Ready to learn how to draw the lewis structure of IO3- ion?
Here, I have explained 6 simple steps to draw the lewis dot structure of IO3- ion (along with images).
So, if you are ready to go with these 6 simple steps, then let’s dive right into it!
Lewis structure of IO3- contains two double bonds 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 3 Oxygen atoms (O). The Iodine atom has 1 lone pair. And the single bonded oxygen atom has -1 formal charge.
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 IO3-).
6 Steps to Draw the Lewis Structure of IO3-
Step #1: Calculate the total number of valence electrons
Here, the given ion is IO3-. In order to draw the lewis structure of IO3- ion, first of all you have to find the total number of valence electrons present in the IO3- 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 IO3-
- For Iodine:
Iodine is a group 17 element on the periodic table.
Hence, the valence electrons present in iodine is 7 (see below image).
- For Oxygen:
Oxygen is a group 16 element on the periodic table.
Hence, the valence electron present in oxygen is 6 (see below image).
Hence in a IO3- 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 IO3- ion = 7 + 6(3) + 1 = 26
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).
Here in the IO3- 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 3 oxygen atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of IO3 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 IO3 molecule.
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 IO3, 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 24 valence electrons of IO3- ion are used in the above structure.
But there are total 26 valence electrons in IO3- ion (as calculated in step #1).
So the number of electrons left to be kept on the central atom = 26 – 24 = 2.
So let’s keep these 2 electrons (i.e electron pair) 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 and the iodine atom is stable.
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 = 2
Bonding electrons = 6
- For Oxygen:
Valence electron = 6 (as it is in group 16)
Nonbonding electrons = 6
Bonding electrons = 2
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charge on iodine is +2 and the formal charge on all the three oxygen atoms is -1.
This indicates that the above lewis structure of IO3 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 pairs from negatively charged oxygen atoms to the positively charged iodine atom to form a double bonds.
Now, in the above structure, you can see that the charges are minimized and the above lewis structure of IO3- is the final stable structure.
There is a -ve charge left on the oxygen atom, which indicates the -1 formal charge on the IO3- ion.
Each electron pair (:) in the lewis dot structure of IO3- ion represents the single bond ( | ). So the above lewis dot structure of IO3- ion can also be represented as shown below.