Ready to learn how to draw the lewis structure of ICl5?
Here, I have explained 5 simple steps to draw the lewis dot structure of ICl5 (along with images).
So, if you are ready to go with these 5 simple steps, then let’s dive right into it!
Lewis structure of ICl5 contains five single bonds between the Iodine (I) atom and each Chlorine (Cl) atom. The Iodine atom (I) is at the center and it is surrounded by 5 Chlorine atoms (Cl). The Iodine atom has 1 lone pair while all the five Chlorine atoms have 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 ICl5).
5 Steps to Draw the Lewis Structure of ICl5
Step #1: Calculate the total number of valence electrons
Here, the given molecule is ICl5 (Iodine pentachloride). In order to draw the lewis structure of ICl5, first of all you have to find the total number of valence electrons present in the ICl5 molecule.
(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 ICl5
- 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 Chlorine:
Chlorine is a group 17 element on the periodic table.
Hence, the valence electron present in chlorine is 7 (see below image).
Hence in a ICl5 molecule,
Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Chlorine (Cl) atom = 7
So, total number of Valence electrons in ICl5 molecule = 7 + 7(5) = 42
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 ICl5 molecule, if we compare the Iodine atom (I) and Chlorine atom (Cl), then the Iodine is less electronegative than chlorine.
So, Iodine should be placed in the center and the remaining 5 chlorine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of ICl5 molecule, put the two electrons (i.e electron pair) between each Iodine atom and chlorine atom to represent a chemical bond between them.
These pairs of electrons present between the Iodine (I) and Chlorine (Cl) atoms form a chemical bond, which bonds the iodine and chlorine atoms with each other in a ICl5 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 ICl5, the outer atoms are chlorine atoms.
So now, you have to complete the octet on these chlorine atoms (because chlorine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the chlorine atoms form an octet.
Also, only 40 valence electrons of ICl5 molecule are used in the above structure.
But there are total 42 valence electrons in ICl5 molecule (as calculated in step #1).
So the number of electrons left to be kept on the central atom = 42 – 40 = 2.
So let’s keep these two electrons (i.e electron pair) on the central atom.
Now, let’s move to the next step.
Step #5: 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 chlorine atom (Cl).
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 = 10
- For Chlorine:
Valence electron = 7 (as it is in group 17)
Nonbonding electrons = 6
Bonding electrons = 2
|Formal charge||=||Valence electrons||–||Nonbonding electrons||–||(Bonding electrons)/2|
So you can see above that the formal charges on iodine as well as chlorine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of ICl5 is the final stable structure only.
Each electron pair (:) in the lewis dot structure of ICl5 represents the single bond ( | ). So the above lewis dot structure of ICl5 can also be represented as shown below.