Lewis Structure of IF4- (With 5 Simple Steps to Draw!)

Ready to learn how to draw the lewis structure of IF4- ion?

Awesome!

Here, I have explained 6 simple steps to draw the lewis dot structure of IF4- 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 IF4- ion contains four 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 4 Fluorine atoms (F). The Iodine atom has 2 lone pairs and it also 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 IF4-).

5 Steps to Draw the Lewis Structure of IF4-

Step #1: Calculate the total number of valence electrons

Here, the given ion is IF4- ion. In order to draw the lewis structure of IF4- ion, first of all you have to find the total number of valence electrons present in the IF4- 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 IF4-

• 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 Fluorine: 

Fluorine is a group 17 element on the periodic table. ^[2]

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

Hence in a IF4- ion, 

Valence electrons given by Iodine (I) atom = 7
Valence electrons given by each Fluorine (F) atom = 7
Electron due to -1 charge, 1 more electron is added
So, total number of Valence electrons in IF4- ion = 7 + 7(4) + 1 = 36

Step #2: Select the center atom

While selecting the center 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 IF4 molecule, if we compare the iodine atom (I) and fluorine atom (F), then iodine is less electronegative than fluorine.

So, iodine should be placed in the center and the remaining 4 fluorine atoms will surround it.

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

Now in the above sketch of IF4 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 IF4 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 IF4, 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 32 valence electrons of IF4- ion are used in the above structure.

But there are total 36 valence electrons in IF4- ion (as calculated in step #1).

So the number of electrons left to be kept on the central atom = 36 – 32 = 4.

So let’s keep these four electrons (i.e 2 electron pairs) on the central atom.

Now, let’s move to the next step.

Step #5: 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 = 8
• 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
I	=	7	–	4	–	8/2	=	-1
F	=	7	–	6	–	2/2	=	0

Let’s keep this charge on the iodine atom in the above lewis structure of IF4 molecule.

As you can see in the above sketch, there is one -ve charge on the iodine atom, which indicates the -1 formal charge on the IF4 molecule.

Hence, the above lewis structure of IF4- ion is the stable lewis structure.

Each electron pair (:) in the lewis dot structure of IF4- ion represents the single bond ( | ). So the above lewis dot structure of IF4- ion can also be represented as shown below.

Related lewis structures for your practice:
Lewis Structure of PBr3
Lewis Structure of NO2-
Lewis Structure of CO3 2-
Lewis Structure of NF2-
Lewis Structure of SO4 2-

---

Article by;