IF5 Lewis structure

IF₅ (iodine pentafluoride) has one iodine atom and five fluorine atoms.

In the IF₅ Lewis structure, there are five single bonds around the iodine atom, with five fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the iodine atom has one lone pair.

Alternative method: Lewis structure of IF₅

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, both iodine and fluorine lie in group 17.

Hence, both iodine and fluorine have seven valence electrons.

Since IF₅ has one iodine atom and five fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of five fluorine atoms = 7 × 5 = 35

And the total valence electrons = 7 + 35 = 42

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 42 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 = 42 ÷ 2 = 21

• Third, determine the central atom

We have to place the least electronegative atom at the center.

Since iodine is less electronegative than fluorine, assume that the central atom is iodine.

Therefore, place iodine in the center and fluorines on either side.

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 21 electron pairs. And five I — F bonds are already marked. So we have to only mark the remaining sixteen electron pairs as lone pairs on the sketch.

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

Always start to mark the lone pairs from outside atoms. Here, the outside atoms are fluorines.

So for each fluorine, there are three lone pairs, and for iodine, there is one lone pair.

Mark the lone pairs on the sketch as follows:

Formal charge

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

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

For iodine atom, formal charge = 7 – 2 – ½ (10) = 0

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

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

Final structure

The final structure of IF₅ contains a central iodine atom connected to five fluorine atoms through single covalent bonds. In this arrangement, the iodine atom utilizes an expanded octet to accommodate twelve valence electrons, consisting of five bonding pairs and one lone pair. Each fluorine atom fulfills its octet by maintaining three lone pairs of its own. This setup is the most stable because it results in formal charges of zero for every atom, representing the most energetically favorable state for the molecule. Consequently, this specific electronic distribution serves as the definitive and most accurate Lewis representation of iodine pentafluoride.

Next: HNO₃ Lewis structure

External video

• IF5 Lewis Structure: How to Draw the Lewis Structure for IF5 – YouTube • Wayne Breslyn

External links

• https://techiescientist.com/if5-lewis-structure/
• https://lambdageeks.com/if5-lewis-structure/
• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/IF5-lewis-structure.html
• https://topblogtenz.com/if5-lewis-structure-molecular-geometry-bond-angle-hybridization/
• https://sciedutut.com/if5-lewis-structure/