IO3- Lewis structure

IO₃^– (iodate) has one iodine atom and three oxygen atoms.

In IO₃^– Lewis structure, there are two double bonds and one single bond around the iodine atom, with three oxygen atoms attached to it. The oxygen atom with double bonds has two lone pairs, the oxygen atom with a single bond has three lone pairs, and the iodine atom has one lone pair.

Also, there is a negative (-1) charge on the oxygen atom with a single bond.

Alternative method: Lewis structure of IO₃^–

Rough sketch

• First, determine the total number of valence electrons

In the periodic table, iodine lies in group 17, and oxygen lies in group 16.

Hence, iodine has seven valence electrons and oxygen has six valence electrons.

Since IO₃^– has one iodine atom and three oxygen atoms, so…

Valence electrons of one iodine atom = 7 × 1 = 7
Valence electrons of three oxygen atoms = 6 × 3 = 18

Now the IO₃^– has a negative (-1) charge, so we have to add one more electron.

So the total valence electrons = 7 + 18 + 1 = 26

Learn how to find: Oxygen valence electrons

• Second, find the total electron pairs

We have a total of 26 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 = 26 ÷ 2 = 13

• Third, determine the central atom

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

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

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

• And finally, draw the rough sketch

Lone pair

Here, we have a total of 13 electron pairs. And three I — O bonds are already marked. So we have to only mark the remaining ten 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 oxygen 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 oxygens.

So for each oxygen, 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 (6) = +2

For each oxygen atom, formal charge = 6 – 6 – ½ (2) = -1

Here, both iodine and oxygen atoms have charges, so mark them on the sketch as follows:

The above structure is not a stable Lewis structure because both iodine and oxygen atoms have charges. Therefore, reduce the charges (as below) by converting lone pairs to bonds.

Convert a lone pair of the oxygen atom to make a new I — O bond with the iodine atom as follows:

Since there are charges on iodine and oxygen atoms, again convert a lone pair of the oxygen atom to make a new I — O bond with the iodine atom as follows:

Final structure

The final structure of IO₃^– features a central iodine atom connected to three oxygen atoms. In this layout, the iodine atom serves as an exception to the octet rule, utilizing an expanded valence shell to accommodate twelve electrons through two double bonds, one single bond, and one lone pair. Each oxygen atom involved in a double bond retains two lone pairs, while the oxygen atom connected by a single bond maintains three lone pairs to fulfill its octet. This arrangement is the most stable because it optimizes the formal charge distribution; the iodine atom and the two double-bonded oxygen atoms carry a formal charge of zero, while the single-bonded oxygen atom carries a formal charge of -1. Accordingly, this specific electronic distribution serves as the definitive and most accurate Lewis representation of the iodate 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 ion.

Next: HOFO Lewis structure

External video

• IO3- Lewis Structure: How to Draw the Lewis Structure for the Iodate Ion – YouTube • Wayne Breslyn

External links

• https://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/IO3–lewis-structure.html
• https://www.chemistryscl.com/general/IO3–lewis-structure/
• https://lambdageeks.com/io3-lewis-structure/
• https://www.numerade.com/ask/question/draw-the-lewis-structure-for-the-io3-ion-before-answering-the-following-questions-iodine-is-the-central-atom-and-all-atoms-obey-the-octet-rule-how-many-bonds-does-the-iodine-atom-form-how-ma-02045/
• https://study.com/academy/answer/for-the-io3-a-draw-the-electron-dot-diagram-and-structural-formula-b-predict-the-shape-c-predict-whether-it-is-polar-or-non-polar-and-justify-your-prediction-indicate-the-positive-and-negativ.html