IF4- Lewis structure

IF₄^– (iodine tetrafluoride) has one iodine atom and four fluorine atoms.

In the IF₄^– Lewis structure, there are four single bonds around the iodine atom, with four fluorine atoms attached to it. Each fluorine atom has three lone pairs, and the iodine atom has two lone pairs.

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

Alternative method: Lewis structure of IF₄^–

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 IF₄^– has one iodine atom and four fluorine atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of four fluorine atoms = 7 × 4 = 28

Now the IF₄^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 28 + 1 = 36

Learn how to find: Fluorine valence electrons

• Second, find the total electron pairs

We have a total of 36 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 = 36 ÷ 2 = 18

• 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 18 electron pairs. And four I — F bonds are already marked. So we have to only mark the remaining fourteen 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 are two 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 – 4 – ½ (8) = -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 IF₄^– has a central iodine atom connected to four fluorine atoms through single covalent bonds. In this configuration, the iodine atom utilizes an expanded octet to accommodate twelve electrons, which includes two lone pairs, while each fluorine atom completes its octet with three lone pairs. This arrangement is the most stable because it minimizes the formal charges of the atoms, resulting in a -1 charge on the iodine atom while each fluorine remains at zero. Consequently, this specific distribution of electrons represents the most accurate and stable Lewis representation for the IF₄^– ion.

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 IF₄^– ion.

Next: SF₃^– Lewis structure

External video

• IF4- Lewis Structure: How to Draw the Lewis Structure for IF4- – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/IF4–lewis-structure.html
• https://lambdageeks.com/if4-lewis-structure/