Ready to learn how to draw the lewis structure of IF5?
Here, I have explained 5 simple steps to draw the lewis dot structure of IF5 (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 IF5 contains five single bonds between the Iodine (I) atom and each Fluorine (F) atom. The Iodine atom (I) is at the center and it is surrounded by 5 Fluorine atoms (F). The Iodine atom has 1 lone pair while all the five Fluorine 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 IF5).
5 Steps to Draw the Lewis Structure of IF5
Step #1: Calculate the total number of valence electrons
Here, the given molecule is IF5 (Iodine pentafluoride). In order to draw the lewis structure of IF5, first of all you have to find the total number of valence electrons present in the IF5 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 IF5
- 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 Fluorine:
Fluorine is a group 17 element on the periodic table.
Hence, the valence electrons present in fluorine is 7 (see below image).
Hence in a IF5 molecule,
Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Fluorine (F) atom = 7
So, total number of Valence electrons in IF5 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 IF5 molecule, if we compare the Iodine atom (I) and Fluorine atom (F), then the Iodine is less electronegative than fluorine.
So, Iodine should be placed in the center and the remaining 5 fluorine atoms will surround it.
Step #3: Put two electrons between the atoms to represent a chemical bond
Now in the above sketch of IF5 molecule, put the two electrons (i.e electron pair) between each Iodine atom and fluorine atom to represent a chemical bond between them.
These pairs of electrons present between the Iodine (I) and Fluorine (F) atoms form a chemical bond, which bonds the iodine and fluorine atoms with each other in a IF5 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 IF5, the outer atoms are fluorine atoms.
So now, you have to complete the octet on these fluorine atoms (because fluorine requires 8 electrons to have a complete outer shell).
Now, you can see in the above image that all the fluorine atoms form an octet.
Also, only 40 valence electrons of IF5 molecule are used in the above structure.
But there are total 42 valence electrons in IF5 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 fluorine atom (F).
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 Fluorine:
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 fluorine are “zero”.
Hence, there will not be any change in the above structure and the above lewis structure of IF5 is the final stable structure only.
Each electron pair (:) in the lewis dot structure of IF5 represents the single bond ( | ). So the above lewis dot structure of IF5 can also be represented as shown below.