Lewis Structure of IO2- (With 6 Simple Steps to Draw!)

I’m super excited to teach you the lewis structure of IO2- in just 6 simple steps.

Infact, I’ve also given the step-by-step images for drawing the lewis dot structure of IO2- ion.

So, if you are ready to go with these 6 simple steps, then let’s dive right into it!

Lewis structure of IO2- contains one double bond and one single bond between the Iodine (I) atom and Oxygen (O) atom. The Iodine atom (I) is at the center and it is surrounded by 2 Oxygen atoms (O). The Iodine atom has 2 lone pairs, one Oxygen atom has 2 lone pairs and the other oxygen atom has 3 lone pairs.

Let’s draw and understand this lewis dot structure step by step.

(Note: Take a pen and paper with you and try to draw this lewis structure along with me. I am sure you will definitely learn how to draw lewis structure of IO2-).

6 Steps to Draw the Lewis Structure of IO2-

Step #1: Calculate the total number of valence electrons

Here, the given ion is IO2-. In order to draw the lewis structure of IO2- ion, first of all you have to find the total number of valence electrons present in the IO2- ion.
(Valence electrons are the number of electrons present in the outermost shell of an atom).

So, let’s calculate this first.

Calculation of valence electrons in IO2-

• For Iodine:

Iodine is a group 17 element on the periodic table. [1]

Hence, the valence electrons present in iodine is 7 (see below image).

• For Oxygen:

Oxygen is a group 16 element on the periodic table. [2]

Hence, the valence electron present in oxygen is 6 (see below image).

Hence in a IO2- ion,

Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Oxygen (O) atom = 6
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in IO2- ion = 7 + 6(2) + 1 = 20

Step #2: Select the center atom

While selecting the atom, always put the least electronegative atom at the center.

(Remember: Fluorine is the most electronegative element on the periodic table and the electronegativity decreases as we move right to left in the periodic table as well as top to bottom in the periodic table). [3]

Here in the IO2- ion, if we compare the Iodine atom (I) and oxygen atom (O), then the iodine is less electronegative than oxygen.

So, iodine should be placed in the center and the remaining 2 oxygen atoms will surround it.

Step #3: Put two electrons between the atoms to represent a chemical bond

Now in the above sketch of IO2- molecule, put the two electrons (i.e electron pair) between each iodine atom and oxygen atom to represent a chemical bond between them.

These pairs of electrons present between the Iodine (I) and Oxygen (O) atoms form a chemical bond, which bonds the iodine and oxygen atoms with each other in a IO2- ion.

Step #4: Complete the octet (or duplet) on outside atoms. If the valence electrons are left, then put the valence electrons pair on the central atom

Don’t worry, I’ll explain!

In the Lewis structure of IO2-, the outer atoms are oxygen atoms.

So now, you have to complete the octet on these oxygen atoms (because oxygen requires 8 electrons to have a complete outer shell).

Now, you can see in the above image that all the oxygen atoms form an octet.

Also, only 16 valence electrons of IO2- are used in the above structure.

But there are total 20 valence electrons in IO2- ion (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 20 – 16 = 4.

So let’s keep these four electrons (i.e two electron pairs) on the central atom.

Now, let’s move to the next step.

Step #5: Check whether the central atom has octet or not

In this step, we have to check whether the central atom (i.e iodine) has an octet or not.

In simple words, we have to check whether the central Iodine (I) atom is having 8 electrons or not.

As you can see from the above image, the central atom (i.e iodine), is having 8 electrons. So it fulfills the octet rule.

Step #6: Final step – Check the stability of lewis structure by calculating the formal charge on each atom

Now, you have come to the final step and here you have to check the formal charge on iodine atom (I) as well as each oxygen atom (O).

For that, you need to remember the formula of formal charge;

Formal charge = Valence electrons – Nonbonding electrons – (Bonding electrons)/2

• For Iodine:
Valence electrons = 7 (as it is in group 17)
Nonbonding electrons = 4
Bonding electrons = 4
• For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2

So you can see above that the formal charges on iodine is +1 and the formal charge on both the oxygen atoms is -1.

This indicates that the above lewis structure of IO2 is not stable and so we have to minimize the charges to get a more stable lewis structure.

This can be done by shifting the lone pair from negatively charged oxygen atom to the positively charged iodine atom to form a bond.

Now, in the above structure, you can see that the charges are minimized and the above lewis structure of IO2- is the final stable structure.

There is a -ve charge left on the oxygen atom, which indicates the -1 formal charge on the IO2- ion.

Each electron pair (:) in the lewis dot structure of IO2- ion represents the single bond ( | ). So the above lewis dot structure of IO2- ion can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of CI4
Lewis Structure of BI3
Lewis Structure of CH3I
Lewis Structure of BrO-
Lewis Structure of SeOF2

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Author
Jay Rana

Jay is an educator and has helped more than 100,000 students in their studies by providing simple and easy explanations on different science-related topics. With a desire to make learning accessible for everyone, he founded Knords Learning, an online learning platform that provides students with easily understandable explanations.