen has one valence electron, while iodine has 7 valence electrons. Therefore, HI has a total of 1 + 7 = 8 valence electrons.

### Does the HI Lewis structure satisfy the octet rule?

In HI Lewis structure, the hydrogen (H) atom forms a single covalent bond with the iodine (I) atom.

The hydrogen atom forms a duet, while the central atom (iodine) forms an octet. Hence, the HI Lewis structure follows the duet/octet rule.

### What’s the shape of the HI molecule according to its Lewis structure?

Hydrogen iodide (HI) is a diatomic molecule. In HI Lewis structure, the hydrogen (H) atom is bonded to the iodine (I) atom through a single covalent bond.

Since there are only two atoms in the HI molecule, there is no other option for the shape besides linear. Hence, the HI molecule has a linear shape according to its Lewis structure.

Next: PF4+ Lewis structure