IF2- Lewis structure

IF₂^– has one iodine atom and two fluorine atoms.

In IF₂^– Lewis structure, there are two single bonds around the iodine atom, with two fluorine atoms attached to it, and each atom has three lone pairs.

Also, there is a negative (-1) charge on the iodine atom.

Alternative method: Lewis structure of IF₂^–

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 IF₂^– has one iodine atom and two fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of two fluorine atoms = 7 × 2 = 14

Now the IF₂^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 14 + 1 = 22

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 22 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 = 22 ÷ 2 = 11

• 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 11 electron pairs. And two I — F bonds are already marked. So we have to only mark the remaining nine 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 atom, there are three lone pairs.

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 – 6 – ½ (4) = -1

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

Here, the iodine atom has a charge, so mark it on the sketch as follows:

Final structure

The final structure of IF₂^– comprises a central iodine atom linked to two fluorine atoms through single covalent bonds. In this configuration, the iodine atom utilizes an expanded valence shell to accommodate ten electrons, which consist of two bonding pairs and three lone pairs. Within this layout, each fluorine atom successfully satisfies the octet rule by maintaining three lone pairs alongside its single shared bond. This arrangement represents the most stable state for the ion because it places the negative formal charge specifically on the central iodine atom, while the fluorine atoms maintain a formal charge of zero. Thus, this specific electronic distribution serves as the definitive and most accurate Lewis representation of IF₂^–.

To complete the representation, draw square brackets around the entire Lewis structure and place a “-” or “-1” sign as a superscript outside the upper right bracket. This notation signifies that the negative charge is a property of the whole ion.

Next: BrF₂^– Lewis structure

External video

• IF2- Lewis Structure: How to Draw the Lewis Structure for IF2 – – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/IF2-lewis-structure.html
• https://lambdageeks.com/if2-lewis-structure/
• https://topblogtenz.com/if2-lewis-structure-molecular-geometry-polar-or-non-polar-hybridization/
• https://homework.study.com/explanation/draw-the-lewis-structure-for-if2-use-only-single-bonds-to-fluorines-and-answer-the-following-questions-a-how-many-valence-electrons-are-there-b-what-is-the-electron-geometry-c-what-is-the-molecular-geometry-d-what-are-the-hybridized-orbitals-e.html
• https://www.chegg.com/homework-help/questions-and-answers/molecular-geometry-if2-1-lewis-structure-q3210366